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

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 $\text{E2}$ elimination because:

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

cis-1-Chloro-3-methylcyclopentane

The structure of the substrate is

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

We see appropriate β-hydrogens at $\text{C2}$ and $\text{C5}$.

Elimination of the hydrogen from $\text{C2}$ forms 3-methylcyclopentene.

Elimination of the hydrogen from $\text{C5}$ forms 4-methylcyclopentene.

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

trans-1-Chloro-3-methylcyclopentane

The structure of the substrate is

Again, we see trans β-hydrogens at $\text{C2}$ and $\text{C5}$.

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

Elimination of the hydrogen from $\text{C5}$ will be faster than from $\text{C2}$ because of its greater distance from the methyl group.

Thus, the major product will be 4-methylcyclopentene.