(A) xrightarrow{KOBr} (B) xrightarrow[KOH]{CH | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $1$. The reaction sequence starts with $(A)$,which is $p$-methylbenzamide $(CH_3-C_6H_4-CONH_2)$.$2$. Hofmann bromamide degradation: $(A) \xrighta

Vedclass · Accepted Answer

p-methylphenyl isocyanide

Vedclass · Answer

(B) $1$. The reaction sequence starts with $(A)$,which is $p$-methylbenzamide $(CH_3-C_6H_4-CONH_2)$.$2$. Hofmann bromamide degradation: $(A) \xrightarrow{KOBr} (B)$. This converts the amide group $(-CONH_2)$ to a primary amine group $(-NH_2)$. Thus,$(B)$ is $p$-toluidine $(CH_3-C_6H_4-NH_2)$.$3$. Carbylamine reaction: $(B) \xrightarrow{CHCl_3/KOH} (C)$. Primary amines react with chloroform and alcoholic $KOH$ to form isocyanides (isonitriles). Thus,$(C)$ is $p$-methylphenyl isocyanide $(CH_3-C_6H_4-NC)$.$4$. Reduction: $(C) \xrightarrow{LiAlH_4} 	ext{Product}$. Isocyanides are reduced to secondary amines $(R-NH-CH_3)$. Therefore,the final product is $N$-methyl-$p$-toluidine $(CH_3-C_6H_4-NHCH_3)$.$5$. The question asks for the structure of $(C)$,which is $p$-methylphenyl isocyanide.

$(A)$ $\xrightarrow{KOBr} (B)$ $\xrightarrow[KOH]{CHCl_3} (C)$ $\xrightarrow{LiAlH_4} \text{Structure of } (C) \text{ is}$

Similar Questions

Which of the following reactions does not yield an amine?

This reaction is an example of:

Rank the following compounds in order of increasing basic strength. (weakest $\to$ strongest)

When $methyl$ $iodide$ is heated with $ammonia$,the product obtained is

An organic compound $A$ on reduction gives compound $B$,which on reaction with trichloromethane and caustic potash forms $C$. The compound $C$ on catalytic reduction gives $N$-methylbenzenamine. The compound $A$ is,

Vedclass Test Series

Exam Paper Generator

Online Exam Module