The Dehydration Of 2 Methylcyclohexanol Essays

1082 Words Nov 13th, 2016 5 Pages
Introduction: The purpose for this experiment is to carry out the dehydration of 2-methylcyclohexanol, which undergoes an E1 mechanism by following Saytzev’s rule to form different product such as 1-methylcyclohexanol, 3-methylcyclohexanol and methylenecyclohexane. The reaction can be dehydrated using sulfuric acid and H3PO4. The dehydration mechanism occurs in several steps like –
1. Converting the OH group into a better leaving group that is deprotonating with H3PO4.
2. Removing a beta-hydrogen besides the methyl group to form a stable product.
3. The most stable carbocation is a tertiary carbon, which means the major product is going to be a tri-substituted alkene (
The overall reaction for the dehydration of 2-methylcyclohexanol is shown below.
Protonation of the Alcohol Group -

Figure 1: Protonation of the alcohol group.

Carbocation Rearrangement -

Figure 2: Carbocation Rearrangement.
Beta-Hydride Elimination to Form the Alkene -

Figure 3: Beta-Hydride Elimination to form the alkene.
Beta-elimination reactions usually undergo a hydride or a methyl shift to form the different products in different amounts. The reaction is also reversible, in order to prevent the reaction from getting converted back into an alcohol, we can use a weaker acid like H3PO4 instead of
H2SO4. Then, the water generated as the side product should be immediately removed using vacuum line, so the water is sucked out from the reaction giving only an alkene…

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