A) formic acid
B) acetic acid
C) propionic acid
D) benzoic acid
Correct Answer: B
Solution :
\[RCOOH\xrightarrow{{}}\underset{59\,\,g}{\mathop{RCON{{H}_{2}}}}\,\xrightarrow{{}}\underset{17\,\,g}{\mathop{N{{H}_{3}}}}\,\] Since, \[17g\] of\[N{{H}_{3}}\]is liberated from\[59g\]of acid amide, the amide has molecular mass of \[59,\,\,i.e.,RCON{{H}_{2}}=59\] \[R+12+16+14+2=59\] \[R+44=59\] \[R=15\] Hence,\[R\]is\[C{{H}_{3}}\]group and thus acid is\[C{{H}_{3}}COOH\](Acetic acid)You need to login to perform this action.
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