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ALKYL & ARYL HALIDES

ALKYL HALIDES Alkyl halides are the derivatives of alkanes which can be obtained by replacing one hydrogen

atom by halogen atom. The general formula of monohalogen derivative of alkanes is CnH2n+1X (where X is any halogen atom). The monohalogen derivatives of alkanes are commonly called as alkyl halides. Common names of alkyl halides are derived from the name of parent hydrocarbon.

IUPAC NOMENCLATURE OF ALKYL HALIDES For writing IUPAC name of alkyl halide we use prefix halo (fluoro, chloro, bromo, iodo)

before the root word. Numbering of principal chain is done from that side which gives least sum to the position of substituents. The position of halogen is written before the prefix by using roman numerals (1, 2, 3 etc).

In case more than one halogen atoms are present, we use di, tri, tetra etc before prefix to

indicate the number of halogen atoms. In case of more than one type of prefixes, they are written in alphabetical order. It must be kept in mind that the di, tri, tetra, etc, are not considered for deciding the alphabetical order. Common names and IUPAC names of some alkyl halides are given as

Alkyl halide Common Name IUPAC Name CH3Cl Methyl chloride Chloromethane CH2Cl2 Methylene dichloride Dichloromethane CHCl3 Chloroform Trichloromethane CCl4 Carbon tetrachloride Tetrachloromethane C2H5Cl Ethyl chloride Chloroethane CH3CHCl2 Ethylidine chloride 1,1−Dichloroethane CH2ClCH2Cl Ethylene dichloride 1,2−Dichloroethane

METHODS OF PREPARATION OF ALKYL HALIDES

(I) FROM ALCOHOLS

(a) By using hydrogen halide

R − OH →HX RX + H2O It must be noted that the HX used should be dry, which is produced, in situ, as follows

2NaCl + H2SO4 →heat 2HCl↑ + Na2SO4



2NaBr + H2SO4 →heat 2HBr↑ + Na2SO4

6NaI + 2H3PO4 →heat 6HI↑ + 2Na3PO4 It may be noted that H3PO4 instead of H2SO4 is used to prepare HI. This is because HI is a

reducing agent and H2SO4 is an oxidising agent. The above substitution reactions proceed via SN1 and SN2 mechanism. Both the above

mechanisms are operative during reaction having competition among them.

R − OH →+H R − O − H

| H

1S

OH

N

2 →− R⊕ →−+X R − X +

(Some rearranged product if possible)

⊕

SN2 X−

R − X + H2O Details about SN1 and SN2 will be discussed later in this topic. (b) By using phosphorous halides ROH + PCl5 → RCl + POCl3 + HCl

3ROH + PCl3 → 3RCl + H3PO3

3ROH + PBr3 → 3RBr + H3PO3

3ROH + PI3 → 3R − I + H3PO3 Phosphorous halides are prepared by treating red phosphorous and halogen. The merit of

using phosphorous halides is that free carbocation is not generated and so rearrangement does not take place.

(c) By using SOCl2 (thionyl chloride)

R − OH + SOCl2 →Pyridine R − Cl + SO2↑ + HCl↑ The usefulness of this method is that there is no side product which has to be separated. The

side products are gaseous SO2 which escape from the reaction mixture and HCl which forms a salt with the base (pyridine) named pyridinium chloride ( −+ClNHC 55 ). The product (R−Cl) is formed with inversion of configuration in the presence of pyridine and with retention of configuration in absence of pyridine.

(II) BY DIRECT HALOGENATION OF HYDROCARBONS

R − H ν→

h2X R − X + HX

Reactivity of above reaction with respect to type of hydrogen to be replaced follows following order

Tertiary hydrogen > Secondary hydrogen > Primary hydrogen As far as the reactivity of halogen is concerned, F2 is most reactive while I2 is least reactive.

Infact reaction with I2 is reversible and is carried out in the presence of some oxidising agents



like HIO3, HNO3 etc. to oxidise HI. Mechanisms of these reactions has been discussed in the topic “alkanes”.

(III) BY HALIDE EXCHANGE (FINKELSTEIN REACTION)

Alkyl iodides can be prepared from alkyl chlorides and alkyl bromides by nucleophilic substitution. This can be achieved by treating them with NaI, using acetone as a solvent. Feasibility of this reaction is due to the solubility of NaI in acetone and more nucleophilic character of I− ion. Moreover, NaCl and NaBr are comparatively insoluble in acetone due to strong ion paring with respect to NaI.

R − Cl eAceton

NaI→ R − I + NaCl

R − Br Acetone

NaI→ R − I + NaBr

(IV) BY ADDITION OF HX TO ALKENES

Alkyl chlorides, bromides and iodides can be prepared by treating an alkene with corresponding hydrogen halides (HCl, HBr and HI). The addition of these compounds to alkene takes place according to Markovnikov's rule as discussed earlier in the topic “alkenes”. The reaction proceeds by electrophilic addition of +H to give more stable carbocation followed by attack of X−. Anti−Markovnikov addition of HBr can be achieved if reaction is carried out in presence of peroxide (H2O2 or benzoyl peroxide). This reaction proceeds by the attack of bromine free radicals (.Br).

CH3 − CH = CH2 →HX CH3 − ⊕

CH − CH3 →−X CH3 − CH − CH3

| X

CH3 − CH = CH2 Peroxide

HBr → CH3 − .CH − CH2 − Br . H CH2 − CH2 − CH2X

(V) FROM SILVER SALTS OF CARBOXYLIC ACID

RCOOAg + X2 → 4CCl RX + AgX + CO2 (X2 = Cl2 or Br2) This reaction is popularly known as Hunsdieker reaction. Mechanism The probable mechanism is the free−radical one in which acyl hypohalite is formed first

which then decomposes into free radicals.

RCOO2Ag + X2 → RCOOX + AgX↓

RCOOX → RCOO• + X•

Step (i): RCOO• → R• + CO2

Step (ii): R• + RCOOX → RX + RCOO• etc. Then steps (i) and (ii) repeated again to get the chain reaction.



In Hunsdieker reaction, Br2 in CCl4 gives better yield than Cl2 in CCl4. The yield and ease of formation of R −X is 1°RX > 2°RX > 3°RX If I2 is used in Hunsdieker reaction in CCl4, then a ester is formed. This is known as

Birnbaum−Simonini Reaction.

2RCOOAg + I2 → R − C − OR + 2AgI + CO2

O

But if acids are used with I2 and red mercuric oxide, alkyl iodides are obtained.

2RCO2H + 2I2 + HgO → 2RI + 2CO2 + HgI2

(VI) PREPARATION OF ALLYLIC OR BENZYLIC HALIDES

(i) Direct halogenation of any aromatic hydrocarbon preferably gives benzylic halide. This is because benzyl radical is resonance stabilized.

CH2 − CH3

ν→

h

X 2

X − CH − CH3

(ii) If we treat alkene with halogen at high temperature or in presence of radiations or any reagent which is able to provide halogen radicals in low concentrations, then allyl halides are formed.

CH3 − CH = CH2 →•X X − CH2 − CH = CH2

CH3 − CH = CH2 + Cl2 Cto ° →

500400

etemperaturhigh CH2 − CH = CH2 | Cl

CH3 − CH = CH2 + Br2 →NBS CH2 − CH = CH2 | Br

CH3 − CH = CH2 → 2SOCl CH2 − CH = CH2 | Cl

(VII) PREPARATION OF POLYHALIDES

(i) Addition of halogen to alkenes produces vicinal dihalides.

CH2 = CH2 + X2 → = )BrorCl(X 222 CH2 − CH2 | | Cl Cl

(ii) Addition of halogen (Cl2 or Br2) to alkynes produces tetrahalodervatives.



HC ≡ CH → 2X CH = CH | | X X

→ 2X

X X | | H − C − C − H | | X X

Mechanism of addition of halogen to alkenes has been discussed in the topic “alkenes”.

(VIII) PREPARATION OF VINYL HALIDES

CH ≡ CH 2CuCl

HCl → CH2 = CH | Cl

The above reaction proceeds via electrophilic addition as follows

CH ≡ CH →+2Cu

CH CH →−Cl

Cu2+

Cu+ | CH = CH | Cl

→+OH3 CH2 = CH

| Cl



GENERAL PHYSICAL PROPERTIES OF THE ALKYL HALIDES Because of small polarity in alkyl halides (carbon−halogen bond is polar due to difference in

electronegativities of the two atoms) as well as greater molecular weights, alkyl halides have considerably higher boiling points than alkanes containing same number of carbon atoms. Branching leads to decrease in surface area due to which Van der Waals forces decrease. So boiling point of isomeric alkyl halides decreases with branching. For similar alkyl halides with different halogen atoms, boiling point increases with molecular mass (alkyl iodides have maximum boiling point whereas alkyl fluorides have minimum boiling point).

