Question

Question: A solution of 5-methylcyclopenta-1,3-diene rearranges at room temperature to a mixture containing 1-methyl-, 2-methyl-, and 5-methylcyclopenta-1,3-diene. (a) Show how both isomeric products are formed from the starting material by a sigmatropic rearrangement involving a C–H bond. (b) Explain why 2-methylcyclopenta-1,3-diene is not formed directly from 5-methylcyclopenta-1,3-diene by a [1,3] rearrangement.

Short answer

Answer

a.

b. The [1,3] thermal shift demands the antarafacial migration, which is forbidden due to the geometric and steric constraints, and hence, 5-methylcyclopenta-1,3-diene does not undergo a [1,3] shift.

Step-by-step solution

Sigmatropic rearrangement

A pericyclic reaction involving the cleavage of a sigma bond, the rearrangement of a pi bond, and the formation of a new sigma bond is known as a sigmatropic rearrangement.

Sigmatropic rearrangement of 5-methylcyclopenta-1,3-diene

The sigmatropic rearrangement of 5-methylcyclopenta-1,3-diene yields the following products:

Sigmatropic rearrangement of 5-methylcyclopenta-1,3-diene

In thermal conditions, the isomerization occurs via two successive [1,5] sigmatropic shifts of the C-H bond.

The thermal pathway of [1,5] shift involves suprafacial migration.

[1,3] rearrangement

The [1,3] thermal shift demands antarafacial migration, which is forbidden due to geometric and steric constraints.

Therefore, 5-methylcyclopenta-1,3-diene does not undergo a [1,3] shift yielding 2-methylcyclopenta-1,3-diene.