Based on the provided reaction scheme,determi | vedclass

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

Question

(A) The reaction scheme shows that compound $A$ reacts with $NaBH_4$ to form a hydroxy-acid derivative. Since $NaBH_4$ reduces ketones but not carboxy

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The reaction scheme shows that compound $A$ reacts with $NaBH_4$ to form a hydroxy-acid derivative. Since $NaBH_4$ reduces ketones but not carboxylic acids,$A$ must be a keto-acid. The product of $NaBH_4$ reduction is a hydroxy-acid with a bicyclic structure. $LiAlH_4$ is a stronger reducing agent that reduces both the ketone and the carboxylic acid group to alcohols. Thus,$B$ is a diol with the structure of a bicyclo[$2.2$.$1$]heptane derivative containing two hydroxyl groups. Specifically,the reduction of the keto-acid yields a diol where the two hydroxyl groups are attached to the bicyclic framework. This structure contains two chiral centers at the bridgehead or substituted carbons. For a bicyclic system with two chiral centers,the number of stereoisomers is $2^n$,where $n$ is the number of chiral centers. Given the rigid bicyclic geometry,there are $2$ possible stereoisomers (the $cis$ and $trans$ forms relative to the bridge).

Based on the provided reaction scheme,determine the total number of stereoisomers of $B$.

Similar Questions

$A$ new carbon-carbon bond formation is possible in

The major product in the following reaction sequence is:

Assertion: $tert$-Butyl methyl ether is not prepared by the reaction of $tert$-butyl bromide with sodium methoxide.
Reason: Sodium methoxide is a strong nucleophile.

From the following compounds,determine the number of compounds that give $(i)$ a red colouration with ceric ammonium nitrate and $(ii)$ a positive iodoform test: $...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module