Question

1-Chloro-2-butene undergoes hydrolysis in warm water to give a mixture of these allylic alcohols. Propose a mechanism for their formation.

Short answer

Answer: The mechanism involved is an SN1 reaction, where a carbocation intermediate is formed followed by a nucleophilic attack by water. A possible 1,2-hydride shift may occur leading to the formation of two different allylic alcohol products.

Step-by-step solution

Identifying the structure of 1-chloro-2-butene

First, let's identify the structure of 1-chloro-2-butene, which is CH3-CH(Cl)=CH-CH3. It is a 4-carbon chain with a double bond between the second and third carbon, and a chlorine atom attached to the second carbon.

Understanding allylic alcohols

Allylic alcohols have a hydroxyl group (OH) attached to an allylic carbon (a carbon that is adjacent to the carbon with a double bond). In this reaction, we expect the allylic alcohol products to be CH3-CH(OH)=CH-CH3 and CH3-CH=CH-CH2(OH).

Identifying the reaction mechanism

Given that the reaction occurs in warm water, it suggests that this is a nucleophilic substitution reaction. Specifically, it is an SN1 reaction because the reactant, 1-chloro-2-butene, is an allylic halide which typically favors an SN1 reaction. In an SN1 reaction, the bond between the carbon and the halogen (Cl) breaks, and the nucleophile (water) attacks the carbocation intermediate to form the product.

Formation of a carbocation intermediate

In an SN1 reaction, the first step is the formation of a carbocation intermediate. The bond between the second carbon and the chlorine atom breaks, generating a carbocation on the second carbon (CH3-CH+=CH-CH3) and a chloride ion (Cl-).

Nucleophilic attack by water

Next, the water molecule (H2O) acts as a nucleophile and attacks the positively charged carbocation. This results in the formation of an oxonium ion intermediate, which is CH3-CH(OH2+)=CH-CH3.

Deprotonation to form the first allylic alcohol

Another water molecule acts as a base and abstracts a proton from the oxonium ion intermediate, resulting in the formation of the first allylic alcohol, CH3-CH(OH)=CH-CH3, and a hydronium ion (H3O+).

Formation of the second allylic alcohol

Since the reaction gives a mixture of allylic alcohols, it is possible that a 1,2-hydride shift occurs in the carbocation intermediate (from step 1) to form an alternative carbocation intermediate on the third carbon (CH3-CH=CH-CH2+). A nucleophilic attack by water, followed by a deprotonation event similar to steps 2 and 3, would then lead to the formation of the second allylic alcohol, CH3-CH=CH-CH2(OH). Thus, the proposed mechanism for the formation of the mixture of allylic alcohols involves an SN1 reaction, with nucleophilic attack by water on the carbocation intermediate and a possible 1,2-hydride shift forming two different products.