Chemistry Reference
In-Depth Information
CHAPTER 4 SOLUTIONS
1. In many S
N
2 reactions, the nucleophile is generated by deprotonation of an organic
acid. For each molecule, chose the base best suited to completely remove the labeled
proton. (Consider pK
a
values and recognize that, in some cases, dianions should be
considered.) Explain your answers.
The pK
a
of the highlighted proton is approximately 20. Therefore, NaOCH
3
(pK
a
of conjugate acid methanol
¼
16) is not a strong enough base. CH
3
Li (pK
a
of
conjugate acid methane
¼
50) will deprotonate this molecule; however, it is too
nucleophilic a base and will predominantly add to the carbonyl to produce a tertiary
alcohol (see Chapter 7). (CH
3
)
2
NLi (pK
a
of conjugate acid dimethylamine
¼
35) is a
bulkier base than CH
3
Li and is, therefore, less nucleophilic and the best base for
this case.
The pK
a
of the highlighted proton is approximately 12. As described in the answer
for Problem 1(a), CH
3
Li (pK
a
of conjugate acid methane
¼
50) will deprotonate
this molecule; however, it is too nucleophilic a base and will predominantly add to
the carbonyls to produce tertiary alcohols (see Chapter 7). While (CH
3
)
2
NLi (pK
a
of conjugate acid dimethylamine
¼
35) is a bulkier base than CH
3
Li and is, therefore,
less nucleophilic, it is also more basic than required for removal of the specified
proton. NaOCH
3
(pK
a
of conjugate acid methanol
¼
16), on the other hand, is a
milder base and, based on pK
a
values, is adequate to fully deprotonate the illustrated
compound.
In this case, the most acidic proton is not the proton of interest. Therefore, it is import-
ant to remember that once the most acidic proton is removed, the resulting enolate
