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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

Predict the product $B$ in the sequence of reaction:
$HC \equiv CH$ $\xrightarrow{30\% \ H_2SO_4, HgSO_4} A$ $\xrightarrow{dil. \ NaOH} B$

Which of the following reactions does not represent the aldol condensation reaction?

The major product $(P)$ in the following reaction is :

Which among the following compounds is a hemiacetal?

Select the option in which the incorrect major product is shown.

Vedclass Test Series

Exam Paper Generator

Online Exam Module