Identify the correct statement for the below given transformation.

Explanation: The reaction proceeds through an elimination $$\mathrm{(E2)}$$ mechanism because a strong base, sodium ethoxide $$\mathrm{(C_2H_5ONa)}$$, is used.

The substrate is a quaternary ammonium salt containing the bulky leaving group:

$$\mathrm{-N^+(CH_3)_3}$$

Quaternary ammonium salts undergo elimination according to Hofmann’s rule due to the large steric bulk of the leaving group.

Hofmann elimination favors formation of the less substituted alkene.

The base can remove a $$\mathrm{\beta}$$-hydrogen from two different carbons:

Removal from the terminal $$\mathrm{CH_3}$$ group gives:

$$\mathrm{CH_3CH_2CH_2CH=CH_2}$$

which is 1-pentene, the less substituted alkene.

Removal from the internal $$\mathrm{CH_2}$$ group gives:

$$\mathrm{CH_3CH_2CH=CHCH_3}$$

which is 2-pentene, the more substituted alkene.

Because of steric hindrance from the bulky ammonium leaving group, abstraction of the more accessible terminal hydrogen is preferred.

Therefore:

Major product $$\mathrm{(A)}$$:

$$\mathrm{CH_3CH_2CH_2CH=CH_2}$$

Minor product $$\mathrm{(B)}$$:

$$\mathrm{CH_3CH_2CH=CHCH_3}$$

These are Hofmann products.