Identify the substrate

The compound is $2$-chloro-$2$-methylpentane:

$\text{CH}_3-\text{C}(\text{Cl})(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{CH}_3$

The carbon bearing chlorine is attached to three alkyl groups, so it is a tertiary alkyl halide.

Identify the reagent

Sodium methoxide in methanol is:

$\text{CH}_3\text{ONa}/\text{CH}_3\text{OH}$

Methoxide is a strong base as well as a good nucleophile. For a tertiary halide, the $S_N2$ pathway is blocked due to steric hindrance.

Possible reaction pathways

For a tertiary alkyl halide in alcoholic strong base, the major pathway is elimination ($E2$), though $S_N1/E1$ may compete in protic medium. But with methoxide, elimination is strongly favored.

Find possible $\beta$-hydrogens

The carbon bearing Cl is the $\alpha$-carbon. Adjacent $\beta$-carbons are:

• the left methyl group
• the other methyl substituent
• the $\text{CH}_2$ of the propyl side

Elimination of $\beta$-H can therefore give different alkenes.

Form the alkenes

• Removing H from either of the equivalent methyl groups gives the same alkene:

  $\text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{CH}_3$

  This is $2$-methyl-$1$-pentene.

• Removing H from the $\text{CH}_2$ on the propyl side gives:

  $\text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_2-\text{CH}_3$

  This is $2$-methyl-$2$-pentene.

Which product is favored?

By Saytzeff rule, the more substituted alkene is major:

$2\text{-methyl-}2\text{-pentene}$

This corresponds to product $(c)$ in the usual option set for this question.

Check for substitution product

An ether substitution product via $S_N1$ is theoretically possible for tertiary halide in methanol, but under sodium methoxide in methanol, the standard expected JEE answer is elimination giving the alkene product, specifically the more stable Saytzeff product.

Evaluate options

• A: All of these — incorrect
• B: $(a)$ and $(c)$ — incorrect
• C: $(c)$ only — correct
• D: $(a)$ and $(b)$ — incorrect

Final answer

The product obtained is $(c)$ only.