Inspite of small polarity, alkyl halides are very less (negligibly) soluble in water, because of their inability to form hydrogen bonds. However alkyl fluoride are comparatively more soluble in water. Their solubility further decreases with increase in the size of alkyl group. They are soluble in typical organic solvents.

The density of alkyl halides is more than that of alkane and decreases from alkyl iodides to alkyl fluorides. Alkyl iodides and alkyl bromides are denser than water whereas monochloro alkanes and monofluoroalkanes are lighter than water. More the number of halogen atoms more will be the density if carbon skeleton remains same.

GENERAL CHEMICAL PROPERTIES OF THE ALKYL HALIDES

Carbon halogen bond in alkyl halide is polar owing to partial positive charge on carbon and partial negative charge on halogen. Due to the positive charge on carbon, any nucleophile can attack the molecule and if it is a stronger nucleophile than halide, it can substitute halide. So nucleophilic substitution reactions are the most common reactions of alkyl halides.

(I) PREPARATION OF ORGANOMETALLIC COMPOUNDS

R − X ether

Mg→ RMgX

(Grignard reagent)

R − X ether→Li R − Li + LiX

(Organo lithium compound) Mechanism:

R − X + Li → [R − X]− + +Li

[R − X]− → R• + X−

R• + Li → R − Li The extra electron in free radical anion occupies an antibonding molecular orbitals. It must be noted that the reactivity of alkyl halides to form organometallic compounds

increases from alkyl fluorides to alkyl iodies. Infact alkyl fluorides do not respond to the reaction and alkyl chloride and aryl halides give considerable yield only if THF (Tetrahydrofuran) is used as a solvent. Further, more the stability of free radical, more will be reactivity of alkyl halide to form organo metallic compound.



Alkyl organometallic compounds are good sources of carbanions (as carbon is more electronegative than the metals) and are very useful in organic synthesis.

(II) REDUCTION OF ALKYL HALIDES

R − X →Reduction R − H The reagents used for reduction of alkyl halide to alkane are Zn and CH3COOH, Zn and

HCl, Zn and NaOH, Zn − Cu couple and alcohol, aluminium amalgam and alcohol etc.

(III) WURTZ REACTION

2R − X + Na →ether R − R Details about this reaction had already been discussed during preparation of alkanes.

(IV) DEHYDROHALOGENATION OF ALKYL HALIDES

R − X → KOHalcoholic Alkene CH3 − CH − CH2 − CH3 → KOHalcoholic CH3 − CH = CH − CH3 + CH2 = CH − CH2CH3 (major) (minor)

X Details of this reaction had been discussed during preparation of “alkenes”. A very important reaction of alkyl halides i.e. Aliphatic nucleophilic substitution is dealt as a

separately heading.

ALIPHATIC NUCLEOPHILIC SUBSTITUTION In nucleophilic substitutions, the attacking reagent (nucleophile) brings an electron pair to the

substrate, which uses this pair to form a new bond and the leaving group comes away with an electron pair. When nucleophile is a solvent, the reaction is called solvolysis.

R − X + Y → R − Y + X Y is a nucleophile which may be neutral or negatively charged while RX may be neutral or

positively charged substrate. R − I + HO− → R − OH + I−

R − I + NMe3 → R− 3MeN+

+ I−

3MeNR+

− + OH− → R − OH + NMe3

3MeNR+− + H2S → R −

+S H2 +NMe3



R − OH + NaX or KX

R − O − R' + NaX

R − SH + KX or NaX

R − S − R' + NaX

RNH2 + HX

R − NHR' + HX

R − C ≡ C — R'

R − NR'2 + HX

R − C ≡ N + R − N C + X−

R − O − N = O + R − NO2 + X−

R − I + X−

aq. NaOH or KOH

or moist Ag2O R'O−Na+

KSH or NaSH

R'S−Na+

NH3

R'NH2

R'2NH

−C ≡ C — R'

I−, acetone

−2NO

R — X

AgCN/KCN

CH−(CO2Et)2 R − CH(CO2Et)2 Before going on for discussion of the mechanism of nucleophilic substitution reactions in

alkyl halides, we must have an idea about the few basic terms. These are mentioned below.

Basicity and Nucleophilicity

Nucleophilicity is defined by a reaction of a :Nu with an electrophilic C and it influences the rate of reaction as reflected by the rate constant, kr. Basicity is the ability to remove H+ from an acid as represented quantitatively by the equilibrium constant kb. Basicity of the nucleophile determines the thermodynamics (equilibrium) of the reaction,

Nu:− + RX NuR + X:− ; nucleophilcity the kinetics (rate). Nucleophilicity Basicity

Nu:− + C − L → Nu − C + :L− B:− + H − A B − H + :A−

In nonpolar and weakly polar aprotic solvents, the salts of Nu:− are present as ion−pairs (or ion−clusters) in which nearby cations (counterions) diminish the reactivity of the anion. Since ion−pairing is strongest with the smallest anion, F− and weakest with the largest anion, I− the reactivity of X− decreases with decreasing size. In polar protic solvents, H−bonding attenuates the reactivity of the anion, more so with the smallest anion, again making it the least reactive. Polar aprotic solvents solvate only cations, leaving free, unencumbered (“naked”) anions. The reactivities of all anions are greatly enhanced but more so for the smallest one because it is the most basic. F− > Cl− > Br− > I− and nucleophilicity and basicity are directly related. Apparently, the inverse relationship of nucleophilicity and basicity demonstrated.

Bases are better nucleophiles than their conjugate acids. Basicity and nucleophilicity are directly related−they both decrease. They are inversely related nucleophilicity increases while basicity decreases.



When the nucleophilic and basic sites are the same atom (here on O), nucleophilicity parallels basicity.

These relative rates are counter to relative basicities. When the atom bonded to the nucleophilic site also has an unshared pair of electrons, For example, G: − Nu:, nucleophilicity of the species increases.

SN1 REACTION

SN1 stands for substitution nucleophilic unimolecular. The ideal version of SN1 consists of two steps. The first step is the slow ionization of the substrate and is rate determining step. The second step is the rapid reaction between the carbocation and nucleophile.

R − X →RDS +R + X− Step 1 (slow) +R + Y− → R − Y Step 2 (fast) The ionization is always assisted by solvent, since the energy of activation required for

breaking of the bond is largely recovered from solvation of ions produced. For example, ionization of t−BuCl in gas phase requires about 150 kcal/mol but in water it requires only 20 kcal/mol. In pure SN1 reactions, solvent molecules assist the departure of leaving group from the front side.

The carbocation generated by first step has an 2sp hybridized carbon i.e. the structure is flat (trigonal planar). Thus nucleophile will attack the carbocation from the front side as well as from the rear side with equal ease, leading to the formation of two isomers, if the chiral carbon is present in the substrate.

STEREOCHEMICAL ASPECTS OF SN1 MECHANISM Let us understand the mechanistic details of SN1 reaction in the following manner:

−−

→Br

.s.d.r

slow

A planar carbocation is formed during the slow rds of the reaction, thus if may be expected

that the nucleophilic attack by OH− (or solvent H2O) would occur from either side of the carbocation leading to the formation of 50/50 mixture of substituted products having same and opposite configuration as that of starting material, i.e. racemisation will occur leading to an optically inactive (±) product.

In practice, the complete racemisation is never observed. It is always accompanied by some degree of inversion. The relative proportion of the two depend on (a) the stability of



carbocation (b) the ability of solvent as nucleophile (c) the ionizing ability of the solvent and (d) the bulkiness of the groups on the substrate.

Till the formation of free carbocation, the nucleophile attacks from the rear side of the compound leading to inverted product while once the carbocation is formed the probability of attack from both sides are equally feasible. As a result, overall inverted product exceeds the retented product.

SN2 REACTION MECHANISM SN2 stands for substitution nucleophilic bimolecular. In this mechanism, there is a backside

attack of the nucleophile. The nucleophile approaches the substrate from a position 180° away from the leaving group. The reaction is a one−step process with no intermediate. The C − Y bond is formed as C − X bond is broken.

Y− + C − X → [Y C X]# → Y − C + X− δ− δ− RDS

Transition State The energy necessary to break the C−X bond is supplied by the simultaneous formation of

C− Y bond. The group X must start leaving as Y comes in, because at no point of time can the carbon atom have more than 8 electrons in its outermost shell.

In the transition state, the carbon is sp2 hybridized with a p atomic orbital available for overlapping of its lobes with an orbital of the incoming :Nu− while the other lobe overlaps with an orbital of the leaving group X−. These overlaps account for the partial bonds drawn in figure. The reaction is initiated by :Nu− beginning to overlap with the small lobe (tail) of the sp3 hybrid orbital bonding with X. In order to provide more bonding volume to give a stronger bond, the tail becomes the larger lobe (head) and the head becomes the tail, inverting the configuration of carbon. The configuration of the original compound is the opposite to that of the compound obtained provided the leaving group and the attacking nucleophile have the same priorities. As :Nu− starts to bond to carbon, it loses some of its full charge and in the transition state has a δ− charge, as does X as it begins to leave as an anion.

In SN2 mechanism, the front side attack has never been observed. In a hypothetical front side attack, both the nucleophile and nucleofuge (leaving group) would have to overlap with the same lobe of p−orbital whereas the back side attack involves the maximum amount of overlap throughout the course of reaction. During the transition state, the three non−reacting groups and the central carbon atom are approximately coplanar.

Nu: C

R

H D

sp3 hybridized

X X C

R

H D

Nu δ−

sp2 hybridized with p orbital

δ−

Nu C

D H

+ :X

R

sp3 hybridized

S Configuration

Transition State R Configuration

(Nu and X of same priority)

#

In the above representation it is clear that the carbon atom is linked to five groups so it is

highly overcrowded. Due to high overcrowding it is easier to make a decision that more bigger the groups attached to carbon, more unstable will be the transition state. Hence steric



factors will have important role in SN2 substitution. Due to this reason the reactivity towards SN2 is as follows.

R | R − C − X | R

<

H | R − C − X | R

<

H | R − C − X | H

<

H | H − C − X | H

SN2 reactions are stereospecific because stereoisomeric reactants give stereochemically

different products. They are also stereoselective because they form exclusively or predominantly only one of a possible pair of enantiomers or one of the possible diastereoisomers.

1. COMPARATIVE DETAILS OF SN1 AND SN2 REACTIONS

SN1 SN2

(a) Number of steps 2 steps (i) R : L →slow +R + : L− (II) +R + : NuH → fast R:Nu + +H

1 step

R : L + : Nu− → R : Nu +: L− OR

R : L + : NuH → R : uHN+ +: L−

(b) Reaction rate & order Rate = k1[RL]; first order Rate =k1[RL][:Nu−]; second order

(c) Molecularity Unimolecular Bimolecular

(d) TS of slow step HNu: ……. Cδ+ ……. δ− L ……. HNu: δ−Nu ……. C ……. :Lδ− (with : Nu−) δ+HNu…….C…….:Lδ−(with :Hnu)

(e) Stereochemistry Inversion and retention (Partial racemization)

Inversion (backside attack)

(f) Reacting nucleophile Nucleophilic solvent; stable R+ may react with added nucleophile

Added nucleophile

(g) Structure of R 3° > 2° > 1° > Me Me > 1° > 2° > 3°

(h) Nature of Leaving group

Weakest base is best leaving group, i.e. I− > Br− > Cl− > F−

Weakest base is best leaving group, i.e. I− > Br− > Cl− > F−

(i) Solvent effect Rate α H−bonding ability and dielectric constant

Depends on charge type. Polar aprotic solvents leaves "freest" most reactive Nu.

(j) Determining factor Stability of R+ Steric hindrance

(k) Rearrangement Observed Not observed, except for allylic

(l) Catalysis Lewis and Bronsted acids: Ag+, AlCl3 and strong HA

No specific catalyst



ILLUSTRATION

CH3 − C − CH2Br CH3

CH3 C2H5O−

SN2

C2H5OH SN1

Identify the products of both reactions. SOLUTION EtOH is polar protic solvent which will favour SN1 reaction for the given alkyl halide

although it is 1° alkyl halide. In first case, C2H5O− will be the nucleophile which is a stronger base also, causing more elimination than substitution. So, elimination product is the major product in first case.

In the second case, it is C2H5OH itself acting as nucleophile, so it is a case of ethanolysis (solvolysis) C2H5OH is a weak base and a weak nucleophile, which causes more of substitution than elimination. Thus substitution product is the major one in second case.

CH3 − C − CH2−Br

solventproticPolar

EtOH → Br− + CH3

CH3 CH3 − C − CH2

CH3

CH3

(Substitution product)

→−3CH~ ⊕

CH3 − C − CH2CH3 CH3

⊕

EtOH −H+

C2H5O− −EtOH

CH3 − C − CH2CH3 CH3

OEt CH3 − C = CH −CH3 CH3

(Elimination product)

(1°)

2. AMBIDENT NUCLEOPHILES

Some nucleophiles have lone pair of electrons on more than one atom and can attack through more than one site. Such nucleophiles are called ambident nucleophiles. In such cases different products due to attack by different sites are possible. Attack by a specific site can be promoted under special conditions. Two well known examples are discussed in detail.

Attack by CN− nucleophile (:−C ≡ N :)

R − X →−CN R − CN + R − NC + X−

nitriles isonitriles In CN−, carbon (negatively charged) will be a soft base as compared to nitrogen. So if the

reaction proceeds via SN1 mechanism, which produces a free carbocation (a hard acid), then attack through nitrogen (hard base) will take place. But if the reaction proceeds via SN2 mechanism (small positively charged carbon is soft acid) then attack through carbon (soft base) will take place. So if we want to increase relative yield of nitriles, we can use NaCN or KCN etc in a less polar solvent which facilitates SN2 substitution. Similarly if we want to increase the yield of isonitriles, we use AgCN. Ag+ has very high affinity for X− so it favours the formation of +R and the reaction proceeds via SN1 mechanism. This will result in attack



by hard base giving R−NC. Further if we compare primary, secondary and tertiary alkyl halides, formation of R − NC should be favoured due to more favourable SN1 substitution in tertiary alkyl halide. But the exception is that tertiary alkyl halides undergo elimination and the yield decreases. This is because CN− is a strong base which can cause elimination reaction easily.

Attack by −2NO nucleophile (−O − N = O)

R − X →−2NO R − O − N = O + R − NO2 + X−

alkane nitrite nitro alkane

In −2NO , oxygen (negatively charged) will be a hard base as compared to nitrogen. So if the

reaction proceeds via SN1 mechanism, then attack through oxygen (hard base) will take place to produce alkane nitrite. But if the reaction takes place via SN2 mechanism then attack through nitrogen (soft base) takes place to give nitro alkane.

If we want to increase the yield of nitro alkane the reaction should proceed via SN2 mechanism. i.e. we can use NaNO2, KNO2 etc. Moreover the yield will be best if we use primary alkyl halide and less polar solvent. Formation of nitrite will dominate if we use tertiary alkyl halide, more polar solvent and AgNO2 because Ag+ has strong affinity for X− and can form a carbocation to force the reaction to proceed via SN1 mechanism. Primary alkyl halide with AgNO2 chiefly gives nitroalkane but if secondary and tertiary alkyl halides are used then AgNO2 will yield nitrite as the major product.

COMPARATIVE DETAILS OF E1 AND E2 REACTIONS E1 reaction E2 reaction

(a) Steps (i)

H − C − C − X → H − C − C+ +

(ii) H − C − C+ →+−H C = C

B:− + H − C − C − X → B : H + C = C + : X

(b) Transition state (i) H − C − Cδ+ Xδ− HS:

(ii)

HS:δ+ H C Cδ+

B:δ− H C C Xδ− HS:

(c) Kinetics First order, unimolecular Rate = k1[RX]; rate of ionization

Second−order, bimolecular

Rate = k2[RX][:B−]

(d) Driving force

Ionization of R − X Attack by B:− on H

(e) Stere− ospecificity

Non−stereospecific Anti elimination

(f) Effect of R Stability of R+ 3° > 2° > 1° RX

Alkene stability (Saytzeff Rule)



(g) Rearr− angement

Common None

(h) Regios− electivity

Saytzeff Usually, Saytzeff but Hofmann with bulky bases (Me3CO−)

(i) Alkyl group

3° > 2° > 1° 3° > 2° > 1°

(j) Base strength

Weak Strong

(k) Leaving group

Weak base favours reaction

I− > Br− > Cl− > F−

Weak base favoures reaction

I− > Br− > Cl− > F−

(l) Catalysis Ag+, AlCl3 No specific catalyst

COMPARATIVE DETAILS OF E2 AND SN2 REACTIONS

Factor Favors E2 Favors SN2

(1) Alkyl group 3° > 2° > 1° 1° > 2° > 3°

(2) Leaving group I− > Br− > Cl− > F− I− > Br− > Cl− > F−

(3) Reagent Strong, bulky Bronsted base Strong nucleophile

(4) Solvent (i) Low polarity (ii) Protic Polar (iii) Aprotic polar

Favored Disfavored Strongly favored

Slightly favored Disfavored Favored

COMPARATIVE DETAILS OF E2 AND SN1 REACTIONS

Factor Favors E2 Favors SN1

(1) R group 3° > 2° > 1° 3° > 2° > 1°

(2) X I− > Br− > Cl− > F− I− > Br− > Cl− > F−

(3) Base Strength Concentration

Strong High

Very weak Low



ILLUSTRATION (a) Give the product from the reaction of Mg/Et2O with (i) BrCH2CH2Br, (ii)

