reaction diagram

In the above dehydration reaction (asked in a competitive examination) I have shown two paths, obviously only one of them is correct.

According to test correct answer is product formed through A but I think other one should be the correct major product :

After the first carbocation forms, it would rearrange; after rearrangement to A by methyl shift, number of alpha hydrogen in A is only 6 (carbocation being 3°) while in B, number of alpha hydrogen is 7 (Hydride shift). B also being 3° carbocation. So yielding more stable carbocation it should rearrange to B and further corresponding Saytzeff product should form.

Also we can't apply migratory aptitude here since there is no scope of back bonding.

So who is right and why ? Please explain.

  • $\begingroup$ You might consider the reversibility of the reaction which would favor path A. $\endgroup$
    – user55119
    Mar 17, 2023 at 23:38

1 Answer 1


So who is right and why?

The person who made the question is correct (OP said according to the test, correct answer is the product formed through inter mediate $\bf{A}$, because that major product, 1,2,3-trimethylcyclohexene is the thermodinamically most stable product (assuming reaction is set under correct condition for thermodynamic control). Since the whole process is reversible except for the last step to give product, there are other posibilities to obtain several minor products mainly based on the conditions applied. Energy barrier to remove those protones should be harder that that to remove tertiary protons in intermediate

The migratory aptitude matters if the intermediate stability increased by the migration. From initial carbo cation, there are two migratory choices: hydride shift or alkide shift, both of which give most stable tertiary carbocations $\bf{B}$ and $\bf{A}$ intermediates, respectively. Although hydride is superior to alkide in migratory aptitude, methide shift is dominate here because it gives the thermodinamically most stable product with tetrasubstithted double bond. There is also possibility that $\bf{A}$ intermediate can rearrange again by shifting another hydride to give a third tertiary carbocation where positive charge is in the Carbon-2 (of 1,2,3-trimethylcyclohexane frame). That gives an additional tertiary proton (one on C1 and the other on C3) to leave giving the identical product.

The disadvantage on $\bf{B}$ intermediate is it has two secondery protons to leave giving the two different products, 2,4,4-trimethylcyclohexene (as shown in OP's scheme) and 1,3,3-trimethylcyclohexene $\bf{A}$. Thus, those products must be produced as minor products.

Note: There is also posibility to have two cyclopetene and cyclopenten products by ring consruction rearrangement based on condition used. I leaved it to OP to envision.


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