The major product in the reaction

We need to find the major product when tert-butyl chloride reacts with potassium tert-butoxide.

Identify the substrate and reagent: Substrate: $$(CH_3)_3C{-}Cl$$ (tert-butyl chloride) — a tertiary alkyl halide.

Reagent: $$K^+ \; ^-OC(CH_3)_3$$ (potassium tert-butoxide) — a strong, bulky base.

Determine the reaction pathway: With a tertiary substrate:

$$\bullet$$ $$S_N2$$ is not possible due to steric hindrance at the tertiary carbon.

$$\bullet$$ $$S_N1$$ could occur but is disfavored because tert-butoxide is a strong base, not just a nucleophile.

$$\bullet$$ The bulky tert-butoxide base strongly favors E2 elimination over substitution.

Determine the elimination product: E2 elimination of HCl from $$(CH_3)_3CCl$$:

$$(CH_3)_3C{-}Cl \xrightarrow{KOC(CH_3)_3} (CH_3)_2C{=}CH_2 + KCl$$

The product is 2-methylprop-1-ene (isobutylene), formed by removal of H from one of the methyl groups and Cl from the tertiary carbon.

Evaluate the options: Option A: t-Butyl ethyl ether — Would require $$S_N$$ reaction, which is not favored here.

Option B: 2-Methyl pent-1-ene — Incorrect carbon count; the substrate has only 4 carbons.

Option C: 2, 2-Dimethyl butane — Would require C-C bond formation, not possible in this reaction.

Option D: 2-Methyl prop-1-ene — Correct. This is the E2 elimination product.

The correct answer is Option D: 2-Methyl prop-1-ene.