Chemistry Reference
In-Depth Information
CHAPTER 6 SOLUTIONS
1. E2 eliminations do not necessarily require acidic protons in order to proceed. Explain
how this can occur.
The orientation of any proton in a trans-periplanar relationship to a given leaving group
is usually enough to allow elimination to occur under basic conditions even when, in the
absence of an electron-withdrawing group, the proton is not acidic enough to be
removed.
2. When CH
3
OCH
2
CH
2
CH
2
Br is treated with magnesium, we get the Grignard reagent
CH
3
OCH
2
CH
2
CH
2
MgBr. However, when CH
3
OCH
2
CH
2
Br is treated with magnesium,
the product isolated is H
2
C
55
CH
2
. Explain this result.
Grignard reagents are carbanions stabilized by a MgBr cation. As with all anionic
species, if a leaving group is situated on an adjacent center, the structure is subject to
an E2 elimination process. Furthermore, CH
3
O
2
is a sufficient leaving group when it
is located adjacent to an anionic center. Therefore, in the case of bromomethoxyethane,
E2 elimination leads to formation of ethylene when the negative charge adopts a trans-
periplanar relationship to the methoxy group.
3. With an understanding of E1 mechanisms, one may realize that under S
N
1 reaction con-
ditions multiple products may form. In addition to the products predicted in Chapter 5
for the following molecules, predict plausible elimination products.
Silver is very efficient at removing halides, resulting in generation of carbocations.
Because protons adjacent to carbocations are acidic and, therefore, participate in
