CH_3-CCl_3 xrightarrow[(2) H^{+}]{(1) KOH_{(a | vedclass

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(D) $Step$ $1$: $CH_3-CCl_3$ reacts with aqueous $KOH$ followed by acidic workup $(H^{+})$ to form $CH_3-C(OH)_3$,which is unstable and loses a water

Vedclass · Accepted Answer

$CH_3-COCH_3$

Vedclass · Answer

(D) $Step$ $1$: $CH_3-CCl_3$ reacts with aqueous $KOH$ followed by acidic workup $(H^{+})$ to form $CH_3-C(OH)_3$,which is unstable and loses a water molecule to form $CH_3-COOH$ (Acetic acid). Thus,$P$ is $CH_3-COOH$.
$Step$ $2$: Acetic acid reacts with $Ca(OH)_2$ to form calcium acetate,$(CH_3COO)_2Ca$.
$Step$ $3$: Calcium acetate upon dry distillation $(\Delta)$ undergoes decarboxylation to give $CH_3-COCH_3$ (Acetone) and $CaCO_3$. Thus,$Q$ is $CH_3-COCH_3$.

$CH_3-CCl_3$ $\xrightarrow[(2) H^{+}]{(1) KOH_{(aq)}}$ $P$ $\xrightarrow[(2) \Delta]{(1) Ca(OH)_2}$ $Q$; $Q$ is:

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