Q29P A bottle of allyl bromide was fo... [FREE SOLUTION]

Chapter 13: Q29P (page 693)

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

Short Answer

Expert verified

(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.