Ch18 Sm Chemistry2e

Solutions Manual

to accompany

Chemistry2nd Edition by Blackman, Bottle, Schmid,

Mocerino & Wille

Prepared by John Hill

© John Wiley & Sons Australia, Ltd 2012

Solutions Manual to accompany Chemistry 2e by Blackman et. al.

© John Wiley & Sons Australia, Ltd 2012 18.2

Chapter 18 - Haloalkanes

Practice exercises 18.1 (a) 1-chloro-3-methylbut-2-ene (b) 1-bromo-1-methylcyclohexane (c) 1,2-dichloropropane (d) 2-chlorobuta-1,3-diene 18.2 (a)

1-bromo-2,3-dimethylbutane

2-bromo-2,3-dimethylbutane (b)

1-chloro-2-methylcyclopropane

1-chloro-1-methylcyclopropane

chloromethylcyclopropane (c)

1-chloropentane

2-chloropentane

Chapter 18: Haloalkanes


3-chloropentane (d)

3-bromo-3,4,4-trimethylpentane



2-bromomethyl-2,3-dimethylpentane

3-bromomethyl-4,4-dimethylpentane

18.3

+Br CH3CH2S-Na+(a) S CH2CH3 + Na+Br-

(b) Br CH3CO-Na++O

OCCH3 + Na+Br-O



18.4

Br

Na+SH-+(a) acetone + NaBrSHSN2

The SH– is a good nucleophile and because the reaction involves a secondary

haloalkane with bromide anion as good leaving group, the reaction follows an SN2 mechanism. The SN2 results in the inversion of stereochemistry at the carbon bearing the leaving group.

Br

SH SH + Br

Mechanism:

(b) CH3CHCH2CH3

Cl

R EnantiomerHCOH

O

formic acidCH3CHCH2CH3

O

R and S Enantiomers

CHO

SN1

H3C CH2CH3

Cl HC CH2CH3

CH3

H+ Cl-

HCO

C CH2CH3CH3

H+ Cl-OH

attack

from above C CH2CH3HCH3

OCH

O

H

C CH2CH3H

CH3

OCH

O

H

C CH2CH3HCH3

OCH

O

C CH2CH3H

CH3

OCH

O

attack

from below

- H+

- H+

Mechanism:

S enantiomer

R enantiomer

+

+

+

+

A SN1 reaction is favoured for a poor nucleophile (formic acid) reacting with a

secondary halide. Formic acid is also an excellent ionising solvent. SN1 reactions result in racemisation.



18.5

(a) CH3

ClNaOCH2CH3

CH3CH2OH+ Na+Cl-

Majorproduct

Minorproduct

(b)

CH2Cl NaOCH2CH3

CH3CH2OH + Na+Cl-

(c)CH3Cl NaOCH2CH3

CH3CH2OH + Na+Cl-+

Equal amounts 18.6 These reactions proceed by an E2 mechanism. E2 mechanisms are favoured when

using secondary or tertiary halides with nucleophiles that are also strong bases like methoxide or ethoxide. If two or more alkenes are possible products, the most stable product is the major product according to Zaitsev’s rule. Trans alkenes are more stable than cis alkenes.

Br

(a) CH3O-Na++methanol

+ CH3OH

Major product

Minor product

(b) +

Cl

CH3CH2O-Na+ethanol

+ CH3CH2OH

18.7 Reaction (a) involves the methoxide anion, which is both a strong base and strong

nucleophile. Secondary halides in polar solvents can undergo either substitution or elimination reactions. SN2 reactions are favoured with strong nucleophiles and E2 is favoured with strong bases, therefore these two reactions are competing in (a). The E2 reaction favours the most stable alkene as the major elimination product, which is the trans isomer.



methanol(a) + CH3O- Na+

Br

Major productby E2 reacton

OCH3

Major productby SN2 reacton

Reaction (b) involves the iodide anion, which is a weak base and a strong

nucleophile. SN2 reactions are favoured when secondary halides react with strong nucleophilic anions that are also weakly basic in weakly ionising solvents. SN2 reactions proceed with complete inversion of stereocentres that are bonded to the leaving group, thus yielding the cis-1-iodo-4-methylcycloheane from trans-1-chloro-4-methylcyclohexane.

