Solutions to Homework #2
Solutions to problems 9.1 -- 9.5, 9.8, 9.11 -- 9.12 can be found in your
"Study Guide and Solutions Manual"
Problem 9.13:
Problem 9.14:
Problem 9.17:
Problem 9.18:
The three alkyl halides are (a), (d), and (e).
For (e) you will receive a single isolated product, namely 4-methylcyclohexene.
Compound (a) will generate two products 3-methylcyclohexene and 4-methylcyclohexene.
Compound (d) will certainly give 3-methylcyclohexene and possibly 1-methylcyclohexene
depending on the stereochemistry at the adjacent methyl group position (remember, you need
trans-antiperiplanar geometry to get E2).
Problem 9.19:
Alkyl halides (a) and (d) are chiral. Only (d) will produce an achiral alkene
(1-methylcyclohexene). HINT: think E2 only.
Alkene (e) is achiral, but produces (R) and (S) 4-methylcyclohexene upon elimination.
Problem 9.20:
Rank (from fastest to slowest SN2): (c) > (e) > (a) > (d) >
(b)
The reason that (e) > (a) > (d) is due to the degree of beta-branching tht might
interfere sterically with substitution.
Problem 9.21:
The 3-bromocyclohexanone will react faster than the 2-bromocyclohexanone because the
intermediate carbanion produced upon removal of the acidic hydrogen is resonance
stabilized. The more stable carbanion intermediate is more stable and will form
faster.
Problem 9.23:
Problem 9.26:
In both cases, under acidic conditions, the OH group that is present at the tertiary
carbon will be protonated (activated) and will leave to generate a carbocation
intermediate. The other alcohol group will likely be protonated, but will not leave
because it will generate the less stable primary or secondary carbocation. In
addition, the tertiary alcohol will lead to the more stable Zaitsev's alkene.
Therefore, regardless of whether the reaction is under kinetic or thermodynamic
control, the indicated product is favored.
Problem 9.28:
Problem 9.30:
