chem 232 fall 2015 uiuc notes

Understanding Reactivity Chemistry is the study of change. Up to this point, the molecular changes we have studied involved internal changes such as conformational rotation about single bonds of alkanes or ring flipping of cyclohexane. None of these changes involved the breaking and making of covalent bonds. The chemical change we will study from this point forward involves changes in electron configuration. Much of the language that was developed in previous lessons will apply to covalent bond changes. Our description of these changes will greatly benefit from the curved arrow notation that we began to study when we discussed resonance. The discussion of intermediates and transition states from the last lesson will be utilized as well.

Our goal in this course is to learn and understand the chemical transformations (i.e., reactions) common to molecular compounds. To achieve this goal, we will analyze reactions by breaking them down into a sequence of “elementary steps”. This sequence of steps is called the reaction mechanism. In order to write mechanisms, you will need a working knowledge of the electronic structure of molecules and the skill in writing curved arrow notation. This lesson will begin to put these skills to use as we learn to write the mechanism for a simple reaction - the proton transfer reaction.

Organizing chemistry around mechanisms allows the process of change to be understood, not just memorized. Once basic principles of reactivity are understood, you’ll learn to reason-by-analogy to rationalize and predict outcomes of reactions that you’ve never previously seen.

“Elementary step”?

Elementary steps describe the process of change between adjoining minima on the MEP of an overall reaction. Many reactions have several minima (i.e., intermediates) located between starting and ending points. Thus, the overall reaction mechanism requires a sequence of elementary steps to describe the entire process.

Each elementary step has: • A reaction arrow (→) connecting the starting and ending structures of that step • Balanced charge • Balanced stoichiometry

Nucleophiles and Electrophiles We will expand the concept of acids and bases to a related classification scheme known as electrophiles and nucleophiles. The word nucleophile derives from the Greek nucleo, for “nucleus,” and philos, for “loving.” The word electrophile is derived from the Greek electros, “electron,” and philos, “loving.” Most elementary steps involve a nucleophile reacting with an electrophile. Electrophiles are electron deficient. They are characterized by partial or fully developed positive charges and by incomplete octets. Electrophiles have low-lying LUMOs. Nucleophiles are electron rich. They are characterized by partial or fully developed negative charges, and by electron lone pairs or pi bonds. Nucleophiles have high-lying HOMOs. It should not surprise you to learn that all acids are electrophiles, and all bases are nucleophiles.

Nu E

Nu E

empty orbital able to accept an electron pair

an electrophilea nucleophile

new bond between nucleophile and

electrophile

Notice that charge is balanced (the sum the formal charges on one side of the reaction arrow is the same as the sum of the charges on the other side)

reaction arrowNu E

Nu E

The same process takes on a sense of "action" when curved arrows are used to illustrate electron flow

filled orbital ready to donate an electron pair

An elementary step between a generic nucleophile, Nu–, and a generic electrophile, E+

Electron Reconfigurations - the Flow of Electrons - Usually Involve Electron Pairs Covalent bond making and breaking are changes in electron configuration. Electron reconfiguration most often takes place as pairs of electrons, not as single electron entities. In fact, most all of the chemistry that we will study involves the “movement” of electrons in pairs. Heterolytic chemistry is the class of reactions that involves electrons moving in pairs, while homolytic chemistry describes the less common mode involving single electron reconfiguration. Organic chemists like to think of the “movement” of electrons as a “flow” of electron density from regions of high to regions of low electron density – that is – from nucleophiles to electrophiles. The movement of electron pairs is represented by curved arrows.

Nu E

Nu E

an electrophilea nucleophileNu E

Nu E

The same process takes on a sense of "action" when curved arrows are used to illustrate electron flow

Proton Transfer - A Single-step Mechanism (building block considerations)

OH

OH

O

+ HO

H

phenol phenolatehydroxide

+water

phenol

hydroxide

water

phenolate

Notice that most of the atoms remain unchanged. Focus is given only to those building blocks that undergo a change in electron configuration. !

Curved Arrows Describe the Electron Reconfigurations in a Chemical Reaction

OH OH

O

+ HO

H

phenol phenolate

sink

source

Each double-headed curved arrow symbolizes the movement of an electron pair that’s involved in the electron reconfiguration. The tail of the first arrow is placed at the source of electrons. This may be a lone pair, pi-bond or sigma bond. The head of the last arrow indicates the destination or sink. The electron sink tends to either be an electronegative atom – that is, an atom that can stabilize negative charge – or an atom with an empty atom-centered orbital (we symbolize empty, atom-centered orbitals with the letter a).

If, in the process of “pushing” an electron pair to an uncharged C, H, N, or O atom, a new bond is formed, then the electron flow must continue – in the same step – and subsequently break one of the existing bonds at the newly bonded atom. This avoids exceeding the allowed electron count of that atom (e.g., octet for C, N or O; duet for H).

The hydroxide anion is the source. A lone pair of electrons on the hydroxide anion forms a new bond with the phenolic hydrogen atom making water. Simultaneously the O-H bond in phenol breaks and this electron pair moves onto oxygen to make a phenolate anion.

“pushing electrons” using curved arrow notation

OH OH

O

+ HO

H

2nd arrow

1st arrow

Proton Transfer Reactions Involve Sigma Bond Breaking and Sigma Bond Making

Empty σ* Filled n

New σ bond

σ"

n

π

σ� "a π�

σ→σ� " σ→a σ→π�"

filled

empty

n→σ� " n→a n→π�"

π→σ� " π→a π→π�"

Curved Arrows Imply the Frontier Orbitals

Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in an elementary step. The tail of the first arrow implies the filled orbital (HOMO) is a non-bonded electron pair, n. The head of the first arrow points to the empty orbital (LUMO, σ*) that accepts the electron pair. The empty orbital may not be obvious until you realize that phenol’s O-H sigma bond breaks, as implied by the tail of the second arrow. Sigma bond breaking would result upon filling σ* with the electron pair from n. The frontier orbitals align coaxially (a σ-type interaction).

[pt]

This symbol means a proton transfer step

An n → σ* σ-type interaction

LUMO / HOMO Applied to Acid / Base Interactions

Low energy LUMO = strong acid (reactive electrophile) High energy HOMO = strong base (reactive nucleophile)

H3O+ H2O HO-

n

σ∗

n

σ∗

n

σ∗

Consider the O-H bond in…

Energy Changes Accompany Each Elementary Step

progress of the reaction

free

ener

gy

(a)

(b)

(c)

(e)

(d)

A Reaction Coordinate Diagram for a One-Step Process

(a) reactants(b) transition state(c) products(d) ΔG˚ Gibbs standard free energy change(e) ΔG≠ free energy of activation

We have already encountered a useful tool that allows us to track energy changes along a complex reaction pathway – the tool was the reaction coordinate diagram used to describe the ring flipping process of cyclohexane. Although cyclohexane’s “reaction” did not involve electron reconfiguration, the minimum energy pathway (MEP) correlated energy and progress of change for a complex set of conformations. Not only does the reaction coordinate diagram and its MEP serve as a way to track energy changes for conformational processes, but it also is a tool for tracking energy of reactions that involve electron reconfiguration.

We are interested only in the minima and maxima on reaction coordinate diagrams. By knowing the energy at these positions, either qualitatively or quantitatively, we will be able to judge the equilibrium position of the reaction, and how fast each step of the reaction takes place.

The essential components of a reaction coordinate diagram for a one-step reaction are shown above. Most important is the MEP on this diagram. The particular MEP on the above diagram might, for example, apply to the proton transfer from phenol to hydroxide. The reactants (a) are higher in energy than the products (c), so there is a favorable driving force (d) to promote the change. This driving force is the Gibbs standard free energy and it’s directly related to the equilibrium constant (Keq) by ΔG˚ = -RT ln Keq where T is the temperature in Kelvin and R is the gas constant.

The transition state (b) is a transient structure caught in the act of undergoing electron reconfiguration. The symbol TS≠ represents the transition state structure. Notice that the energy associated with TS≠ is higher than the reactants. This maximum on the energy profile represents a barrier that resists the transformation of reactants to products. The higher the barrier is, the slower the rate of the reaction. The coefficient, k, is proportional to the reaction rate (k and ΔG≠ are related by k = 2.084x1010·T·e-(ΔG≠/1.986T)).

MEP on a 1-step reaction coordinate diagram

Activated-Rate Theory Getting from reactants to products requires energy, e.g., energy to surmount a transition state barrier or energy to become an unstable intermediate on the MEP. Where does this energy come from?

In order to surmount barriers on the MEP, reactants must somehow acquire energy from their surroundings. This energy might come from heat, light, electrical potential, or mechanical force. The barrier is thus only a temporary one, because as soon as the reactants acquire sufficient energy from their surroundings, electron reconfiguration can take place. In the case of proton transfer, the barrier is not very high and is easily surmounted by the thermal energy (i.e., random, Brownian motion) available to the reactants at room temperature.

The energy transduction network (diagram at the right) reminds us that potential energy of one form (e.g., electrical potential) can be converted into another form (e.g., chemical potential). When molecules acquire energy from their surroundings, they can surmount the barriers on an MEP and reactions will proceed. This is the basic idea behind the activated-rate theory which describes how fast reactants transform into products. The theory says that reactants must become activated to their transition state structures by acquiring energy from their surroundings.

Structure of the Transition State The transition state (TS≠) is a transient species lying on the reaction pathway that is undergoing the act of electron reconfiguration. But what does the structure of the TS≠ look like? We can generate a representation of the transition state structure using curved arrow notation as a guide. In representing the TS≠, lone pairs are not shown because the number of lone pairs often changes as part of the electron reconfiguration. Bonds being broken and bonds being made are represented as dashed lines. Changes in formal charge are important. In the TS≠ charges on one particular atom may be growing or diminishing and such charges are non-integer (i.e., they are partial charges). Partial charges, developing or diminishing, are represented as δ+ (for positive charge) and δ� (for negative charge).

OH OH

O

+ HO

H

OH

OHδ−

δ−

transition state structure

diminishing chargedevelopingcharge

developing bond

diminishing bond

signifies transition state structure

For the proton transfer reaction shown below, notice the connection between the curved arrow notation and the representation of the TS≠. The tail of each curved arrow either corresponds to a lone pair whose electron density is being consumed as the reaction unfolds, or a bond that is being broken. The head of each arrow either corresponds to a lone pair being produced, or a bond being made. The changes in charge are determined by looking at the location of charge in the reactant and in the product. Each atom associated with a changing charge is given by a δ+ or a δ�. The entire structure is placed in square brackets with a double dagger (≠) in the upper right sided to signify that the structure is transient.

Activated-rate theory says that the energy of a TS≠ must be known to predict reaction rates. Although TS≠ energies can accurately be calculated with modern computational tools like WebMO, TS≠’s cannot be studied experimentally because they only exist transiently. For building chemical intuition, a qualitative analysis may be better than sophisticated computation; thus, it is worth-while to have some rules to help us rationalize the relationship between TS≠ energy and structure. In the TS≠, bonds are being made and bonds are being broken. How far along are the bond-formation and bond-breaking processes at the point the transition state is reached? Are bonds nearly fully developed or just beginning to form? Are bonds fully broken, or are they just beginning to weaken? The Hammond postulate helps to answer these questions. The Hammond postulate tells us that for any two minima connected via a TS≠, the structure of the transition state will more closely resemble the minimum that it is closer to in energy. This might sound complicated, but in reality there are only two possibilities to consider. Let’s examine them with reference to the diagram.

Case 1: the beginning state is higher than the ending state – In this case Hammond’s postulate tells us that the TS≠ will more closely resemble the beginning state. The energy of the TS≠ must be closer to the beginning state. There’s just no other way to draw the diagram. We call this TS≠ “early” since the structure of the transition state has not evolved far from its starting point. Case 2: the ending state is higher than the beginning state – In this case Hammond’s postulate tells us that the TS≠ will more closely resemble the ending state. The energy of the TS≠ must be closer to the energy of the ending state. We call this TS≠ “late” since the structure of the transition state has nearly evolved to its ending point.

Hammond’s Postulate: Transition State Structure & Energy

The Reaction Coordinate Diagram for Proton Transfer

The complete reaction coordinate diagram includes the structures of the reactants, products and a representation of the TS≠. The rate of the reaction in the forward direction is determined by ΔG≠

f while the rate of the reaction in the backwards direction is determined by ΔG≠

b. Based on the relative heights of ΔG≠

f and ΔG≠

b, notice that the forward rate is faster than the backward rate. Recall that Keq is given by the ratio of the forward to backward rate coefficients. The driving force for the reaction is determined by ΔG˚.

OH

OH

O

+

HO

H

Keq+

free

ener

gy


OH

OH

O

HO

H

OH

HO

H

δ−

δ−

ΔG≠f

ΔG≠b

ΔG˚

Keq > 1 for this proton transfer step (soon we will be able to predict this result, but for now just accept the fact). Since Keq > 1, the reaction is thermodynamically downhill; case 1 of Hammond’s Postulate applies. The TS≠ is early, and the TS≠ structure more closely resembles the reactants.

Summary

1.  Curved arrow notation shows the changes in electron configuration for each step of a reaction mechanism. !

2.  Curved arrows indicate the partial bonds formed or broken in the transition state (e.g., an arrow tail drawn from a bond represents a bond that is breaking; an arrow head drawn between two atoms represents a bond being made).!

3.  The sum of the formal charges on each side of a reaction arrow must balance. !

4.  The atoms in transition state structures may have developing or diminishing charge. Developing or diminishing charge in transition states is represented by the partial charge symbol (e.g., δ+ or δ–). If the formal charge on a particular atom changes from one side of the reaction arrow to the other, that atom is assigned a partial charge in the transition state.

Structure-Reactivity Relationships A change in structure corresponds to a chemical reaction. Reactivity is the term we use to describe the potential of a structure to undergo chemical change. Some structures are more prone to react than others. Structures that are likely to undergo chemical change are said to be reactive. Structures that are resistant to chemical change are said to be unreactive. We sometimes choose to speak about the complementary property known as stability. Structures that are highly reactive are generally unstable. Structures that are unreactive are generally stable.

We want to view molecular structure from the perspective of potential to undergo chemical change. A goal is to establish trends that will allow you to examine a structure and make predictions about its chemical characteristics. Just like a physician diagnoses patients, chemists learn to associate certain features of structure with the tendency to react. From these trends emerge the structure-reactivity relationships.

Potential energy (specifically, chemical potential) is the link between structure and reactivity. The greater a structure’s chemical potential, the greater is its reactivity. Understanding the link between structure and reactivity will help us decide which, among several possible changes, are most reasonable (e.g., pathways having intermediates with exceedingly high energies are not likely to be reasonable and should therefore be avoided, especially if a pathway with lower energy intermediates is available).

structure chemical potential reactivity

Analysis of Chemical Potential Analysis of Chemical Potential: Chemical potential makes sense in the context of a chemical reaction because we can compare what’s on the two sides of a reaction arrow (→) and determine the difference in chemical potential. By comparing what’s on one side with what’s on the other, it becomes possible to know if the chemical change is favored or not. Does the left-hand-side or the right-hand-side have greater chemical potential? The favored direction will be from high to low chemical potential.

An analysis of chemical potential can be done for individual elementary steps, or an overall chemical reaction, as the problem demands. The things that contribute to chemical potential - the things that need to be compared - are:

• Bond energy changes • Strength (reactivity) of electrophiles and /or acids • Strength (reactivity) of nucleophiles and/or bases

• Weak bonds • Low energy LUMO = strong acid (reactive electrophile) • High energy HOMO = strong base (reactive nucleophile) • Permanent charges

What Factors Contribute to High Chemical Potential?

A → B

free

ener

gy

GB

GA

ΔG˚ = GB - GA

Chemical potential for the reaction A → B

Factors Contributing to Charge Stability

____________________________ - All other things being equal, greater charge delocalization leads to greater stability.

________________________________ - All other things being equal, if negative charge is localized on an atom, the greater the s-character of that atom’s hybrid orbitals, the greater the stability (for negative charge: sp is more stable than sp2, which in turn is more stable than sp3). Conversely, if positive charge is localized on an atom, the lesser the s-character of that atom’s hybrid orbitals, the greater the stability (for positive charge: sp3 is more stable than sp2, which in turn is more stable than sp).

(2) Delocalization

(3) Hybridization

_______________ Like charges repel; opposite charges attract. (4) Coulombic

Charged species are commonly encountered in chemical reactions, either as starting components, end products, or as intermediates and TS≠ along a reaction pathway. There is considerable energy associated with charged species; consequently, the relative stability of charged atoms often provides important information about chemical reactivity. The four factors listed below are some of the most important aspects of structure that contribute to charge stability.

________________________________ - The electronegativity and size of charged atoms contribute to its ability to stabilize charge. (1) Atom type and periodic table trends

Bond Energy Changes Bond dissociation energies are a useful way to estimate chemical potential. Values for bond dissociation energies are summarized in the table below. Note that by convention, all bond energies are assigned a negative value. In working these calculations, be sure to note all bond changes that take place on both sides of the reaction arrow (→). An example is provided on the next slide. Hint: Drawing the implied hydrogen atoms is often VERY helpful!

Average Bond Energies (kcal mole-1)

H C N O F Si S Cl Br I

104 99 93 111 135 76 83 103 87 71 H 83a 73b 86c 116d 72 65 81 68 52 C 39 53e 65 46 N 47 45 108 52 48 56 O 37 135 F

aC=C 146, C≡C 200 53 91 74 56 Si 60 61 52 S

bC=N 147, C≡N 213 dIn CF4 58 Cl 46 Br

cC=O 176 (aldehydes) e In nitrites and nitrates 36 I 179 (ketones) All bond energies are assigned a negative value by convention.

For a more extensive list see: http://www.jhu.edu/chem/lectka/Bond%20Strengths.html

Bond Energy Changes: Example Calculation

C CH

H H

HH Br+ C C

H

HH

H

HBr

To determine the net change in bond energy, we first must determine which bonds undergo change. For the reaction:

the H–Br and C=C double bond are broken. The energy of the C=C double bond is not entirely lost, however, as the double bond becomes a C–C single bond. Thus, we will take this change into account by assuming that a C=C bond is broken and a C–C bond is made. Other new bonds include C–H and C–Br. The energies of all bonds broken and all bonds made are then summarized as shown in the table below. The reaction is favorable by 17 kcal/mol as a result of the increased net bond strength. Note that this calculation is for bond enthalpies (ΔH), not ΔG.

Bonds changes (reactants) Bonds changes (products) bond Energy (kcal/mol) bond Energy (kcal/mol) C=C -146 C-C -83 H-Br -87 C-H -99

C-Br -68 net -233 net -250

Overall change = bonds made – bonds broken = (-250) – (-233) = -17 kcal/mol

Recognizing Nucleophiles and Bases

HO Cl

H2O CH3NH2

C CR

RR

R

The usual order of energy levels

σ* π* a n π σ

ener

gy

Nucleophiles donate electrons from their highest occupied MO

The signatures of a nucleophile • an electron rich atom, especially as revealed by negative formal charge • a nonbonded pair of electrons (a lone pair) • an electron pair in a pi bond • a partial negative charge revealed via resonance contributor(s) • strong bases tend to be strong nucleophiles • any species with a high-lying HOMO

Typical nucleophiles

Nucleophilicity and HOMO Energy Level HO

MO

Ene

rgy

Leve

l

Nu 1 Nu 2 Nu 3 Nu 4

Consider the series of nucleophiles Nu 1 to Nu 4 with HOMO energies as shown above!

LUMO Energy of E+!

Given the LUMO energy of E+ as indicated above:

Which frontier orbital interaction will be the strongest?

Which electron pair is the most “energized”?

Which nucleophile has the highest nucleophilicity ?

There’s a good analogy between nucleophiles and bases. Just as some bases are stronger than others, some nucleophiles are stronger than others. Whereas we speak of basicity to describe the strength of bases, we speak of nucleophilicity to describe the strength of nucleophiles.

The usual order of energy levels

σ* π* a n π σ

ener

gy

Recognizing Electrophiles and Acids

The signatures of electrophiles and acids • an electron deficient atom especially as revealed by an incomplete octet (duet for

H) of electrons (e.g., carbocation, boron, H+) • a partial positive charge owning to a polarized H-X or C-X bond (polarization

results when H or C is bound to electronegative atom “X” or an atom X+ that bears a positive formal charge such as [H–NH3]+)

• a partial positive charge revealed via resonance contributor(s) • a weak bond usually involving a pair of heteroatoms e.g., X–Y bonds like Br–Br,

I–Cl, -O–O- • strong acids tend to be strong electrophiles • any species with a low lying LUMO

Electrophiles accept electrons into their lowest unoccupied MO

H R CR

RH B

H

H

O

R Rδ+

δ−H3C O

O CH3

Cl Cl

Typical electrophiles

Electrophilicity and LUMO Energy Level

Consider the series of electrophiles E+ 1 to E+ 4 with LUMO energies as shown!

There’s a good analogy between electrophiles and acids. Just as some acids are stronger than others, some electrophiles are stronger than others. Whereas we speak of acidity to describe the strength of acids, we speak of electrophilicity to describe the strength of electrophiles.

REACTIVITY CONCISELY SUMMARIZED: Continuing the acid/base analogy further, strong nucleophiles will generally react with strong electrophiles just as strong acids and strong bases always react. Such favorable reactions are expected from small frontier orbital HOMO-LUMO energy gaps. Weak nucleophiles and weak electrophiles are not likely to react at all; the frontier orbital gap is too wide in this case. A weak electrophile is likely to react only if it encounters a strong nucleophile; a weak nucleophile is likely to react only if it encounters a strong electrophile.

Given the HOMO energy level of Nu as indicated:

Which frontier orbital interaction will be the strongest?

Which empty orbital is most easily accessed?

Which electrophile has the highest electrophilicity?

LUM

O E

nerg

y Le

vel

E+ 1 E+ 2 E+ 3 E+ 4

HOMO Energy of Nu!

Recognizing Acids and Bases Acids and bases will be one of two types as illustrated by the two equations below.

Acids: • Must have a hydrogen atom to donate • The most acidic hydrogen is often bound to a heteroatom (N, O, S, X) • In rare (but important!) cases, the acidic hydrogen is bound to carbon • Will be neutral or positively charged

Bases: • Must have a lone pair to accept a hydrogen • Will be neutral or negatively charged

Ka

Ka

H––X H X+acid proton base

H––X H X+acid proton base

Type I

Type II

Henderson-Hasselbalch eqn.

Acid-Base Equilibria: Concepts & Definitions Many protonation and deprotonation steps are reversible. The equilibrium can be expressed in terms of pKa. Let’s review the meaning of pKa:

HA + H2O H3O+ + A- Keq Keq =

[H3O+][A-] [HA][H2O]

Assuming the [H2O] to be constant (55.5 M) i.e., [H2O]>>[H3O+], we can define a new constant _________________ which allows us to drop water from the above equilibrium expression. This is the basis of the widely used pKa convention:

Ka = Keq[H20]

HA H+ + A- Ka Ka =

[H+][A-] [HA]

log Ka = log [H+] + log [A-] [HA] Or

-log Ka = -log [H+] + log [A-] [HA]

pKa pH = + log [A-] [HA]

definition

pH

pKa = 5.2 HA A- "

Relative conc. vs. pH

Determining the Position of Equilibria

Keq

Keq can be calculated by analyzing this reaction as a sum of two acid / base equilibria

Ka(1)

Ka(2)

pKa ____ Relevant equilibria _______________

-2.5

-1.7

Keq = ––––– 1 Ka

(1) · Ka

(2) pKa = - log Ka 10-(-1.7) 10-(-2.5)

= 0.16 Keq = –––––

Based on the magnitude of Keq, can we conclude that protonated alcohol is a reasonable species?

Acid-Base Equilibria in Organic Chemistry 1.  Each side of the equation will have one proton donor and one proton acceptor (i.e., one acid

and one base will be on each side of the equation). 2.  Charge must be balanced. 3.  Stoichiometry must be balanced. 4.  Draw all electron lone pairs on all structures to see what bonding takes place in the proton

transfer process. In cases where the proton donor is a C-H group, the implied hydrogen atom must be explicitly drawn.

5.  Some molecules have more than one acidic (or basic) site. Use the rules of charge stability to determine the location to add (or remove) a proton. If necessary, check your guess with the pKa calculator.

6.  The position of the equilibrium is predicted (qualitatively) by consideration of charge stability. If necessary, the pKa calculator can be used to check your guess.

7.  The equilibrium arrows (�) indicate the reaction is reversible. A longer arrow is drawn toward the species favored at equilibrium.!

8.  Curved arrows are added to show the flow of electrons.!

H3C N

CH3

H

HH

+H

OH or

circle one

+

ACE Organic Requires Arrows to Be Drawn Carefully STEP 1 - To begin, click on the electron source. If the source is a lone pair, click on the atom to which the lone pair belongs. The cursor is correctly positioned when you see the blue

STEP 2 - Move the cursor to the electron pair’s new location.

STEP 3 - Prior to releasing the mouse button, blue ( ) will indicate the implied position of the new bond. Release the mouse only when the ( ) are correctly positioned.

More About ACE Arrow Conventions

STEP 4 - Check each arrow after releasing the mouse button. Move the cursor to the arrow’s head. The blue rectangle shows the implied location of the new bond.

STEP 5 - The second arrow begins from a bond (the electron source is a bond). In this case, look for the blue ( ) to signify the position of the second arrow’s tail.

Final ACE Arrow Conventions

STEP 6 - The second arrow ends on an atom. After releasing the mouse, check that it looks like this. Here the blue rectangle shows the bond source.

STEP 7 - Once the arrows are drawn, finish drawing the structures. Take advantage of ACE shortcuts. For example, double click on the structure and copy it to the product box. Adjust the structure as required, but don’t draw it from scratch.

Commonly Used Acids and Bases Proton transfer steps (protonation or deprotonation reactions) are an extremely common way to initiate chemical reactions. Because of this, chemists frequently add acids or bases (often in sub-stoichiometric quantities) to execute a variety of useful reactions. What do some commonly used acids and bases look like in practice? We’ve collected the common ones for you on this page.

HCl

H2SO4

HNO3 nitric acid

sulfuric acid

hydrochloric acid

strong inorganic acids

organic acids

S

O

O

OHH3C

p-toluenesulfonic acid (PTSA, TsOH)

CH3COOHacetic acid

F3C S

O

O

OH

trifluoromethanesulfonic acid(triflic acid, TfOH)

sodium hydrideNaHstrong nitrogen bases

NaNH2

lithium diisopropylamide(LDA)

sodium amide

CH3

NLi

CH3H3C

CH3

weak nitrogen bases oxygen bases

NaOH

NaOMe

NaOEt

KOtBu

sodium hydroxide

sodium methoxide

sodium ethoxide

potassium tert-butoxide

N

pyridine(Py)

N

Et

Et Et

triethylamine(TEA)

strong bases

CH3CH2CH2CH2Li

butyllithium(nBuLi)

Conventions and Terms Used in Writing Chemical Equations Chemists have developed a loose style for writing chemical transformations as equations. In a typical equation, one or more reactants (also called substrates) are combined with a reagent(s), and compounds are usually dissolved in a solvent. Reactions are often performed under conditions of heat (Δ) or light (hν). The conditions (solvent, temperature, etc.) are commonly listed above or below the reaction arrow. Sometimes reagents are on the left side of the reaction arrow, while other times reagents are listed above or below the reaction arrow. Very often the stoichiometry is not balanced. The following example is illustrative.

Sometimes transformations are performed by subjecting the substrate to a sequence of reaction conditions. In such a case, the conditions are listed numerically above and / or below the reaction arrow, as shown.

O

1) H2N-NH22) KOH, Δ

OCH3

O

LiI

DMF , ΔO

O

OCH3

O

+ LiIDMF , Δ

O

O

LiI is the reagent. Dimethylformamide (DMF) is the solvent. Here LiI is written to the left of the reaction arrow.

Here LiI is written above the reaction arrow.

Cyclohexanone is first treated with H2NNH2 followed by KOH and heat.

Common organic solvents: http://organicdivision.org/organic_solvents.html

1) Acid-Base - hydrogen swap

Classes of Chemical Transformations This page summarizes the main classes of chemical transformations. The purpose of presenting this overview is to provide an organizational framework as an outline of upcoming studies. In learning new reactions you will benefit from recognizing to which of the seven classes the new reaction belongs.

2 & 3) Oxidation & Reduction

4) Substitution - replace C’s substituent (-X) with another (-Y), neither being -H

X

C

Y

CY + X+

neither X nor Y being hydrogen

Yreplaces

X

5) Elimination - loss of XY elements with concomitant pi bond formation 6) Addition - gain of XY elements with concomitant loss of pi bond

reductionX

C

H

Coxidation

Hreplaces

X

Xreplaces

H

change in the number of C-H bonds in relation to the number of C-X bonds

X

CZ

Z

YC + XY

elimination

addition

form π bond-XY

break π bond+XY

base

H

X

H

YY + X+

baseacid acid

7) Rearrangement - isomerization process (no atoms lost or gained); results in new bonding connectivity (one of many examples shown as there is no generic representation).

RCH3

RR stands for a generic "residue"

Z = C, N, O

Substitution at sp3 Carbon: Examples

Nu + CH3–Br CH3–Nu Br+

Substitution at sp3 Carbon: More Examples

Nu + CH3–Br CH3–Nu Br+

Two Mechanistic Pathways for Substitution One step mechanism: [SN2] = [AN] + [DN]

Two step mechanism: [SN1] = [DN] then [AN]

Here bond making and breaking take place at the same time - a concerted process. Any one-step process is its own elementary step. So we’ll call the elementary step that results from concerted combination of [AN] + [DN] the [SN2] step, which stands for nucleophilic, bimolecular substitution. It’s bimolecular because two species must come together. We need to re-consider the frontier orbitals involved in this new elementary step. The tail of the first arrow implies the filled orbital (HOMO) is a non-bonded electron pair, n. The head of the first arrow points to the empty orbital (LUMO, σ*) that accepts the electron pair. The empty orbital may not be obvious until you realize that the C-L sigma bond breaks, as implied by the tail of the second arrow. Sigma bond breaking would result upon filling σ* with the electron pair from n. The frontier orbitals align coaxially (a σ-type interaction).

[SN2]

This symbol means bimolecular nucleophilic substitution

[SN2] involves σ-bond making

& breaking An n → σ* σ-type interaction

C L C + LNuC

Nu

[DN] [AN]

Nu CNu LC L+ +

The second pathway is a sequential combination of [DN] followed by [AN]. This sequential, two-step mechanistic pathway is called unimolecular nucleophilic substitution, or [SN1]. The slowest step is the unimolecular [DN]. process, hence the “unimolecular” in the name. The most important thing to realize is that the [SN1] pathway will only be available if a stable carbocation can form.

Substitution Mechanisms Require New Elementary Steps: Nucleophile Dissociation [DN] & the Nature of Leaving Groups

Empty a

Filled σ [DN] involves sigma bond

breaking

σ#

n

π

σ� #a π�

σ→σ� # σ→a σ→π�#

filled

empty

n→σ� # n→a n→π�#

π→σ� # π→a π→π�#


Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in an elementary step. The tail of the arrow implies the filled orbital (HOMO) is σ electron pair. The head of the arrow points to an atom on the leaving group (L) suggesting that an atom-centered empty orbital (LUMO, a) accepts the electron pair. The process produces no new bonding interactions. A good leaving group will have a high energy σ and low lying a.

[DN]

This symbol means a dissociation of a nucleophile

[DN] involves a σ → a “no-bond” interaction

C L C L+-L is the leaving group

Substitution Mechanisms Require New Elementary Steps: Nucleophile Association [AN]

Empty a

Filled n [AN] involves sigma bond

making

σ#

n

π

σ� #a π�

σ→σ� # σ→a σ→π�#

filled

empty

n→σ� # n→a n→π�#

π→σ� # π→a π→π�#


Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in an elementary step. The tail of the arrow implies the filled orbital (HOMO) is non-bonded electron pair, n. The head of the arrow points to an empty orbital on the carbocation, suggesting that an atom-centered empty orbital (LUMO, a) accepts the electron pair. A good nucleophile will have a high energy n.

[AN]

This symbol means an association of a nucleophile

[AN] involves an n → a σ-type interaction

is the nucleophile Nu

CNu + Nu C

Bimolecular Nucleophilic Substitution: Simultaneous Bond Making and Bond Breaking The ____ pathway ( ) Substitution, Nucleophilic, Bimolecular [SN2]

The TS≠ of the rate-determining step involves the association of ___ molecules (bimolecular) two

δ� δ� HO Br+ HOBrHO

http://csi.chemie.tu-darmstadt.de/ak/immel/misc/oc-scripts/animations.html?structure=sn2

Unimolecular Nucleophilic Substitution: Bond Breaking Precedes Bond Making

The ____ pathway ( ) Substitution, Nucleophilic, Unimolecular SN1

δ+ δ�

The TS≠ of the rate-determining step involves just ___ molecule (unimolecular) one

CH3 CCH3

CH3Br

slowstep

CH3 CCH3

CH3Br H

OH

CH3 CCH3

CH3OH

H

HO

H

faststep

faststep

CH3 CCH3

CH3OH +

HO

H HH3O+

Comparison of [SN1] and [SN2] Pathways

[SN2] ---- [SN1] –––

first step is the RDS for [SN1]

http://www.chemtube3d.com/

favors SN1

favors SN2

Structure-Reactivity and the Haloalkane

Carbocation stability

Access to the site of reaction

Governed by electronic factors

Governed by steric factors

Recognizing Leaving Groups (L)

3.2

-7

-9

-10

___ pKa

Incr

easi

ng a

cidi

ty H

–X

Incr

easi

ng b

asic

ity X

–

Incr

easi

ng le

avin

g ab

ility

Leaving group reactivity correlates to basicity. weaker better The ______ the base, the ______ its leaving ability.

This general trend holds for [SN2], [SN1] and other reactions that we’ll study

Why Does L Correlate to Basicity? In the TS≠ (RDS), the leaving group develops negative charge; thus, the ability of a group to function as a leaving group is related to the stability of the corresponding anion

δ+ δ� δ� δ� LNu

TS≠ for SN2

C L

TS≠ for SN1

From Paula Y. Bruice Organic Chemistry 4th Ed. Pearson

The List of Good Leaving Groups is Short

Increasing reactivity

> > >> > ~

Not only is it important to recognize what qualifies as a good -L, but it is equally important to recognize what is not a good -L

Nucleophile Reactivity in

crea

sing

nuc

leop

hilic

ity good

moderate

poor

Nucleophilicity and Basicity

In a series of nucleophiles with the same nucleophilic atom, the stronger the base, the greater the nucleophilicity.

nucleophile

conjugate acid

pKa 4-5 15.7 17

< <

Increasing Nucleophilicity

Nucleophilicity of Neutrals vs. Anions


In a series of nucleophiles with the same nucleophilic atom, _______ nucleophiles are stronger than neutrals anionic

<

<

<

Nucleophilicity of Nucleophiles in the Same Row

row 2


basicity

< < <

< < row 3

Within a row, nucleophilicity increases from right to left and tracks with________

(regardless of solvent type) basicity

Solvents Influence Nucleophilicity Nucleophilicity also depends on the _______ solvent

solvation

In polar _______ solvents (DMSO, acetone, acetonitrile, DMF) relative nucleophilicities parallel relative basicities.

The interaction between solvent and nucleophile: ________

Reason: Cations are effectively solvated but anions are poorly solvated and thus the nucleophile is not shielded by solvent.

DMSO

aprotic

From Paula Y. Bruice Organic Chemistry 4th Ed. Pearson

Nucleophilicity and Solvent In polar ______ solvents, the solvent shell is tightly held around the nucleophile due to _______________

protic hydrogen bonding

The smaller the ion, the tighter the solvent shell. Thus, for nucleophiles of different size (moving up or down in the periodic table) the ______ the size, the ______ the nucleophile (opposite to base strength).

larger better

Modified from Paula Y. Bruice Organic Chemistry 4th Ed. Pearson

Nucleophilicity of Nucleophiles in Different Rows

Increasing Nucleophilicity solvent

Polar aprotic

Polar protic

basicity

< < <

< < <

Summary

1.  Substitution reactions at 1˚ and 2˚ (but not 3˚) C(sp3) centers usually proceed by the SN2 mechanism under basic or neutral conditions. Substitution at C(sp2) centers rarely (if ever) occurs by the SN2 mechanism.!

2.  Leaving group ability: !•  The weaker the base, the better is the leaving ability.!•  F–, OH–, RO– (except epoxides), H–, and carbanions are almost never

leaving groups in [SN1] or [SN2] reactions.!3.  Nucleophilicity (an aprotic solvent like DMSO): !