(b) +

acetone

Cl

Na+ I-

I

Review questions 18.1 (a) 1,1-difluoroethene (b) 3-bromocyclopentene (c) 1,6-dichlorohexane (d) 2-chloro-5-methlyhexane (e) dichlorodifluoromethane (f) 3-bromo-3-ethylpentane 18.2 (a) (S)-2-bromobutane (b) trans-1-bromo-4-methyl-cyclohexane (c) 3-chlorocyclohexene (d) (E)-1-chlorobut-2-ene (e) (R)-2-bromo-2-chlorobutane (f) meso-2,3-dibromobutane 18.3 (a)

(b) (c)

B r

C l H

B r B r H H



(d)

(e) (f)

18.4 (a) (b) (c) (d) CH2Cl2

(e) CHCl3 (f) (g) 18.5 2-Iodooctane and trans-1-chloro-4-methylcyclohexane are 2˚ haloalkanes.

(a) isobutyl chloride: 1˚ haloalkane

(b) 2-iodoctane: 2˚ haloalkane

B r

C l C l

B r

C l

B r

I

C l

C l



(c) trans-1-chloro-4-methylcyclohexane: 2˚ haloalkane

18.6 (a)

(b)

3-chloro-1,1-dimethylcyclohexane (c)



(d)

1-bromo-2-cyclohexyl-2-methylpropane

1-bromo-4-tert-butylcyclohexane

1-bromo-3-tert-butylcyclohexane

1-bromo-2-tert-butylcyclohexane 18.7 (a) CH2Cl2 (b)

(c) CH3CH2OH (d)

(e)



18.8 Alkyl groups decrease the polarity of solvents and therefore the order of increasing

polarity is: CH3CH2OH < CH3OH < H2O 18.9 The carbonyl group of acetone is a polar functional group, so acetone is the most

polar of the three. The oxygen atom of diethyl ether adds polarity to this solvent compared to the hydrocarbon pentane. The order of increasing polarity is:

pentane < diethylether < acetone 18.10 (a) OH–

(b) OH–

(c) CH3S– 18.11 (a) True: both nucleophile and haloalkane are reacting in the rate-determining

step.

(b) True: backside attack at the substitution centre results in inversion of configuration.

(c) True: SN2 reactions result in an inversion of stereochemistry at the

substitution centre, therefore, optical activity is retained in the product. Therefore, the product is optically active and rotates plane-polarised light to the same extent, but in the opposite direction of the original.

(d) False: steric crowding influences the order of reactivity in SN2 reactions. As

the number of substituent groups on the substitution centre increases, the reaction rate decreases. For SN2 reactions, the order of reactivity is: methyl > 1° > 2° > 3°.

(e) False: all nucleophiles are Lewis bases and therefore must have an unshared

pair of electrons. Moderate nucleophiles such as ammonia (NH3) react in SN2 reactions, even though they do not carry a negative charge.

(f) True: the nucleophile is involved in the rate-determining step; therefore, as

nucleophilic strength increases, the rate also increases. 18.12

(a)

BrNaOCH2CH3

CH3CH2OH + Na+Br-



(b) ClNaOCH2CH3

CH3CH2OH + Na+Br-

Majorproduct

Minorproduct

(c)

ClNaOCH2CH3

CH3CH2OH + Na+Br-

Majorproduct

Minorproduct

(d)Br NaOCH2CH3

CH3CH2OH+ Na+Br-

Majorproduct

Minorproduct

18.13

Cl

(a)-HCl

Major Zaitsev products: cis-trans isomers are possible

Br(b)

-HBr

Major Zaitsev products: cis-trans isomers are possible

(c) Cl-HCl No cis-trans isomers are possible

Cl(d) -HCl No cis-trans isomers are possible



18.14

(a) CH2Br -HBr CH2

(b) CH3CHCH2CH=CH2

CH3

CH3CHCH2CH2CH2BrCH3

-HBr

18.15 The carbocation that needs to be formed for an SN1 reaction mechanism is too

unstable for this to be a viable route to the product. 18.16 Potassium tert-butoxide is both a strong base and a hindered nucleophile.

