The stable form of trans-1,4-dimethylcyclohex | vedclass

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Question

(C) In $trans-1,4-$dimethylcyclohexane,the two methyl groups are on opposite sides of the cyclohexane ring. In the chair conformation,for the $trans$

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(C) In $trans-1,4-$dimethylcyclohexane,the two methyl groups are on opposite sides of the cyclohexane ring. In the chair conformation,for the $trans$ isomer,one methyl group must be axial and the other equatorial,or both must be equatorial. The conformation where both methyl groups are in the equatorial position is the most stable because it minimizes $1,3-$diaxial interactions. Looking at the options,the structure where both methyl groups are in equatorial positions represents the most stable form of $trans-1,4-$dimethylcyclohexane.

The stable form of $trans-1,4-$dimethylcyclohexane is represented as:

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