•  The stronger the base, the stronger the nucleophile.!4.  Nucleophilicity (in a protic solvent like methanol):!

•  Neutrals are poorer nucleophiles than anions.!•  Comparing nucleophilic atoms that are in the same row, the stronger

the base, the stronger the nucleophile.!•  Comparing nucleophilic atoms that are in the same column, the larger

the atom, the stronger the nucleophile.!


Elimination Across sp3 Carbons Next we examine a class of elimination reactions to produce carbon-carbon multiple bonds. We recently studied substitution reactions at sp3 carbon atoms. Elimination and substitution are competing reactions; some mechanisms of elimination and substitution have common intermediates.



X

C

Y

CY + X+


Yreplaces

X


reductionX

C

H

Coxidation

Hreplaces

X

Xreplaces

H


X

CZ

Z

YC + XY

elimination

addition

form π bond-XY

break π bond+XY

base

H

X

H

YY + X+

baseacid acid


RCH3


Z = C, N, O

Elimination Across Carbon-Carbon Bonds: Three Examples

CH3 C BrCH3

CH3+ Br

CC

H H

CH3 CH3 HO

H H+H2O, Δ

CC

HH

HH

BrCH2CH3

HOH+ +CC

H

HH

CH2CH3KOH KBr

O OH O

KOHHOH+

The Relative Stability of Carbon-Carbon Double Bonds

Alkenes are hydrocarbons that contain the double bond functional group. Alkenes are also called olefins and they are said to be unsaturated compounds meaning that their carbon atoms do not have the maximum number of allowed substituents (4 for carbon). As explained on the following page, heats of hydrogenation can be used to determine stability for alkenes that differ by their degree of alkene substitution. The general conclusion is that the more alkyl substituents that are bonded to sp2 carbons of an alkene, the more stable is the double bond.

Heats of Hydrogenation and an Experimental Determination of Alkene Stability

The addition of hydrogen to carbon-carbon double bond is known as hydrogenation. This reduction reaction transforms alkenes into alkanes. The conditions and mechanism of the reaction do not concern us at this time. We are going to use this reaction to experimentally determine the relative stability of alkenes.

CH3

CH3

CH3 CH3

transisomer

cisisomer

ΔH˚

C CCH3 CH3

H

H

H

H

H2 + H2 +

-27.6 kcal/mol-28.6 kcal/mol

1.0 kcal/mol

butane

The heat produced by a reaction is accurately measured by a calorimeter. Let’s compare the hydrogenation of cis and trans 2-butene. Both reactions yield the same product, butane. The enthalpy change for the trans isomer is –27.6 kcal/mol whereas for the cis isomer the enthalpy change is –28.6 kcal / mol. Since both reactions yield the same product, we know that the energy difference comes from the difference of the starting alkenes. The diagram illustrates this result, and shows that the trans isomer is more stable than the cis isomer by about 1.0 kcal/mol. The origin of this stability difference is a steric interaction between the methyl groups. You can use MarvinSketch to convince yourself of the close proximity of the methyl groups in the cis isomer.

Elimination Mechanisms Require New Elementary Steps: Electrophile Dissociation From an Adjacent Carbocation

Empty a

Filled σ � [DE] involves σ-bond breaking, π-bond making

Curved Arrows Imply the Frontier Orbitals Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in the elementary step. The tail of the arrow implies the filled orbital (HOMO) is an electron pair in a σ-bond. The head of the arrow points between two carbon atoms and suggests a new π-bond. An empty orbital on the carbocation is available to accommodate this electron pair implying that the LUMO is an atom-centered, empty orbital (a). The HOMO (σ-orbital) is beside (and nearly parallel to) the LUMO (a) allowing for a π-type interaction.

[DE]

This symbol means dissociation of an electrophile

[DE] is a σ → a π-type interaction

-E is the electrophile (-E is often an -H, hydrogen)

C CE

E + C C

σ$

n

π

σ� $a π�

σ→σ� $ σ→a σ→π�$

filled

empty

n→σ� $ n→a n→π�$

π→σ� $ π→a π→π�$

Elimination Mechanisms Require New Elementary Steps: Beta Elimination [Eβ]

Empty σ* Filled n � [Eβ] involves

π-bond making σ-bond breaking

Curved Arrows Imply the Frontier Orbitals Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in this elementary step. The tail of the first arrow implies that the filled orbital (HOMO) is a nonbonded electron pair, n. The head of the first arrow points between two carbon atoms suggesting in a new π-bond. An empty, low-lying σ*-orbital from the adjacent C-L bond is available to accommodate this electron pair suggesting the LUMO is σ*. The HOMO, n, is beside (and nearly parallel to) the LUMO, σ*, allowing for a π-type interaction.

[Eβ]

This symbol means beta elimination

[Eβ] is an n → σ* π-type interaction

-L is a leaving group

σ$

n

π

σ� $a π�

σ→σ� $ σ→a σ→π�$

filled

empty

n→σ� $ n→a n→π�$

π→σ� $ π→a π→π�$

The alpha carbon (Cα) is defined as the carbon bearing the leaving group, -L. Many leaving groups (-L) participate in [Eβ], even poor ones (e.g., -OH). The new π-bond is formed across the Cα-Cβ bond.

C YL

C Y + Lβα

C CL

C C + Lβα

There are Three Mechanistic Pathways for Elimination

[E1]

[E1cb]

[E2]

[DN]

[pt]

[DE]

[Eβ]

[pt] + [DN] (concerted)

[pt] then [Eβ] (two-step)

[DN] then [DE] (two-step)

The three elimination pathways differ in the timing of the leaving group’s departure. In the unimolecular elimination, [E1], the leaving groups departs in the first step via a [DN] process. In the [E1cb] (conjugate base variant of the unimolecular elimination), the leaving group departs in the second step via a [Eβ]. In the [E1cb] mechanism, a strong base must be present, the leaving group will generally be poor, and the beta hydrogen must be acidic. In the bimolecular elimination pathway, [E2], the leaving group leaves at the same time as the beta hydrogen is loss. This is a one-step, concerted process with [pt] + [DN] happening in unison.

H

L

CH

L

H

L

B:δ−

δ−

pathwayHL

B B–H + L

B B–H L

β$α$

[E2] is a Concerted Elimination Pathway All bond breaking happens simultaneously with pi-bond making. The ___ pathway ( ) Elimination, Bimolecular [E2]

β α

δ�

δ�

The TS≠ for the rate determining step involves the association of ___ molecules (bimolecular).

2

Many frontier orbitals are involved in the [E2] pathway. There’s an n → σ� σ-type interaction between ethoxide and the beta hydrogen. There’s a σ → σ� π-type interaction between the beta C–H and the C–X bond. There’s a σ → a no-bond interaction leading to leaving group dissociation.

http://www.chemtube3d.com/index.html

Regiochemistry of [E2] Elimination Reaction

major minor

β'α' α'

[E2]: Double bond character is _______________ in the TS≠ so the relative stability of possible C=C determines the major product.

highly developed

δ�

δ�

δ�

δ�

Product Ratios for [E2] Elimination Reactions Generally Follow Stability of Developing Double Bond in TS≠

Label the energy parameter that determines the product ratio

[E1] Is a Two-Step Elimination Pathway Bond breaking precedes bond making. The ___ pathway ( ) Elimination, Unimolecular

[E1]

β"

α"

The [SN1] and [E1] mechanisms proceed through a ________ intermediate common

http://www.chemtube3d.com/Elimination%20-%20E1.html

Regiochemistry of [E1] Elimination Reactions

The major product is the more substituted alkene. Why?

As before, this has to do with the stability of the developing double bond in the TS≠

ΔΔG≠

δ+

δ+

δ+

δ+

Consider the Developing Double Bond Character in Each TS≠

[E2] Regiochemistry Exceptions (When the Less Substituted Product is Major)

 A bulky base is used •  The leaving group is ___ •  A conjugated double bond can form

–F

major

major


•  A bulky base is used  The leaving group is ___ •  A conjugated double bond can form

–F

When the leaving group is a poor one (e.g., -F), deprotonation precedes C-L bond breaking and negative charge begins to accumulate on carbon. In these cases, the reaction mechanism resembles the [E1cb] pathway. These reactions follow the pathway of the most stable carbanion (carbanion stability increases as the number of C-H increases).


•  A bulky base is used •  The leaving group is ___  A conjugated double bond can form

–F

major

major

http://www.chemtube3d.com/Elimination%20-%20E2.html

Summary

E2 Promoting Factors

check the boxes corresponding to the conditions that favor the E2 mechanism

strong base (CH3O–) weak base (CH3OH)� �

CH3CH2 XCH3

HCX

CH3 CH3 CX

CH3

CH3

� � �small base (HO–) large (bulky) base ((CH3)3CO–)

� �no heatheat (�)

� �

Br

1.  C(sp3)–X electrophiles can undergo β-elimination reactions as well as substitution. The β-carbon is the carbon adjacent to the electrophilic C atom. Electrons in the Cβ–H bond form a π bond with the electrophilic C atom. "

2.  Under basic conditions, elimination proceeds by the [E2] mechanism. Under neutral or acidic conditions, elimination proceeds by the [E1] mechanism."

3.  Like the [SN2] mechanism, the [E2] mechanism is a concerted process. A reaction is said to be concerted if bond breaking and bond making occur simultaneously.

http://wps.prenhall.com/wps/media/objects/725/742549/0044f.html

Stereochemistry of the [SN2] Pathway

The SN2 reactions proceeds with _______ of configuration. Stereospecific or Stereoselective?

inversion

Stereospecific Reaction: A reaction in which the stereochemistry of the reactant completely determines the stereochemistry of the product without any other option.

Stereoselective Reaction: A reaction in which there is a choice of pathway, but the product stereoisomer is formed preferentially because its reaction pathway is more favorable than the others available.

Stereochemistry of the [SN1] Pathway

The nucleophile will attack both faces of the planar carbocation with equal probability resulting in a racemic mixture.

planar carbocation

(achiral)

Starting with a single enantiomer


The [E2] reaction pathway has very specific requirements: There must be ________________ among the orbitals in which σ bonds are made/broken and π bonds are made.

continuous overlap

Stereochemistry of the [E2] Pathway

Conformational Requirements of [E2] Elimination

(1) –X and –H oriented anti-periplanar

(2) –X and –H oriented syn-periplanar

For steric reasons, the preferred pathway for [E2] elimination is through (1). The double bond geometry depends on which pathway is followed. Consider the following example:

The carbon atoms smoothly rehybridize from sp3 to sp2

http://www.chemtube3d.com/EliminationE2stereospecific.html

Here’s another example:

[E2] Pathway in Cyclic Compounds

In cyclic compounds, anti-periplanarity demands that the groups to be eliminated occupy _____ positions. axial

This requirement has important consequences on rates and regiochemistry of elimination reactions involving substituted cyclohexanes

[E2] Elimination in Substituted Cyclohexanes Explain the reason for the difference in relative rates and regiochemistry in the following examples:

[E2] Reactions in Substituted Cyclohexanes: Example A

Two anti-periplanar pathways are possible from the most stable chair conformation. The observed product follows the pathway leading to the more substituted alkene

http://www.chemtube3d.com/EliminationE2regioselectiveA.html

[E2] in Substituted Cyclohexanes: Example B

An anti-periplanar arrangement exists only in an unstable conformation (slow rxn rate) Only one anti-periplanar pathway exists (regiochemistry)

No anti-periplanar

relationship exists ring-flip

http://www.chemtube3d.com/EliminationE2regioselectiveB.html

[E2] Pathways in Acyclic Compounds

The major product eliminates via a pathway with the bulky groups on opposite sides.

The carbon atoms smoothly rehybridize from sp3 to sp2

trans

cis

Which -H is this?


Stereochemistry and the [E1] Pathway There is no anti-periplanar requirement for the [E1] pathway. The carbocation intermediate undergoes rotation about the single bond.

Summary • Orbital alignment requirements in the [SN2] and [E2] pathways have

stereochemical consequences. • The → * sigma-type interaction of the [SN2] pathway results in

inversion of configuration at carbon. • In the [E2] pathway, the C –H bond and the C –X bonds break

simultaneously; these bonds must be parallel to one another to allow the orbitals involved to be properly aligned.

⇒ Acyclic compounds can orient the two bonds in two ways: synperiplanar (eclipsing at a 0˚ dihedral angle) or antiperiplanar (staggering at a 180˚ angle). The antiperiplanar conformation is generally lower in energy and the reaction usually proceeds through this conformation.

⇒ In cyclic compounds, there are much greater restrictions on conformational flexibility. In six-membered rings, the antiperiplanar requirement for [E2] elimination is satisfied when both the leaving group and the adjacent H atom are axial.

• In principle, any lone pair can act as either a base or a nucleophile toward C(sp3)–X. Thus, it is sometimes difficult to decide if a reaction will follow the [E2] or [SN2] pathway. The competition between substitution vs. elimination depends on two factors: (1) the nucleophilicity vs. basicity of the lone-pair and (2) the identity of the alkyl halide substrate (1˚, 2˚, 3˚). The table summarizes the behavior.

Factors determining substitution vs. elimination under basic conditions Alkyl halide good nucleophile poor base good nucleophile good base poor nucleophile good base

1˚ SN2 SN2 E2 2˚ SN2 SN2 < E2 E2 3˚ E2 or no reaction E2 E2

Examples of good nucleophiles and poor bases: Br–, I–, RS– Examples of good nucleophiles and good bases: RO–, R3N, RC≡C– Example of poor nucleophile and good base: t-BuO–

Carbocation Intermediates Carbocations are common intermediates in organic chemistry. We have encountered carbocations in [SN1] substitution and [E1] elimination. Although carbocations are never present in very high concentrations (because they are unstable and highly reactive), they do open up a reaction pathway through which important chemistry ensues.

C

•6-electron species•trigonal planar geometry•sp2 hybridized•empty p orbital on carbon

carbocation

CHHH

methyl cation

bonds directly attached to the carbocation do not share their electron density

CH2CHH

ethyl cation

bonds directly attached to a carbocation substituent do share their electron density

H

least stable carbocation stabilized by hyperconjugation

Carbocations have a carbon-centered empty p-orbital

Stability of Carbocation Intermediates Certain carbocations are more easily formed than others. Only if the carbocation can be formed easily does it open a viable pathway. We thus need to understand how carbocation stability relates to structure, and how this structure relates to reactivity. A simple rule predicts stability for carbocations: the more nonhydrogen substituents attached to a carbocation, the greater is its stability.

Increasing stability

π-type σ → a Resonance Interaction Stabilizes Carbocations

http://www.chemtube3d.com/index.html

Why are substituted carbocations more stable than unsubstituted carbocations? The carbocation is a 6-electron species; any atom (or bond) that supplies this electron deficient carbon with electron density increases its stability. Electrons in bonds that are directly attached to the carbocation do not offer any help because the filled and empty orbitals are perpendicular and incapable of overlapping. However, electrons in bonds that are one atom away from the carbocation center are able to donate some of their electron density through a σ → a (π-type) interaction.

C

•6-electron species•trigonal planar geometry•sp2 hybridized•empty p orbital on carbon

carbocation

CHHH

methyl cation

bonds directly attached to the carbocation do not share their electron density

CH2CHH

ethyl cation

bonds directly attached to a carbocation substituent do share their electron density

H

least stable carbocation stabilized by hyperconjugation

The σ→a filled→empty orbital interaction of hyperconjugation

Carbocation Intermediates Undergo Rearrangement Reactions via 1,2-Shifts

We have encountered carbocations in [SN1] substitution and [E1] elimination. Any reaction that proceeds through carbocation intermediates may undergo a type of rearrangement reaction known as the 1,2-shift (also called 1,2-rearrangement). Recall from the seven classes of reactions that a rearrangement is a reaction with only a change in bonding connectivity. Typical groups undergoing 1,2-shifts are hydrogen (hydride shift), methyl (methyl shift), and C-C bonds in rings (ring expansion). These reactions are driven by formation of a more stable carbocation. The reactions may be reversible. It is possible for hydride and methyl migrations to take place over longer distances. However, when longer migrations do occur, they usually take place via a sequence of successive 1,2-shifts, not by a single-step, long-distance pathway.

HCH3CCH3

C CH

HCH3

HH3C C

CH3

C CH

HCH3

H

H

a secondary carbocation

a tertiary carbocation

HCH3CCH3

C CH

CH3

Ha secondary carbocation

H

1,2-shift 1,2-shift

a long-distance migration is the result of

successive 1,2-shifts

Carbocation Rearrangements Require a New Elementary Step: The 1,2-Rearrangement [1,2R]

Empty a

Filled σ � [1,2R] involves σ-bond breaking, σ-bond making

Curved Arrows Imply the Frontier Orbitals Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in the elementary step. The tail of the arrow implies the filled orbital (HOMO) is an electron pair in a σ-bond. The head of the arrow points to the electron deficient carbon. An empty orbital on the carbocation is available to accommodate this electron pair implying that the LUMO is an atom-centered, empty orbital (a). The HOMO (σ-orbital) is beside (and nearly parallel to) the LUMO, a, allowing for a π-type interaction (σ-type overlap is not possible due to bonding constraints).

[1,2R]

This symbol means a 1,2-rearrangement

[1,2R] involves a σ → a π-type interaction

-Y is the group undergoing the 1,2-shift -Y is typically -H, (hydride shift) or -CH3

σ$

n

π

σ� $a π�

σ→σ� $ σ→a σ→π�$

filled

empty

n→σ� $ n→a n→π�$

π→σ� $ π→a π→π�$

The 1,2-shift involves migration of a sigma bond (its electrons and the attached atoms to that bond) from a carbon that is adjacent to the carbocation onto the carbocation. It is illustrated for the generic group labeled “Y”. The 1,2-numbering reflects the fact that the migration takes place from atom labeled “1” to the adjacent atom labeled “2”.

CY

C C CY

1 2 1 2

An Example of Ring Expansion CCH3

CCH3

HCCH3

CCH3

H≡ C

C

CH3HCH3

a strained 4-membered ring less strained

5-membered ring

[1,2R]

The way it looks in ACE Organic…

Step 1

Step 2

Step 3 Step 4 Step 5

either way is fine

Rearrangements May Intervene in the [SN1] Pathway!

