# Reaction of alc. KOH with alkyl halide

The following question was given in FIITJEE study material:

At first, I thought it should be elimination. But, soon I realized that there is no heating, plus a primary alkyl halide is used. So, "no elimination".

Then, I thought it would proceed through $$\ce{S_N2}$$, and hence (B) should be correct, as it involves concerted mech.. But, again I was wrong. Why is it so?

The correct answer given is:

(D)

• My hint would be "enolate".
– Karl
Jan 19 '20 at 17:13
• Just as @Karl mentioned, use the base to abstract the $\alpha$ $\ce{H}$, and form an enolate and subsequently use it as a nucleophile to substitute the $\ce{Cl}$. thus forming the 3 membered ring. Jan 19 '20 at 17:26
• Ok, now I got it. Thank you, Karl and SirArthur7 :) Jan 19 '20 at 17:29
• @RahulVerma How do you have to change the reaction to get (C)?
– Karl
Jan 19 '20 at 20:55
• If you're asking about the actual question, then I would better attribute this reaction to NGP (neighboring group participation). SNi is a bit rare and specific to chlorination of alcohols with thionyl chloride, or the decomposition of alkyl chloroformates, though basically both involve the same concepts. But NGP includes other cases too, so I feel its safe to use it. I'd like someone else to contradict my belief if there are any. Jan 20 '20 at 14:25

## 1 Answer

The solvent used is a polar protic solvent which heavily solvates the OH- radical thereby decreasing it's nucleophilicity as the solvent forms a cage around the radical anion .Even though the basic strength of the anion is also decreased by solvation still the decrease in nucleophilicity is dominant as it is more difficult for a large solvated anion to attack at a carbon atom than to sneakily abstract an acidic hydrogen and hence the enolate is formed which then , proceeding through the SN NGP (neighbouring group participation) mechanism eliminates the Cl ion forming a cyclopropane.

Hope that this helps.