Consequently H+ abstraction leading to alkene formation is favoured over substitution.

18.17 The major product is substitution to give 2-ethoxypropane. Elimination to give

propene is a minor side reaction. 18.18 The tertiary haloalkane, 2-bromo-2-methylpropane, is sterically hindered, but sodium

hydroxide is a strong nucleophile and a strong base. This combination leads primarily to E2 elimination to give the alkene. Steric hindrance favours abstraction as the nucleophile is unable to form a new bond. The increased numbers of C-C bonds in tertiary alkanes such as 2-bromo-2-methylpropane, favours alkene formation (from Zaitsev’s rule).

18.19 (a) SN1 (tertiary substrate)

(b)

(c) (d) No. The rate is not dependent on the concentration or strength of the

nucleophile.

(e)



18.20 (a) SN2

(b)

(c) (d) Yes. The reaction rate would double.

(e) Review problems 18.21



Br2

light

1-chloromethyl-1,3,3-trimethylcyclopentane

(b)Br

Br

Br2-bromo-1,1,3,3-tetramethylcyclopentane

1-bromo-2,2,4,4-tetramethylcyclopentane

Br2

heat

3,4-diethyl-1-bromohexane

(c)

Br

Br

Br



Br2

light

3-ethyl-1-bromopentane

(d)



Br

Br

Br



18.22

acetone(a) +Na+I- CH3CH2CH2Cl CH3CH2CH2I NaCl

ethanol+(b) BrNH3 NH3 Br

ethanol(c) + CH2=CHCH2ClCH3CH2O-Na+ CH2=CHCH2OCH2CH3 + NaCl

18.23

Cl

CH3CO- Na+O

ethanol(a) +

O

O + Na+Cl-

CH3CHCH2CH3

ICH3CH2S-Na+

acetone(b) +

S+ Na+I-

acetone(c) +CH3

CH3CHCH2CH2Br Na+I- + Na+Br-

I

acetone+(d) (CH3)3N CH3I (CH3)4N + I

+methanol

(e) CH2Br CH3O- Na+ CH2OCH3 + Na+Br-

ethanol+(f) H3C Cl CH3S- Na+ H3C

SCH3

+ Na+Cl-

(g) + ethanolNH CH3(CH2)6CH2Cl N

(CH2)7CH3

H+ + Cl

+(h) CH2Cl NH3 ethanol

CH2NH3+

+ Cl



18.24 (a) A 2˚ halide, a moderate nucleophile and a moderately ionising solvent all

favour an SN2 mechanism. (b) A 2˚ halide, with ethyl thiolate as a good nucleophile that is a weak base and

a weakly ionising solvent all favour an SN2 mechanism. (c) A 2˚ halide, with iodide as a good nucleophile that is a weak base and a

weakly ionising solvent all favour an SN2 mechanism. (d) Methyl halides can only undergo SN2 reactions because their extremely

unstable carbocations preclude SN1 reactions. Trimethylamine is a moderate nucleophile and acetone is a weakly ionising solvent, further supporting an SN2 mechanism.

(e) A 1˚ halide, with a good nucleophile like methoxide favours an SN2

mechanism. (f) A 2˚ halide, with methyl thiolate as a good nucleophile that is a weak base

and a moderately ionising solvent all favour an SN2 mechanism. (g) Piperidine is a moderate amine nucleophile and ethanol is a moderately

ionising solvent. SN2 mechanisms are favoured with moderate nucleophiles reacting with 1˚ haloalkanes.

(h) Ammonia is a moderate nucleophile and ethanol is a moderately ionising

solvent. An SN2 mechanism is favoured in this case because the haloalkane is primary.

18.25 When comparing nucleophilicity of atoms in the same row of the periodic table, as in

reactions (a) and (b), nucleophilicity increases with basicity of the atom (oxygen is less basic than nitrogen). For reaction (c), when comparing atoms in the same column of the periodic table, nucleophilicity increases from top to bottom; therefore, sulfur is more nucleophilic than oxygen.