2˚ carbocation 3˚ benzylic carbocation

Encountering New Chemical Reactions

The steps of writing reaction mechanisms • make sure the overall stoichiometery is balanced • draw implicit hydrogen atoms wherever they might be important (e.g., on carbon atoms of reactants

and products near to where the bond changes takes place) • list the bonds made and broken • identify nucleophiles, bases, electrophiles and acids; repeat for each step of the mechanism • draw curved arrows to show the flow of electrons • make sure charge is balanced for each step of the mechanism • find a reasonable pathway that leads to the products • look for hidden reactivity by drawing resonance contributors for each intermediate • work backwards if you get stuck

Write mechanisms for the following reactions. Use clear, carefully drawn structures and precise curved arrow notation for all steps. Provide distinct structures for all intermediates. Show resonance contributors and electron lone pairs where necessary.!

CH3

Cl

CH3H3CCH3

CH3

CH3

H3C CH3Br

CH3

CH3

CH3OH, Δ

CH3OH, Δ

Reactions at sp3 Carbon: Leaving Groups Other Than Halogen

In this lesson we will reason by analogy to predict the outcome of new reactions (e.g., predictions of leaving group ability)

Breaking C–O bonds: –OH is a poor leaving group and does NOT directly participate in [SN1], [SN2], [E1] or [E2] reactions.

Why is –OH a poor leaving group?

Why –OH is a Poor LG

-10.0

-9.0

15.7

Protonation of an Alcohol Improves the Leaving Group Ability

___ _____________ ____________ LG Conjugate Acid pKa of

Conjugate Acid

H2O H3O+ -1.7

Protonated Alcohol as a Leaving Group

Reactions of Protonated Ethers [SN2] or [SN1] pathway depending on: •  steric accessibility of nucleophile •  stability of carbocation

If a stable carbocation can form, the reaction will follow the [SN1] pathway (otherwise [SN2])

[SN2]

[SN1]

Examples - Nucleophilic Substitution

Examples - Water Elimination (dehydration)

Activated Alcohols Based on PBr3

PBr3 is an electrophile δ�

δ� δ�

δ+

Nucleophilic attack on phosphorus by oxygen

–OH, a poor leaving group is converted into this good leaving group

Activated Alcohols Based on SOCl2

Nucleophilic attack on sulfur by oxygen

δ�

δ� δ� δ+

Activated Alcohols Based on Sulfonate Esters

sulfonyl chloride

sulfonate ester

-1.0

Common Sulfonyl Chlorides

δ+ δ�

(tosyl abbreviated Ts)

(mesyl abbreviated Ms)

(triflyl abbreviated Tf)

Ethers and Epoxides

Like alcohols, ethers are poor leaving groups. They can be activated by acids or _______ (epoxides are strain-activated ethers)

a 3-membered cyclic ether

strain

pKa = ? -3.6

Protonation

Summary: Converting Alcohols,

Ethers and Amines

into Leaving Groups

CH3CH2O HBrPBr

Br CH3CH2O PBr

CH3O HOS ClCl CH3O S

Cl

CH3CH2CH2O H

S OOCl

OS

C N

OH H2SO4

O SO

OOHH O

HH

OHOP ClClCl

O PCl

ClO

Leaving Groups

+a bromophosphite

+a chlorosulfite

+a tosylate

+

+

CH3 O CCH3

CH3

CH3

H+ CH3 O CCH3

CH3

CH3H

CH3CH2CH2NH2 CH3–I+ K2CO3 NCCH3

CH3

CH3

CH

HCH3

H

H Δ

http://wps.prenhall.com/wps/media/objects/725/742646/0048f.html!Complete the diagrams (use the link for help)

Energetics of Epoxide Reactions

Epoxides react faster because their ground state is ___________ relative to ethers (by about ___ kcal·mol-1)

destabilized 25

25

ΔG≠

ΔG≠

ΔΔG≠ ~ 25 kcal·mol-1

Epoxides are more reactive than ethers because the strain is released when the ring opens

Acid-Catalyzed Ring Opening ([pt] then [SN2])

δ+

δ+ δ+

•  [SN2]-like in that nucleophilic attack follows backside displacement •  [SN1]-like in that the regiochemistry places the nucleophile on the

more substituted carbon

δ+

[pt]

[SN2]

Regiochemistry Rationalization

Because there is some ____________ character in the TS≠, attack occurs at the carbon better able to bear a partial positive charge.

carbocation

protonated intermediate

2nd best resonance contributor

Where’s the Site of Electrophilicity? Two Pathways NOT Followed

Option 1 [AN] at O+

Option 2 [SN2] at O+

[SN2] at carbon (NOT oxygen) is most reasonable. Why? On which atom of C–O+ is σ* largest? Draw an MO diagram to illustrate these differences.

NuCH3

H

O

CH3

H

O NuHH

oxygen exceeds octet

NuCH3

H

O H

CH3

H

O NuH

charges are not reasonable

the nucleophile does NOT attack the positively charged oxygen via [AN] as there is no empty orbital centered on this atom."

Ring Opening Under Nucleophilic (Basic) Conditions

[SN2] backside attack at less hindered carbon

Many Nucleophiles React with Epoxides

Summary of Epoxide Reactivity

Ring Opening of Epoxides

O H2SO4 (cat.)CH3OH

OHOCH3

OCH3

OH

OH

2˚ 1˚

OO

H–O–SO3H

H–OCH3

acidic conditions

basic conditions

O CH3ONaCH3OH

OHOCH3

OCH3

OH

OO

H–O–CH3

product

Complete the diagrams (use the link for help) http://wps.prenhall.com/wps/media/objects/725/742646/AADFQQH0.html

Carbon-Based Nucleophiles

A special class of carbon-centered nucleophiles are _____________ compounds organometallic

δ�" δ+"M is less electronegative than C

The carbon acts as an _______ nucleophile anionic

< < <

Increasing stability

These organometallic compounds are strongly basic

Preparation and Reactions of Carbon Nucleophiles

Organolithium

Grignard (Organomagnesium)

New C–C bond

Summary 1) Here’s an important general rule to be followed in writing all reaction mechanisms. If a reaction is executed

under acidic conditions, no strong bases can be present. Any negatively charged species must be a weak base (e.g., Cl–). If a reaction is executed under basic conditions, no strong acids can be present. Any positively charged species must be a weak acid.

2) Under basic conditions, a hydroxyl group cannot be substituted directly because HO– is too poor of a leaving group. The hydroxyl group must first be transformed into a derivative that is a suitable leaving group.

3) Under acidic conditions, the R3C–X+ bond can ionize spontaneously to give X: and R3C+. The likelihood of ionization depends on the stability of the carbocation product, the leaving group and the solvent.

4) Under acidic conditions, a lone pair nucleophile bearing a proton (e.g., H2O, ROH, RCO2H, RNH2) is always deprotonated after it adds to the electrophile (e.g., carbocation).

5) Under acidic conditions, epoxides open by attack of the nucleophile onto the protonated ring. The nucleophile attacks the most highly substituted carbon.

6) Under basic conditions, epoxides open by nucleophilic attack onto the least substituted carbon (the reaction resembles an SN2 reaction; the leaving group is a poor one, but the process is driven by the release of ring strain).

7) If carbocations are intermediates, rearrangement is possible.

Electrophilic Addition Reactions - Adding XY Across Unsaturated Carbon-Carbon Bonds




X

C

Y

CY + X+


Yreplaces

X


reductionX

C

H

Coxidation

Hreplaces

X

Xreplaces

H


X

CZ

Z

YC + XY

elimination

addition

form π bond-XY

break π bond+XY

base

H

X

H

YY + X+

baseacid acid


RCH3


Z = C, N, O

This page summarizes the main classes of chemical transformations. Having studied the process of elimination, we are now set to examine the reverse reaction - addition of the elements “XY” across carbon-carbon multiple bonds.

Examples of Addition Reactions to Carbon-Carbon Double Bonds

CHCH3H2C + H2O HH

H2CHC CH3

OH

Br BrCH2Cl2

+BrBr

a non-nucleophilic solvent (i.e., inert)

cyclopentene

HH a vicinal dibromide (or vic-dibromide for short)

Cl ClH2O

+OHCl

a nucleophilic solvent participates in the reaction

cyclopentene

HH + HCl

a chlorohydrin

CH3

+ H–BrCH3

HHBrH

2121

Addition Mechanisms Require a New Elementary Step: Association of an Electrophile with a π-bond [AE]

Empty a

Filled π

[AE] involves π-bond breaking and

σ-bond making

σ#

n

π

σ� #a π�

σ→σ� # σ→a σ→π�#

filled

empty

n→σ� # n→a n→π�#

π→σ� # π→a π→π�#


Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in an elementary step. For [AE] the tail of the arrow implies the filled orbital (HOMO) is an electron pair in a pi-bond. The head of the arrow points between one carbon atom of the π-bond and the E+ suggesting that an atom-centered empty orbital (LUMO, a) accepts the electron pair with σ-type orbital interaction (note: the HOMO for [AE] will always be π but the LUMO will vary according to the specific E+ involved).

[AE]

This symbol means an electrophile association with a π-bond

[AE] involves a π → a σ-type interaction

E+ is the electrophile

The Electrophilic Pathways for C=C Addition (two variations of [AdE2])

When E+ has no lone pair (e.g., H+)

When :E+ has a lone pair (e.g., :Br+)

Two-step mechanism: [AdE2] = [AE] then [AN]

Two step mechanism: [AdE2] = [AE] then [SN2]

Depending on whether or not the electrophile has an electron lone pair will determine the nature of the intermediate resulting from the [AE] step. If the electrophile has a lone pair, then a 3-membered intermediate will form in order to satisfy the octet rule. If no lone pair exists on E+, then [AE] results in a carbocation.

C C

E

[AE]C C

E [AN]C

E

NuCNu

C C

E

[AE] [SN2]C

E

NuCNu

EC C

Addition of H-Br When the Electrophile Has No Lone Pair,

[AE] Gives a Carbocation Intermediate


C CH

H H

HH Br+ C C

H

HH

H

HBr

C CHHH

HHBr

intermediate

[AE] [AN]

[AE] Drawn in ACE

Clean in 2D

Draw the 1st arrow

After the 2nd arrow

Check the 1st arrow

[AN] Drawn in ACE

Draw the arrow

Check the arrow

http://aceorganic.pearsoncmg.com/epoch-plugin/public/mechmarvin.jsp

Reaction Coordinate Diagram for the [AdE2] of HBr to Ethylene

free

ener

gy


H2C CH2+

HBr CH3 CH2Br

C CHHH

HH Brintermediate

C CHHH

HH

Brδ−

δ+

C C HHH

HHBr

δ+

δ−

Unsymmetrical Alkenes CH3 CH3

H HBr

CH3

H BrH

+HBr

major minoran unsymmetrical

alkene

this carbon bears no hydrogen atoms

this carbon bears one hydrogen atom

The first step of the mechanism involves the formation of a carbocation intermediate. Because the alkene is unsymmetrical, there are two possible carbocations that can form. The relative stability of these carbocation intermediates (technically, the transition state leading to these intermediates) determines which product is formed the fastest. Hammond’s postulate allows us to predict the relative energy of the TS≠s leading to the two possible intermediates. Draw a representation of the transition state structures for both modes of reactivity. Your representations should show that one begins to develop character of a secondary carbocation, while the other develops character of a tertiary carbocation. The TS≠ leading to the secondary carbocation is a higher energy species comparied to the TS≠ leading to the tertiary carbocation intermediate. Thus, the secondary carbocation pathway represents a higher barrier to reaction. In contrast, the TS≠ leading to the tertiary carbocation is a lower energy species, and thus represents a lower barrier to the reaction pathway. The reaction coordinate diagram illustrates this logic and helps explain why the major observed product results from attack of bromide on a tertiary carbocation. Start from the reactants. Notice the pair of diverging pathways, one leading to the major product and the other leading to the minor product.

Diverging Reaction Pathways


The relative energies of these TS≠ can be determined because: 1)  Both steps are examples of a “late” TS≠ 2)  The structure and energy of each TS≠ will track according to

the carbocation intermediate to which it is en route

Acid-Catalyzed Addition of Water or Alcohols to Alkenes

Notice first, that the stoichiometry is balanced. Water is probably the solvent of this reaction, and in this case, it participates in the reaction, too. Such behavior where the solvent gets incorporated into the product is fairly common in organic chemistry. In contrast to water, the H+ above the equilibrium arrow does not enter into the stoichiometry. However, H+ does play an important role. If you were to combine water and an alkene without any acid, no change will take place for a long time, at least not at lower temperatures. These observations point to the notion that the reaction is catalyzed by H+, meaning that acid enhances the rate of the reaction, but does not show up in the stoichiometry. So, when you see H+ or HO– listed above the arrow, but not enter the stoichiometry, think about acid or base catalyzed reactions. Complete the mechanism, reasoning by analogy to the [AdE2] mechanism for H-Br addition to an alkene.

C CH

CH3 H

H CH3 CH3

OHHH2O +

HH

H2O

step 1

step 2

step 3

Step 1 combines neutral reactants and H+. The charge at the end of step 1 must be +1.

Step 2 combines the positively charged intermediate from step 1 and neutral H2O. The 2nd intermediate must also have a +1 charge.

Balance charge at every step

[AE] [AN]

[pt]

Rationalizing Br2 and Cl2 As Electrophiles Reagents such as Cl2, Br2, I2, and ICl are electrophiles that add to alkenes. We will see that these reactions follow the [AdE2] mechanism with a slight variation, as a consequence of the lone pair on these electrophiles. First, how can these reagents be rationalized as electrophiles? The answer is that they have weak bonds and low lying empty σ* orbitals. The resonance picture below may help build your intuition about structure and reactivity. Very often the second best resonance contributor provides insight about reactivity.

Addition of Br2 to an Alkene When the Electrophile Possesses a Lone Pair,

[AE] Gives a “Halonium Ion” Intermediate

[AE]

CH2Cl2+

BrBrBr Br

BrBr

BrBr

HH

This carbocation does not satisfy the octet rule, yet there is a way to do so.

Br

H HBr

This bromonium ion satisfies the octet rule. It's true that it is highly strained and has a positive charge on an electronegative atom; nonetheless, this is a better structure than an electron deficient carbocation.

[AN]

[SN2] Here bromonium ion formation is written as [AE] then [AN]

CH2Cl2 (DCM) a non-nucleophilic solvent

[AE] of Br2 to a Carbon-Carbon π- Bond It is possible to proceed directly to the bromonium ion from Br2 and alkene by combining [AE] and [AN] into one step. This reflects the fact that the carbocation may not be an intermediate on the MEP for this reaction. The curved arrow notation might seem complex, but shows all of the electron reconfiguration that accompanies this process. The overall process is [AE], an association of the electrophile, Br2, with C=C π-bond. The curved arrows imply a pair of filled-empty (frontier) orbitals for this step: •  π → σ* •  n → π* Expressed in words, we’ve combined the alkene’s HOMO with Br2’s LUMO (π → σ*), and we’ve combined Br2’s HOMO with the alkene’s LUMO (n → π*).

+ Br Br

Br

H H

curved arrow notation for direct formation of bromonium ion

Br

[AE]

What it looks like in ACE Organic…

Clean in 2D

Draw the 1st arrow

After the 2nd arrow

Draw the 3rd arrow

After the 3rd arrow


n → π*

π → σ*

A Bromonium Ion Behaves Much Like a Protonated Epoxide

Br

H H

the bromide nucleophile does NOT attack the positively charged Br

BrH H

BrBr

[AN] at Br+ is not reasonable because there’s no low lying empty atom-centered orbital. [SN2] at carbon (NOT Br+) is most reasonable.

Reason by analogy

Br

H HBr

Br

H HBr

the electrophilic sites of the bromonium ion are revealed by the second best resonance contributors

Br

H HBr

best contributorsecond best contributor second best contributor

Addition Reactions to Alkynes Alkynes contain carbon-carbon triple bonds. The triple bond is an electron dense region and it should therefore not be surprising that it has good nucleophilic characteristics. In studying the reactions of alkynes, we will reason-through new transformations by drawing analogies to the mechanisms and concepts that we have previously learned. The success of applying established ways to a new class of compounds illustrates the power of the mechanism-based approach for studying reactivity.

CH3CH2C CH + H–Br CH3CH2C CBr

a vinylic cation intermediate

CH3CH2C CBr

H

H

H

H

[AdE2]

[AE] [AN]


The Vinylic Cation

CH3CH2C CH2 CH3CH2CH CHmore stable less stable

A two-EPD carbocation (sp-hybridized)

The vinylic cation possesses a positively charged carbon atom having only two electron pair domains (EPDs); the carbon bearing the positive charge is thus sp-hybridized. The addition of the electrophilic H+ takes place so as to place the positive charge on the most highly substituted carbon, just as for alkenes. As before, the more highly substituted the vinylic cation, the greater its stability. This preference determines the regioselectivity of the addition. As a consequence, the bromide is located on the more highly substituted carbon (i.e, the internal carbon). It turns out that the vinyl cation is much less stable (i.e., much more reactive) than the corresponding alkyl cation. One reason is because sp hybrid orbitals, consisting of 50% s character, are less well-suited to stabilize positive charge compared to sp2 hybrid orbitals that have only 33% s character (the s orbital resides closer to the positively charged nucleus).

R C HH

RCH C–H

primary alkyl cation primary vinylic cation

3-EPDs(sp2 hybridized)

2-EPDs(sp hybridized)

Relative Stability of Various Carbocations

The order of stability for various carbocations is provided in the chart below. It can be seen that the secondary vinyl cation is not even as stable as the primary alkyl cation.

Successive Addition Since the addition product to an alkyne is a vinyl bromide – a type of alkene – it should not be surprising that subsequent reaction involving further addition is possible. In fact, successive addition tends to be the normal mode of reactivity for alkynes, as addition to the alkene is generally faster than the initial alkyne addition.

+ H–BrCH3CH2C CBr

H

HCH3CH2C C

Br

H

HH

Br

a vinyl bromide (an alkene)

a geminal dibromide

One reason why the subsequent reaction is faster is because of the special stability of the carbocation intermediate that results when an electrophile adds to a vinyl bromide. The lone pairs on the bromo group can donate to the electron deficient carbocation. Although this places a positive charge on the electronegative bromo group, all atoms have an octet of electrons making this a very important resonance contributor.

CH3CH2C CBr

H

HH CH3CH2C C

Br

H

HH

the most important resonance contributor (each atom has an

octet of electrons)

n → a π-type resonance interaction

Addition of Br2 to Alkynes

Halogens add to alkenes in a way that is similar to alkenes. An analogous bromonium ion is formed in this case as well, and the subsequent attack of bromide anion results in trans addition. However, this initially formed product usually undergoes a second addition to give a tetrabromoalkane.

Br2 + CH3CH2C CCH3

trans alkene

Br

CH3CH2 Br

CH3

a second additionusually occurs

CH3CH2C CCH3

Br

Br Br

Br

a tetrabromide

Br

CH3CH2 CH3

Bra bromonium ion

intermediate

The Addition of Water to Alkynes

CH3H3C21

H3CCH3

O

H H

21

Bonds Made Bonds Broken

C1=OC2-H

C1≡C2

H-OC2-H H-O

+ H–O–H H2SO4

The complete mechanism is outlined in a “discussion problem”. Try to work through this on your own.

The addition of water to an alkyne initially results in an enol. The enol, whose name is derived from alkene + alcohol, is generally not the form that is most stable. Enols tend to undergo a process known as tautomerization, an equilibrium reaction which shifts the position of a double bond together with a C-H, N-H or O-H bond (in this case, an O-H). A bond energy calculation will show why the equilibrium favors the keto form. What we are after right now is not the knowledge that this reaction is a tautomerization, but to understand the mechanism of how tautomerization occurs.

The acid-catalyzed tautomerization mechanism is included with the “discussion problem”. You can fully understand this mechanism using elementary steps that you previously learned.

CHCH3H3C

OH HH3C

CH3

O

HHa ketonean enol

tautomerization

1 1 2 2

Acid-Catalyzed Tautomerization Mechanism

To begin, it should not be surprising that the enol is a nucleophilic π-bond capable of undergoing [AE]. This π-bond nucleophile finds an electrophilic partner in H+ (from H2SO4). The new C-H bond must be made at carbon atom C2, consistent with the need for a new C2-H bond. The resulting carbocation is stabilized by resonance with the lone pair of oxygen (Box 4). This resonance contributor has an octet of electrons on every atom. Moreover, this resonance contributor has the necessary C=O bonding of the final product. All that remains to realize the final product is dissociation of a proton, breaking the final O-H bond in our bonds made / bonds broken list. Notice how H2SO4 is regenerated each time it is used, consistent with its use as a catalyst.

Addition of Hydrogen to Alkenes and Alkynes The addition of hydrogen to an organic compound is called hydrogenation. An example is shown. Hydrogenation is an addition reaction and it is also a reduction reaction (there is an increase in the number of C-H bonds without an increase in C-X bonds).

H Hhydrogen

oleic acid (an unsaturated fatty acid)OH

O

H H

OH

O

H HH H

steric acid (an saturated fatty acid)

The addition of H2 occurs only in the presence of a metal catalyst (Pd, Pt or Ni) that is adsorbed onto a finely divided inert solid, such as charcoal. The catalyst 10% Pd on carbon is common. The reaction takes place at the metal’s surface but we won’t study the details.

Pd-C

Alkynes Usually Undergo Successive Hydrogen Addition

CH3CH2C CH H2Pt / C CH3CH2CH C

H

H

H2Pt / C CH3CH2CH2 C

H

HH

CH3CH2C CCH3H2

Lindlar'scatalyst CH3CH2

H H

CH3a cis alkene

Hydrogen adds to alkynes in an analogous way as with alkenes. The initially formed alkene generally continues on to the saturated alkane under the reaction conditions.

The addition of H2 to triple bonds is somewhat faster than to alkenes; thus, it is possible to stop the reaction at the alkene using a type of “poisoned” or deactivated catalyst known as Lindlar’s catalyst. Lindlar’s catalyst is stereoselective for the formation of the cis alkene.


The Alkyne’s Terminal Hydrogen Is Acidic C C CH3CH3 C C HCH3

an internal alkyne a terminal alkynethe triple bond ends in a hydrogen atom

Compared to other C-H bonds, the terminal hydrogen of a terminal alkyne is acidic. Yes, it is a weak acid, but comparatively speaking, it is acidic. Deprotonation results in a carbon-centered anion that is sp hybridized. This anion is called an acetylide anion. As stated earlier, sp hybrid orbitals are more effective at stabilizing negative charge than sp2 or sp3 orbitals. It is the stability of the anion that makes the C-H of a terminal alkyne more acidic than other C-H bonds.

C C HCH3 C CCH3H +

compared to other C–H bonds this one is acidic

pKa = 25

CH2 CH

H

pKa = 44

H + CH2

H

an acetylide anion

a vinylic anion

Homolytic vs. Heterolytic Fragmentation Most organic transformations involve the movement of electron pairs (heterolytic reactions). There are a few important addition reactions, however, in which the electron reconfiguration involves the movement of single electrons. Whereas heterolytic bond cleavage leads to ion pairs, homolytic bond cleavage results in unpaired electrons – or free radicals. Some weak bonds have a tendency to fragment homolytically (e.g., peroxides, halogens). Chemists use a slight variation of curved arrow notation to show the movement of single electrons. For single electron movement, “fishhook” arrows, i.e., single headed arrows are drawn. The difference in these two notations is illustrated below for heterolytic vs. homolytic fragmentation of Br2.

How it looks in ACE…

http://aceorganic.pearsoncmg.com/epoch-plugin/public/mechmarvin.jsp

The “fishhook” arrow tool.

Comparing Homolytic and Heterolytic Addition When peroxides are heated, they fragment homolytically. This behavior can be used to initiate addition reactions that proceed via radical intermediates. A consequence of the different reaction mechanism is that the regiochemistry of addition varies from that seen for the heterolytic case, as the comparisons below illustrate.

Stability of Carbon-Centered Radicals In order to rationalize the reactivity trends seen above, we must first understand how radical stability depends on structure. The radical is a half-occupied electron pair domain (“open shell”). The geometry at carbon is approximately planar, and the single electron can be assumed to occupy the p-orbital of an sp2-hybridized atom. Since the carbon atom has 7 electrons, one short of the desired octet, it is electron deficient, and stability as a function of substitution follows the same trend found for carbocations.

The Radical Chain Mechanism A chain mechanism has repeating characteristics. Species produced at the end reenter the mechanism to cause it to start all over. Radical reactions are often of this type. Chain mechanisms have three events in common: initiation events, propagation events and termination events. For radical chain mechanisms, initiation events are what create the radicals. Propagation events are what produce products and these are the steps that repeat in what are known as cycles, (like the links of a chain, hence the name chain mechanism). A single radical from an initiation event may yield several cycles of propagation. Since each cycle produces product, a single radical can account for several equivalents of product. The termination events consume radicals and shut down the propagation cycles.

HBr +

O O

Brheatn-butyl bromide

tert-Butyl peroxide

The thermally induced bond homolysis of peroxides is a common way to initiate radical reactions. The oxygen-oxygen bond of peroxides is a weak covalent bond of 47 kcal/mol and hence is readily susceptible to thermal dissociation. The result is a pair of oxygen-centered radicals.

Initiation

Thermal homolysis of peroxides results in a pair of oxygen-centered radicals. The next step in initiation involves HBr. A newly formed oxygen-centered radical abstracts a hydrogen atom from H-Br resulting in a bromine-centered radical and a hydroxyl group. The bromine-centered radical then enters the propagation cycle.

RO OR RO2a peroxide

RO H Br RO H Br+

Propagation The first step of the propagation cycle involves the addition of the bromine-centered radical to an alkene. This step creates a carbon-centered radical, and it determines the isomer that is formed (i.e., n-butyl vs. sec-butyl bromide). The formation of the most stable carbon-centered radical is what governs the outcome of this crucial step. By having the bromine-centered radical add to the terminal position of the alkene, a secondary carbon-centered radical results. This pathway is lower energy than the addition that leads to a primary radical.

The second half of propagation involves the carbon-centered radical abstracting a hydrogen atom from another molecule of HBr. This regenerates a bromine-centered radical along with the alkene addition product. The bromine-centered radical is ready to reenter a new chain propagation cycle.

Br+CH3 BrH

H Br CH3 BrHH

Termination Termination events are the ways that radicals annihilate themselves. All propagation steps involve a radical combining with a non-radical species. Through combination of a pair of radicals, the radicals will disappear. Because the concentration of radicals tends to be very low, these events are relatively rare. Some examples of termination reactions are shown below. All of them involve the combination of two radicals, and they produce non-radical products.

CH3 BrH

Br BrBr Br

Br CH3 BrHBr

BrBr

CH3 BrH

CH3 BrH

Stereoisomers and Addition Reactions Many addition reactions have important stereochemical consequences and as we have previously seen, new details about a reaction mechanism will be unveiled when we examine the stereochemical relationship between reactants and products. For example, the hydrogenation to alkenes normally results in both hydrogen atoms adding to the same side, or face of the planar double bond. Same-face addition of atoms to a double bond is known as syn addition. An example of syn addition is illustrated for the hydrogenation of cis-2,3-dideuterio-2-pentene (deuterium, D is the nuclear isotope of hydrogen with an atomic mass of 2). Syn addition can take place from either the bottom or top. Because both top and bottom face addition are equally possible, an equimolar mixture of an enantiomer pair results. An equamolar mixture of enantiomers is known as a racemic mixture. Note that different products result from the trans alkene.

Anti vs. Syn Stereoselectivity Although hydrogenation generally follows syn addition, special conditions exist that favor anti addition. In anti addition, one hydrogen atom adds to one face while the other adds to the other. The hydrogenation conditions shown in the reaction below exhibit anti stereoselectivity. You should draw the implied hydrogen atoms in the two products, showing their proper stereochemial orientation. The “anti” and “syn” products are diastereomers. The enantiomer of the anti product is not shown but is produced with equal preference to the anti isomer that is shown (i.e., of the 73% anti product, half of this amount is one enantiomer and half is the other). The syn product is a meso molecule. Can you find the plane of symmetry?

CH3

CH3

CH3

CH3anti

CH3

CH3syn

73% 27%

+H2Pd / Al2O3CH3CO2H

these specialized reaction conditions favor the anti

addition product

The specific set of conditions written above and below the reaction arrow describe to a chemist important details about how the reaction was performed. In this particular case, palladium metal was deposited on solid aluminum oxide and the reaction was conducted in a solvent of acetic acid (CH3CO2H) in the presence of gaseous hydrogen. Please note that it is NOT important for you to memorize these specialized conditions or even understand the mechanistic details of how these conditions produce anti products. What is important is that you have the ability to examine a reactant / product pair and determine the mode of addition (syn vs. anti).

Same Face Addition = Syn Opposite Face Addition = Anti

By comparing the stereochemistry of the products and reactant, it can be seen that the anti diastereomer arises from the addition of hydrogen atoms to opposite faces of the double bond, while the syn product results from same face addition.

Hydrogen atoms add to the opposite faces of the double bond

Hydrogen atoms add to the same face of the double bond

Halogens Add Stereospecifically to Alkenes A mechanism that constrains the stereochemical outcome of a reaction is said to be stereospecific. Such a close relationship between mechanism and stereochemistry means that it is often possible to gain valuable insight into reaction mechanisms by studying their stereochemical features. This kind of a relationship is found for the addition of halogens to alkenes. In fact, the evidence that a bromonium ion was involved in alkene addition came from studies of the reaction stereochemistry. Consider the reactions of trans and cis 2-pentene shown below. The product has two stereocenters so up to four different stereoisomers are possible (there are no meso molecules in this case). Notice, however, that trans and cis substrates each give rise to unique subsets of the possible stereoisomers. The mechanism must account for these stereochemical observations.


trans 2-penteneBr2, CH2Cl2

Br

Br Br

Br

+

enantiomeric pair

cis 2-penteneBr2, CH2Cl2

Br

Br Br

Br

+

enantiomeric pair

Rationalizing the Stereoisomers From Bromine Addition to Trans 2-Pentene

The addition of bromine to alkenes is a stereospecific reaction. In particular, the analysis below will show that this reaction is stereospecific for anti addition. The fact that the reaction follows anti addition provides important details of how the reaction mechanism must proceed. Trans 2-pentene gives rise to the bromonium ion shown below. The observed product is a mixture of enantiomers. These stereoisomers, being the only 2 of all the possible stereoisomeric products, can only come about if the Br– nucleophile always attacks from the opposite side of Br+, as expected for n → σ* interaction of an [SN2] step involving a bromonium ion intermediate. The different enantiomers arise from Br– attack at the two different carbon atoms. The anti relationship of the bromine atoms can be seen in the Newman projection.

[SN2]

[SN2]

Rationalizing the Stereoisomers From Bromine Addition to Cis-2-Pentene

bromonium ion from cis-2-pentene!

Bromine addition to cis-2-pentene produces a different set of enantiomers. However, the exact same mechanism discussed for the trans case also accounts for the stereoisomers produced in the cis case.

A Non-Stereospecific Example

CH3CH3CH2

CH3CH2 CH3

CH2CH3

CH2CH3

H3C Br

H3C H

CH2CH3

CH2CH3

Br CH3

H CH3

CH2CH3

CH2CH3

Br CH3

H3C H

CH2CH3

CH2CH3

H3C Br

H CH3

HBr

enantiomers

enantiomers

diastereomersdiastereomers diastereomers

cis-3,4-dimethylhex-3-ene

Reactions that proceed by non-stereospecific mechanisms will result in a much greater array of stereoisomers. Consider the reaction of HBr adding to an alkene (recall that the addition of HBr to an alkene proceeds via either a cationic or radical intermediate). As a specific case, the addition of HBr to cis-3,4-dimethylhex-3-ene results in all four possible stereoisomers as shown below. Those isomers related as enantiomers must be formed in equal amounts. Those isomers related as diastereomers may be present in unequal amounts. Think through the mechanism and decide if trans-3,4-dimethylhex-3-ene will produce the same stereoisomer mixture as cis (will the proportions of the various isomers be the same for cis vs. trans?).

Pathways Involving Carbocation and Radical Intermediates Are Not Stereospecific

The uniqueness of the bromonium ion intermediate in its ability to stereospecifically govern the stereochemical outcome of reactions is appreciated when we compare its behavior to that of pathyways that proceed via carbocation or radical intermediates. In these cases, there is generally equal probability that the nucleophile or radical will react on either side as shown. Thus, carbocation and radical intermediates are not-stereospecific and unless there is another stereocenter in the molecule, non-stereoselective. Only if a stereocenter exists elsewhere in the carbocation or radical intermediate will there be a stereoselective (not stereospecific) preference. Typically, even when another stereocenter is present, the stereoselectivity induced from this remote site will only result in a small bias.

Substitution at Aromatic Carbon




X

C

Y

CY + X+


Yreplaces

X


reductionX

C

H

Coxidation

Hreplaces

X

Xreplaces

H


X

CZ

Z

YC + XY

elimination

addition

form π bond-XY

break π bond+XY

base

H

X

H

YY + X+

baseacid acid


RCH3


Z = C, N, O

We previously studied substitution reactions at sp2 carbonyl carbon and learned that the process goes by variations of nucleophilic acyl substitution. Here we examine substitution at sp2 aromatic carbon. These substitution reactions often follow the electrophilic aromatic substitution pathway, although 3 other pathways are possible in certain cases.

Aromatic Substitution: Four Examples

+

HNO3

N+H2SO4

HO

NOH

OO O

H OH

I. Electrophilic Aromatic Substitution

II. Nucleophilic Aromatic Substitution Cl

NO2

NO2

NaOHH2O, 100 ˚C

ONO2

NO2

H

Aromatic Substitution: Four Examples

IV. Substitution via benzyne intermediate

III. Substitution via arenediazonium ion

CH3

Cl

Na NH2

NH3 (l), -33 ˚C

CH3

NH2

CH3

NH2

NaCl+ +

Four Mechanistic Pathways for Aromatic Substitution: I. Electrophilic Aromatic Susbstitution

[AE]

H

+ E

E

+ H

EH EH EH

EH

[DE]

resonance stabilized cationic intermediate

this filled orbital shows a σ → a π-type interaction

The LUMO shows positions of electron deficiency

MO pictures of the cationic intermediate

Four Mechanistic Pathways for Aromatic Substitution: II. Nucleophilic Aromatic Susbstitution

[AdN] [Eβ]

resonance stabilized anionic intermediate

LG

Nu

Nu

EWG EWG

LG

LGNu LGNu LGNu

EWG EWG EWG

+ +

Nu

LG

The HOMO shows positions of surplus electron density

MO picture of the anionic intermediate

NN

HH

HH H

N N

HH

HH

H

NuH

H

HH

H

Nu

Nu N2+ +

Four Mechanistic Pathways for Aromatic Substitution: III. Substitution via Arenediazonium Ion

[AN] [DN]

an aryl cation The LUMO shows positions of electron deficiency

Two views of the aryl cation’s LUMO

the unsubstituted carbon of the aryl cation approaches sp hybridization and distorts the ring’s geometry

side

top

Four Mechanistic Pathways for Aromatic Substitution: IV. Substitution via Benzyne Intermediate

[E2]

Cl

NH2

H

HH

HH

H

HH

NH2

H

NH2

H

NaNH2, NH3 (l), -33 ˚C NaCl+

NH3 + Cl–

NH2

NH2

NH2H

[AdN]

[pt]

The LUMO shows positions of electron deficiency

Two views of the benzyne’s LUMO

side

top benzyne

Substituents Greatly Influence Reactivity and the Stability of the Various Intermediates on the Substitution Pathways Does the substituent make benzene a better nucleophile or better electrophile?

Substituents that interact through σ bonds influence reactivity by what are known as ________ effects. Withdrawing substituents accept π-electron density into an appropriately-aligned, low-energy σ* orbital; in contrast, donating substituents contribute electron density from an appropriately-aligned, high-energy σ orbital. This behavior is revealed below in the frontier orbital pictures. For -CF3 (left), the LUMO (but not the HOMO) shows an interaction between the substituent and the π-system.. Thus, -CF3 lowers the LUMO energy and makes the ring more electrophilic. For -CH3 (right), the HOMO (but not the LUMO) shows an interaction between the substituent and the π-system. Thus, a C-H σ bond of -CH3 contributes electron density to the HOMO making the ring a better nucleophile.

inductive

CF F

FH

CH HH

reference

electron withdrawing

electron donating

LUMO LUMO

HOMO HOMO

Substituents with Non-bonding Pairs Donate Electron Density Through an n → π� (π-type) Resonance Interaction

NH H

NH H

NH H

NH H

HOMO LUMO

top top side side

Substituents adjacent to a benzene ring can donate electron density from an appropriately-aligned, high-energy n orbital. This behavior is revealed below in the frontier orbital pictures of aniline. Notice that nitrogen is planar (not pyramidal) for maximum n → π� overlap. The HOMO (but not the LUMO) shows an interaction between the substituent and the π-system (indicated by the block arrows). Thus, nitrogen’s lone pair contributes electron density to the HOMO making the ring a better nucleophile. This is predicted by the resonance contributors. Those atoms sharing the negative charge contribute most to the HOMO and are the most nucleophilic positions. http://wps.prenhall.com/wps/media/objects/725/743034/0081f.html!

Substituents with Low-Lying π* Accept Electron Density Through a π → π� (π-type) Resonance Interaction

Substituents adjacent to a benzene ring can accept π electron density into an appropriately-aligned, low-energy π* orbital. This behavior is revealed below in the frontier orbital pictures of benzaldehyde. Notice that the carbonyl and ring atoms are coplanar for maximum π-type overlap. The LUMO (but not the HOMO) shows an interaction between the substituent and the π-system (indicated by the block arrows). Thus, the C=O lowers the LUMO energy and makes the ring more electrophilic. This is predicted by the resonance contributors. Those atoms sharing the positive charge contribute most to the LUMO and are the most electrophilic positions. http://wps.prenhall.com/wps/media/objects/725/743034/0082f.html!

H O H O H O H O

HOMO LUMO

top side top side

Rationalizing Substituent Effects for Electrophilic Aromatic Substitution Reactions

EWG

+ E

EWG

E

A

H

+ E

H

E

B

EDG

+ E

EDG

E

Cprogress of reaction

free

ene

rgy

A B C A B Cproducts

starting substrates

RDS

RDS

RDS

TS≠

rese

mbl

es

the

inte

rmed

iate

Since the benzene ring is the __________, the reaction is accelerated by EDGs nucleophile

Comparing Intermediates For Rings with an EWG vs. EDG

EDG activating

deactivating O H

+ EA

O H

E H

O H

E H

NCH3 CH3

+ EC

NCH3 CH3

E H

NCH3 CH3

E H all atoms have a complete octet

EWG

Substituents and Regiochemistry in Electrophilic Aromatic Substitution

X

+ E

XE

XX

EE

or orH

X is EWG or EDG ortho meta para

•  All activating substituents (EDG) and weakly deactivating halogens are __________ directors ortho / para

meta •  All substituents more deactivating than the

halogens are ____ directors (i.e., the EWGs)

Substituent - Reactivity Correlation Chart

Ortho / para directing

From P. Y. Bruice Organic Chemistry (5th Ed.)

Summary • The π bonds in aromatic compounds behave as nucleophiles and are thus reactive toward electrophiles. In the first step of electrophilic aromatic substitution, the aromatic ring attacks an electrophile in an [AE] fashion to give a delocalized carbocation intermediate."

• In the second step of electrophilic aromatic substitution, the carbocation undergoes [DE] with loss of H+ from the same carbon to which the electrophile added, re-forming the benzene ring (the net result is substitution at this C)."

• The electronic nature of substituents on an aromatic ring has profound effects on the rate of reactivity and the regiochemistry of electrophilic aromatic substitution reactions."

• Draw your curved arrows correctly. Arrows should not show the H+ flying off into space! If H+ loss via [DE] bothers you, show a weak base making a new bond to hydrogen at the same time the C–H bond is cleaved (technically now a [pt] step). Often the base is assumed to be present but need not be shown." E

H

H

E

H

E

H

H

B

CORRECT CORRECT WRONG

E

H

H

[DE] [pt]

Which best represents the transition state for the first step in the electrophilic aromatic substitution pathway?" A" B" C" D!

Eδ+

δ+

E

H

H

δ+

δ+

E

H

H

δ+

δ+δ+

δ+ H Eδ+

δ+

A B C D

Electrophiles for Electrophilic Aromatic Substitution Bromination

+ Br2FeBr3

Br

+ HBr

Br BrBrFe

Br Br+ Fe Br

Br

BrLewis Acid

Lewis BaseBr Br

[AN]

Electrophiles for Electrophilic Aromatic Substitution Other Halogenations

Chlorine is similar to bromine

For iodine, the following conditions are used

Here the active electrophilic species is presumed to be† ________

+ Cl2FeCl3

Cl

+ HCl

†J. Chem. Soc. B, 1971, 2264 - 2268, DOI: 10.1039/J29710002264

OHNIO

+ I2HNO3

I

+ HI

Nitration by Electrophilic Aromatic Substitution

+

HNO3

N+H2SO4

HO

NOH

OO O

H OH

The active electrophilic species is O N Onitronium

ion

(NO2 )

How does it form?

NO

O O

HO S

O

OO HH+ N

O

O O

H H

O SO

OO H

O N O

+

H2O +[Eβ]

[pt]


Electrophiles for Electrophilic Aromatic Substitution Sulfonation

+S

+ H2OH2SO4O O

OH

The active electrophilic species is

How does it form?

O SHO

O

O SO

OO HH+ O S

O

OO H+

H2O +

O SO

OO HH O S

O

OOHH

H

O SHO

O

[pt]

[DN]


Friedel-Crafts Acylation O

R

R

An acyl group has the structure

An alkyl group has the structure

H

H

+

+

acyl chloride

RO

Cl

acid anhydrideR O R

O O

1) AlCl32) H2O

1) AlCl32) H2O

R O


The Acylium Ion is the Electrophile in Friedel-Crafts Acylation Reactions

The active electrophile for Friedel-Crafts Acylation is the acylium ion

How does it form?

R C O

R C Oacylium ion

AlCl4

R CO

Cl +ClAl

Cl ClR C

OCl Al

ClCl

Cl

ion pairdissociation /

formation

[AN]

[DN] (or [Eβ])

[AN] (or [AdN])

For the acylium ion, which resonance contributor is more important? Why?

All atoms have octet of electrons

Friedel-Crafts Alkylation

HR Cl+ AlCl3

R+ HCl

The active electrophilic species is the carbocation R+

Example: H

+ Clisobutylchloride

tert-butylbenzene

+ HCl

CH3 CHCH3

CH2–ClClAl

Cl Cl CH3 CCH3

CH2–ClH

AlCl

ClCl

isobutyl chloride -aluminum chloride complex

CH3

CCH3 CH3 AlCl4

H

H

+

Mechanism of Friedel-Crafts Alkylation

[1,2R] + [DN]

[AE]

[AN]


General Considerations About Ortho vs. Para Substitution

+ HNO3H2SO4 NO2

NO2

+

1 : 1

+ HNO3H2SO4 NO2

NO2

+

1 : 4.5

Ortho and para substitution have similar electronic considerations. ______ factors can have an influence on the ortho / para product ratio. Steric

Electrophile Orientation With Disubstituted Benzenes

With disubstituted benzenes: •  First consider the contributions of the individual substitutents and determine if

they direct to a common position to produce a major product. •  When substituents are in competition, strongly activating substituents will

dominate weakly activating or deactivating ones.

CH3

+ HNO3H2SO4

CH3NO2

NO2 NO2p-nitrotoluene 2,4-dinitrotoluene

Summary • Electrophiles preferentially attack ortho or para to electron-donating substituents such as (RO–, R2N–, RCONH–) and halo substituents, whereas electrophiles attack meta to electron-withdrawing groups such as carbonyl, –CN, –NO2, and –SO3H.!

• The directing ability of a group is closely related to its ability to stabilize or destabilize the intermediate carbocation.!

Give the major product of these reactions.

CH2CH3

FeBr3+ Br2

CH2CH3

Br

CH2CH3

BrBr

CH2CH3

BrBr

CH2CH3

Br

CH2CH2Br

A B C D E

Br

OCH3

+ HNO3H2SO4

NO2

OCH3

Br

NO2

Br

OCH3

NO2

Br

OCH3

NO2

A B C D

Anilines are Prepared by Reduction of the Nitro Group

Anilines are commonly formed by the reduction of the nitro group. The reaction is accomplished by the reagents shown below. Given that the nitro group is easily introduced via electrophilic aromatic substitution, the synthetic sequence:

Nitration → reduction → diazotization → substitution is a powerful route to many substituted arenes.

NO2 NH2

Sn, HCl

NO2 NH2

Pd/C, H2

Substitution Reactions via Arenediazonium Salts

Mechanism of Diazotization

nitrosonium ion ORnitrosyl cation

O N Osodiumnitrite

H–H

O N OHnitrous

acid

H

–HO N O

H

H

N ON O

–H2O

NH2NH

HN O

NH

N O NH

N OH

H

–H

H

–HN N O

H

H

–H

H–H

N N OH

HN N

–H2O

[pt] [pt] [Eβ] or

[DN]

[AN] or [AdN]

[pt]

[pt] [pt]

[pt]

[Eβ]

NN

HH

HH H

N N– Nu

HH

HH

H

NuH

H

HH

H

aryl cation

Mechanistic Considerations: Substitution of Arenediazonium Ions

Although conveniently thought of as proceeding via nucleophilic capture of the aryl cation via an [AN] step, evidence suggests that many of these reactions may involve radical chain mechanisms (for example, see: J. Am. Chem. Soc. 72, 1950, 3013).

[DN] [AN]

In these reactions the leaving group leaves before the Nu adds (compare to electrophilic aromatic substitution mechanism)

Benzene as an Electrophile: Nucleophilic Aromatic Substitution

This reaction goes by an ___________________ pathway, analogous to nucleophilic acyl substitution (recall that substitution reactions at sp2 carbon do not follow the [SN2] pathway).

addition / elimination

ClNO2

NO2

NaOHH2O, 100 ˚C

ONO2

NO2

H

Mechanism of Nucleophilic Aromatic Substitution

ClNO2

NO2

ONO2

NO2

H

OH N

N

addition step(slow step)

Cl OHN

N

Cl OH

O

O

N

N

Cl OHNCl OH

NO O

elimination step(fast step)

Cl

[AdN]

[Eβ]

O

O

O

O

OO OO

OO

O

O

Substrate Requirements for Nucleophilic Aromatic Substitution

ClNO2

NO2

•  EWG to activate ring (lower LUMO energy)

The EWG must be located ______ or _____ positions. A variety of nucleophiles can be used. The incoming group must be a stronger base than the outgoing leaving group.

ortho para

•  Good leaving group (–F best in these reactions)

Substitution that Involves Elimination at Benzene

This reaction goes by an _________________ pathway and it involves a high-energy intermediate known as _______

elimination - addition benzyne

OCH3

Na NH2NH3 (l), -33 ˚C

OCH3

NH2

NaBr+Br

OCH3

Na NH2NH3 (l), -33 ˚C

OCH3

NH2

NaBr+Br

OCH3

Br

Na NH2NH3 (l), -33 ˚C

OCH3

NH2

OCH3

NH2

NaBr+ +

Elimination−Addition (Benzyne) Reactions of Bromoanisole Isomers with Sodium Amide in Liquid Ammonia

Published in: John D. Roberts; J. Org. Chem. 2009, 74, 4897-4917. DOI: 10.1021/jo900641t Copyright © 2009 American Chemical Society

NaNH2 is a very strong base. The elimination process follows the [E1cb] pathway. [AdN] to benzyne is favorable because a triple bond in a 6-membered ring causes considerable strain.

Functional Groups With C=O and Related Carbon-Heteroatom Multiple Bonds

aldehyde

N N

NH2H2N

NC

O

OH

O

O

O

O

Cl

O

S

O

NH2

O

O

O

H

O S

ketone thioketone

imine nitrile urea (or guanidine)

carboxylic acid

carboxylic ester

carboxylic anhydride

amide thioester acid chloride

α

β

α,β�unsaturated carbonyls

The acyl group

Chemistry of the Carbonyl Functional Group

Electrophilic site: nucleophiles attack here

Basic site: protons add here

α

Acidic hydrogen: deprotonation creates a nucleophile

Leaving group (substitution reactions)

Types of Carbonyl Reactions I. Addition reactions

Here the carbonyl serves as an ___________ electrophile

The elements of “Nu-H” add across the C=O double bond Nu

+ H OC Nu

HOC

II. Substitution reactions Here the carbonyl serves as an ___________ and “X” is a ____________. electrophile

leaving group

The “X” group is replaced by the “Nu” group. This is substitution at an sp2 carbon.

O

R X

Nu O

R Nu

X

Overall result is substitution of -H at the α-carbon by -E.

O

H H

E O

H E

Hα

Here the carbonyl serves as a ___________ following deprotonation. nucleophile

III. Reactions at the α-carbon

Resonance Contributors and the LUMO Reveal a Consistent Picture of Carbonyl Polarity

δ+

δ–

The molecular orbital model is consistent with the above resonance picture. The LUMO (π*) of formaldehyde is shown; the π* lobe on carbon is much larger than on oxygen. In other words, carbon contributes more to π* than does oxygen. Nucleophiles will be attracted to the largest lobe of the LUMO. Conclusion: Carbon (not oxygen) is the electrophilic site in carbonyl groups.

OC HH

formaldehyde

2nd best resonance contributor

Structure-Reactivity and the Carbonyl Group

Reason:

Better electrophile

Steric factors

Reason:

better electrophile

Electronic factors; LUMO energy level

Aldehydes are more reactive than ketones.

The Carbonyl Oxygen is Protonated Under Acidic Conditions

The carbonyl is converted into a more powerful electrophile by ___________ the carbonyl oxygen. protonating

OC HH O

C HH

H

π

π∗

π

π∗

Characteristic of a more powerful electrophile, the protonated carbonyl has a lower lying π* (LUMO) energy than the neutral carbonyl.

add H+

π* energy is lowered

On the Bascisity of Carbonyl Oxygen Which carbonyl is the stronger base?

Which of these protonated forms (A or B) is more stable? Why? __________________

Circle the stronger base

delocalized charge

A pKa = -7.2 (stronger acid)

B pKa = -6.5 (weaker acid)

On the Acidity of α-Hydrogen fre

e en

ergy

reaction progress+ B

A

BC CO

CO

HC C

O

H+ B

Cα-H acidity correlates to the stability of the conjugate base. The more stable the conjugate base, the stronger the acid.

weaker acid

stronger acid

Negative charge better delocalized in B than A

pKa Values of Some Carbon Acids


From Paula Y. Bruice Organic Chemistry 4th Ed.

Other Nucleophiles Derived From Carbonyl Groups

The enol is the tautomeric form of the carbonyl group

Mechanism, pH, and Charge: General Considerations

•  Under acidic conditions: the carbonyl oxygen will be protonated first. DON’T draw negatively charged intermediates under acidic conditions (charged intermediates should only bear ___ charge). +

•  Under basic conditions: the carbonyl oxygen will become negatively charged. DON’T draw positively charged intermediates under basic conditions (charged intermediates should only bear ___ charge). –

These are good rules to follow when you write any organic reaction, but they are especially common in carbonyl chemistry (We will see specific examples in the next several lessons).

Oxidation of Alcohols to Form Carbonyls

1. Oxidation of secondary alcohols to form ketones

Reagents: PCC (pyridinium chlorochromate), PDC (pyridinium dichromate), chromic acid (seen as H2CrO4 or K2Cr2O7/H2SO4/H2O)

2. Oxidation of primary alcohols to form aldehydes

Reagents: PCC (pyridinium chlorochromate), PDC (pyridinium dichromate)

3. Oxidation of primary alcohols to form carbox. acids

Reagent: Chromic acid (seen as H2CrO4 or K2Cr2O7/H2SO4/H2O)

Mechanism of Chromic Acid Oxidation of 2-Propanol

1. Condensation of alcohol and chromic acid (couple elementary steps)

2. Proton-transfer/beta-elimination

3. A series of redox reactions converts the chromium from the 4+ oxidation state to the 3+ oxidation state

Summary • Tautomers are isomers in which a double bond and a hydrogen atom change their locations. The most important kinds of tautomers are carbonyl-enol tautomers. Tautomerization is a chemical equilibrium that occurs very rapidly in acidic or basic media; it should not be confused with resonance (resonance is not an equilibrium process).

• An organic acid, HA, is much more acidic when the lone pair of the conjugate base can be stabilized by resonance (e.g., PhOH is more acidic than EtOH). A C-H bond is acidic when the lone pair of the conjugate base (a carbanion) can be delocalized into a carbonyl group, and even more so when it can be delocalized into two carbonyl groups.

• Carbonyl compounds are some of the most important organic compounds. To help understand their behavior, remember these pKa values:

O

O CH3H

O

N CH3H

O

C CH3H

H HH

pKa = 5pKa = 16 pKa = 20

α

O

CH2

CH3

OH

CH2 CH3

> 99 %< 1 %

H

Nucleophilic Addition Reactions - Adding XY Across Unsaturated C=Z and C=C Bonds




X

C

Y

CY + X+


Yreplaces

X


reductionX

C

H

Coxidation

Hreplaces

X

Xreplaces

H


X

CZ

Z

YC + XY

elimination

addition

form π bond-XY

break π bond+XY

base

H

X

H

YY + X+

baseacid acid


RCH3


Z = C, N, O

We previously studied the electrophilic addition of “XY” across carbon-carbon multiple bonds. Here we examine the nucleophilic variations of addition. We will consider C=C and C=Z π-bonds (where Z is a heteroatom such as O or N).

Addition Mechanisms Require a New Elementary Step: Nucleophile Addition to a C=O π-bond [AdN]

Empty π*

Filled n [AdN] involves

σ-bond making and π-bond breaking

σ#

n

π

σ� #a π�

σ→σ� # σ→a σ→π�#

filled

empty

n→σ� # n→a n→π�#

π→σ� # π→a π→π�#


Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in an elementary step. The tail of the first arrow implies the filled orbital (HOMO) is non-bonded electron pair. The head of the arrow points between Nu– and a carbon atom from the π-bond. The tail of the second arrow shows that a π-bond breaks suggesting that π* accepts the electron pair via σ-type orbital interaction.

[AdN]

This symbol means nucleophile addition to a π-bond

[AdN] is a n → π* σ-type interaction

C ONu

+δ+

δ− C ONu

pathwayHNu

C ONu

H

CO

C ONu

H

C ONu

HC O

HNu

There are Two Mechanistic Pathways for Nu Addition

[AdN2]

[pt] [AdN]

[AdN] then [pt] (two-step, basic

conditions)

The two addition pathways differ in the timing of the proton transfer and nucleophile addition. Both are two-step mechanisms and both are called [AdN2] (bimolecular nucleophilic addition to a polarized π-bond). The bimolecular [AdN] step is rate-determining for both. In the top case, the two steps are often actually performed sequentially (i.e., in the laboratory, a proton source is subsequently introduced to “quench” the initially formed adduct). For the top case nucleophilic addition is under basic conditions and an acid then neutralizes the initially formed alkoxide adduct. In the bottom case, reversible proton transfer takes place followed by [AdN]. Unlike the top case, this two-step process occurs without changing conditions (no intervention on the part of the chemist is needed).

