CHAPTER 15 Practice Exercises 15.1 15.3 (A) C9H18

Home , Cyclododecane, Methylcyclopropane

CHAPTER 15

Practice exercises

15.1

15.3 (a) C9H18: isobutylcyclopentane (b) C11H22: sec-butylcycloheptane (c) C6H2: 1-ethyl-1-methylcyclopropane

15.5 (a) 3,3-dimethyl-1-pentene (b) 2,3-dimethyl-2-butene (c) 3,3-dimethyl-1-butyne

15.7 (a) (E)-1-chloro-2,3-dimethyl-2-pentene (b) (Z)-1-bromo-1-chloropropene (c) (E)-2,3,4-trimethyl-3-heptene

15.9

cis,trans-2,4-heptadiene cis,cis-2,4-heptadiene

15.11 (a)

2-iodopropane

(b)

1-iodo-1-methylcyclohexane

15.13 CH3 CH3 + HI I

Step 1: Protonation of the alkene to give the most stable 3˚ carbocation intermediate:

slow step + CH3 HI CH3 I rate-determining step

Step 2: Nucleophilic attack of the iodide anion on the 3˚ carbocation intermediate to give the product:

CH + 3 CH3 I I

15.15 H2O CH3 CH3 + H3O OH

Step 1: Protonation of the alkene to give the most stable 3˚ carbocation intermediate:

slow step H + + CH3 HOH CH3 O H rate-determining H step

Step 2: Nucleophilic attack of the water on the 3˚ carbocation intermediate to give the protonated alcohol:

H H O+ H + CH3 O H CH3

Step 3: The protonated alcohol loses a proton to form the product:

H + O H H CH3 + O + H3O CH3 H OH

15.17 (a) 2-phenyl-2-propanol (b) (E)-3,4-diphenyl-3-hexene (c) 3-methylbenzoic acid or m-methylbenzoic acid

Review questions

15.1 A hydrocarbon is a compound composed only of hydrogen and carbon atoms.

15.3 In saturated hydrocarbons, each carbon is bonded to four other atoms, either hydrogen or carbon atoms. Unsaturated hydrocarbons have carbon atoms that have a double or triple bond to another carbon atom.

15.5 (a) (b)

(e) (f)

15.7 (CH2)2CH3

(a) CH3(CH2)4CH(CH3)2 (b) CH3(CH2)2CH(CH2)2CH3

(CH2)2CH3 (c) CH3(CH2)2C(CH2)4CH3

CH3

15.9 (a) different compounds (b) constitutional isomers (c) constitutional isomers (d) different compounds (e) constitutional isomers (f) constitutional isomers

15.11 (a) 2-methylpentane (b) 2,5-dimethylhexane (c) 3-ethyloctane (d) 2,2,3-trimethylbutane (e) isobutylcyclopentane (f) 1-tert-butyl-2,4-dimethyl-cyclohexane

15.13 (a) 1,3-dimethylbutane The longest chain is pentane. The IUPAC name is 2-methylpentane.

(b) 4-methylpentane The pentane chain is numbered incorrectly. The IUPAC name is 2-methylpentane.

(d) 2-ethyl-3-methylpentane The longest chain is hexane. The IUPAC name is 3,4-dimethylhexane.

(e) 2-propylpentane The longest chain is heptane. The IUPAC name is 4-methylheptane.

(f) 2,2-diethylheptane The longest chain is octane. The IUPAC name is 3-ethyl-3-methyloctane.

(g) 2,2-dimethylcyclopropane The ring is numbered incorrectly. The IUPAC name is 1,1-dimethylcyclopropane.

(h) 1-ethyl-5-methylcyclohexane The ring is numbered incorrectly. The IUPAC name is 1-ethyl-3-methylcyclohexane.

15.15 No, because alkanes do not have rings or C—C double bonds and so all conformations are usually interconvertable by rotation about a C—C single bond. However, there are some extremely crowded molecules which are locked into specific conformers. These are called ‘conformational isomers’ or ‘locked conformational isomers’ and are not correctly defined by cis-trans nomenclature.

15.17 There are two enantiomers of the trans-1,2-dimethylcyclopropane..

CH3

CH2CH3

ethylcyclopropane cyclopentane methylcyclobutane

CH H CH3 CH3 CH3 3

CH3 H H H CH3 1,1-dimethyl- cis-1,2-dimethyl- trans-1,2-dimethylcyclopropane cyclopropane cyclopropane

15.19 (a) trans-2-methylhex-3-ene (b) 2-methyl-3-hex-3-yne (c) 2-methylbut-1-ene

(d) 3-ethyl-3-methylpent-1-yne (e) 2,3-dimethylbut-2-ene (f) cis-pent-2-ene

15.21 (a) 2-isobutylhept-1-ene (b) 1,4,4-trimethylcyclopentene (c) 1,3-cyclopentadiene (d) 3,3-dimethylbut-1-yne (e) 2,4-dimethylpent-2-ene (f) oct-1-yne (g) 2,2,5-dimethylhex-3-yne (h) 3-methylpent-1-yne

15.23 (a) 2,2-dimethylhex-3-yne (b) oct-2,5-diyne (c) 3,6-dimethylhept-2-ene-4-yne (c) hept-1,4-diyne

15.25 (a) Correct name: but-2-ene.You must select the longest carbon chain containing the ‘ene’ functional group.