(a) HOCH2CH2NH2 + CH3I ethanol HON

CH3

H H

++ I

(b)N

O

H

+ CH3Iethanol

N

O

H CH3

++ I



(c) HOCH2CH2SH + CH3I ethanol HOS

CH3+ HI

18.26 (a) False: formation of the carbocation is the higher activation energy which

controls the rate of reaction. Once formed, the carbocation reacts directly with the nucleophile.

(b) False: the carbocation formed is planar and has lost all stereochemistry.

Addition of the nucleophile to the carbocation can occur from either side leading to the two stereochemically different enatiomers, usually as a racemic mixture.

(c) True: the carbocation formed is planar and has lost all stereochemistry.

Addition of the nucleophile to the carbocation can occur from either side leading to the two stereochemically different enantiomers, usually as a racemic mixture.

(d) True: methyl carbocations are so difficult to form that they have never been

detected in solution. (e) False: although this statement is generally true, for SN1 reactions the

carbocation is flat and so steric bulk on the three groups does not normally impact on the rate at which the nucleophile can approach the charged carbon atom. The rate-determining step Is the rate at which the starting material loses the leaving group to form the carbocation.

(f) False: again the rate of formation of the carbocation governs the rate of the

reaction. 18.27

(a) +ethanol

Cl

CH3CHCH2CH3 CH3CH2OH

S enantiomer

OCH2CH3

OCHCH3

Racemic mixture

+ HCl

(b) + methanolCH3OHCl

CH3

OCH3+ HCl



acetic acid+(c) CH3COHCH3CCl

CH3

CH3

O

O

O+ HCl

(d)methanol

+ CH3OHBr

OCH3

OCH3

+ HBr

Racemic mixture 18.28 (a) Chloride is a good leaving group and the resulting 2˚ carbocation is a

relatively stable intermediate. The most important factor influencing the reaction towards SN1 is ethanol, which is a poor nucleophile and a moderately ionising solvent.

(b) Chloride is a good leaving group and thus results in a very stable 3˚

carbocation intermediate. An equally important factor influencing the reaction towards SN1 is methanol, which is a poor nucleophile and a good ionising solvent. 3˚ Haloalkanes almost exclusively undergo SN1 reactions with poor nucleophiles.

(c) Chloride is a good leaving group and acetic acid is a strongly ionising

solvent/poor nucleophile. 3˚ Haloalkanes exclusively undergo SN1 reactions with poor nucleophiles. All these factors strongly favour SN1 reactions.

(d) Methanol is a good ionising solvent and a poor nucleophile. Bromide is a

good leaving group resulting in a relatively stable 2˚ carbocation, all favouring an SN1 reaction.

18.29 The order of reactivity for haloalkanes in SN1 reactions is:

3° > 2° > 1° > methyl (a)

(b)

(c)



18.30

18.31 All of the reaction products shown can be produced from the same carbocation

intermediate that results from the ionisation of the carbon-halogen bond. The reaction proceeds in an ionising solvent, there are no good nucleophiles present and the 3˚ carbocation is a very stable intermediate. All of these factors are very favourable for SN1 and E1 reaction mechanisms.

SN1 reaction mechanism with ethanol as a nucleophile:

C

CH3

CH3

H3C Clslow

C

CH3

CH3

H3C

C

CH3

CH3

H3C HOCH2CH3 C

CH3

CH3

H3C OCH2CH3

H+

OH

HCH3COCH2CH3

CH3

CH3

+ Cl

+ H3O+

SN1 reaction mechanism with water as a nucleophile:

C

CH3

CH3

H3C Clslow

C

CH3

CH3

H3C

C

CH3

CH3

H3C C

CH3

CH3

H3C O

H+

OH

HCH3COH

CH3

CH3

+ Cl

+ H3O+OH

HH

E1 reaction mechanism:

C

CH3

CH3

H3C Cl C

H2C

H3CClCH3

H

OH

HCH3C

CH3

CH2

slow+ H3O+



18.32 The reaction mechanism involves the formation of a carbocation in the rate-

determining step. The addition of the more polar water to ethanol makes the solvent mixture more polar and more ionising than pure ethanol. Thus, as the percentage of water in the ethanol increases, carbocations form by ionisation of the carbon-chloride bond more easily in the water-ethanol solvent mixture relative to pure ethanol.

