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Organic Chemistry

Chapter 11

Organic ChemistryChapter 11
Problem 11-1
(R)-1-Methylpentyl acetate, CH3CO2CH(CH3)CH2CH2CH2CH3
Problem 11-2
(S)-2-Butanol
Problem 11-4
(a)
1-Iodobutane
(b)
1-Butanol
(c)
1-Hexyne
(d)
Butylammonium bromide
Problem 11-5
(a)
(CH3)2N−
(b)
(CH3)3N
(c)
H2S
Problem 11-6
CH3OTos > CH3Br > (CH3)2CHCl > (CH3)3CCl
Problem 11-7
Similar to protic solvents
Problem 11-8
Racemic 1-ethyl-1-methylhexyl acetate
Problem 11-9
90.1% racemization, 9.9% inversion
Problem 11-11
H2C ═ CHCH(Br)CH3 > CH3CH(Br)CH3 > CH3CH2Br > H2C ═ CHBr
Problem 11-12
The same allylic carbocation intermediate is formed.
Problem 11-15
(a)
Major: 2-methyl-2-pentene; minor: 4-methyl-2-pentene
(b)
Major: 2,3,5-trimethyl-2-hexene; minor: 2,3,5-trimethyl-3-hexene and 2-isopropyl-4-methyl-1-pentene
(c)
Major: ethylidenecyclohexane; minor: cyclohexylethylene
Problem 11-16
(a)
1-Bromo-3,6-dimethylheptane
(b)
4-Bromo-1,2-dimethylcyclopentane
Problem 11-17
(Z)-1-Bromo-1,2-diphenylethylene
Problem 11-18
(Z)-3-Methyl-2-pentene
Problem 11-19
Cis isomer reacts faster because the bromine is axial.
Problem 11-20
(a)
SN2
(b)
E2
(c)
SN1
(d)
E1cB
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