Neopentyl chloride on reaction with ethanolic KOH is likely to:(a)Neopentyl alcohol(b)pentylene(c)2-methyl-2-butene(d)undergo no reaction

Neopentyl chloride (also known as 2-chloro-2-methylbutane) is a tertiary alkyl halide. When it reacts with ethanolic KOH, it typically undergoes an elimination reaction rather than a substitution reaction. This is because ethanolic KOH is a strong base that favors elimination reactions (E2 mechanism) over substitution reactions (SN2 mechanism), especially with tertiary alkyl halides.

In the elimination reaction, the base abstracts a proton from a carbon adjacent to the carbon bearing the halide, leading to the formation of a double bond and the loss of the halide as a leaving group.

Here’s how the reaction proceeds:

1. **Base Action:** The strong base (ethanolic KOH) abstracts a proton from the β-carbon (the carbon next to the one bearing the chloride).
2. **Formation of Double Bond:** This leads to the formation of a double bond between the α-carbon (the carbon with the chlorine) and the β-carbon.
3. **Elimination of Halide:** The chloride ion is eliminated as a leaving group.

For neopentyl chloride:

\[ \text{(CH}_3\text{)}_2\text{C(Cl)CH}_2\text{CH}_3 \]

The elimination reaction will produce:

\[ \text{(CH}_3\text{)}_2\text{C=CHCH}_2\text{CH}_3 \]

which is **2-methyl-2-butene**.

Thus, the correct answer is:

**(c) 2-methyl-2-butene**.