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Which of the following reaction will NOT give primary amine as the product?
medium
Amines
2023
chemistry
C H 3 C O N H 2 ā B r 2 / K O H {CH}_3{CONH}_2 \xrightarrow{Br_2 / KOH} C H 3 ā CON H 2 ā B r 2 ā / K O H ā Product
C H 3 C N ā ( i i ) H 3 O + ( i ) L i A l H 4 {CH}_3CN \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4} C H 3 ā CN ( i ) L i A l H 4 ā ( ii ) H 3 ā O + ā
C H 3 N C ā ( i i ) H 3 O + ( i ) L i A l H 4 {CH}_3NC \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4} C H 3 ā NC ( i ) L i A l H 4 ā ( ii ) H 3 ā O + ā
C H 3 C O N H 2 ā ( i i ) H 3 O + ( i ) L i A l H 4 {CH}_3CONH_2 \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4} C H 3 ā CON H 2 ā ( i ) L i A l H 4 ā ( ii ) H 3 ā O + ā
Explanation To solve this problem, we need to analyze each reaction to determine if it produces a primary amine.
\newline Let's examine each option step-by-step:
\newline ā¢ Option 1: C H 3 C O N H 2 ā B r 2 / K O H {CH}_3{CONH}_2 \xrightarrow{Br_2 / KOH} C H 3 ā CON H 2 ā B r 2 ā / K O H ā
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Product
Product
Product
Product
This is the Hofmann bromamide degradation reaction.
ā¢ In this reaction, an amide is converted to a primary amine with one less carbon atom.
ā¢ The product is
C H 3 N H 2 , CH_3NH_2, C H 3 ā N H 2 ā , which is a primary amine.
ā¢ Option 2:
C H 3 C N ā ( i i ) H 3 O + ( i ) L i A l H 4 {CH}_3CN \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4} C H 3 ā CN ( i ) L i A l H 4 ā ( ii ) H 3 ā O + ā Product
This involves the reduction of a nitrile using
L i A l H 4 . LiAlH_4.
\newline L i A l H 4 ā . ā¢ The nitrile is reduced to a primary amine.
ā¢ The product is
C H 3 C H 2 N H 2 , CH_3CH_2NH_2, C H 3 ā C H 2 ā N H 2 ā , which is a primary amine.
ā¢ Option 3:
C H 3 N C ā ( i i ) H 3 O + ( i ) L i A l H 4 {CH}_3NC \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4} C H 3 ā NC ( i ) L i A l H 4 ā ( ii ) H 3 ā O + ā Product
This involves the reduction of an isocyanide (isonitrile).
ā¢ The reduction of an isocyanide typically leads to the formation of a secondary amine.
ā¢ Therefore, the product is not a primary amine.
ā¢ Option 4:
C H 3 C O N H 2 ā ( i i ) H 3 O + ( i ) L i A l H 4 {CH}_3CONH_2 \xrightarrow[(ii) H_3O^+]{(i) LiAlH_4} C H 3 ā CON H 2 ā ( i ) L i A l H 4 ā ( ii ) H 3 ā O + ā Product
This involves the reduction of an amide using
L i A l H 4 . LiAlH_4.
\newline L i A l H 4 ā . ā¢ The amide is reduced to a primary amine.
ā¢ The product is
C H 3 C H 2 N H 2 , CH_3CH_2NH_2, C H 3 ā C H 2 ā N H 2 ā , which is a primary amine.
Conclusion:
ā¢ Option 3 does not produce a primary amine; it produces a secondary amine.
Therefore, the correct answer is Option 3.