[AdN] [pt]

[AdN2] [pt] then [AdN]

(two-step, acidic conditions)

alkoxide adduct

Addition Involving Carbon Nucleophiles Grignard reagents

Synthesis of primary alcohols

Reacts like a carbanion Nu:–


Acetylide Anions As Carbon Nucleophiles

sodium acetylide

C C HCH3 C CCH3H +

compared to other C–H bonds this one is acidic

pKa = 25

CH2 CH

H

pKa = 44

H + CH2

H

an acetylide anion

a vinylic anion

Hydride Reduction of Ketones and Aldehydes


hydride Hydride is a strong base and nucleophile

Addition of H-CN to a Ketone or Aldehyde: Cyanohydrin Formation

a cyanohydrin

For aldehydes and most ketones, the position of the equilibrium favors cyanohydrin formation. For aryl ketones and sterically hindered aliphatic ketones, starting materials are favored.

pKa = 9.3


Nucleophilic Addition to α,β�Unsaturated Carbonyl Compounds

1 2 3 4

Draw resonance structures that reveal the two electrophilic sites

• 1,2-addition is called ______ addition • 1,4-addition is called ________ addition

direct conjugate

[AdN] Also Applies to Polarized C=C π-Bonds

Empty π*

Filled n [AdN] involves

σ-bond making and π-bond breaking

[AdN]

This symbol means nucleophile addition to a π-bond

EWG is an electron withdrawing group that polarizes the π-bond

and lowers the LUMO energy

C CEWG

Nu+

δ+δ−

C CEWG

Nu

[AdN] is a n → π* σ-type interaction

LUMO (side view) LUMO (top view)

H

HOH

H

αβ

Orbital calculations (see below) show that Cβ makes a significant contribution to π*

Analogous Pathways for Nu Addition to Polarized C=C

[pt] [AdN]

Nucleophilic addition to a polarized carbon-carbon π-bond takes place by analogous pathways. The top case is just like carbonyl addition except that [AdN] occurs at the Cβ site. The initially formed enolate anion is resonance stabilized (via electron donation into the withdrawing group (e.g., C=O)). For the bottom case, the proton first adds to the carbonyl oxygen. Then, [AdN] occurs at the Cβ site giving rise to an enol. The enol is usually unstable and undergoes spontaneous tautomerization to produce the addition product.

[AdN] [pt]

[taut]

enol

enolate anion

HNu

C OC

C CNu

H

C CNu

Nu

C

C OC C

Nu

C O

H

C CC O

HC O

H

C CNu

C O

H

O

+ K C NCH3CO2H

OH

HH

CN

Li

H HO

Direct vs. Conjugate Addition


•  Nucleophiles that are weak bases tend to give conjugate addition products.

•  Nucleophiles that are strong bases tend to give direct addition products.

Understanding Direct vs. Conjugate Addition (Kinetic vs. Thermodynamic Control)

• Conjugate addition _____ irreversible • Direct addition ___________reversible • Direct addition is _________ than conjugate

is may be

faster

Whether or not direct addition is reversible depends on the ability of the -Nu to act as a good ___________________. (More about this in the next lesson)

leaving group

CH

CH

C

O

CH

CH

C

OH

Nu

HC CH2 C

O

Nu

fast

slow

thermodynamic product

CH

CH

C

O

CH

CH

C

OH

Nu

HC CH2 C

O

Nu

fast

slow

kinetic product

Summary • Alkenes and alkynes that are substituted with electron-withdrawing groups such as carbonyl, nitro, or sulfonyl groups are electrophilic. Many nucleophiles react with these alkenes to give conjugate addition (i.e., 1,4-addition) products.!

• The enol resulting from 1,4-addition undergoes spontaneous tautomerization to give alkene addition as the net result.!

• Nucleophiles that are weak bases such as alcohols, thiols, amines, and halides, add reversibly to α,β-unsaturated carbonyls and tend to give the thermodynamic product (i.e., typically the product of conjugate addition). Nucleophiles that are strong bases add irreversibly and give direct addition (i.e., 1,2-addition) products.!


1,2- vs. 1,4-Additions to α,β-Unsaturated Carbonyl Compounds

O+

nucleophile

CH3CH2 MgBr H3O+

C N H3O+

HH3O+

strong or weak base(circle one)


+


+

Substitution at the Carbonyl Carbon (Acyl Substitution)




X

C

Y

CY + X+


Yreplaces

X


reductionX

C

H

Coxidation

Hreplaces

X

Xreplaces

H


X

CZ

Z

YC + XY

elimination

addition

form π bond-XY

break π bond+XY

base

H

X

H

YY + X+

baseacid acid


RCH3


Z = C, N, O

We previously studied substitution reactions at sp3 carbon and concluded that they proceed either by the [SN1] or the [SN2] pathway. Here we examine substitution at the sp2 carbonyl carbon (nucleophilic acyl substitution reactions). We’ll see that substitution at sp2 atoms generally proceed by pathways much different than those at sp3 atoms.

Nucleophilic Acyl Substitution: Three Examples

CH3 O CH3 CO

Cl+ CH3 CO

OCH3+ Cl

OH ++

OHCO

CH3OCH3C

OCH3

esterification

CH3

OHH

H

CH3 NHCH3 O CH3

O O

O CH3

O+

CH3 NHCH3

OCH3 NH2 CH3 NH2

H+

H

Three Mechanistic Pathways for Acyl Substitution

[pt]

[AdN]

All three pathways include [AdN] and subsequently [Eβ] steps (i.e., addition followed by elimination, proceeding through tetrahedral intermediate [TI]). The three paths differ by the role of [pt]. Which pathway is best will depend on the pH, the nucleophile strength, and the leaving group quality. A rough generalization is that the top pathway is for basic, the middle for neutral, and the bottom for acidic conditions. Slight variations on these pathways will be encountered.

[AdN]

[pt]

[pt]

[Eβ]

[Eβ]

[Eβ] [AdN]

CO

L

H

C OL

Nu

HC O

L H

Nu

C ONu

C OL

NuL

H

Nu H

NuH

C OL

NuHC O

L

NuB BH

base

C ONu

L

C OL

Nu

HC O

Nu

H

H

H

LH[pt]

Acyl Substitution Requires a Variation of the [Eβ] Step: Beta Elimination Across the C-Z: Bond

Empty σ* Filled n � [Eβ] involves

π-bond making σ-bond breaking

Curved Arrows Imply the Frontier Orbitals Curved arrows indicate the HOMO-LUMO pair (Frontier Orbitals) involved in this elementary step. The tail of the first arrow implies that the filled orbital (HOMO) is a nonbonded electron pair, n. The head of the first arrow points between C-Z: suggesting a new π-bond. An empty, low-lying σ*-orbital from the adjacent C-L bond is available to accommodate this electron pair suggesting the LUMO is σ*. The HOMO (n) is beside (and nearly parallel to) the LUMO (σ*) allowing for a π-type interaction.

[Eβ]

This symbol means beta elimination

[Eβ] is an n → σ* π-type interaction

-L is a LG; Z: is O or N

σ%

n

π

σ� %a π�

σ→σ� % σ→a σ→π�%

filled

empty

n→σ� % n→a n→π�%

π→σ� % π→a π→π�%

[Eβ]

We first encountered [Eβ] when we studied the [E1cb] elimination pathway (top equation). In the second equation, [Eβ] is now shown for the generalized case where Z: may be an oxygen or nitrogen atom. Many leaving groups (-L) participate in [Eβ], even poor ones (e.g., -OH). This step is the reverse of [AdN].

C YL

C Y + Lβα

C CL

C C + Lβα

C ZL

C Z + Lβ α

Nucleophilic Acyl Substitituon: Addition-Elimination Through a Tetrahedral Intermediate

[AdN] [Eβ]

O

R Cl

O

R OCH3+ Cl

tetrahedralintermediate

[TI]

R Cl

O OCH3

eliminationaddition

CH3O

Here’s a related mechanism on the formation of amides via acid chlorides.


Equilibrium Position and the RDS Depend on Relative Basicity of Nu and X

free

ener

gy

reaction progress

[TI]

R CO

LNu+ + L

R CO

Nu free

ener

gy

reaction progress

R CO

LNu+ + L

R CO

Nu

relative basicities:Nu > L

relative basicities:L > Nu

[TI]

R CO

Cl

leavinggroup

derivative

Cl

R CO

O CO

R

O CO

R O R

R CO

O RR CO

OH

OH NH2

R CO

NH2acyl chloride acid anhydride esteracid amide

(conjugate acid pKa)

(-2.2) (4.8) (16)(15.7) (35)

Base-Catalyzed Ester Hydrolysis

Note that under basic conditions, all intermediates are either neutral or negatively charged. All steps are reversible, but the last step lies very far to the side of the carboxylate. This nearly irreversible step makes formation the reverse process (formation of esters from acids under basic conditions) nearly impossible.

[AdN]

[Eβ]

carboxylate

carboxylic ester

carboxylic acid

addition

elimination

CH3 CO

OCH3 + OH CH3 CO

OCH3OH

CH3 CO

O–H OCH3+CH3 CO

O + CH3OH

Ester Formation via the Acid Anhydride (Near Neutral Conditions)

These reactions are often conducted in presence of a weak base such a pyridine. The base assists in proton transfer steps.

[AdN]

[pt]

[Eβ]

acid anhydride

ester

pyridine

Acid-Catalyzed Ester Formation via Carboxylic Acid

Note that under acidic conditions, all intermediates are either neutral or positively charged. All steps are reversible so formation and hydrolysis are related as the forward and backward directions of these pathways. For ways to manipulate the equilibrium see the link below.


[AdN] [Eβ]

carboxylic ester

carboxylic acid

Manipulating the Equilibrium http://wps.prenhall.com/wps/media/objects/725/743131/0085f.html!

Acid-Catalyzed Amide Hydrolysis

OHNH2CO

R OHCO

RH+

NH2CO

R

H

NH2CO

R

H

OH H

NH2CO

R

H

OH

NCO

R

H

O

H

H

OHCO

R

H+ NH3

–H H

addition elimination

H

+ H–ClHO

H H

H

+ NH4Cl

NH2CO

R

H

Amides are stable functional groups but under forcing conditions can be made to hydrolyze to carboxylic acids. The acid-catalyzed process is used to hydrolyze peptides and proteins for amino acid composition and sequence determination.

Interconversion of Carboxylic Acid Derivatives D

ecre

asin

g re

activ

ity!

For the reagents needed to accomplish these interconversions, see the following link: http://wps.prenhall.com/wps/media/objects/725/743131/0087f.html

Mechanism of Acid Chloride Formation

thionyl chloride

chlorosulfite

Summary Many carbonyl compounds including esters, acyl chlorides, and acid anhydrides have leaving groups attached to the carbonyl C, and many reactions proceed with substitution of this leaving group by a nucleophile.!

Substitutions at the carbonyl C usually occur by an addition-elimination mechanism. The nucleophile (Nu–) adds to the electrophilic C of the carbonyl group to make a tetrahedral intermediate. The leaving group (L–) then leaves in the elimination step to give a new carbonyl compound. Note that either L– or Nu– may be expelled from the tetrahedral intermediate. Which one that is expelled depends on their relative leaving group abilities and the reaction conditions. Expulsion of Nu– gives back the starting material (which is not very productive!).!

(1) What products are observed from the following reaction (the oxygen of methanol is labeled with the 18O isotope)?!

  A!  B!  C!  D!  E!  F!

Acyl Substitution Followed by Addition

CO

L

HNu

C ONu

L

C ONu

Nu

H

Nu

acyl substitution addition

The carbonyl product resulting from acyl substitution may be susceptible to further attack by the nucleophile. These sequential processes follow pathways that have previously been discussed. Whenever substitution is slower than addition, stopping the reaction at the end of the substitution reaction can be nearly impossible. The result will be two equivalents of nucleophile attaching to the carbonyl carbon (the first by substitution, the second by addition). In general, the substitution-addition sequence takes place for very strong nucleophiles such as hydrides and Grignard reagents. A source of H+ must intentionally be added by the chemist to quench the reaction after formation of the addition adduct (i.e., in chemical equations this is often written in “H+ workup”).

Examples of Acyl Substitution Followed by Addition

+ OCO

CH3CH2

CH3COH

CH3CH2

HO

CH3 MgBr CH3+following H+ workup

+LiHAlH4 OCO

CH3CH2+

CH3

HO

HCOH

CH3CH2

Hfollowing H+ workupCH3

NH2

O1) LiAlH4

NH2

H H

2) H+ workup

Substitution-Addition Mechanism for Grignard Nucleophiles

2 CH3 MgBr + OCO

CH3CH2

CH3COH

CH3CH2CH3H

OH

+

OCO

CH3CH2CH3

[MgBr]

CH3CO

CH3CH2

O+CH3 MgBr CH3C

OCH3CH2

CH3

[MgBr]

Hadd

[AdN]

[Eβ]

[AdN]

[pt]

Substitution-Addition Mechanism for Hydride Nucleophiles & Esters

+ OCO

CH3CH2

HCOH

CH3CH2HH

OH

+

OCO

CH3CH2H

HCO

CH3CH2

O+

HCO

CH3CH2H

Hadd

H AlH

HH

Li

H AlH

HH

Li

Li

Li

Li

[AdN]

[Eβ]

[AdN]

[pt]

Substitution-Addition Mechanism: LiAlH4 with Primary or Secondary Amides

[AN]

[Eβ]

[pt]

[AdN]

[AdN]

Substitution-Addition Mechanism: LiAlH4 with Tertiary Amides

[AN]

[Eβ]

[pt]

[AdN]

[AdN]

+

add H+

R N

O

R CH

HN

AlH

HH H

R CH

NO

CH3

CH3

AlH

HH

tertiary amide

R C NH CH3

CH3

CH3

CH3

CH3

CH3low temp.hydrolysis

H3O

R CO

H

AlH

HH H

AlH

H H

OAl

H

HH

For tertiary amides, the elimination step is not assisted by loss of H2 (there’s no N-H available). Thus, the initial tetrahedral intermediate is fairly stable. If kept cold, hydrolysis of this TI yields aldehydes. At higher temperature, elimination takes place to the iminium ion which undergoes further hydride addition.

an aldehyde

an iminium ion This TI is relatively

stable to [Eβ]

CO

NH

CO

NH

HO H

R

RCN R

H

H

CO

C

NR

CO

C NH

HO H

R

R

H

R

H

α

H

α CC NR

Rα

Substitution of the Carbonyl Oxygen of Ketones and Aldehydes by Nitrogen: Imines and Enamines

addition

When primary amines are combined with ketone and aldehyde groups, the carbonyl C=O is replaced by the C=N (imine) group. The mechanistic sequence is addition followed by elimination of water (the second N–H proton is necessary for imine formation). When secondary amines are combined with aldehydes and ketones, there is no second proton on nitrogen to permit imine formation. In this case, elimination of water is achieved via proton transfer from a H–Cα bond (assuming that the aldehyde or ketone substrate has at least one α-hydrogen). The functional group that results is called an enamine.

addition

imine

enamine

1˚ amine

2˚ amine

elimination and [pt] from N

elimination and [pt] from Cα

Examples of Imine and Enamine Formation

primary amine

O

+NH2 H

N+

HO H(cat.)

N+H

H

O

α N

H

HH Hα

HO H+

H(cat.)

secondary amine

enamine

imine

Other Amino Reagents that Produce Derivatives of Aldehydes and Ketones

H2N OHhydroxylamine

Reagent H2N R Derivative Name Structure

oxime NOH

H2N NHRhydrazine hydrazone N

NHR

H2NHN

semicarbazide

CO

NH2 semicarbazone NNH C

ONH2

Mechanism of Imine Formation O

+NH2

HN

+ H2O

O NH

HH

NH

HO

H

H

HO NH H

HNH H2O

[pt]

[pt]

[pt]

[AdN]

[Eβ]


N+H

H

O

H

N

H

α

H2O

N

H

HH

H

H

H

OH

N

OH

H H

N

OHH

H

H2O+

H

NH

α

Mechanism of Enamine Formation


[pt]

[AdN] [Eβ]

[pt]

[pt] (or [DE])

Hydride Reduction of Imines Produces Amines: Reductive Amination

CH3

ONH3

NaBH(OAc)3pH 6

CH3

H NH2

CH3

NH2 excess C=ONaCNBH3

pH 6 CH3

NCH3 CH3

Imines and iminium ions are reduced by hydride sources to make amines. The process is known as reductive amination. The hydride reducing reagents in the above equations are carefully chosen so as to remain stable under the mildly acidic (pH=6) conditions and to selectively react with protonated imines rather than ketones.

OC

RN

H H

NC

+H2O

H+ + H–hydride

NC H

HRR

CO

OH

CO

OH

HO H

R OH

R

RCO

OR

R

CO

H

OH

CO

HOH

HO H

R OH

R

RCO

HOR

Raldehyde

ketonehemiketal ketal

hemiacetal acetal

Substitution of the Carbonyl Oxygen in Ketones and Aldehydes with Two C–O Bonds: Acetals and Ketals

Addition of one molar equiv alcohol (R–OH) to aldehydes and ketones gives hemiacetals and hemiketals, respectively. Hemiacetals and hemiketals are not usually isolated. Elimination of water followed by addition of a second molar equiv of R-OH gives the corresponding acetals and ketals. Acetal and ketal derivatives are isolable compounds. Note: several [pt] steps assist the elimination of water.

addition

elimination then addition

addition

elimination then addition


Examples of Acetal and Ketal Formation


a cyclic ketal

OCH3C CH3

HOOH+

H3C CH3

OCO + H2O

OC

H3C H+ + H2O

H3C HH3CO

COCH32 CH3OH

a ketone a diol

an aldehyde two molar equiv

of an alcohol an acetal

Removal of Water to Shift Equilibria Reactions such as ester, imine, enamine, and acetal formation are reversible equilibria that generate water as a byproduct. The equilibria constants for these reactions are near unity, meaning that at equilibrium approximately equal quantities of reactants and products are present. To obtain high yields, the reaction must be driven to the right. Le Chatelier's principle suggests a useful strategy for driving these equilibria is to remove water. The apparatus below is called a Dean-Stark trap and is a practical device for removing water to shift chemical equilibria. Benzene is a commonly used solvent for these reactions. When heated to boil, benzene vapor forms an azeotropic mixture with the water that is generated in the reaction.

