instructor: prof. duncan wardrop time/day: t, 4:30-8:15 p.m. september 9, 2010

University of Illinois at Chicago

UICCHEM 494 Special Topics in Chemistry

Instructor: Prof. Duncan Wardrop

Time/Day: T, 4:30-8:15 p.m.

September 9, 2010

Welcome to CHEM 494

UICUniversity of

Illinois at Chicago

2CHEM 494, Fall 2012 SlideLecture 1

Course Website

http://www.chem.uic.edu/chem494

Syllabus

Course Policies

Other handouts

Announcements (Course News)

Course Calendar

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Keys to Success

1. Attend all lectures and discussion sections.

2. Don’t fall behind. Organic chemistry is easy, but each topic

builds on the previous.

3. Don’t memorize. Organic chemistry is conceptual.

4. Always ask yourself “Why” for everything you read or hear.

You may not always find the answer, but just asking will help

you to find connections and remember more.



Introduction

Origins and Examples of Organic Chemistry

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Vitalism

Living

vs.

Nonliving.

•posses vital force•compounds derived from are “organic” (coined by J. Berzelius, 1807)

•could not be synthesized in the laboratory

•termed “inorganic”

•derived from nonliving

matter

•can be synthesized in the

lab

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Wohler Synthesis Debunks Vitalism

ammonium cyanate(inorganic)

urea(organic)

AgOCN(aq) + NH4Cl(aq) AgCl(s) + NH4OCN(aq)

F. Wohler(Germany, 1828)

N

H

H

HH O C NO

CNH2H2N

Expected Product Actual Product

X

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Self Test Question

What is the minimum requirement, today, for a chemical substance to be classified as organic?

A. derived from living matter

B. contains carbon

C. cannot be synthesized

D. bought at Whole FoodsE. combustion yields SO2

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Vitalism Lives On. . .

CH3

OH

terpinen-4-olsold as tea tree oil

or Melaleuca oil

fructose“fruit sugar”also in HFCS

HO O

OH

OH

OH

OH

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Organic Chemistry Everywhere

H3C O

O

3-methylbutyl acetate(bananas)

O

O

CH3

OH

methyl salicylate(oil of wintergreen)

OHO

O

O

CH3

aspirin(analgesic)

O

O

NH3C

cocaine(analgesic)

O

O

N

procaine, "Novocaine"(analgesic)

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10CHEM 494, Fall 2012 Slide

Lecture 1

Natural Products

Order: coleoptera

Family: coccinellidae

Order: Tetraodontoformes

NHN

H2N

OH

OHHO

OO

OHO

(-)-tetrodotoxin

N

C6H13

NCS

fasicularin

3

6 7

3'

A

BC

NH

O

(-)-adalinine

C5H11

O

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Lecture 1

Pharmaceuticals

N

F3C

CH3

S NH2

O

O

Tamiflu(influenza)

Celebrex(arthritis)

Morphine(analgesic)

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Lecture 1

Optics - Transition Lenses?

NC

O

N

NC

N

O

does not absorb visible

light

does absorb visible light

UVlight

heat



Atomic StructureA General Chemistry

Review

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Lecture 1

particle chargemass

molar masssymbol

proton positive1.6726 x 10-24 g1.0073 g/mol p

neutron neutral1.6750 x 10-24 g1.0087 g/mol n

electron negative9.1096 x 10-38 g

5.486 x 10-4 g/mol

e–

Atomic Composition

+++

-

-

-

+

Bohr Model-

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Lecture 1

Atomic Number & Mass Number

+++

-

--

XA

Z

A = protons + neutronsZ = protons

Li6

3For neutral molecules, the

number of electrons equals the number of protons.

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Lecture 1

Self Test Question

How many neutrons are in the following atom:

A. 14B. 6C. 8D. 20E. cannot be determined

C14

6

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Lecture 1

Tenets of Schrӧdinger Equation

• electrons have wave properties• wave equation gives energy of electron at a

location• solutions to wave equation are wave functions (Ψ);

a.k.a orbitals• probability distribution = Ψ2 (Heisenburg

uncertainty principle)

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Lecture 1

Probability Distribution vs. Boundary Surface

Probability distribution (Ψ2)(“electron cloud”)

Boundary Surface (1s orbital)

(where the probability = 90-95%)

1s 1s

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Lecture 1

Wave Function Values:Quantum Numbers

quantumnumber

principle orbital magnetic spin

symbol n l ml ms

values 1, 2, 3. . . 0, 1, . . . n-1(ml = −l, −l+1 ...

