what product will form from dehydration of 2-methylcyclohexanol?
http://www.thecatalyst.org/experiments/Moores/Moores.html
2-methylhexene
To determine the product formed from the dehydration of 2-methylcyclohexanol, we need to understand the reaction and the mechanism involved. Dehydration is a type of elimination reaction in which a molecule of water (H2O) is removed from an alcohol.
In the case of 2-methylcyclohexanol (C8H16O), it will undergo dehydration to form an alkene. The reaction typically occurs in the presence of an acid catalyst, such as sulfuric acid (H2SO4).
The dehydration of 2-methylcyclohexanol follows the E1 mechanism, which involves a unimolecular elimination process. Here are the steps involved:
1. Protonation: The acid catalyst, such as sulfuric acid (H2SO4), donates a proton (H+) to the hydroxyl group (OH) of 2-methylcyclohexanol, forming a more reactive species called a carbocation.
2. Rearrangement (if applicable): In some cases, carbocations may undergo a rearrangement to form a more stable carbocation. The stability of carbocations follows the order: tertiary > secondary > primary.
3. Deprotonation: A molecule of water (H2O) acts as a base and removes a proton (H+) from the carbocation, forming an alkene (C=C double bond).
In the case of 2-methylcyclohexanol, only one product is expected to form. The elimination of water from 2-methylcyclohexanol produces 3-methylcyclohexene (C7H12) as the main product.
The reaction can be represented as follows:
2-methylcyclohexanol ⟶ 3-methylcyclohexene + H2O
It is worth noting that in some cases, a minor product called 1-methylcyclohexene may also be formed due to the possibility of carbocation rearrangement.