Hydrogenation of C_6H_5CHOHCOOH over Rh-Al_2O | vedclass

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

Question

(B) The hydrogenation of $C_6H_5CHOHCOOH$ (mandelic acid) over a $Rh-Al_2O_3$ catalyst is a selective reaction that reduces the aromatic benzene ring

Vedclass · Accepted Answer

$C_6H_{11}CHOHCOOH$

Vedclass · Answer

(B) The hydrogenation of $C_6H_5CHOHCOOH$ (mandelic acid) over a $Rh-Al_2O_3$ catalyst is a selective reaction that reduces the aromatic benzene ring to a cyclohexane ring while leaving the carboxylic acid $(-COOH)$ and hydroxyl $(-OH)$ groups intact.The reaction is as follows:$C_6H_5CHOHCOOH + 3H_2 \xrightarrow{Rh-Al_2O_3} C_6H_{11}CHOHCOOH$Thus,the product formed is $C_6H_{11}CHOHCOOH$ (cyclohexylmandelic acid).

Hydrogenation of $C_6H_5CHOHCOOH$ over $Rh-Al_2O_3$ catalyst in methanol gives:

Similar Questions

In the given reaction: $CH_3COCH_3 \xrightarrow{\text{Conc. } H_2SO_4, \text{Distillation}} [X]$. $[X]$ will be:

In the reaction,the major product formed is:

Explain the bromination of benzene with its mechanism.

Nitrobenzene can be prepared from benzene by using a mixture of conc. $HNO_3$ and conc. $H_2SO_4$. In the nitrating mixture,$HNO_3$ acts as a

Which active species is involved in the nitration of benzene with concentrated nitric acid and concentrated sulfuric acid?

Vedclass Test Series

Exam Paper Generator

Online Exam Module