Determine the structure and name of $A$ and $B$ in the following reaction sequence:
$(A)$ $\xrightarrow{NaOH, -H_2O} (B)$ $\xrightarrow{NaOH, CaO, \Delta} CH_3CH_3$
$(A)$ $\xrightarrow{NaOH, -H_2O} (B)$ $\xrightarrow{NaOH, CaO, \Delta} CH_3CH_3$
(N/A) The given reaction sequence is for the preparation of alkanes via decarboxylation.
Step $1$: Propanoic acid $(CH_3CH_2COOH)$ reacts with $NaOH$ to form sodium propanoate $(CH_3CH_2COONa)$ and water.
Thus,$(A) = CH_3CH_2COOH$ (Propanoic acid) and $(B) = CH_3CH_2COONa$ (Sodium propanoate).
Step $2$: Sodium propanoate undergoes decarboxylation with soda-lime $(NaOH + CaO)$ upon heating to yield ethane $(CH_3CH_3)$.
Step $1$: Propanoic acid $(CH_3CH_2COOH)$ reacts with $NaOH$ to form sodium propanoate $(CH_3CH_2COONa)$ and water.
Thus,$(A) = CH_3CH_2COOH$ (Propanoic acid) and $(B) = CH_3CH_2COONa$ (Sodium propanoate).
Step $2$: Sodium propanoate undergoes decarboxylation with soda-lime $(NaOH + CaO)$ upon heating to yield ethane $(CH_3CH_3)$.
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