Q 35 E - Organic Chemistry (Mcmurry)

Question

Question:When the alcohol below is treated with ${\mathbf{POCI}}_{\mathbf{3}}$and pyridine, the expected elimination product is formed. However, when the same alcohol is treated with ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$, the elimination product is 1,2-dimethylcyclopentene. Propose a mechanism for each pathway to account for these differences.

Step-by-Step Solution

Answer

a) 

b)

Step 1: E1 elimination

It is the process of removal of the halide to form an alkene. It is a single-step process without the formation of an intermediate.

E2 elimination is not able to form alkene from alcohol because removal of hydroxyl is difficult. So it first reacts with ${\mathrm{POCI}}_3$ to form halide. Then the removal of halide occurs.

Step 2: Formation of a compound on reacting with phosphorous trichloride

The alcohol hydroxyl group reacts with the ${\mathrm{POCI}}_3$   to form the dicholorophosphate intermediate, followed by the E2 elimination in the presence of the amine base and pyridine, which abstracts proton. At the same time, dichlorophosphate is leaving, and alkene formation occurs as shown:

                                Formation of compound A

Step 3: Formation of the compound on reacting with sulphuric acid

The reaction of alcohol with sulfuric acid leads to the removal of the hydroxyl group by the loss of the water molecule, and the formation of carbocation occurs. 

A four-membered ring is less stable than a five-membered ring, so ring expansion occurs, and the four-membered ring is rearranged into five-membered rings.

 Then, the abstraction of the proton occurs, which leads to the formation of alkene, as shown below: