$P$ and $Q$ are isomers of dicarboxylic acid $\mathrm{C}_{4}\mathrm{H}_{4}\mathrm{O}_{4}$. Both decolorize $\mathrm{Br}_{2}/\mathrm{H}_{2}\mathrm{O}$. On heating, $\boldsymbol{P}$ forms the cyclic anhydride. Upon treatment with dilute alkaline $\mathrm{KMnO}_{4}$, $P$ as well as $\boldsymbol{Q}$ could produce one or more than one from $\boldsymbol{S}, \boldsymbol{T}$ and $\boldsymbol{U}$. $\boldsymbol{S}$ $\boldsymbol{T}$ $\boldsymbol{U}$ Compounds formed from $\boldsymbol{P}$ and $\boldsymbol{Q}$ are, respectively

(B) Optically inactive $\boldsymbol{S}$ and optically inactive pair $(\boldsymbol{T}, \boldsymbol{U})$

(A) Optically active $\boldsymbol{S}$ and optically active pair $(\boldsymbol{T}, \boldsymbol{U})$

(C) Optically active pair $(\boldsymbol{T}, \boldsymbol{U})$ and optically active $\boldsymbol{S}$

(D) Optically inactive pair $(\boldsymbol{T}, \boldsymbol{U})$ and optically inactive $\boldsymbol{S}$

General Organic Chemistry — JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 Question with Solution

P

and

Q

are isomers of dicarboxylic acid

C_{4} H_{4} O_{4}

. Both decolorize

Br_{2} / H_{2} O

. On heating,

P

forms the cyclic anhydride. Upon treatment with dilute alkaline

KMnO_{4}

P

as well as

Q

could produce one or more than one from

S, T

and

U

S

T

U

Compounds formed from

P

and

Q

are, respectively

Answer: (B)

P

and

Q

are isomers of

C_{4} H_{4} O_{4}

that decolorize

Br_{2}

(unsaturated).

P

forms a cyclic anhydride on heating, so

P

is maleic acid (cis-butenedioic acid).

Q

is fumaric acid (trans-butenedioic acid).

Oxidation with dilute alkaline

KMnO_{4}

Maleic acid ( $P$ ) → meso-tartaric acid ( $S$ ), optically inactive (has a plane of symmetry)
Fumaric acid ( $Q$ ) → racemic mixture of D- and L-tartaric acid ( $T, U$ ), optically inactive as a pair

Both products are optically inactive.

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option