0 ... l−1, l)+1/2 or -1/2

examples/abbreviatio

ns

l= 0, sl = 1, pl = 2, dl = 3, f

s: 1 orbitalp: three orbitalsd: five orbitals

f: seven orbitals

•Shrodinger equation → wave function (orbital, Ψ)

•many solutions for Ψ, energies of electrons in atom

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Lecture 1

s-Orbitals• spherically symmetric

• possible for n≥1

• 1 s-orbitals for each value of n

• 1s = no nodes, 2s = 1 nodes, 3s = 2 nodes, etc.

• probability of finding s-electron at nodal surface = 0

• s-orbital energy increases with increasing nodes (as n increases)

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Lecture 1

p-Orbitals• shaped like dumbells

• not possible for n = 1 (n≥2)

• 3 p-orbitals for each value of n; they are degenerate (equal in energy)

• wave function changes sign at the nucleus (node)

• probability of finding p-electron at nodal plane = 0

• higher in NRG than s-orbitals of the same shell

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Lecture 1

Relative Energies of Orbitals

1s

2s2p 2p 2p

3s3p 3p 3p

n = 1

3d 3d 3d 3d 3d

n = 2

n = 3/4

4s

En

erg

y

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Lecture 1

Order of Orbitals from Periodic Table

1s2s 2p3s 3p4s 3d 4p5s6s7s

4d5d6d

5p6p7p

4f5f

Read left to right starting at the top left; just like a

typewriter.

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Lecture 1

Electron Configuration: Filling Orbitals

1s

2s2p 2p 2p

3s3p 3p 3p

n = 1

3d 3d 3d 3d 3d

n = 2

n = 3/4

1. Aufbau principle: fill lowest energy orbitals first.

2. Pauli exclusion principle: two electrons with same quantum numbers cannot occupy a single orbital. Electrons must have opposite spins in the same orbital.

3. Hund’s rule: degenerate orbitals filled singly first.

spin (ms = +1/2)

spin (ms = –1/2)

4s

En

erg

y

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Lecture 1

Self Test Question

Which of the following orbitals is highest in energy?

A. 3fB. 3pC. 2sD. 3sE. 2d

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Lecture 1

Self Test Question

Which is the correct electron configuration for carbon?

A. 1s2, 2s8

B. 1s2, 2s4

C. 1s2, 2s2, 2s2

D. 1s2, 2s2, 2p2

E. none of the above



Bonding and Molecular Structure

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Lecture 1

What is a Covalent Chemical Bond?

Forces involved:

•electron-electron repulsion

•nucleus-nucleus repulsion

•electron-nucleus attraction

valent = bondingcovalent = electrons shared between two

nuclei

electron-electron repulsion < electron-neutron attraction

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Lecture 1

Three Models of BondingLewis Model:

•atoms gain, lose or share electrons in order to achieve a closed-shell electron configurations (all orbitals fully occupied)•closed-shell for row 2 = 8 electrons (octet rule)•only valence electrons (in outermost shell) are involved in bonding

Atomic StructuresMolecular

Lewis StructuresMolecular

Lewis Structures

H–H

F–F

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Lecture 1

Three Models of BondingValence Bond Model:

•in-phase overlap of two half-filled orbitals•in-phase = constructive interference•increases probability electrons between two nuclei

H H H–HBoundar

y Surfaces

Electrostatic Potential

maps:red =

negativeblue = postive

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Lecture 1

Three Models of BondingMolecular Orbital (MO) Model:

•combine atomic orbitals (AO) of all atoms, then extract molecular orbitals•number of atomic orbitals in equals the number of molecular orbitals out•combination possibilites:•additive = produces bonding molecular orbitals•subtractive = produces antibonding molecular orbitals

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Lecture 1

Molecular Orbital Diagram

1s AO 1s AO

antibonding (σ*) MO

bonding (σ) MO

nodal planee- probability = 0

= no bond

H H

Rules Still Apply:1. Aufbau2. Pauli3. Hund

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Lecture 1

Self Test Question

Which is the correct Lewis dot structure for CH4O?