BrCH2CH2CH2Br and (iii) BrCH2CH2CH2CH2Br. (b) Discuss the difference in behavior of (ii) and (iii) in part (a). (c) Give the product of the reaction of CBr4 + MeLi + cyclohexene. SOLUTION (a) (i) H2C = CH2 (elimination of vic X’s), (ii) cyclopropane, (iii) BrMg(CH2)4MgBr (b) Unlike compound (ii), compound (iii) does not react intramolecularly because a 4−

membered ring does not form as readily as does a 3−membered ring. (c) CBr4 + MeLi

exchange

metalenlogha → [Br3CLi] + MeBr,

Br3CLi → Br2C: + LiBr Dibromocarbene The carbene then adds to the alkene in the typical fashion giving

Br

Br

7,7−Dibromo[4.1.0]−bicycloheptane

ILLUSTRATION Give the organic products of the following reactions

(a) n−PrBr+ N −O − →acetone

O

(b) i−PrBr + [ SC ≡ N ]− (isocyanate) →acetone

(c) EtBr + [ SSO3]2− (thiosulfate) →acetone (d) ClCH2CH2CH2I + CN− (One mole each) → (e) H2NCH2CH2CH2CH2Br →

+−H SOLUTION The nucleophiles in (a), (b) and (c) are ambident since they each have more than one

reactive site. In each case, the more nucleophilic atom reacts even though the other atom may bear a more negative charge.

(a) n−PrNO2 (b) i−PrSCN (c) [EtSSO3]− (d) ClCH2CH2CH2CN. I− is a better leaving group than Cl−.

(e)

N

H

. When the nucleophilic and leaving groups are part of the same molecule, an

intramolecular displacement occurs if a three, a five− or a six−membered ring can form.



ARYL HALIDES

Aryl halides are compounds containing halogen attached directly to an aromatic ring. The general formula of aryl halides is ArX, where Ar is phenyl, substituted phenyl or aryl groups.

m-bromonitrobenzene Chlorobenzene Iodobenzene

Cl I Br

NO2

p-chloronitrobenzene o-bromobenzoic acid

Cl

NO2

CO2H Br

An aryl halide is not just any halogen compound containing an aromatic ring. For example,

benzyl chloride is not an aryl halide, because halogen is not directly attached to the aromatic ring. Its structure and properties resemble substituted alkyl halide.

The aryl halides differ a lot from alkyl halides in their preparation and properties. Aryl halides

are comparatively unreactive towards the nucleophilic substitution reactions, which are characteristic of the alkyl halides. However, the presence of certain groups on the aromatic ring greatly increases the reactivity of aryl halides towards nucleophilic substitution. In absence of such groups, reaction can still be brought about by very strongly basic reagents or under drastic conditions. Nucleophilic aromatic substitution follow two very distinct pathways.

(1) The bimolecular displacement mechanism−involving the carbanion as intermediate and applicable to activated aryl halides.

(2) The elimination addition mechanism−involving the remarkable intermediate called benzyne and applicable to unactivated and deactivated aryl halides.

Vinyl halides are the compounds in which halogen is attached directly to a doubly bonded

carbon.

— C = C — X

A vinyl halide show an interesting parallelism to aryl halides. Both are unreactive towards nucleophilic substitution reactions, for basically the same reason. Moreover, this low reactivity is caused partly by the partial double bond character of the carbon-halogen bond.

STRUCTURE AND REACTIVITY OF ARYL AND VINYL HALIDES The low reactivity of aryl and vinyl halides towards nucleophilic displacement has been

attributed to two different factors.



(a) delocalization of electrons by resonance and (b) differences in σ bond energies due to difference in hybridization of carbon.

Cl

(I)

Cl

(II)

Cl

(III)

+ H

Cl

(IV)

+

H

Cl

(V)

+ H

Chlorobenzene is considered to be hybrid of not only the two Kekule structures I and II, but

also structures, III, IV, and V, in which chlorine is joined to carbon by a double bond. In structures III, IV, and V chlorine bears a positive charge and the ortho and para positions bear a negative charge.

In a similar way, vinyl chloride is considered to be a hybrid of structure VI and structure VII

(in which chlorine is joined to carbon by a double bond). In (VII), chlorine bears a positive charge and C-2 bears a negative charge. Other aryl and vinyl halides have structures exactly analogous to these.

CH2 = CH − Cl

(VI)

CH2 − CH = Cl ⊕

(VII) Contribution from III, IV and V, and from VII stabilizes the chlorobenzene and vinyl chloride

molecules, and gives a double bond character to the carbon−chlorine bond. Carbon and chlorine are thus held together by something more than a single pair of electrons, and the carbon chlorine bond is stronger than if it were a pure single bond. The low reactivity of these halides towards nucleophilic substitution is due to resonance stabilization of the halides. This stabilization increases the Eact for displacement, and thus slows down reaction.

In alkyl halides, the carbon holding halogen atom is sp3 hybridized. In aryl and vinyl halides,

carbon is sp2 hybridized, the carbon−halogen bond is shorter and stronger and thus the

molecule is more stable. Chlorobenzene and bromobenzene have dipole moments of only 1.7 D, and vinyl chlorides

and vinyl bromides have dipole moments of only 1.4 D. This is consistant with the resonance picture of these molecules. In the structures that contain double bonded halogen (III, IV, V and VII) there is a positive charge on halogen and a negative charge on carbon, to the extent that these structures contribute to the hybrids, they tend to oppose the usual displacement of electrons towards halogen. Although there is still a net displacement of electrons towards halogen in aryl halides and in vinyl halides but it is less than other organic alkyl halides.



PREPARATION OF ARYL HALIDES (I) FROM DIAZONIUM SALTS

C6H6 42

3

SOH

HNO → C6H5NO2HCl/Sn

.redn → C6H5NH2 C0

HONO

° → C6H5N2

+

C6H5F

C6H5Cl

C6H5Br

C6H5I

−4BF ,∆

CuCl, ∆

CuBr, ∆

I−, ∆

(II) BY HALOGENATION OF ARENES OR SUBSTITUTED ARENES ArH + X2 → acidLewis ArX + HX where X2 = Cl2 or Br2 Lewis acid = FeCl3, AlCl3, Tl (OAc)3 etc. For example,

NO2

→ 32 FeCl,Cl

NO2

Cl

NHCOCH3

→ 2Br

NHCOCH3

Br Acetanilide p−bromo acetanilide

(Major product)

m−chloronitrobenzene

(III) FROM ARYLTHALLIUM COMPOUNDS ArH + Tl(OOCCF3)3 →− HCOCF 23 ArTl(OOCCF3)2

∆→KI ArI

Aryl thallium trifluoro acetate This method is applicable for the preparation of aryl iodides only.



For example,

→ 33 )OOCCF(Tl

p-iodotoluene Toluene

CH3

∆→KI

CH3

I

→ 33 )OOCCF(Tl

CO2H

∆→KI

I CO2H

Benzoic acid o-iodobenzoic acid

The preparation of aryl halides from diazonium salts is more important than direct halogenation for several reasons. First of all, fluorides and iodides, which can seldom be prepared by direct halogenation, can be obtained from the diazonium salts. Secondly, where direct halogenation yields a mixture of ortho and para isomers, the ortho isomer, atleast, is difficult to obtain pure. On the other hand, the ortho and para isomers of the corresponding nitro compounds, from which the diazonium salts ultimately come, can often be separated by fractional distillation. For example, the o- and p-bromotoluenes boil only three degrees apart but the corresponding o- and p-nitrotoluenes boil, sixteen degrees apart.

GENERAL CHEMICAL PROPERTIES OF THE ARYL HALIDES LOW REACTIVITY OF ARYL AND VINYL HALIDES

An alkyl halide is conveniently detected by the precipitation of insoluble silver halides when it is warmed with alcoholic AgNO3. The reaction occurs instantaneously with tertiary alkyl or benzyl bromides, and within five minutes or so with primary and secondary bromides. But bromobenzene or vinyl bromide can be heated with alcoholic AgNO3 for days without the slightest trace of AgBr being detected.

− C = C − X

OH−

OR−

ArH,

AlCl3

X

Aryl Halide

or

Vinyl Halide

Ag+

CN−

NH3 No reaction under

normal conditions.

