In the reaction sequence C_6H_5-CH(OH)-CH_2-C | vedclass

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Question

(C) $(i)$ $NaOBr$ acts as an oxidizing agent and a haloform reagent. It oxidizes the secondary alcohol to a ketone: $C_6H_5-CH(OH)-CH_2-C(=O)-CH_3 \xr

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$C_6H_5-C(=O)-CH_3$ and $CHBr_3$

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(C) $(i)$ $NaOBr$ acts as an oxidizing agent and a haloform reagent. It oxidizes the secondary alcohol to a ketone: $C_6H_5-CH(OH)-CH_2-C(=O)-CH_3 \xrightarrow{NaOBr} C_6H_5-C(=O)-CH_2-C(=O)-CH_3$. Then,the methyl ketone group undergoes the haloform reaction: $C_6H_5-C(=O)-CH_2-C(=O)-CH_3 \xrightarrow{NaOBr} C_6H_5-C(=O)-CH_2-COO^{-} + CHBr_3$. $(ii)$ Acidification converts the salt to a $\beta$-keto acid: $C_6H_5-C(=O)-CH_2-COO^{-} \xrightarrow{H^{+}} C_6H_5-C(=O)-CH_2-COOH$. $(iii)$ Heating $(\Delta)$ causes the $\beta$-keto acid to undergo decarboxylation: $C_6H_5-C(=O)-CH_2-COOH \xrightarrow{\Delta} C_6H_5-C(=O)-CH_3 + CO_2$. The final products are acetophenone $(C_6H_5-C(=O)-CH_3)$ and bromoform $(CHBr_3)$.

In the reaction sequence $C_6H_5-CH(OH)-CH_2-C(=O)-CH_3 \xrightarrow{(i) NaOBr, (ii) H_2O/H^{+}, (iii) \Delta}$ product,the product will be:

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