A. H C

H

H O H

O

CH H

HH

B.

C

H

O

H

H H

H C O

H

H

HC.

D.

E. none of the above

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Lecture 1

Structural Bond-line Condensed

•format varies; typically most covalent bonds are drawn out•bonds are lines•only lone-pair (nonbonding) electrons are drawn as dots

•only atoms written are those that are not C or H bound to C•intersection of two lines is C•terminus of a line is -CH3 group•# H atoms is assumed for C•chains drawn as “zig-zag”

•all or most covalent bond lines are omitted•groups are separated by parentheses (infers group is attached to carbon with available valency on left or right)

Structural, Bond-line and Condensed Formulas

CH3CH2CH2CHOCC

CC

H

O

HH

HHH

H H

CC

CH

HH

HOH

H H

H OHCH3CH(OH)CH3

CC

CC

HH

HH

H H

HHH

C

HH

O

OH

O

O

H (CH3)2CHCH2CH2CO2H

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Lecture 1

VSEPR: Quick Review

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Lecture 1

Self Test Question

Which bond-line drawing correctly represents the following condensed formula?

A.

B.

C.

D.

CH3CH(OCH2CH3)CH2Br

O

Br

O

Br

O

Br

O Br

Br

OE.



Functional Groups

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Lecture 1

Memorize These Functional Groups NOW

Table 4.1, page 140

functional group: a defined group of atoms with a specific connectivity •responsible for properties•predictable reactivity•well-defined nomenclatureMemorizing: “What is the minimum number of atoms needs to define a functional group and in what order are they bonded?” Flashcards!

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39

CHEM 494, Fall 2010 SlideLecture 2

Formal Charge - Example 1My logic (no memorizing equations):

1.Recognize that opposing charges cancel each other2.Determine difference in protons vs. electrons by asking:a.How many valence (outer shell) electrons does the atom have?b.How many valence (outer shell) electrons does the atom “want” (groups #)?1.If atom has more than it “wants,” negatively charged.2.If atom has less than it “wants,” positively charged.

O16

8--

-

+

+++

+ +

-

1s

2s

2p

-

-

-

-

-

-

+

+Has? =

Wants? =

Difference? =

8

6

2 extra

2–

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Formal Charge - Example 2My logic (no memorizing equations):

1.Recognize that opposing charges cancel each other2.Determine difference in protons vs. electrons by asking:a.How many valence (outer shell) electrons does the atom have?b.How many valence (outer shell) electrons does the atom “want” (groups #)?If atom has more than it “wants,” negatively charged.If atom has less than it “wants,” positively charged.

Has? =

Wants? =

Difference? =

5

6

1 less“formal” = count one electron for

each bond

H

O

H

H

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Shortcut for Some 1st and 2nd Period Atoms•For each additional bond

= +1•For each “missing” bond = –1

AtomBonds Forme

dGroup

H 1 I

C 4 IV

N 3 V

O 2 VI

F (X) 1 VII

H3CC

CH3

CH3

H3C

O

O

N

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Self Test Question

What is the formal charge on the carbon atom in red?

A. -2B. -1C. 0D. +1E. +2

C C

H

H

H

H

H

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Self Test Question

According to VSEPR, what is the bond angle between 2 H-atoms in methane (below)?

A. 105ºB. 109.5ºC.107ºD.180ºE. 120º

H

CH

HH?



Section: 1.8, 1.11You are responsible for Section 1.10.

Structure and Bonding: Resonance

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The Contradictory Case of Ozone

electrostatic potential map

red = negativeblue = positive

Microwave spectroscopy shows that ozone is symmetrical (C2v)

Lewis Structuresuggests ozone is non-symmetrical

OO

O

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Curious Case of Benzene

C–C bond length: 150 pmC=C bond length: 134 pm

H

H

H

H

H

H

Predicted Actual

All bonds = 140 pm

resonance

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Resonance Provides the SolutionProblems:

1.Lewis structures fail to describe actual atom electron densities for molecules with more than one possible electron distribution.

2.Lewis formulas show electrons as localized; they either belong to a single atom (lone pair) or are shared between two atoms (covalent bond).

3.Electrons are not always localized; delocalization over several nuclei leads to stabilization (lower energy).

