CH_3-CO-CH_2-COOH xrightarrow{Delta} X; X can | vedclass

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Question

(C) The starting material is acetoacetic acid $(CH_3-CO-CH_2-COOH)$.Upon heating $(\Delta)$,$\beta$-keto acids undergo decarboxylation to form a keton

Vedclass · Accepted Answer

$CH_3-C \equiv CH \xrightarrow{Hg^{2+}, H_2O}$

Vedclass · Answer

(C) The starting material is acetoacetic acid $(CH_3-CO-CH_2-COOH)$.Upon heating $(\Delta)$,$\beta$-keto acids undergo decarboxylation to form a ketone.$CH_3-CO-CH_2-COOH \xrightarrow{\Delta} CH_3-CO-CH_3   CO_2$.Thus,$X$ is acetone $(CH_3-CO-CH_3)$.Now,let us evaluate the options:$(A)$ $CH_3-CH_2-CH_2-Cl \xrightarrow{LiAlH_4} CH_3-CH_2-CH_3$ (Propane,not acetone).$(B)$ $CH_3-CH_2-COOH \xrightarrow{P   HI} CH_3-CH_2-CH_3$ (Propane,not acetone).$(C)$ $CH_3-C \equiv CH \xrightarrow{Hg^{2 }, H_2O} CH_3-CO-CH_3$ (Acetone is formed via hydration of propyne).Therefore,the correct option is $C$.

$CH_3-CO-CH_2-COOH \xrightarrow{\Delta} X$; $X$ can also be formed by

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