Question

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?

Answer 1

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"` elimination because:

• The substrate is a secondary alkyl halide.
• The substrate has at least one β-hydrogen: possible elimination.
• The nucleophile (`"OH"^"-"`) is a strong base: possible elimination.
• There is (presumably) a high base concentration: favouring `"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 `"C2"` and `"C5"`.

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

Elimination of the hydrogen from `"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 `"C2"` and `"C5"`.

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

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

Thus, the major product will be 4-methylcyclopentene.