H3C N

O

CH3

CH3

H3C N

O

CH3

CH3

H3C N

O

CH3

CH3

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Physical Proof of Resonance in Amides

C-N (amide) Bond Length = 133 pm

C-N (amine) Bond Length = 147 pm

C=N (imine) Bond Length = 129 pm

H3C N

O

CH3

CH3

H3C N

O

CH3

CH3

NH

HH

N

CH3H3C

H

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A Question of Arrows

Equilibrium between distinct species

Reaction from one species to another

“Movement” of electronsfrom donor to acceptorIndicates that 2 species are contributing resonance structures

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Resonance

Solutions:

1.Actual molecule is considered a resonance hybrid (weighted average) of contributing Lewis structures. (Note: double headed arrow does NOT indicate interconversion.)

2.Dashed-line notation is sometimes used to indicate partial bonds.

3.Not all structures contribute equally.

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Rules of Resonance

• How to draw resonance structures• How to distinguish between a resonance

structure and a unique electron configuration

• How to predict the major contributing structure

You are responsible for the “Rules of Resonance” (Pg. 27)

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Another Way to ThinkAbout Resonance Contributors

Dürer's Rhinoceros (1515 A.D.)

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Curved Arrow Notation

curved arrows show the

movement of electrons; never

atoms

electronsatoms

H3C N

O

CH3

CH3

resonance: electrons in a covalent bond

moving out to an atom

H3C N

O

CH3

CH3

resonance: lone pair of electrons

moving in between two

atoms to form a new covalent

bond

HO

H H H3O+

bond making: lone pair of

electrons forming a new bond to another atom

H3C O

OH

H3C O

O

+ H

bond breaking: electrons in a

bond leaving to most

electronegative atom

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Preview: Curved Arrows in Reaction Mechanisms

H3C O

O

HH

OH

+H3C O

O

+H

OH

H

bond making

bond breaking

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Another Way of Viewing Arrow-Pushing

Remember that curly arrows originate from e-donors (nucleophiles) and terminate at e-acceptors (electrophiles)

From an M.O. perspective curly arrows originate from filled orbitals (nucleophiles) and terminate at empty orbitals (electrophiles) - bonds are formed from the mixing of filled and empty orbitals

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Molecular Orbital Diagram

1s AO 1s AO

antibonding (σ*) MO

bonding (σ) MO

nodal planee- probability = 0

= no bond

H H

Rules Still Apply:1. Aufbau2. Pauli3. Hund

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Self Test Question

Which of the following is not descriptive of resonance as it pertains to organic chemistry?

A. stabilization of negative chargeB. movement of atomsC. delocalization of electronsD. involves bond-breakingE. involves bond-making



How Structure Affects Physical Properties:

Acid Strength

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Arrhenius Acids and Bases

Acid: Dissociates to provide protons (H+) in water.

Base: Dissociates to provide hydroxide (OH–) in water.

strong acid: ionizes completely

weak acid: does not ionize completely

strong base: ionizes completely

weak base: does not ionize completely

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Brønsted-Lowry Acids & Bases

Acid: Proton (H+) donorBase: Proton (H+) acceptor

• definition does not depend on dissociation in water• broader definition• conjugate acid and base differ by only one proton• most often used definition in this course

B : + H A B + AH :

Base Acid Conjugate Acid Conjugate Base

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Molecular Orbital Picture

B H A B +H:

σ* (H-A)n (B) σ (B-H)

A :

n (A)

Anti-Bonding M.O. Bonding M.O.

H-A Bond Breaks!

Non-Bonding M.O. Non-Bonding M.O.

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Water Can Act as a Base or Acid Under Brønsted-Lowry Definition

OH

HClH+

HO

H

H

Cl–+

NH

HO

H

HH

NH

H

OH–+

water (base)

HCl (acid)

hydronium ion (conjugate acid)

chloride ion (conjugate base)

amide (base)

water (acid)

ammonia(conjugate acid)

hydroxide ion (conjugate base)

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Acid Strength is Measured by the Acid Dissociation Constant (pKa)

pKa = –log10 Ka

H A + AH :

Acid Proton Anion

Note that the value of K reflects the relativethermodyamic stability of acid and anion

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pKa Scale

HCl

H3O+

NH3

H3C

O

OH

OH

CH3CH2O H

H2O

acid pKa

hydrochloric

-3.9

hydronium -1.7

acetic acid 4.7

pyridinium 5.2

phenol 10

water 15.7

ethanol 16

ammonia 36

ethane 62

NH3

CH3CH3

Take home message: the larger the Ka value, the smaller the pKa value,the stronger the acid.

