Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The major product of the reaction is:
seo images

(A)
seo images

(B)
 
seo images

(C)
seo images

(D)
seo images


seo-qna
Last updated date: 27th Jul 2024
Total views: 405.3k
Views today: 11.05k
Answer
VerifiedVerified
405.3k+ views
Hint: The organic reaction of the conversion of a primary amine into the diazonium salt is known as the diazotization reaction. This is a method by which we can convert the amine group into a diazo group which is easily attacked by the other reagent. Here sulphuric acid maintains the acidic medium.

Complete Solution :
The primary amine group undergoes the diazotization reaction. We have been provided with a reaction, we need to complete the reaction and the major product formed,
So, for that:
- In the given reaction diazotization reaction is taking place. Here, the compound is treated with the $\text{ NaN}{{\text{O}}_{\text{2 }}}$ solution. The nitrosation of amine takes place with nitrous acid in presence of acidic medium produced the diazonium salt.
- The obtained Diazonium salts are important synthetic intermediates that can undergo coupling reactions to form azo dyes and electrophilic substitution reactions to introduce functional groups.
So, in this reaction due to diazotization $N{{H}_{2}}$ group leaves as $-{{N}_{2}}$.
seo images

The diazotization reaction does not affect the hydroxyl group. But only converts the $N{{H}_{2}}$ to the$-{{N}_{2}}$.
- In the next step, the product undergoes the Then methyl group shift. These shifts make the carbocation more positive than the former. Then oxygen stabilizes the carbocation by +M effect and ketone formation takes place. Reactant loses the nitrogen gas followed by the 1,2, methyl shift which proceeds via the most stable more positively charged carbocation.
seo images

The product on the deprotonation gives the product which is 3, 3-methyl butanone.
So, the correct answer is “Option B”.

Note: Note that, diazotization reaction is advised to carry out at low temperature $\text{ ( 0 }-{{5}^{0}})C\text{ }$. This is because high temperatures may result in products and give phenol. Here reaction proceeds through the 1, 2 methyl shift as it goes via the most stable carbocation pathway. Whenever there is a possibility of formation of strong carbocation the reaction proceeds via rearrangement.