The typical reaction of alkyl halides, is nucleophilic substitution. R − X +:Z− → R − Z + :X−



where Z = OH−, OR−, NH3, CN−, −2NH , ROH, H2O etc.

It is typical of aryl halides that they undergo nucleophilic substitution only with extreme difficulty. Except for certain industrial processes where very severe conditions are feasible, one does not ordinarily prepare phenols (ArOH), ethers (ArOR), amines (ArNH2), on nitriles (ArCN) by nucleophilic attack on aryl halides. The aryl halides cannot be used in the Friedel-Crafts’s alkylation reaction just like alkyl halides, which can be used.

However, aryl halides do undergo nucleophilic substitution readily if the aromatic ring

contains, in addition to halogen, certain other properly placed groups. The presence of electron withdrawing groups like −NO2, −CF3 at ortho or para position to the halogen atom makes the aryl halides more susceptible to nucleophilic attack.

The reactions of unactivated aryl halides with strong bases or at high temperature proceed via the intermediate benzyne. The Dow’s process used for the manufacture of phenol involves benzyne intermediate.

In aromatic ring to which halogen is attached can of course, undergo the typical electrophilic aromatic substitution reactions like nitration, sulphonation, halogenation, Friedel−Craft alkylation. Halogen is unusual in being deactivating, yet ortho and para-directing.

(I) FORMATION OF GRIGNARD REAGENT ArBr + Mg → etherdry ArMgBr ArCl + Mg → furantetrahydro ArMgCl (II) ELECTROPHILIC AROMATIC SUBSTITUTION Although halogen is deactivating but still it directs the incoming electrophile to ortho and

para position. For example,

Cl

+ HNO3

Cl NO2

→ 42 SOH +

Cl

NO2

Cl Cl CH3

3

3

AlCl

ClCH → +

Cl

CH3

Cl Cl Br

→ Fe/Br2 +

Cl

Br



(III) NUCLEOPHILIC AROMATIC SUBSTITUTION 1. Bimolecular SNAr Mechanism ArX + Z− → ArZ + X−

For facile reaction Ar must contain strongly electron withdrawing groups at ortho and/or para position to the halogen atom. The reaction involves the formation of intermediate as carbanion.

Cl NO2

atm10,C300

OHHC 52

° →

OC2H5 NO2

Cl NO2

atm5,C150

OHHC 52

° →

OC2H5 NO2

NO2 NO2

Cl NO2

atm2,C75

OHHC 52

° →

OEt NO2

NO2 NO2

O2N O2N

Reaction proceeds by carbanion intermediate. The rate of the reaction increases with the

increase in number of electron withdrawing groups, since the carbanion formed would be stabilized more. The mechanism of the reaction is of addition−elimination type.

Step: 1

X NO2

)RDS(slowNu →

−

Nu NO2

X Nu NO2

X Nu N

X O

O

Nu N

X O−

O

Step: 2

Nu NO2

fastX→

−

Nu NO2

X



It can be seen that the presence of −NO2 group at ortho or para position would facilitate to disperse the negative charge of the carbanion, thus stabilizing it and making the reaction to occur fast.

2. Benzyne or Elimination−Addition Mechanism

ArX + :Z− → basestrong ArZ + :X− Ring not activated toward bimolecular displacement.

Cl

conditionsDrasticpressurehigh

C300,NaOH → °

O−Na+

→+H

OH

Cl

3

2

NH.liqin

KNH →

NH2

Cl

pressurehigh

C200,OCu,NH 23 → °

NH2

Unactivated and deactivated aryl halides undergo nucleophilic substitution by elimination-

addition (benzyne) mechanism. Unactivated aryl halide means either no substitutent is present or the presence of electron withdrawing group at meta position. Deactivated aryl halides mean the presence of electron releasing groups at any position. Rate determining step of the reaction is the formation of benzyne intermediate.

Br

stepslowNH,Br

NH

3

2

−− −

−

→ fast

NH: 2 →−

NH2

H Benzyne

fast,NH

NH

2

3−− →

NH2

For example,

OCH3 Br

HBr

NH2

− →−

OCH3

→ 3NH

OCH3

NH2 (a) →

−2NH:

Cl 14

HCl

NH2

− →−

NH2

(b) NH2

14

14

14



Chlorobenzene with chlorine bonded to C14 gives almost 50% aniline having NH2 bonded to C14 and 50% aniline with NH2 bonded to normal C12 atom. In benzyne mechanism, the

formation of product is governed by the inductive effect and resonance effect plays insignificant role.

CH3 Cl

3

2

NH,Cl

NH

−− →−

→ 3NH NH2

(c) →−2NH

CH3 CH3

OCH3

Cl 3

2

NH,Cl

NH

−− →−

→ 3NH NH2

(d) →−2NH

OCH3 OCH3

CH3

Cl 3

2

NH,Cl

NH

−− →−

→ 3NH NH2

(e) →−2NH

CH3 CH3

OCH3

Cl

3

2

NH,Cl

NH

−− →−

→ 3NH

NH2

(f) →−2NH

OCH3 OCH3

CH3

Cl

3

2

NH,Cl

NH

−− →−

→ 3NH NH2

(g) →−2NH

CH3 CH3

(IV) FITTIG REACTION It is an extension of wurtz reaction and consists of heating an ethereal solution of

bromobenzene with metallic sodium.

Br + 2Na + Br → + 2NaBr

(V) ULLMANN BIARYL SYNTHESIS Iodobenzene on heating with copper in a sealed tube forms biphenyl.

I + 2Cu + I → + 2CuI

Chloro− or bromobenzene fails to undergo this coupling reaction unless some strong

electron withdrawing group (e.g. NO2) is present in the ortho or para position.



SOLVED OBJECTIVE EXAMPLES

EXAMPLE 1

R − Cl + Nu:− → R − Nu + Cl− Reactivity order of different nucleophiles (C2H5O−, CH3COO−, OH−) is in order (a) C2H5O− < CH3COO− < OH− (b) CH3COO− < OH− < C2H5O− (c) C2H5O− < OH− < CH3COO− (d) CH3COO− < C2H5O− < OH− SOLUTION If the nucleophilic atom is same, then nucleophilicity parallels basicity. As the acidic order

follows the order ROH < H2O < CH3COOH Thus the basic strength order would be C2H5O− > OH− > −

23COCH and same is the order of nucleophilicity.

∴ (b) EXAMPLE 2

HO H

Et

Me H Cl

Et

Me

II

OH ←−

I

OH →−

H

Et

Me OH + HO H

Et

Me

Steps I and II are (a) both SN1 (b) both SN2 (c) I−SN1, II − SN2 (d) I−SN2, II−SN1 SOLUTION In step (I), there is racemization (formation of product with retention and inversion of

configuration both) while in step (II), there is inversion of configuration hence step (I) is SN1 and step (II) is SN2.

∴ (c) EXAMPLE 3

CH3

H H

Br * 2S

OH

N →−

CH3

H HO

H *

If the configuration at the chiral carbon in the reactant is R, the configuration of chiral carbon (asterik marked) in the product would be

(a) R (b) S (c) R and S both (d) none SOLUTION In all SN2 reactions, inversion is observed if the leaving group and attacking nucleophile

have same priorities. In the given case, it is so. ∴ (b)



EXAMPLE 4

CH3 − C − CH2Br

CH3

CH3

OHHC

1S

52

N → Product

This major product would be

(a)

CH3 − C − C2H5

CH3

OC2H5 (b)

CH3 − C = CH − CH3

CH3

(c) both are correct (d) none is correct SOLUTION

The intermediate formed in the reaction is a stable 3° carbocation

CH3 − C − CH2CH3

CH3

⊕

(after rearrangement). Since EtOH is a weak base and a weak nucleophile, it can not cause elimination but can cause nucleophilic attack to give the product (a).