1 3 2 H C 4 Not: 3 3 2 1 but-2-ene 1-methylpropene

(b) Correct name: pent-2-ene. You must number the chain to give the first carbon of the double bond the lowest possible number.

2 4 4 1 5 5 Not: 2 3 1 3

pent-2-ene pent-3-ene

(c) Correct name: 1-methylcyclohexene. By default, the first carbon of the double bond in a ring is given the number 1. In this, case this can also be the ring carbon with the methyl group attached.

1 2 Not: 2 1

2-methylcyclohexene 1-methylcyclohexene

(d) This name is not incorrect even though there is no indication of where the double bond is positioned. The indication of the two methyls on the third carbon means that the double bond is attached to the first carbon. Technically, the correct name is 3,3-dimethylpent-1-ene.

(e) Correct name: 4-hexyne. As with (b) the chain carbons are numbered to give the first carbon of the double bond the lowest possible number. 7 6 5 4 3 2 1 CH 3CH 2 C H 2 C CCH2CH3

(f) Correct name: 2-isopropyl-2-butene. The longest chain containing the function group is a pentene.

5 4

3 1 2

15.27 (a) No (b) Yes (c) Yes (d) No (e) Yes (f) No

(b)

15.29 For alkenes to exist as a pair of cis-trans isomers, both carbons of the double bond must have two different substituents. Thus, only (b) and (d) can exist as a cis or trans isomer.

H3C Br H3C H (b) H H H Br cis-bromoprop-1-ene trans-bromoprop-1-ene

(d)

15.31 Molecules (a) and (c) do not show cis-trans isomerism.

(b) (d)

15.33 (a) Alkenes that do not show cis-trans isomerism are:

pent-1-ene2-methylbut-2-ene 2-methylbut-1-ene 3-methylbut-1-ene

(b) Alkenes that do show cis-trans isomerism are:

trans-pent-2-ene cis-pent-2-ene

CH CH CH3 2 3 H3C CH3

cyclopentane methylcyclobutane ethylcyclopropane 1,1-dimethyl cyclopropane

(d) Cycloalkanes that do show cis-trans isomerism are:

H3C CH3 H3C CH3 H H H H cis-1,2-dimethylcyclopropane trans-1,2-dimethylcyclopropane

15.35 (a) CH33 CHCH is more stable.

CH3

(b) CH323 C CH CH is more stable. +

15.37 CH3 CH C=CH + 6O 4CO + 4H O 3 2 2 2 2

15.39 H2 (a) H2CCH2 CH3CH3 Ni

H O 2 (b) H2CCH2 CH3CH2OH H2SO4

Br2 (d) H2CCH2 BrCH2CH2Br

HCl (e) H2CCH2 CH3CH2Cl

15.41

H2 / Pd/C

H2 Lindlar H2 catalyst Na/NH3

Br2

Br Br

15.43

15.45 (a) 1-bromo-2-chloro-4-ethylbenzene (b) 4-iodo-1,2-dimethylbenzene Br Cl

O2N NO2 OH

NO2

(e) p-cresol (f) 2,4-dichlorophenol

OH OH Cl

(g) 1-phenylcyclopropanol (h) styrene (phenylethene)

(i) m-bromophenol (j) 2,4-dibromoaniline

NH2 OH Br

Br Br

(k) isobutylbenzene (l) m-xylene

15.47

isopropylbenzene

Cl OH (1) /AlCl3 (2) /H2SO4 (3) /H2SO4

Review problems

15.49 (1) Alkanes are less dense than water. (2) As alkane molar mass increases, density increases. (3) Constitutional isomers have similar densities.

15.51 Boiling points of unbranched alkanes are related to their surface area: the larger the surface area, the greater the strength of the dispersion forces and the higher the boiling point. The relative increase in molecular size per CH2 group is greatest between CH4 and CH3—CH3 and becomes progressively smaller as molecular mass increases. Therefore, the increase in boiling point per CH2 group is greatest between CH4 and CH3—CH3, and becomes progressively smaller for the higher alkanes.

15.53 (a) No (b) Yes (c) Yes (d) It is a liquid. (e) It is less dense than water.

15.55 The three structures with molecular formula C2H2Br are 1,1-dibromoethane, cis-1,2- dibromoethane and trans-1,2-dibromoethane. The dipole moment of the C—Br bond is shown in the structures below by a solid arrow. trans-1,2-dibromoethane has no dipole moment because the two C—Br dipoles point in opposite directions and therefore cancel each other out. However, 1,1-dibromoethane and cis-1,2-dibromoethane both have a net dipole moment (in the direction shown by the dotted arrow) since the two C—Br dipoles do not point in opposite directions.

Br Br Br H Br H + CC CC CC H H Br H H Br No dipole moment 1,1-dibromoethane cis-1,2-dibromoethane trans-1,2-dibromoethane

15.57 (a)

CH3 CH3 CH3

CH3 CH CH3 CH3 CH H CH3CHCH2CH CH2 CC CC 4-methylpent-1-ene H H H CH3 cis-4-methylpent-2-ene trans-4-methylpent-2-ene

(b)

(c)

15.59

15.61 (a) 2,2-dichlorobutane

(b) butan-2-ol

15.63 A hydrocarbon of formula C5H8 must have a triple bond, or two double bonds, or a combination of a double bond and a ring. Since it only reacts with one mole of H2 indicates that the compound is an alkyne. Lindlar’s catalyst gives a cis alkene, which in turn is brominated to give a dibromoalkane which has two chiral carbons. Only the final possibility shown below meets this last criterion.

H H Br H H H Br OR H H Br H H H Br

H Br Br H H H

15.65 (a) The formation of the more stable tertiary carbocation intermediate proceeds at a faster rate, therefore 2-methylbut-2-ene reacts faster than trans-but-2-ene. I +

trans-but-2-ene secondary 2-iodobutane carbocation

I + 2-methylbut-2-ene tertiary 2-iodo-2-methyl carbocation butane (major product)

(b) The formation of the more stable tertiary carbocation intermediate proceeds at a faster rate, therefore 1-methylcyclohexene reacts faster than cyclohexene.

I +

cyclohexene secondary iodocyclohexane carbocation

CH3 CH3 + CH3 I

1-methylcyclohexene tertiary 1-iodo-1-methyl carbocation cyclohexane (major product)

15.67 The first step in the reaction is the protonation of the alkene to generate a carbocation. The reaction path that produces the more stable carbocation occurs at a faster rate, thus producing the observed regioselectivity.

CH CH 3 HI 3 (a) CH3CCHCH2CH3 CH3CCH2CH2CH3 I

CH3 H2O CH3 (b) CH3CCHCH2CH3 CH3CCH2CH2CH3 H SO 2 4 OH

15.69 (a)

(b)

(c)

15.71 Chlorine undergoes an anti-addition to alkenes.

Cl Cl Cl2 Cl2 (a) (b) Cl Cl

CH3 H C Cl 3 Cl Cl Cl2 2 Cl CH (c) (d) 2 Cl

15.73 Both cis- and trans-hex-3-ene yield hexan-3-ol upon acid-catalysed hydration because they both form the same carbocation intermediate that leads to hexan-3-ol according to the partial mechanism:

H + H OH OH + H + H OH

15.75

H2O (a) OH + H3O

CH3 CH3 H2O (b) OH + H3O CH CH3 3

OH H2O (c) CH3 + H3O CH 3

H2O OH (d) H3C H3C + H3O

15.77 Step 1: Protonation of the alkene gives a stable 3˚ carbocation intermediate:

CH 3 H CH3 + CH3C CH2 H O CH3 H3C C + CH3OH CH3

Step 2: Nucleophilic attack of methanol on the carbocation intermediate yields a protonated ether intermediate.

CH3 CH3 CH3 + CH3 H3C C + O H3C C O H H CH3 CH3

Step 3: Proton loss by the protonated ether intermediate yields the ether and regenerates the acid catalyst.

CH3 CH CH CH3 H + 2 3 + H3C C O O H3C C O CH3 H O CH3 H CH CH3 H 3

15.79 The skeleton and location of the double bonds in hydrocarbon A are identified from the structure of the brominated product:

2 mol Br2 Br Br Br Br Hydrocarbon A (C H ) 5 8 1,2,3,4-tetrabromo-2-methylbutane 2-methylbuta-1,3-diene

15.81 Br Br2 (a) Br

OH H2O (b) H2SO4

Br HBr (c)

H2 ( d) Pt

15.83 HBr Br (a)

H O (b) or 2 OH H SO 2 4

Br Br2 (c)

15.85 Resonance accounts for the three equivalent structures of naphthalene:

15.87

H H NH (a)(b) H H (c) O - 8 π electrons - Non-aromatic - 14 π electrons - 10 π electrons - Not planar - Planar - Non-aromatic - Aromatic

H B O O (d)(e) (f)

- 6 π electrons - 6 π electrons - 8 π electrons - Planar - Not a 2p orbital - Non-aromatic - Aromatic on every atom - Non-aromatic

Although compound (b) contains 14 π electrons (a Hückel number), a closer inspection of the molecule reveals that the protons inside the ring are held too close together and force the ring to be non-planar. Molecular modelling of structure (b) confirms its non-planarity and it therefore non-aromatic.

