An optically active compound 'X' having molec | vedclass

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

Question

(C) The molecular formula $C_4H_8O_3$ and the evolution of $CO_2$ with $NaHCO_3$ indicate the presence of a carboxylic acid group $(-COOH)$.Since the

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The molecular formula $C_4H_8O_3$ and the evolution of $CO_2$ with $NaHCO_3$ indicate the presence of a carboxylic acid group $(-COOH)$.Since the compound is optically active,it must contain a chiral center.Reduction of the carboxylic acid group with $LiAlH_4$ converts it into a primary alcohol $(-CH_2OH)$.Let us analyze option $A$: $CH_3-CH(OH)-CH_2-COOH$. Upon reduction,it gives $CH_3-CH(OH)-CH_2-CH_2OH$ (butane$-1,3-$diol). This molecule has a chiral center at the $C-3$ position,so it is optically active.Let us analyze option $B$: $CH_3-CH_2-CH(OH)-COOH$. Upon reduction,it gives $CH_3-CH_2-CH(OH)-CH_2OH$ (butane$-1,2-$diol). This molecule has a chiral center at the $C-2$ position,so it is optically active.Let us analyze option $C$: $HO-CH_2-CH(CH_3)-COOH$. Upon reduction,it gives $HO-CH_2-CH(CH_3)-CH_2OH$ ($2$-methylpropane$-1,3-$diol). This molecule is achiral because it has a plane of symmetry passing through the $C-2$ atom.Therefore,the compound $'X'$ is $HO-CH_2-CH(CH_3)-COOH$.

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

$2CH_3COOH \xrightarrow{P_2O_5} ........;$ What will be the product of this reaction?

The major products obtained in the reaction of oxalic acid with conc. $H_2SO_4$ upon heating are

$A$ key step in the hydrolysis of acetamide in aqueous acid proceeds by nucleophilic addition of

The product $(Q)$ is

The distinguishing test between methanoic acid and ethanoic acid is

Vedclass Test Series

Exam Paper Generator

Online Exam Module