What are the major products for reaction with cis-1-chloro-3-methylcyclopentane and trans-1-chloro-3-methylcyclopentane with "OH"^(-)OH in water at high temperature?

1 Answer
Sep 16, 2017

The cis isomer will form 3- and 4-methylcyclopentene in roughly equal amounts. The trans isomer will form 4-methylcyclopentene as the major product.

Explanation:

What type of reaction?

I predict an "E2"E2 elimination because:

  • The substrate is a secondary alkyl halide.
  • The substrate has at least one β-hydrogen: possible elimination.
  • The nucleophile ("OH"^"-"OH-) is a strong base: possible elimination.
  • There is (presumably) a high base concentration: favouring "E2"E2.
  • Water is a polar protic solvent: favouring elimination.
  • The high temperature favours elimination.

cis-1-Chloro-3-methylcyclopentane

The structure of the substrate is

CisCis

Elimination requires a trans arrangement of the β-hydrogen and the leaving group.

We see appropriate β-hydrogens at "C2"C2 and "C5"C5.

Elimination of the hydrogen from "C2"C2 forms 3-methylcyclopentene.

3-Methyl3-Methyl

Elimination of the hydrogen from "C5"C5 forms 4-methylcyclopentene.

4-Methyl4-Methyl

These two isomers would probably be formed in roughly equal amounts.

trans-1-Chloro-3-methylcyclopentane

The structure of the substrate is

TransTrans

Again, we see trans β-hydrogens at "C2"C2 and "C5"C5.

However, elimination will be slower in each case because of steric hindrance by the methyl group.

Elimination of the hydrogen from "C5"C5 will be faster than from "C2"C2 because of its greater distance from the methyl group.

Thus, the major product will be 4-methylcyclopentene.