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A compound C (molecular formula , $$ C_{2}H_{4}O_{2} $$ ) reacts with Na-metal to form a compound R and evolves a gas which burns with a pop sound. Compound C on treatement with an alcohol A in presence of an acid forms a sweet smelling compounds S (molecular formula $$ C_{3}H_{6}O_{2} $$). On addition of NaOH to C, it also gives R and water, S on treatement with NaOH solution gives back R and A.
Identify C, R, A, S and write down the reactions involved.


Solution

As Compound C evolve a gas which burn with pop sound, i.e. the gas is $$CO_2$$. Therefore, C is Ethanoic acid.
R- Sodium salt of ethanoic acid (sodium acetate) and gas evolved is hydrogen
A-Methanol
S-Ester (Methyl acetate)
(a) $$ \underset{(c)}{2CH_{3}COOH}+2Na\rightarrow \underset{(R)}{2CH_{3}COONa}+H_{2} $$
(b) $$ \underset{(c)}{CH_{3}COOH}\overset{Conc.H_{2}SO_{4}}{\rightarrow}\underset{(s)}{CH_{3}COOCH_{3}}+H_{2}O $$
(c) $$ \underset{(c)}{CH_{3}COOH}+NaOH \overset{Conc.H_{2}SO_{4}}{\rightarrow}\underset{(R)}{CH_{3}COONa}+H_{2}O $$
(d) $$ \underset{(c)}{CH_{3}COOCH_{3}}+NaOH\rightarrow \underset{(R)}{CH_{3}COONa}+\underset{(A)}{CH_{3}OH} $$

Chemistry
NCERT
Standard XII

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