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Magazine Chapter 27: Q 46. (page 1102)
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鈥揌 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.
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