Suppose the following reaction: CH_3-CH(CH_3) | vedclass

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(B) The reaction sequence is as follows: $1.$ Carboxylic acid reacts with methylamine to form an ammonium salt: $CH_3-CH(CH_3)-COOH + CH_3-NH_2 \right

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$2^o$ Amine

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(B) The reaction sequence is as follows: $1.$ Carboxylic acid reacts with methylamine to form an ammonium salt: $CH_3-CH(CH_3)-COOH + CH_3-NH_2 \rightarrow CH_3-CH(CH_3)-COO^- NH_3^+CH_3$ ('$A$'). $2.$ Heating the salt results in dehydration to form an $N$-substituted amide: $CH_3-CH(CH_3)-COO^- NH_3^+CH_3 \xrightarrow{\Delta} CH_3-CH(CH_3)-CONH-CH_3$ ('$B$'). $3.$ Reduction of the amide with $LiAlH_4$ yields a secondary amine: $CH_3-CH(CH_3)-CONH-CH_3 \xrightarrow{LiAlH_4} CH_3-CH(CH_3)-CH_2-NH-CH_3$ ('$C$'). Since the nitrogen in '$C$' is bonded to two carbon atoms,it is a $2^o$ amine.

Suppose the following reaction:
$CH_3-CH(CH_3)-COOH + CH_3-NH_2$ $\rightarrow 'A'$ $\xrightarrow{\Delta} 'B'$ $\xrightarrow{LiAlH_4 (excess)} 'C'$
The final product '$C$' will be:

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Suppose the following reaction: $CH_3-CH(CH_3)-COOH + CH_3-NH_2$ $\rightarrow 'A'$ $\xrightarrow{\Delta} 'B'$ $\xrightarrow{LiAlH_4 (excess)} 'C'$ The final product '$C$' will be: