Identify A, B,and C in the following reaction | vedclass

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(C) The reaction sequence is as follows:$1$. Isopropyl chloride $(CH_3-CHCl-CH_3)$ reacts with aqueous $NaOH$ (nucleophilic substitution) to form Isop

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$CH_3-CH(OH)-CH_3, CH_3-CO-CH_3, CH_3-COONa$

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(C) The reaction sequence is as follows:$1$. Isopropyl chloride $(CH_3-CHCl-CH_3)$ reacts with aqueous $NaOH$ (nucleophilic substitution) to form Isopropyl alcohol $(A)$: $CH_3-CHCl-CH_3 \xrightarrow{NaOH} CH_3-CH(OH)-CH_3 (A)$.$2$. Isopropyl alcohol $(A)$ on dehydrogenation with $Cu$ at $573 \ K$ gives Acetone $(B)$: $CH_3-CH(OH)-CH_3 \xrightarrow{Cu/573 \ K} CH_3-CO-CH_3 (B)$.$3$. Acetone $(B)$ undergoes the iodoform reaction with $NaOI$ to give Sodium acetate $(C)$ and Iodoform $(CHI_3)$: $CH_3-CO-CH_3 \xrightarrow{NaOI} CH_3-COONa (C) + CHI_3$.Thus,the correct sequence is $A = CH_3-CH(OH)-CH_3$,$B = CH_3-CO-CH_3$,and $C = CH_3-COONa$.

Identify $A, B$,and $C$ in the following reactions:
Isopropyl chloride $\xrightarrow{NaOH} A$ $\xrightarrow{Cu/573 \ K} B$ $\xrightarrow{NaOI} C + \text{Iodoform}$

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Identify $A, B$,and $C$ in the following reactions:Isopropyl chloride $\xrightarrow{NaOH} A$ $\xrightarrow{Cu/573 \ K} B$ $\xrightarrow{NaOI} C + \text{Iodoform}$