In compound (c), only one of the electron pairs on oxygen is in a 2p orbital, the other electron pair is in an sp2 hybridised orbital, perpendicular to the π bonds. The nitrogen lone pair in compound (c) is also in a 2p orbital for a total of 10 π electrons in a planar ring making it aromatic.

In compound (d), the boron atom contributes an empty 2p orbital, giving an aromatic compound with six π electrons in a seven-membered ring with seven 2p orbitals. Compound (f), is similar to compound (c) where the oxygen has one electron pair in a 2p orbital and the other in an sp2 hybridised orbital with a total of eight π electrons, giving a non-aromatic compound.

15.89 Two monochloronaphthalenes are possible when naphthalene is treated with Cl2/AlCl3:

Cl Cl Cl2

AlCl3 15.91

AlCl3 2 + CH2Cl2 CH2 + HCl

Step 1: Formation of Lewis acid-Lewis base complex:

Cl Cl + ClCH2 Cl Al Cl ClCH2 Cl Al Cl Cl Cl

Step 2: Nucleophilic attack of benzene on the electrophilic Lewis acid-Lewis base complex and the formation of a resonance-stabilised carbocation (two more resonance structures exist):

Cl CH2 Cl + - ClCH2 Cl Al Cl + AlCl4 Cl H +

Step 3: Deprotonation of the carbocation to give benzyl chloride, HCl, and AlCl3:

CH2 Cl Cl

Cl Al Cl CH2 Cl + AlCl3 + HCl H Cl +

Step 4: Formation of Lewis acid-Lewis base complex between benzyl chloride and AlCl3:

Cl Cl

CH2 Cl Al Cl CH2 Cl Al Cl Cl Cl

Step 5: Dissociation of the complex to give a resonance-stabilised benzyl cation (only – one out of five contributing structures is shown) and AlCl4 :

Cl Cl + CH2 Cl Al Cl CH2 Cl Al Cl Cl Cl

Step 6: Nucleophilic attack of the second molecule of benzene on the benzylic cation to form another resonance-stabilised carbocation (only one structure is shown):

+ CH2 CH2

H +

Step 7: Deprotonation of the carbocation intermediate to regenerate the aromatic ring (diphenylmethane), HCl, and AlCl3:

Cl CH2 Cl Al Cl CH2 + AlCl3 + HCl Cl H +

Additional exercises

15.93 A molecular model of cyclohexane shows hindered rotation about the C—C single bonds, allowing the possibility of cis-trans isomers in disubstituted six-membered rings. The much larger cyclododecane ring is flexible enough to allow unrestricted rotation about the C—C single bonds, thus no cis-trans isomerisation exists.

CH3 CH3

CH3 CH3 CH3 CH3 cis-1,2-dimethyl- trans-1,2-dimethyl- 1,2-dimethyl- cyclohexane cyclododecane cyclohexane

15.95 o 109.5 120o 180oH H (a) (b) CH2OH (c) H CCC C (d) 120o H 120o o 120

15.97 109.5o o 109.5o 109.5 OH Br (a)120o (b) (c)180o (d) o Br 109.5o 120

15.99 (a) Carbons 1 and 3 are sp2 hybridised. Carbon 2 is sp hybridised. (b)

π y-axis 2pz-2pz

H H H H π 2py-2py z-axis

Each of the terminal carbons is sp2 hybridised and thus has three sp2 orbitals oriented 120˚ apart. Two of these sp2 hybrid orbitals form sigma bonds with hydrogen atoms and the third forms a sigma bond with the central carbon. Each terminal carbon has one 2p

orbital (C1 has a 2pz orbital and C3 has a 2py orbital) that each form a π bond with the 2pz and 2py orbitals on the central carbon atom. Because these two 2p orbitals on the centre carbon atom are at right angles to each other, the 2p orbitals on the terminal carbons must be at right angles to each other to permit full orbital overlap.

15.101 Molecules of the trans isomer can ‘pack together’ more tightly than those of the cis isomers, which means that the dispersion forces controlling intermolecular bonding are greater for the trans than for the less symmetrical cis isomer.

COOH COOH COOH COOH COOH COOH

The trans isomer of octadec-9-enoic acid can pack together closely and so it has a higher melting point (44–45° C) than the cis isomer.

COOH

15.103 (a)

OH O enzyme catalyst + H2O2 + 2H2O + heat

OH O

(b) Yes