= Solution
Elimination from A gives
$$
\boxed{\mathrm F=CH_2=C(CH_3){-}C{\equiv}CH},\qquad \mathrm{C_5H_6}.
$$
Removing a beta hydrogen from either equivalent methyl group while losing chloride creates the same conjugated enyne. The strongly basic \b[ethoxide], $\mathrm{CH_3CH_2O^-}$, promotes this <E2 elimination> from the tertiary substrate more strongly than neutral ethanol does. Ethanol instead favors the intended solvolytic capture, although some elimination can still compete.
The formula drawn and used in the rate table is ethoxide. The name “ethanoate” would mean acetate, $\mathrm{CH_3CO_2^-}$, a different and resonance-stabilized anion; it should not be silently treated as the same base.
In Reaction 2 the primary benzylic carbon favors rapid substitution. The adjacent aromatic ipso carbon has no beta hydrogen for ordinary side-chain E2 elimination, so the analogous alkene-forming path is unavailable. This does not assert that all benzyl compounds are immune to every side reaction.
Back to article page