18.33 The reacting carbon in an SN2 reaction transition state is sp2 hybridised, with the

substituents bonding to sp2 hybridised orbitals and the incoming nucleophile and leaving groups each partially bonded to an unhybridised p orbital. The molecular shape of this transition state is trigonal bipyramidal.

CNu L

18.34 Haloalkenes fail to undergo SN1 reactions because the alkenyl carbocations produced

through ionisation of the carbon-halogen bond are too unstable. The electron-rich π-bonds of alkenyl halides repel nucleophiles in SN2 reaction.

18.35

(a)

Br CNNaCN NaBr

(b) Br CNNaCN NaBr

(c)Br

O

ONaOCCH3

ONaBr

(d) Br NaSH SH NaBr

(e) Br NaOCH3 OCH3NaBr

(f) Br NaOCH2CH3 O NaBr



(g)Br

NaSHSH

NaBr

18.36

(a) Br + 2NH3 NH2 + NH4+Br-

(b) CH2Br + 2NH3 CH2NH2 + NH4+Br-

(c) Br + CH3COO

Na+ + NaBrOCCH3

O

(d) SBr S-Na+

+ NaBr

(e)OCCH3Br

+ CH3COO

Na+

O

+ NaBr

(f) OBr O-Na+

+ NaBr

18.37 When elimination reactions can give two or more possible alkenes, Zaitsev’s rule

predicts that the most stable alkene (the most substituted alkene) will be the major product.

(a) Cl or

Cl

KOH

(b)

CH2Cl KOH

(c) Cl KOH



(d)Cl

orCl

KOH

(e)

ClKOH

18.38 (a)

HO OH

Cl

SN2 OHHO

(b)

Cl

HO H

OCH2CH3E2 HO

(c) SN2 reactions occur with a backside attack by the nucleophile on the carbon

bearing the leaving group. In trans-4-chlorocyclohexanol, the nucleophilic atom is created with the deprotonation of the hydroxyl proton. The alkoxide nucleophile is now perfectly oriented to initiate a backside attack. The alkoxide nucleophile from cis-4-chlorocyclohexanol is situated on the same side as the carbon-halogen bond and therefore, cannot undergo a backside attack to complete the substitution.

Cl

OHCH3CH2O

Cl

O O

18.39

(a) ClNaOCH2CH3

CH3CH2OH

(b) Br

HBr



Cl OH(c)

NaOCH2CH3

CH3CH2OH

H2O

H2SO4

(d) Br NaOCH2CH3

CH3CH2OH

(e)Br NaOCH2CH3

CH3CH2OH

OH

OH

OsO4

ROOH

(f)Br Br

Br

NaOCH2CH3

CH3CH2OH

Br2

18.40 (a) 3-chloro-3-methylhexane

(b) 3-bromohexane

18.41 Primary haloalkanes react with bases/nucleophiles to give predominantly substitution

products. With strong bases, such as hydroxide ion and ethoxide ion, a percentage of the product is formed by an E2 reaction, but it is generally small compared with that formed by an SN2 reaction. With strong, bulky bases, such as the tert-butoxide ion, the E2 product becomes the major product.



INaOH

HO

Majorproduct

minorproduct

(a)

INaNH2 H2N

Majorproduct

minorproduct

(b)

Sodium hydroxide and sodium amide are very strong bases and also very good nucleophiles. Some elimination occurs, but the major product arises from substitution.

INaCN

NC

Onlyproduct

(c)

Sodium cyanide is a strong nucleophile but a weak base, which favours substitution over elimination.

INaOOCCH3 O

Onlyproduct

(d)

O

H3C

Sodium acetate is a weak base and so little elimination occurs. It is a good nucleophile and so the predominant product is the substitution of the iodine atom by the acetate group.

INaI I

Onlyproduct(racemate)

(e)

Sodium iodide is a very weak base and so no elimination occurs. The predominant product is the substitution of the iodine atom by another iodine atom so that a racemic mixture results.

INaOC(CH3)3

majorproduct

(f)



The tert-butoxide amion is a strong base which favours elimination. It is also sterically bulky which hinders substitution. The major product is propene with only minor amounts of the ether formed.

18.42 The second method is more efficient because the alkyl halide (methyl iodide) is not

sterically hindered. The first method is not efficient because it employs a tertiary alkyl halide, and SN2 reactions do not occur at tertiary substrates.

Additional exercises 18.43

CH3CO- K+ CH2ClDMSO

CH3COCH2 +(a) + KClCH3

CH3

CH3

CH3Major product (SN2)

CH2O-K+ CH3CCl CH3COCH2(b) + + KClCH3

CH3

CH3

CH3

Minor product

DMSO

Reaction (a) gives the best yield of ether product because it involves favourable

conditions for an SN2 reaction, involving a 1˚ halide substrate and a good nucleophile. The substrate also precludes E2 reactions with the absence of β-protons. Reaction (b) predominately undergoes an E2 reaction with a strong base deprotonating the β-proton on a sterically hindered 3˚ halide to produce an alkene as the major product.

CH2O H2C C

H CH3

CH3

ClE2 H2C C

CH3

CH3

CH2OH

Major product (E2)

18.44 (a) The formation of bond (2) gives a higher yield of the ether because an SN2

reaction between a nucleophile and a primary halide yields less E2 product than the reaction between nucleophile and secondary halide, as in the formation of bond (1).



(2)O

(1)Bond (1):Br

O-Na+

Bond (2):O-Na+

Br

(b) The formation of bond (1) gives exclusively the ether product by an SN2. No

E2 reaction is possible because methyl bromide lacks the required β-proton. Attempts to form bond (2) results predominately in an E2 reaction yielding an alkene as the major product.

O(1)

Bond (1):

Bond (2):

O-Na+ CH3Br

Br CH3O-Na+ O(2)

Only product

Minorproduct

Majorproduct

(c) The formation of bond (1) gives a higher yield of the ether because an SN2

reaction between a nucleophile and a primary halide yields less E2 product than the reaction between nucleophile and secondary halide, as in the formation of bond (2).

(1)O

(2)Bond (1):

Bond (2):

O-Na+

Br

BrO-Na+

18.45

Cl CH2CH2 O H OH Cl CH2CH2 O H2C CH2

O



18.46 Hydroxyl is not the leaving group under acidic conditions. Instead, the hydroxyl is protonated by the strong acid, HBr, then leaves as water, which is a better leaving group than hydroxide.

OH HBr Br

Step 1: Protonation of the hydroxyl by HBr.

OH OH2H Br

+

Br

Step 2: Loss of water to form the 3˚ carbocation.

OH2+

++ H2O

Step 3: Nucleophilic attack of the bromide anion on the carbocation to form the

haloalkane.

+Br

Br

18.47 Under these SN2 conditions, the bromide nucleophile inverts the stereochemistry of

the stereocentre. After 50% of the starting material has reacted, racemisation is attained.

18.48 The negative charge on phenoxide is delocalised by resonance over four different

atoms and 5 resonance structures. This leaves the electron pairs less available for nucleophilic reactions relative to cyclohexanoxide, which has no electron delocalisation.

O O O

OO

18.49 Although alkoxides are poor leaving groups, the release of strain in the three-



(c)CH3

Cl

HCl CH3

Clor

membered ring is a driving force behind ring-opening of epoxides by nucleophiles. 18.50 All of the following reactions take place via the Markovnikov addition of a hydrogen

halide to an alkene.

(a)

Br HBr Br

(b) CH3CCH2CH2CH3

CH3

Bror

HBr Br

18.51

Cl(a) HCl

CH3CH2CH=CH2 CH3CH2CHCH3(b)I

HI

(c) CH3CH=CHCH3 CH3CHCH2CH3

ClHCl

(d) Br

CH3CH3HBr

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