Question

A bottle of allyl bromide was found to contain a large amount of an impurity. A careful distillation separated the impurity, which has the molecular formula C₃H₆0. The following ¹³CNMR spectrum of the impurity was obtained:

1. Propose a structure for this impurity.
2. Assign the peaks in the¹³CNMR spectrum to the carbon atoms in the structure.
3. Suggest how this impurity arose in the allyl bromide sample.

Short answer

(a) and (b).

The peak at 115 is a = CH₂, the peak at 138 is = CH - and the peak at 63 is deshielded aliphatic CH₂ . The formula has changed from C₃ H₅ Br to C₃ H₆Br so OH has replaced the Br.

(c).The impurity arose because of the presence of water, and allyl bromide gets easily hydrolyzed by water by S_N1 process.

Step-by-step solution

Step-1. Explanation of parts (a) and (b):

Since allyl bromide was the starting material, it is expected that an allyl group must be present in the impurity. The peak at 115 is a = CH₂ , the peak at 138 is = CH - and the peak at 63 is deshielded aliphatic a = CH₂ . The formula has changed from C₃ H₅ Br to C₃ H₆ so OH has replaced the Br.

Chemical shift values in ¹³C-NMR spectrum

Step-2. Explanation of part (c):

The impurity arose because of presence of water and allyl bromide gets easily hydrolyzed by water by S_N1 process. S_N1 process involves the formation of carbocation as the intermediate and since, allylic carbocation is stable, thus reaction follows S_N1 pathway and in second step, water which is a impurity acts as a nucleophile and attacks at the carbocation and forms the product.

Hydrolysis of allyl bromide