# 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?

## 1 Answer

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.)

• I would humbly request the contributor to make an edit. The formation of carbocation by removal of H2O is actually a slow step. It is infact the RDS and should proceed slow as is characteristic of any. The addition of the H+ forming the protonated alcohol is the fast step that you've confused with. Jan 6 at 2:22
• but the water is such a nice leaving group that it leaves very fast and forms the cation. That is what I meant. Jan 6 at 2:29
• No, the slow step is carbocation formation. The rest are even faster. Jan 6 at 11:16
• @Desai, please stop using mod flags for everything. Those are meant for exceptional issues. If the answer is wrong, you can comment, or downvote. No need to flag for moderator intervention, or else we'd literally have to fact-check every answer on this site. Jan 6 at 11:17
• OK, let's clarify a few things. (1) The "loss of water" refers to exactly the same step as the carbocation formation. What I'm saying is that that step, whatever you want to call it (water leaving or carbocation being formed), is the slow step. (2) You're still forming an unstable intermediate (a carbocation), which necessarily means that this step represents a large increase in energy, hence higher activation energy, hence slower. Water being the leaving group merely makes that slightly easier, but not to the point where it can be faster than the other steps in the mechanism. [...] Jan 6 at 11:29