An optically active compound 'X' having molec | vedclass

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

Question

(A) The compound $X$ is $CH_3-CH(OH)-CH_2-COOH$ ($3-$hydroxybutanoic acid).It is optically active due to the presence of a chiral center at the $C-3$

Vedclass · Accepted Answer

$CH_3-CH(OH)-CH_2-COOH$

Vedclass · Answer

(A) The compound $X$ is $CH_3-CH(OH)-CH_2-COOH$ ($3-$hydroxybutanoic acid).It is optically active due to the presence of a chiral center at the $C-3$ position.It evolves $CO_2$ gas when treated with $NaHCO_3$ because it contains a carboxylic acid group $(-COOH)$.Upon reduction with $LiAlH_4$,the carboxylic acid group is reduced to a primary alcohol,yielding $1,3-$butanediol $(CH_3-CH(OH)-CH_2-CH_2OH)$.Since $1,3-$butanediol is achiral (it does not possess a chiral center as the $C-3$ carbon is no longer chiral in the product),it satisfies all the given conditions.

An optically active compound $'X'$ having molecular formula $C_4H_8O_3$ evolves $CO_2$ with $NaHCO_3$. $'X'$ on treatment with $LiAlH_4$ gives an achiral compound. Then $'X'$ is:

Similar Questions

The strongest of the four acids listed below is

Which of the following represents the correct order of the acidity in the given compounds?

Number of compounds from the following which will not dissolve in cold $NaHCO_3$ and $NaOH$ solutions but will dissolve in hot $NaOH$ solution is $........$.

$Methyl$ propanoate on hydrolysis with dil $NaOH$ forms a salt that on further acidification with conc. $HCl$ forms:

If the rate of the reaction,$R-CO-Z + Nu^- \rightarrow R-CO-Nu + Z^-$ is fastest,then $Z$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module