-log10(1) = 0-log10(10) = -1-log10(100) = -2

[H+][A-][HA]-log10pKa =

HA H A–+acid conjugate

base

dissociation

-log10(1/10) = 1-log10(1/100) = 2

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Structure Affects Acid Strength

1. Electronegativity of H-Donating Atom •dominant effect for same periods (rows)•more electronegative conjugate base = more stable conjugate base = Ka lies further to right•Alternative reasoning: H of conjugate acid (HA) becomes more positive with increasing electronegativity of A

conjugate base

acid

CH3 NH2 OH F

least stable (highest energy)

conjugate base =

weakest acid

most stable (highest energy)

conjugate base =

strongest acid

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Structure Affects Acid Strength

C C O

H

H

H

H

H

H C C O

H

H

H

H

H

+ H

C C O

H

H

F

F

F

H C C O

H

H

F

F

F

+ H

Ka = 1 x 10-16

pKa = 16

Ka = 5 x 10-12

pKa = 11.3

1. Electronegativity (cont.) •inductive effect of electronegative substituents also stabilizes conj. base•the more electronegative the group, the greater the stabilization•the closer the electronegative group, the greater the inductive effect •inductive effect = structural effects that are transmitted through bonds

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Self Test Question

Which of the following carboxylic acids (-CO2H) is the most acidic (lowest pKa)?

A.

B.

C.

D.

E.

CC

O

O

HCl

H H

CC

O

O

HH

H H

CC

O

O

HC

H HC

Cl

H H

H H

CC

O

O

HF

H H

CC

O

O

HCl

H H

R O

O

carboxylate anion(conjugate base)

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Structure Affects Acid Strength2. Bond Strength

•dominant effect for same groups (columns)•Similar sized bonding orbitals = better overlap = stronger bond•Stronger H–A bond = weaker acid

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Self Test Question

Which of the following is the strongest acid?

A. H2O

S

O

B. H2S

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Structure Affects Acid Strength3. Resonance (delocalization of e–s in conj. base): •delocalization of e–s in conj. base = increased stability (lower NRG) of conjugate base•more stable conj. base = larger Ka = smaller pKa = stronger acid

Ka = 1 x 10-16

pKa = 16

Ka = 2.0 x 10-5

pKa = 4.7

H3CC

OH

H H

H3CC

O

H H+ H

H3CC

O

O

H3CC

O

O

H3CC

OH + H

H3CC

O

OO

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Acid-Base Equilibria: Determining the Direction of Acid-Base Reactions

You must identify the ACID on each side of the equilibrium:

pKeq = pKa (acid left) - pKa (acid right)

Keq = 10-[pKa (acid left) - pKa (acid right)]

• remember: p = -log10• this equation works for any acid-base reaction; doesn’t

matter which way equilibrium is written

OH

C

O

OHO+

KeqO

C

O

OHHO+

Keq = 10-[10 - 6.4]

phenol(pKa = 10)

hydrogen carbonate(pKa = 10.2)

phenoxide anion(pKa = NA)

carbonic acid(pKa = 6.4)

= 10-[3.6]

= 2.5 x 10-4

since Keq <1, then equilibrium lies to the left

acid acidbasebase

Example:

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Finally, What About Strengths of Bases?

Question: Is ammonia or pyridine a stronger base?

Solution: 1. Determine which conjugate acid of each base is the weakest.2. The weaker the conjugate acid, the stronger the conjugate base.

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Molecule of the Week...α-AmanitinBe Careful What You Eat (and Who

You Marry)!

Amanita phalloides

α-Amanitinα-Amanitin (red) boundto RNA polymerase II

Read more about the death of Claudius......

Tiberius Claudius Caesar Augustus Germanicus (10 B.C.–A.D. 54)

Amanita caesarea

instructor: prof. duncan wardrop time/day: t, 4:30-8:15 p.m. september 9, 2010

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