∴ (a) EXAMPLE 5 (I) CH3CH2CH2CH2Cl (II)

CH3CH2 − CH − CH3

Cl

(III)

CH3 − CH − CH2Cl

CH3 (IV) (CH3)3C − Cl

Increasing tendency for SN1 and SN2 reaction is (A) SN1: I < III < II < IV (B) SN2: IV < II < III < I (a) A and B both are correct (b) only A (c) only B (d) both incorrect SOLUTION For SN1, order of reactivity is 1° < 2° < 3° For SN2, order of reactivity is 1° > 2° > 3° ∴ (a) EXAMPLE 6 In the reaction of p−chlorotoluene with KNH2 in liq. NH3, the major product is (a) o−Toluidine (b) m−Toluidine (c) p−Toluidine (d) p−Chloroaniline SOLUTION

Cl

CH3 Cl,NH

KNH

3

2

−− →

CH3

→−2NH

CH3

NH2 →+ 3NH

CH3

NH2

m−toluidine

∴ (b)



EXAMPLE 7 A hydrocarbon (A) having molecular weight 70 gives a single monochloride but three

dichlorides on chlorination in the presence of ultra violet light. The hydrocarbon (A) is (a) 2−pentene (b) cyclopentane (c) 2−methyl−2−butene (d) methylcyclobutane SOLUTION Since the hydrocarbon (A) gives only a single monochloride, it implies that all the C − H

bonds must be of the same type. So the given hydrocarbon should be cyclopentane.

light

uv/Cl2 →

Cl

light

uv/Cl2 →

Cl Cl

+

Cl Cl

+

Cl

Cl (Molar mass = 70)

∴ (b) EXAMPLE 8 In the following groups, (I) −OAc (II) −OMe (III) −OSO2Me (IV) −OSO2CF3 the order of leaving group ability is (a) (I) > (II) > (III) > (IV) (b) (IV) > (III) > (I) > (II) (c) (III) > (II) > (I) > (IV) (d) (II) > (III) > (IV) > (I) SOLUTION The anions of strong acids, i.e. weak conjugate base always act as good leaving groups

particularly in polar solvent. Since the decreasing order of acidic strength is following. F3CSO3H > CH3SO3H > CH3COOH > CH3OH The order of leaving group ability is F3C −

3SO > CH3−3SO > CH3COO− > H3CO−

∴ (b) EXAMPLE 9

Br

OMe

→ 2NaNH (A). The product (A) is

(a)

Br

NH2

(b)

Br

NH2

(c)

OMe

NH2

(d)

OMe

NH2



SOLUTION Since the reactant is deactivated aryl halide (due to the presence of OCH3 group which has

greater +R effect than its −I effect), so in the presence of strong nucleophile like NaNH2 it undergoes nucleophilic substitution by benzyne mechanism.

Br

H

OMe

3

2

NH

NH

− →−

OMe 1

2

3 →−2NH

NH2 OMe

+ NH2

OMe

(more stable) NH3

NH2

OMe

∴ (d) EXAMPLE 10

Et

Br Me

Me

?OH%20

EtOH%80

2

→

The major product of the following reaction is

(a)

Et

OEt Me

Me

(b)

Et

OH Me

Me

(c)

Et

Me − C = CH2 (d)

Me

Me − C = CH − CH3 SOLUTION The solvent employed is a mixture of 80% EtOH and 20% H2O but out of ethanol and

water, water is more polar protic solvent. So for the given 3° alkyl halide, the ionization is caused more by H2O then by EtOH. This means that the 3° carbocation formed will be solvated more H2O than by EtOH, so water being closer to the reaction site attacks to give nucleophilic substitution product as alcohol. H2O is a weak base, not suitable for elimination.

∴ (b)



SOLVED SUBJECTIVE EXAMPLES EXAMPLE 1

CH3 − C − CH2Br CH3

CH3 C2H5O−

SN2

C2H5OH SN1

Identify the products of both reactions. SOLUTION In SN2 reaction, the rate determining step involves the simultaneous bond making and

bond breaking through a transition state. Hence there is no scope of rearrangment and the product in neopentyl ethyl ether.

CH3 − C − CH2Br + C2H5O− →slow →fast CH3

CH3 CH3 − C − CH2OC2H5

CH3

CH3 neopentyl ethyl ether

SN1 reaction involves the formation carbonium ion in the rate determining step. It can undergo rearrangement for better stability as follows.

CH3 − C − CH2Br slow

1SN→ CH3 − C − CH2⊕

CH3

CH3 CH3

CH3

1, 2 methyl shift

CH3 − C − CH2CH3 ← OHHC 52 CH3 − C − CH2CH3

tert−pentyl ethyl ether

1° carbonium ion

CH3 CH3

OC2H5 ⊕

3° carbonium ion (more stable)

−H+

CH3 − C = CH − CH3

CH3

2−methyl−2−butene

Hence, the rearranged product will be obtained. Since the carbonium ion is sterically hindered, elimination will occur at a faster rate than

the nucleophile attack by C2H5OH. So the alkene will be the major product.



EXAMPLE 2 When CH3 − CH = CH − CH2Cl reacts with alcoholic KCN, a mixture of isomeric

products is obtained. Explain. SOLUTION It can undergo both SN1 and SN2 reaction. By SN2 reaction only one product is formed.

But by SN1 reaction, intermediate is carbonium ion, which may lead to rearrangment for better stability.

CH3 − CH = CH − CH2Cl + CN− slow

2SN→

(A)

fast

CH3 − CH = CH − CH2CN

CN

CH3 − CH = CH − CH2CN

SN1 slow

CH3 − CH = CH − CH2 →← CH3 − CH − CH = CH2 ⊕

⊕

CN− CN−

CH3 − CH − CH = CH2 (B) (A)

δ−Cl ……… C ……… δ−CN

CH

CH

CH3

H H Transition state

2° Carbonium ion (more stable)

(minor) (major)

Thus we get two isomeric products using SN1 reaction. EXAMPLE 3 Give the major product (with proper explanation) when following halogen compounds

are treated with sodium ethoxide?

(a)

CH3

CH3 − CH − CHCH3

Br (b)

CH3

Cl

(c) CH2Br

CH3

SOLUTION (a)

CH3

CH3 − CH − CH − CH3 1S

slow

N

→ CH3 − CH − CH − CH3 Br CH3

⊕

2° carbonium ion (A)

→ shifthydride2,1 CH3 − CH2 − C − CH3 →−

52HOC CH3 − CH2 − C − CH3 CH3

OC2H5

3° carbonium ion (more stable)

(minor)

⊕

CH3



CH3 CH3 − CH − CH − CH3

inationlime

H→+− CH3 − CH = C − CH3

CH3

⊕

Due to elimination (major)

(A)

Since sodium ethoxide is a strong base, it has a tendency to abstract proton to give alkene rather than a nucleophilic attack.

(b)

CH3

2° carbonium ion (A) Cl

→slow CH3

⊕

→ shifthydride2,1 CH3 ⊕

3° carbonium ion

CH3

OC2H5 (minor) CH3

(major)

H

C2H5O−

H

⊕ CH3

−H+

(c) CH2Br

CH3 →slow

CH2⊕

CH3 → shiftmethyl2,1

CH2CH3 ⊕

H H

1° carbonium ion 3° carbonium ion (more stable)

−H+

→

−52HCO

CH2CH3

OC2H5

CH2CH3

(major)

(minor) EXAMPLE 4 Identify (A), (B), (C), (D) (E) and (F) in the following series of reactions.

ν→h

Br2 (A) → KOH.aq (B) →Na (C)

alc. KOH

(D) →NBS (E) →+ )C( (F)



SOLUTION

ν→h

Br2

Br

→ KOH.aq

OH

→Na

ONa

(A) (B) (C)

alc. KOH

→NBS

Br

(D) (E)

ONa

+

Br

synthesis

Williamson →

O

(C) (E) (F) EXAMPLE 5 1−Fluoro−2,4−dinitrobenzene is highly reactive toward nucleophilic substitution

through an SNAr mechanism. What product would be formed when 1−fluoro−2,4−dinitrobenzene reacts with each of the following reagents?

(a) CH3CH2ONa (b) C6H5NH2 (c) NH3 (d) CH3CH2SNa SOLUTION

(a)

F NO2

NO2

+ CH3CH2ONa →

OCH2CH3 NO2

NO2

(b)

F NO2

NO2

+ NH3 →

NH2 NO2

NO2



(c)

F NO2

NO2

+ C6H5NH2 → no reaction

Since lone pair on N−atom of aniline will not be available for nucleophilic attack as it is in resonance with benzene ring.

(d)

F NO2

NO2

+ CH3CH2SNa → no reaction

Due to large size of S−atom, the overall nucleophile becomes a bulky one and attack of it one the given substrate given a sterically hindered species which is not stable at all. Hence, the substitution will occur.



Objective Problems 1. The order of reactivities of the following alkyl halides for a SN2 reaction is (a) RF > RCl > RBr > RI (b) RF > RBr > RCl > RI (c) RCl > RBr > RF > RI (d) RI > RBr > RCl > RF 2. Which compound has the lowest boiling point? (a) n-hexyl chloride (b) n-butyl chloride (c) sec-butyl chloride (d) t-butyl chloride 3. The rate law for the reaction is given by Rate = k' [RCI] RCl + NaOH (aq) → ROH + NaCl The rate of reaction will be (a) doubled on doubling the concentration of sodium hydroxide. (b) halved on reducing the concentration of alkyl halide to one-half. (c) decreased on increasing the temperature of the reaction. (d) unaffected by increasing the temperature of the reaction. 4. The compound (A) in the reaction is CH2 = CH − CH2 Cl + OH− → (A). (a) ethyl chloride (b) ethyl alcohol (c) ethylene and methyl alcohol (d) allyl alcohol 5. Which of the following anion has the highest nucleophilicity in polar protic solvent? (a) F− (b) OH− (c) −

3CH (d) −2NH

6. Which of the following compound shows greater reactivity towards SN1 reaction? (a) CH3Cl (b) C2H5Cl (c) CH2 = CHCl (d) CH2 = CH−CH2Cl

7.

Cl

H3C CH3

on treatment with aqueous KOH give

(a)

Cl

H3C CH3

OH

(b)

OH

CH3

CH3

(c)

H3C OH CH3 (a)

Cl

H3C CH3

OH



8. The order of relative reactivity of the given halides towards SN2 reaction is (a) PhCH2Cl > PhCHCl(CH3) > PhCCl(CH3)2 (b) PhCH2Cl < PhCHCl(CH3) < PhCCl(CH3)2 (c) PhCHCl(CH3) > PhCH2Cl > PhCCl(CH3)2 (d) PhCHCl(CH3) > PhCCl(CH3)2 > PhCH2Cl 9. In the reaction ClCH2CH2CH2Br + KCN

Δ

EtOH.Aq →

(1 mole) (a) CNCH2CH2CH2Br (b) ClCH2CH2CH2CN (c) CNCH2CH2CH2CN (d) CNCH2CH = CH2 10. 1−Phenyl ethyl chloride undergoes alkaline hydrolysis to give 1−phenyl ethanol with (a) complete inversion of configuration. (b) racemization plus some inversion. (c) retention of configuration. (d) complete racemization. 11. What is the product of following reaction?

Cl

Br OEt

)equivalent1(Mg

2

→ ?

(a) MgCl

Br (b)

MgBr

Cl

(c)

(d) None of these

12. Which one of the following compound is most rapidly hydrolysed by SN1 mechanism? (a) CH3CH=CHCl (b) ClCH2CH = CH2

(c) (C6H5)3CCl (d) C6H5CH2Cl 13. In the reaction, CH3 − CH = CH2

− →

OH/OH)2(

HB)1(

22

62 (A) →HBr (B) →KCN (C). The compound (C) is

(a) CH3CH2CH2Br (b) CH3CH2CN (c) CH3CH2CH2CN (d) CH3CH2COOH 14. In the reaction, CH3CH2CH2Br

OHHC

NaOH

52

→ (A) →HBr (B) → OH/OAg 22 (C). The compound (C) is

(a) Propene (b) Propyne (c) 1-Propanol (d) 2-Propanol

15. The product which is formed in the reaction Me3CCH2OH →HBr (a) Me3CCH2Br (b) Me2CHCH = CH2 (c) Me3C−CH3 (d) Me2C(Br)CH2Me



16. Compared to ethyl chloride, vinyl chloride is less reactive in SN reactions because (a) hyperconjugative effect is high in ethyl chloride. (b) resonance effect is very high in ethyl chloride. (c) in vinyl chloride, the chlorine atom is attached to sp2 carbon atom and in ethyl

chloride, it is attached to sp3 carbon atom. (d) there is partial double bond character between carbon and chlorine atom in vinyl

chloride due to resonance. 17. In the following reaction,

H EtO acetone

NaI→

Ph

H Ph

Br

the structure of the product is

(a)

I H

EtO H Ph

Ph

(b)

Br H I H

Ph

Ph

(c)

I

EtO H Ph

Ph

H

(d)

Br H

I Ph

Ph H

18. The halide which undergoes nucleophilic substitution most readily is (a) p-H3CC6H4Cl (b) o -H3COC6H4Cl (c) p-ClC6H4Cl (d) p-O2NC6H4Cl 19. An SN2 reaction at an asymmetric carbon of a compound always gives (a) an enantiomer of the substrate. (b) a product with opposite optical rotation. (c) a mixture of diastereomers. (d) a single stereoisomer. 20. In the reaction, CH3 − CH − CH3

Br → KOH.alc (A)

Peroxide

HBr→ (B) acetone

NaI→ (C)

The compound (C) is (a) CH3CH2CH2I (b) (CH3)2CHI (c) CH3 − CHI − CH2I (d) CH3CH = CH − I 21. Which of the following compound give 2−iodopropane with HI? (a) CH3CH2CH3 (b) CH3CH = CH2 (c) CH2OHCH(OH)CH2OH (d) C2H4



22. Which represents the correct decreasing order of leaving ability of groups? (a) −

23COCCl > −23COCH > PhO− > −

3PhSO (b) −23COCCl > −

3PhSO > −23COCH > PhO−

(c) −3PhSO > −

23COCCl > −23COCH > PhO− (d) PhO− > −

3PhSO > −23COCCl > −

23COCH 23. An unknown alkyl halide (A) reacts with alcoholic KOH to produce a hydrocarbon

(C4H8). Ozonolysis of the hydrocarbon affords one mole of propionaldehyde and one mole of formaldehyde. Suggest which organic structure among the following is the correct structure of the above alkyl halide (A)?

(a) CH3(CH2)3Br (b) CH3CH(Br)CH(Br)CH3 (c) CH3CH2CH(Br)CH3 (d) Br(CH2)4Br 24. The product of the following reaction is

+ CHBrClF →+−KCO)CH( 33 ?

(a)

H K⊕ (b)

Br

(c)

Cl

(d)

F

25. When (CH3)3CCl and (CD3)3CCl are allowed to undergo SN1 reaction separately. It is

observed that (a) (CH3)3CCl undergoes reaction faster than (CD3)3CCl. (b) (CD3)3CCl undergoes reaction faster than (CH3)3CCl. (c) Both undergoes reaction at the same rate. (d) Both undergoes reaction by SN2 mechanism.



Subjective Problems 1. An aromatic compound (A) C8H9Br reacts with Na+CH−(CO2C2H5)2 gives (B). Compound

(B) on refluxing with dil. H2SO4 gives (C) which on vigorous oxidation gives (D). The compound (D) is a dibasic acid but on heating does not give an anhydride. It forms a mononitro compound (E) in which all the substituents are at equidistant from one another. Give the structures of (A) to (E) with proper explanations.

2. Show how a benzyne intermediate accounts for the formation of m−MeOC6H4NH2

(C) from either o−MeOC6H4Br (A) or m−MeOC6H4Br (B).

3. What are compounds (A) to (C) ? Cyclohexene → 4CCl,NBS (A),C6H9Br

HCOCHCOK)CH(

23

33 → (B),C6H8 → = 32 CHCOCHCH

(C),C10H14O OH,Zn)ii(

O)i(

2

3 →

CHO

CHO

COCH3

4. Two isomeric organic compounds (A) and (B) (M.F. C4H8Br2) on hydrolysis gave two

compounds (C) and (D) of formula C4H8O. Both (C) and (D), on oxidation gave two acids (E) and (F) of formula C4H8O2. Both acids on heating with soda lime gave propane. Identify (A) to (F).

5. A compound (A) (C4H9Br) gives with hot alcoholic potash, three isomeric compounds (B),

(C) and (D). The minor product (D) on ozonolysis yields methanal as one of the products. What product(s) may be obtained from (B) and (C) on ozonolysis ? Write the structures of (A), (B), (C) and (D).

6. An organic compound (A), C4H8Br2 on heating with alcoholic KOH gives (B) of formula

C4H6, which decolourises Br2/CCl4 and cold dilute KMnO4 but does not react with ammonical AgNO3 and ammonical CuCl solutions. It on partial reduction by Na in liquid NH3 gives (C) while with Lindler’s catalyst gives (D) of formula C4H8. Both (C) and (D) on ozonolysis followed by oxidation gives two molecules of ethanoic acid. Identify the compounds (A), (B), (C) and (D).

7. Account for the rapid rate of ethanolysis of ClCH2OCH2CH3, although the substrate is a

1° halide.

8. An organic compound (A) (M.F. C7H15Cl) on treatment with alcoholic caustic potash gives a hydrocarbon (B) (M.F. C7H14). Compound (B) on reductive ozonolysis produced acetone and butyraldelyde. Compound (B) on addition of HBr in presence of benzoyl peroxide produced compound (C) which on treatment with KOH in tertiary butyl alcohol



gave compound (D). Oxidative cleavage of compound (D) gave (E) and (F). Identify (A) to (F).

9. 1,4−Pentadiene reacts with excess of HCl in presence of H2O2 to give compound (X). Compound (X) on reaction with Mg in dry ether produced compound (Y). Compound (Y) on treatment with excess of ethyl acetate followed by hydrolysis produced compound (Z) (M.F. C9H16O2). Identify (X), (Y) and (Z).

10. An organic compound C5H9Br (A) which readily decolorizes bromine water and KMnO4

solution, gives C5H11Br(B) on treatment with Sn/HCl. The reaction of (A) with NaNH2 produces (C) with evolution of ammonia. (C) neither reacts with sodium nor forms metal acetylides but reacts with Lindlar catalyst to give (D) and on reaction with Na/liquid NH3 produces (E). Both the compounds (D) and (E) are isomeric. Give structures of (A) to (E) with proper reasoning.

11. Two isomeric alkyl bromides (A) and (B), C5H11Br yield the following results in the

laboratory. (A) on treatement with alcoholic KOH gives (C) and (D), C5H10 (C) on ozonolysis gives formaldehyde and 2−methyl−propanal. (B) on treatment with alcoholic KOH gives (D) and (E), C5H10. Deduce structures of (A), (B), (C), (D) and (E). Ignore the possibility of geometrical and optical isomerism.

12. A white precipitate was formed slowly when silver nitrate was added to compound (A) with molecular formula C6H13Cl. Compound (A) on treatment with hot alcoholic potassium hydroxide gave a mixture of two isomeric alkenes (B) and (C) having formula C6H12. The mixture of (B) and (C) on ozonolysis furnished four compounds.

(i) CH3CHO (ii) C2H5CHO (iii) CH3COCH3 (iv) CH3 − CH(CH3) − CHO What are the structures of (A), (B) and (C) ? 13. Compound (A), C6H12 is treated with chloride in the presence of CCl4 to form compound

(B), C6H12Cl2. (B) is treated with alc. KOH followed by NaNH2 resulting in the formation of compound (C) C6H10. (C) is treated with hydrogen gas over platinum and forms a compound identified as 2−methylpentane. Compound (C) does not react with an ammonical solution of silver nitrate. On ozonolysis (A) gives two aldehydes (D) and (E). (E) is identified as acetaldehyde. From the above information, deduce the structural formulae of (A), (B), (C) and (D).

14. An unsaturated organic compound (A), M.F., C9H9Cl decolourises Br2 water and when

heated with NaNH2 in liquid NH3 produced another unsaturated compound (B) which does not exhibit geometrical isomerism, having M.F., C9H8. The compound (B) on treatment with dilute sulphuric acid containing mercuric sulphate give a compound C, which on vigorous oxidation with acidified KMnO4 produces 1,4-benzenedicarboxylic acid. Identify (A), (B), (C) and show how (C) is formed from (B). Treatment of compound (C) with methyl magnesium iodide followed by acidification produces a compound which on dehydration followed by catalytic hydrogenation produces D. Identify (D). An isomer (E) of (B) on reduction with Na in liquid ammonia produces a compound (F) which exihibits trans- stereochemistry. Write the structures of (E) and (F). What reagent would you use for converting (E) to (G) which is a geometrical isomer of (F). Give the structure of (G).



15. When bromobenzene is monochlorinated, two isomeric compounds (A) and (B) are obtained. Monobromination of (A) gives several isomeric products of molecular formula C6H3ClBr2, while monobromination of (B) yields only two isomers (C) and (D). Compound (C) is identical with one of the compounds obtained from the bromination of (A), however (D) is totally different from any of the isomeric compounds obtained from the bromination of (A). Give the structures of (A), (B), (C) and (D).



ANSWERS

Objective Problems

1. d 2. d 3. b 4. d 5. c

6. d 7. c 8. a 9. b 10. b

11. c 12. c 13. c 14. d 15. d

16. d 17. b 18. d 19. d 20. a

21. b 22. c 23. a 24. d 25. b

Subjective Problems

1.

CH2Br

CH3

→−+

2)COOEt(CHNa

CH2 − CH

CH3

CO2Et

CO2Et

(A) (B)

dil. H+/H2O

CO2H

CO2H NO2 42

3

SOHHNO ←

CO2H

CO2H

←Oxidation

CH2CH2CO2H

CH3

(E) (D) (C) 2. Both isomers react through the same intermediate benzyne (D). It is the only benzyne that

can arise from (A) but one of the two possibilities from (B).

Br

OMe

+

−

− →H

NH: 2

OMe

→−−Br

OMe Br

(A) (D)



more stable carbanion (E)

Br

OMe

+

−

−HNH: 2

OMe

Br

OMe

Br

→−−Br (D)

(less stable carbanion)

(B)

The negative charge in the carbanion is better stabilized when on 2C , nearer to the

electron−withdrawing MeO, as in (E) and not on 4C . Consequently, little or no benzyne forms with the triple bond between 3C and 4C . (C) results when NH3 bonds to 3C of (D), the negative charge again ending up on 2C .

more stable adduct

OMe

NH3

(D) → 3NH solventNH

withtransferH

3

→

OMe

NH2 +

(C)

3.

→ 4CCl,NBS

Br

(A)

HCOCHCOK)CH(

23

33 →

(B)

→ = 32 CHCOCHCH

CH − COCH3

CH2

ZnOH/O 23 →

CHO

CHO

COCH3

(C)

4. (A) & (B) : CH3 −CH − CHBr2

| CH3

and CH3CH2CH2CHBr2

(C) & (D) : CH3 −CH − CHO | CH3

and CH3CH2CH2CHO

(E) & (F) : CH3 −CH − COOH | CH3

and CH3CH2CH2COOH



5. (A) : CH3 −CH2 − CH−CH3 | Br

(B) & (C) :

C = C

CH3 H

CH3

CH3

H

and C = C

CH3 H

CH3

CH3

H

(D) : CH3 − CH2 − CH = CH2

(B) & (C) →Oxidation CH3CHO 6. (A): CH3 − CH − CH − CH3

Br Br

(B): CH3 − C ≡ C − CH3

(C): CH3 − C = C − CH3

H

H (D): CH3 − C = C − CH3

H H

7. The rapidity of this SN1 reaction is attributed to the stability of a +C bonded to —O—••

••.

The empty p atomic orbital on +C can overlap sidewise with a filled p atomic orbital on O, thereby delocalizing and stabilizing the positive charge. The +C then reacts with EtOH, giving an ether.

CH3CH2 − O − CH2Cl → Cl− + CH3CH2 − O − +

2CH CH3CH2 − O = CH2

+ +− →

H

OHHC 52 EtO − CH2 − OEt

8. (A):

CH3 Cl | | CH3 − C − CH2 − CH2CH2CH3 or CH3 − CH − CH −CH2CH2CH3 | | Cl CH3

(B): CH3 − C = CH− CH2CH2CH3 | CH3

(C):

Br | CH3 − CH − CH −CH2CH2CH3 | CH3

(D): CH3 −CH − CH= CH − CH2CH3 | CH3

(E) & (F):

O || CH3 −CH − CHO & H − C − CH2CH3 | CH3



9. (X): CH3 − CH − CH2 − CH − CH3

| | Cl Cl

(Y): CH3 − CH − CH2 − CH − CH3 | | MgCl MgCl

(Z):

O O || || CH3 − C − CH − CH2 − CH − C − CH3 | | CH3 CH3

10. (A):

Br Br | |

CH3 − CH2 − CH = C − CH3 or CH3 − CH2 − C = CH − CH3

(B):

Br Br | |

CH3 − CH2 − CH2 − CH − CH3 or CH3 − CH2 − CH − CH2 − CH3

(C): CH3 − CH2 − C ≡ C − CH3 (D):

C = C

CH3 H

CH3

H

CH3CH2

(E):

C = C

H

CH3

H

CH3CH2

11. (A): CH3 −CH − CH−CH3 | | Br CH3

(B):

Br | CH3 −CH2 − C−CH3 | CH3

(C): CH2 =CH − CH−CH3

| CH3

(D): CH3 − CH = C − CH3 | CH3

(E): CH3 − CH2 − C = CH2 | CH3

12. (A): CH3 − CH − CH − CH2CH3 | | CH3 Cl

(B) & (C): CH3 − C = CH − CH2 − CH3 | CH3

and

CH3 − CH − CH = CH − CH3 | CH3



13. (A): CH3 − CH − CH = CH − CH3 | CH3

(B): CH3 − CH − CH − CH − CH3 | | | CH3 Cl Cl

(C): CH3 − CH − C ≡ C − CH3 | CH3

(D): CH3 − CH − CHO | CH3

14. (A):

CH3

C = CH2 | Cl

(B):

CH3

C ≡ CH

(C):

CH3

C − CH3

O

(D):

CH3

CH − CH3 | CH3

(E):

C ≡ C − CH3

(F):

C = C

CH3

H

H

(G):

C = C

CH3

H

H

(E) quinoline

BaCOPd,H 32 → − (G)

15. (A):

Cl

Br

(B):

Cl

Br

(C):

Br

Cl Br

(D):

Br

Cl

Br

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