# Major product in dehydration of alcohol

Consider the dehydration of the following molecule with conc. $$\ce{H2SO4}$$ at $$443 \text K$$:

The possible products are:

The first one is the Zaitsev product and the second one Hofmann product.

Which compound is the major product?

There are 2 factors to consider here:

1. Zaitsev product is more stable than Hofmann product due to hyperconjugation
2. Anti elimination is more favorable than syn elimination due to steric requirements

Obviously both the factors clash with each other, so I couldn't come to a definite conclusion. Can you help me out?

The major product is the Zaitsev one i.e 1-methylcyclohexene. The OH needs to be protonated first in order for it to leave and moreover it is a very fast leaving group (also we have the case that $$\ce{HSO_{4}^{-}}$$ or $$\ce{H_{2}O}$$ are very weak bases and considering that only elimination reactions occur tells us that it must follow an E1 elimination). When the OH is protonated and it leaves, we form a secondary carbocation, but that can be further more made stabilized via an hydride shift from the adjacent carbon producing a tertiary carbocation. Now either of the base $$\ce{HSO_{4}^{-}}$$ or $$\ce{H_{2}O}$$ can capture the hydrogen from the adjacent carbon leading to the desired Zaitsev product. (Also the stereochemistry isn't necessary to worry about here, as the reaction proceeds through E1.)