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Question
In the above reactions product \(A\) and product \(B\) respectively are
\( NH _{2} \stackrel{ KOBr }{\longrightarrow} \)
(major product)
\( NH _{2} \stackrel{ LiAlH _{4}}{ H _{2} O ^{+}} \underset{\text { (major product) }}{ B } \)
In the above reactions product \(A\) and product \(B\) respectively are
\( NH _{2} \stackrel{ KOBr }{\longrightarrow} \) (major product) \( NH _{2} \stackrel{ LiAlH _{4}}{ H _{2} O ^{+}} \underset{\text { (major product) }}{ B } \)
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Solution
The correct option is C
Product A:
When an amide is treated with hypobromite (NaOBr), degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction. The primary amine thus formed contains one carbon less than the number of carbon atoms in that amide.
Product B:
Second is reduction reaction. LiAlH4 is a strong reducing agent. It will reduce amide group to amine. The product will have same number of carbon as of reactant.
Hence, option (d) is correct.
Product A:
When an amide is treated with hypobromite (NaOBr), degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction. The primary amine thus formed contains one carbon less than the number of carbon atoms in that amide.
Product B:
Second is reduction reaction. LiAlH4 is a strong reducing agent. It will reduce amide group to amine. The product will have same number of carbon as of reactant.
Hence, option (d) is correct.
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