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Problem 7-3
C16H13ClN2O
Problem 7-4
(a)
3,4,4-Trimethyl-1-pentene
(b)
3-Methyl-3-hexene
(c)
4,7-Dimethyl-2,5-octadiene
(d)
6-Ethyl-7-methyl-4-nonene
Problem 7-6
(a)
1,2-Dimethylcyclohexene
(b)
4,4-Dimethylcycloheptene
(c)
3-Isopropylcyclopentene
Problem 7-7
(a)
2,5,5-Trimethylhex-2-ene
(b)
2,3-Dimethylcyclohexa-1,3-diene
Problem 7-9
Compounds (c), (e), and (f) have cis–trans isomers.
Problem 7-10
(a)
cis-4,5-Dimethyl-2-hexene
(b)
trans-6-Methyl-3-heptene
Problem 7-11
(a)
−CH3
(b)
−Cl
(c)
−CH = CH2
(d)
−OCH3
(e)
−CH = O
(f)
−CH = O
Problem 7-12
(a)
−Cl,  −OH,  −CH3,  −H
(b)
−CH2OH,  −CH = CH2,  −CH2CH3,  −CH3
(c)
−CO2H,  −CH2OH,  −C N,  −CH2NH2
(d)
−CH2OCH3,  −C N,  −C CH,  −CH2CH3
Problem 7-15
(a)
2-Methylpropene is more stable than 1-butene.
(b)
trans-2-Hexene is more stable than cis-2-hexene.
(c)
1-Methylcyclohexene is more stable than 3-methylcyclohexene.
Problem 7-16
(a)
Chlorocyclohexane
(b)
2-Bromo-2-methylpentane
(c)
4-Methyl-2-pentanol
(d)
1-Bromo-1-methylcyclohexane
Problem 7-17
(a)
Cyclopentene
(b)
1-Ethylcyclohexene or ethylidenecyclohexane
(c)
3-Hexene
(d)
Vinylcyclohexane (cyclohexylethylene)
Problem 7-19
In the conformation shown, only the methyl- group C−H that is parallel to the carbocation p orbital can show hyperconjugation.
Problem 7-20
The second step is exergonic; the transition state resembles the carbocation.
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