Question

trans-3,4-Dimethylcyclobutene can open by two conrotatory paths to give either (2E,4E)-2,4-hexadiene or (2Z,4Z)-2,4-hexadiene. Explain why both products are symmetry-allowed, and then account for the fact that only the 2E,4Eisomer is obtained in practice.

Short answer

2E,4E isomer is obtained in practice because it is most stable among two.

This is most stable due to the fact of not having steric hinderance, but in case of 2Z,2Z isomer, there is much steric hinderance

Step-by-step solution

Tran-form can open by conrotatory paths.

In the trans isomer of 3,4-dimethylcyclobutene, if clockwise rotation is followed, the trans(E,E) isomer of diene is formed.

Conrotation of C₃ and C₄results in the formation of the E,E diene from the trans-cyclobutene.

Both product are symmetry allowed.

In the trans isomer of 3,4-dimethylcyclobutene, if clockwise rotation is followed, the trans(E,E) isomer of diene is formed.

Conrotation of C₃ and C₄results in the formation of the E,E diene from the trans-cyclobutene.

It has been observed that by disrotation of that groups attached to C₃ and C₄rotated in opposite directions (one clockwise and one counter clockwise as shown in figure) in such case trans cyclobutene isomer yields(E,Z)-2,4 hexadiene; therefore he opposite results from what is actually observed. Therefore, the reaction was experimentally observed are not disrotatory, nor a mixture of dis and conrotatory