Question

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields

(A)- a and c
(B)- c only
(C)- a and b
(D)- all of these

Answer 1

Reaction in tertiary alkyl halide may proceed via substitution (${{S}_{N}}1$) or elimination depending on the nucleophile or base. However, both substitution and elimination products may form depending on the stability of carbocation or the alkene formed as a result of elimination.

Complete answer:
The structure of 2-chloro-2-methylpentane is shown below:

2-chloro-2-methylpentane is a tertiary (${{3}^{o}}$) alkyl halide.
Unimolecular substitution reaction (${{S}_{N}}1$) involves the formation of a carbocation and is generally carried out in polar solvent.
Reaction of 2-chloro-2-methylpentane with sodium methoxide ($C{{H}_{3}}{{O}^{-}}N{{a}^{+}}$) in methanol ($C{{H}_{3}}OH$) medium satisfies the conditions for ${{S}_{N}}1$ reaction and Thus, tertiary carbocation, which is very stable, is formed. Nucleophile ($^{-}OC{{H}_{3}}$) attacks the ${{3}^{o}}$ carbocation to complete the substitution reaction.

Now, 2-chloro-2-methylpentane contains $\beta$- hydrogen (i.e. hydrogen on the carbon adjacent to the carbon bearing chlorine), so there is a possibility of elimination reaction. Also, ${{3}^{o}}$ alkyl halide is sterically crowded and hinders the nucleophile attack. Then, $^{-}OC{{H}_{3}}$ acts as a base to abstracts a proton from the $\beta$-position and forms an alkene.
Since, there is more than one $\beta$-hydrogen available, the stability the alkenes formed is determined by Zaitsev’s rule, which states that the alkene containing more number of alkyl groups attached to carbon-carbon double bond is more stable and hence, is the major product in elimination.

Based on the above discussion, we conclude that all the three products are possible.

Hence, the correct option is (D).

Note:
Keep in mind that $^{-}OC{{H}_{3}}$ acts as a nucleophile in ${{S}_{N}}1$ reaction and as a base in elimination. Formation of ${{3}^{o}}$ carbocation favours ${{S}_{N}}1$ and availability of $\beta$-hydrogen favors elimination in 2-chloro-2-methylpentane.