Manipulating the Equilibrium http://wps.prenhall.com/wps/media/objects/725/743131/0085f.html!

The azeotropic vapor phase is relatively rich in water content. When this vapor condenses it separates into immiscible liquid fractions, one rich in benzene and the other rich in water. In the receiver arm, the liquid benzene phase floats above the more dense water fraction. The benzene-rich fraction spills back into the pot, ready again to carry over more water. The water-rich phase sinks to the bottom of the receiver arm, withdrawn as needed. The Dean-Stark trap works because of the benzene-water density difference and because benzene and water are more compatible as vapors than they are as liquids.

Mechanism of Ketal (Acetal) Formation

[pt]

[AdN]

[pt]

[AdN]

[pt]

[Eβ]

[pt]

O

CH3 CH3

O

CH3 CH3

H

HOCH3

OCO

CH3CH3

H

H CH3

OCO

CH3CH3

H

CH3

OCO

CH3CH3

H

CH3

H

CO

CH3CH3

CH3H2O

HOCH3

OCO

CH3CH3

CH3

CH3

H

+ 2 CH3OHCH3 CH3

CH3O OCH3+ H2O

elimination

addition

addition

H

H

H

H

The Wittig Reaction: Substitution of =O by =C

O

+ Ph3P CH2+

CH2

OPh3PO

Ph3Pa betaine an oxaphosphetane

Ph3P=O

PhPPhPh

CH2

PhPPhPh

CH2

a phosphonium ylide

Ph =

(carbon nucleophile)

[AN]

This elimination involves breakdown of the 4-membered ring

[AdN]

Phosphonium ylides (powerful nucleophiles) add to aldehyde or ketone carbonyl groups via normal [AdN]. This tetrahedral intermediate, a betaine, (BAY-tah-een) has no leaving group, yet it’s not stable. Because phosphorous has a high affinity for oxygen, the oxyanion attacks the positively charged P via [AN] to make a 4-membered ring (called a phosphaoxetane). This strained phosphaoxetane reacts further to split into Ph3P=O and an alkene. This elimination is a little funky, but you should be able to draw a reasonable set of curved arrows (and identify the frontier orbitals).

This net substitution proceeds by an addition-elimination sequence.

Formation of Ylides

Ph3P + CH3 I[SN2]

Ph3P C HH

HI

methyltriphenylphosphoniumiodide

CH3CH2CH2CH2 LiCH3CH2CH2CH3

butyllithium(strong base)

butanePh3P CH2 +

+ LiI

triphenylphosphine

[pt]


Summary • Primary amines (RNH2) add reversibly to ketones or aldehydes to give imines (Schiff bases) and related compounds via the intermediate hemiaminals. The position of the equilibrium depends on the structure of the amine and the carbonyl compound. With alkylamines, the equilibrium favors the carbonyl compound, but it can be driven to the imine by removal of water. With hydrazines (R2NNH2) and hydroxyl and alkoxylamines (RONH2), the equilibrium greatly favors the hydrazone, oxime, or oxime ether.!

• Secondary amines (R2NH) can form hemiaminals and iminium ions, but they cannot form imines. Removal of the α-hydrogen from iminium ions gives enamines. The equilibrium can be driven to favor the enamine by removal of water from the reaction mixture.!

• Water and alcohols add reversibly to aldehydes and ketones under either acidic or basic conditions to give hydrates, hemiacetals, or hemiketals. Only under acidic conditions can further reaction take place to give acetals and ketals. !

• Acetals and ketals are usually thermodynamically unfavorable, so these reactions are usually driven to completion by removing H2O from the reaction mixture.!

Because of the acidity of the α-hydrogen, ketones exist in another, generally less stable form known as ____________

The Carbonyl’s Alpha C-H is Acidic

CH3 CO

CH2 CH3

α−hydrogens

α−carbons

O

CH2

CH3

OH

CH2 CH3

> 99 %< 1 %

H

keto tautomer enol tautomer

enol tautomer

Recall, the larger the pKa, the weaker the acid. The data below shows that of all the C–H bonds, the carbonyl Cα–H is the strongest acid. The unusually high acidity of this C-H bond gives it some very important reactivity (e.g., deprotonation of Cα-H makes the alpha position a carbon nucleophile).

bondCH3CH2O H

CH3 CO

CH2 H

C CCH3 H

H2C CH

H

CH3CH2 H

16

20

25

44

51

pKa

Interconversion Between Tautomeric Forms is Catalyzed by Acid or Base

RCH CH

OR

HO

H H+ RCH C

H

OR

H

OH H

RCH CO

R

H

H3O+

Acid-catalyzed mechanism of keto-enol interconversion

enol tautomer keto tautomer

Base-catalyzed mechanism of keto-enol interconversion

RCH CH

OR

HO

RCH CO

R RCH CO

R

H

OH

H+

HO+enolate anion

enol tautomer

keto tautomer

LDA is a Base Used to Form Enolate Anions

Strong organic bases such as LDA (Lithium DiisopropylAmide) can be used to drive the ketone-enolate equilibrium completely to the enolate side. LDA is a strong base that is useful for this purpose. The steric bulk of its isopropyl groups makes LDA non-nucleophilic. Even so, it’s a strong base. LDA is prepared by the deprotonation of diisopropyl amine using a very strong base such as n-butyl lithium as shown.

NH

H N+

diisopropyl amine

CH3CH2CH2CH2

pKa = 35

Li NLi

LDA

CH3CH2CH2CH3+

pKa = 50n-BuLi

Enolate Equilibria are Acid-Base Reactions To which side does the equilibrium lie?

CH3 CO

CH3pKa = 20+ NaOH CH2 CCH3

O Na + H2OpKa = 15.7

Keq = 5 x 10-5

CH3 CO

CH3pKa = 20+ LDA CH2 CCH3

O Li +

pKa = 35

Keq = 1 x 1015

NH

Using a strong enough base, quantitative enolate formation is feasible.

Thermodynamic vs. Kinetic Control of Enolate Formation

Which enolate is more stable? ____ Which enolate is formed faster? ____ A

B

OCH3 + LDA

slight excess

OCH3

OCH3

+A B

99% 1%

H H H

H

H

HH0 ˚C

OCH3 + LDA

slight excess

OCH3

OCH3

+A B

10% 90%

H H H

H

H

HH0 ˚C

The indicated difference in reaction conditions determines whether deprotonation is reversible or irreversible.

Summary - How the Various Carbonyl Species Interconvert Under Acidic or Basic Conditions!

RR'

O

baseR"OH

RR'

O OR"

RR'

OH

base

RR'

O

base RR'

OH

RR'

HO OR" + H+

– H2O

RR'

OR"

RR'

R"O OR"

acetal / ketal

enolenolate

hemiacetal / hemiketal

base

carbonyl

+ R"OH– H+

+ R"OH– H+

+H+

+H+ COLOR KEY•intermediates generated under basic conditions are blue•neutral compounds are black•intermediates generated under acidic conditions are red•pay careful attention to which side of the equilibrium arrow (i.e., forward or backward direction) the base or acid lies

[AE]

[pt]

[pt] then [DN]

[AN] then [pt]

[AN] then [pt]

-H+

[DE]

[pt] then [AdN]

The enolate anion is as an ________ nucleophile (two nucleophilic sites, carbon and oxygen). Electrophiles generally add to carbon rather than oxygen. The net result of a reaction at C-alpha is substitution (the electrophile, E+, replaces the carbonyl’s alpha hydrogen).

Enolate and Enol Nucleophiles

RCH CH

OR

HO

RCH CO

R RCH CO

RE RCH C

E

OR

ambident

enolate anion

α�substituted product

The enol is nucleophilic too

•  Under basic conditions, the nucleophilic form will be the ____________ •  Under acidic conditions, the nucleophilic form will be the ____ enol

enolate anion

α-substituted product

RCH CO

R

H

E

RCH CO

R

H

E

–HRCH C

OR

E

enol tautomer

Alkylation of Enolate Anions (Enolates are Ambident Nucleophiles)

O

+ LDA

OLi

OSi(CH3)3

OCH3CH3 I

(CH3)3Si Cl"O"-alkylation

"C"-alkylation

+ LiCl

+ LiI

These are [SN2] reactions

As the HOMO indicates, enolates are ambident nucleophiles, being reactive at both Cα and oxygen. Consistent with the largest lobe of the HOMO being on Cα, most electrophiles react at carbon; some very powerful electrophiles (and oxophilic electrophiles) react mainly at oxygen (e.g., silicon is strongly oxophilic).

enolate HOMO

Alkylation via Enamine Nucleophiles

Enamine nucleophiles are more mild than enolates:

•  Less basic •  More C-alkylation •  Mono-alkylation products only

N

O

Br+N

O

Br

O

HCl / H2Ohydrolysis

iminiumbromide

an enamine from cyclohexanone

[SN2]

N

CH3 CO

Cl+N

CH3

O

O

CH3

O

HCl / H2O

Cl

NH2

Acylation via Enamine Nucleophiles

1 2 3

[AdN] then [Eβ]

a 1,3-dicarbonyl

an enamine from cyclohexanone

Enamine nucleophiles are mild enough to give high yields of acylation products without interference from “over acylation” or addition products. Hydrolysis of the iminium ion involves the [AdN] addition of water, [pt], and [Eβ] elimination.

an iminium

ion

Halogenation of the α-Carbon The Enol as a Nucleophile; Br2 as an Electrophile

Because the conditions are acidic, all intermediates are neutral or positively charged. The [AE] step is most easily seen by drawing an alternative resonance contributor. Can you draw it?

[AE]

CH2

O

+ Br2 CH3CO2H CH2Br

O

+ H–Br

CH2

OH

enolformation

Br Br+ CH2Br

OH

H

Br

H

α-bromo substituted carbonyl

α" α"

α-Halogenation Under Basic Conditions

However, successive halogenation steps are faster than previous ones, since the electron-attracting halogens increase the acidity at the α-carbon

C CH O

+ OH C CO+ X X C C

X

O+ X

α-Halogenation can also be conducted under basic conditions with the enolate anion as the nucleophile.

R CO

CH3+ + R C

OCBr3

++3 Br2 3 OH 3 Br 3 H2O

R CO

CBr3 + OH R CO

O + CHBr3

The product is unstable to the reaction conditions. It reacts with hydroxide to give a carboxylate salt (the Haloform Reaction).

The Michael Reaction (Conjugate addition with an α-Carbon Nu)

Nu HCC

O

R+ CC

O

RNuH

HHα

β

We have seen that nucleophiles can add across α,β-unsaturated double bonds, (conjugate addition). When the nucleophile is an α-carbon nucleophile, this reaction is called _______ addition. Michael

O+

O O O

Nu

1 2 3 4 5

•  Draw a (----) line across the new C–C bond in the product •  Identify the nucleophilic carbon •  Assign partial charges to atoms of the electrophile to indicate their state of polarization

in the reactant •  Number the carbonyl carbons in the product and use this to recognize “Michael adducts”

δ+$δ�$

O

H CH3 CH3

O O

H HCH3 CH3

O O

O H

+ OH

CH3 CH3

O O

HCH3 CH3

O O

O H

OH H

OH

HH

HH

H

OH

H2O

OH

H

H H

Michael Reaction: Example and Mechanism

[pt]

[AdN]

[pt]

For the acid-catalyzed version of this reaction see: http://www.chemtube3d.com/

Reactants for the Michael

Reaction

O

H CH3 CH3

O O

H H

CH3 OCH3

O O

H H

CH3C

O

H H

EtO OEt

O O

H H

N

CH3

N

O

CH3

O

OCH3

O

NH2

CN

NO

O

α,β−unsaturated compound nucleophile

β-diketone

β-diketoester

β-diketonitrile

β-diester

enamine

aldehyde

ketone

ester

amide

nitro

nitrile

Summary

The conjugate base of a carbonyl compound, an enolate, is generated under basic conditions. Enolates are nucleophilic at the α-carbon (and at oxygen). Enolates are better nucleophiles than enols. They react with alkyl halides (alkylation) by the [SN2] mechanism. They react with α,β-unsaturated carbonyl compounds to give conjugate addition products (another version of the Michael reaction).!

Enolates can be prepared quantitatively by deprotonation with a strong base such as lithium diisopropylamide (LDA).!

Enols are good nucleophiles that can be generated under acidic conditions. Under acidic conditions, carbonyl compounds are thus electrophilic at the carbonyl C and nucleophilic at the α-carbon (and also at oxygen).!

Enols react with electrophiles (e.g., bromine) leading to net substitution of one α-hydrogen by the electrophilic atom(s).!

Enols are particularly reactive toward electrophilic alkenes such as α,β-unsaturated carbonyl compounds to give conjugate addition products (the Michael reaction). This is one of the mildest, most versatile, and most efficient methods for forming C-C bonds.!

The Aldol Addition Reaction

In this reaction, two molecules of an _________________ combine to produce a β-hydroxycarbonyl. One molecule serves as the nucleophile (enol or enolate) and the other is the electrophile which undergoes carbonyl addition. The reaction is acid or base catalyzed.

aldehyde or ketone

New C–C bond

α# β#

Base-Catalyzed Aldol Addition

H CO

CH2

HH C

OCH3

+ H CO

CH2

HCOH

CH3

H CO

CH2

Na OH

H CO

CH2enolate anion

H CO

CH2

HCO

CH3

+ H OH

a β-hydroxyaldehyde

HO

H

OH

H CO

CH3

OH

[pt] [pt]

[AdN]

Under basic conditions the intermediates bear negative charges

Acid-Catalyzed Aldol Addition

H CO

CH2H

H CO

CH3+ H C

OCH2

HCOH

CH3

H CO

CH3H COH

CH2 H CO

CH2HCOH

CH3

H OH

H

H+

H

H OH

HH O

H

H

H OH H OH

[pt]

[AdN]

[pt] [taut]

enol Under acidic conditions the intermediates bear positive charges

Dehydration of β-Hydroxy Carbonyl Compounds Is Common in Aldol Reactions

HCO

CHCHOH

CH3 HCO

CHCHCH3H

warm acid or base

HCOH

CHCHOH

CH3enol tautomer

HCO

CHCHO

CH3

HCO

CHCHCH3

H

H H

HO

H

H

H

OH H +

H2O

HO

H

H

H2O

H2OThe acid-catalyzed dehydration is shown. The base-catalyzed dehydration follows the [E1cb] pathway.

[Eβ]

[pt]

[taut]

[pt]

Mixed Aldol Reactions

Mixed aldol reactions involve two different carbonyl compounds. Up to four products are possible (two “self-addition” and two “crossed-addition” products).

Mixed aldol reactions between an aldehyde with no α-hydrogens and a ketone generally give good yields of a single product

•  Only enolate possible is from the ketone •  Aldehydes are better electrophiles than ketones

Possible Products from a Mixed Aldol Reaction

Under conditions where enolate formation is reversible, a complex mixture results.

Crossed-Aldol Using Preformed Enolates

If the enolate is preformed and added to the aldehyde, the desired crossed-aldol product will be obtained in high yield.

Intramolecular Aldol Reaction

Another example of a “successful” mixed aldol reaction is when both enolate anion and carbonyl group to which it adds are in the same molecule

Because of the greater stability of 6-membered rings compared to 4-membered rings, only a single product is formed in this intramolecular aldol reaction.

The Claisen Condensation In this reaction, two molar equiv of an ____ combine to produce a β-ketoester. One equiv serves as the nucleophile (enolate) and the other is the electrophile which undergoes addition and elimination. The reaction is driven to product by the final deprotonation step.

ester

+O CH3

O EtO

O

O

CH3

OO CH2

O

HH

O

O

CH3

O

a β-ketoester

Mechanism of the Claisen Condensation

+ O CH3

OEtO

O

O

CH3

O

O CH2

O

O CH2

O

O

OCCH3

OOEt

O CH2

O

O

O

CH3

O

H H

H

EtO

a tetrahedral intermediate

EtOH

O CH3

OEtO

EtOH

EtO[pt]

[pt]

[AdN]

[Eβ]

an ester enolate

pKa = 15.9

pKa = 10.7

The position of the equilibrium for the first three steps lies toward the starting ester. The overall reaction is driven to completion by the acid-base reaction between the ethoxide and β-ketoester.

Mixed Claisen Condensation

Like mixed aldol reactions, mixed Claisen condensations are useful if differences in reactivity exist between the two esters as for example when one of the esters has no α-hydrogen. Examples of such esters are:

excess

β-Ketoacids are Easily Decarboxylated

O

O

OH

3-oxobutanoic acid

warm O+ CO2

O OH

Oa cyclic 6-membered TS≠

OH

+OCO

keto-enoltautomerization

Decarboxylation is the loss of ____ from the carboxyl group of an acid. Most carboxylic acids are quite resistant to moderate heat. Exceptions are carboxylic acids that have a carbonyl group __ to the carbonyl. These carboxylic acids are derived from Claisen condensations (following ester hydrolysis).

CO2

β

Alkylation of β-Dicarboxylic Esters (Malonic Ester Synthesis)

O O

O O

malonic ester 2) R–Br (alkylation)3) H / H2O / heat (hydrolysis & decarboxylation)

COHO

RCH2

O O

O OO O

O O

R

HO OH

O O

R

hydrolysis2)

1)

heat

1) CH3CH2O (deprotonation)

–CO2

Other synthetic plans take advantage of the ease at which decarboxylation occurs when a carbonyl group is in the β position relative to the carboxylic acid. Here is a general synthetic route to carboxylic acids.

Alkylation of β-Ketoesters (Acetoacetic Ester Synthesis)

OEt

O O1) EtO Na2)

Br

OEt

O O

3) NaOH / H2O4) HCl / H2O

OH

O O

heat

O

+ CO2

acetoacetic ester

Here is a general synthetic route to methyl ketones.

The product of acetoacetic ester synthesis is a substituted acetone.!

Functional Group Relationships and Retrosynthesis

Michael

Aldol

Aldol + dehydration

Claisen

Acetoacetic acid synthesis

Malonic acid synthesis

Learning to recognize compounds as being products of a specific organic reaction is the process of retrosynthetic analysis - working a synthetic plan backwards.