JEE Advanced 2019 Paper 2 Chemistry: Paragraph 2

(P) Hexa-2,4-diene has two stereogenic double bonds with identical ends, so the E,Z and Z,E forms are the same compound: EE, ZZ, EZ $=3$, so P→3. (Q) Pentane-2,3,4-triol has stereocentres C2 and C4 (identical environments) and C3, a pseudoasymmetric centre. This gives $4$ stereoisomers: two meso forms (r and s at C3) and one enantiomeric pair, so Q→4. (R) There is one double bond (2 forms) and two unlike stereocentres (4 forms): $2\times4=8$, so R→6. (S) The double bond (E or Z) and two identical stereocentres: for each geometry RR, SS and meso, so $2\times3=6$, giving S→5. The correct code is P→3, Q→4, R→6, S→5; the others reflect treating P as $2\times2$, treating S as unsymmetrical ($8$), or Q as unsymmetrical ($8$).

Optically active stereoisomers are the total number minus the achiral ones. (P) All three isomers (EE, ZZ, EZ) of hexa-2,4-diene have a mirror plane in the molecular plane, so $0$ are active, P→1. (Q) Of the $4$ stereoisomers of pentane-2,3,4-triol, two are meso (achiral) and two are enantiomers, so $2$ are active, Q→2. (R) With no element of symmetry, all $8$ stereoisomers are chiral, so R→6. (S) Of the $6$ stereoisomers, the E-meso and Z-meso forms are achiral; the other $4$ (RR and SS for each geometry) are optically active, so S→4. The correct code is P→1, Q→2, R→6, S→4. Distractors count all four triol isomers as active, count all three hexadiene isomers as active, or count the meso forms of S as active.