Why would the reaction of the trans isomer yield a different product? Does it have to do with axial positions

Whe cis-1-iodo-2-methylcyclohexane is treated with potassium tert-butoxide, the major product is 1-methylcyclohexene. However when the corresponding trans isomer is subjected to the same conditions, the reaction progresses at a slower rate and the major product is 3-methylcyclohexene, a disubstituted alkene. Explain.

This is my reason for why the reactions is slower: The reaction progresses at a slower rate with the trans-1-iodo-2-methylcyclohexane (when treated with potassium tert-butoxide) because it is in the diaxial conformation which is of higher energy. When you look at the more accurate drawing of the compound to observe where exactly the iodine and methyl are you can tell the diaxial conformation of the trans and find that it makes sense for the reaction to take longer. Axial addition is slower.

I can't figure out when it yields 3-methylcyclohexene instead of 1-methylcyclohexene though, please explain so i can learn this and understand :)

Thanks in advance! :)

Check it check it check it. Very easy to understand.

For the 1st one, the hydrogen bound to the carbon with the methyl group is trans (anticoplanar) with the iodine, which is a requirement for E2 rxns. Therefore 1-methylcyclohexene can be formed. However, for the 2nd one, the hydrogen is cis with the iodine so it must use the hydrogen from the other carbon which is trans, which forms 3-methylcyclohexene

No problem :) feel free to ask me anything else.

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