JEE Main 2026 · 21 January · morning · Single correct · +4 / −1
A hydrocarbon ' P ' (C₄H₈) on reaction with HCl gives an optically active compound ' Q ' (C₄H₉Cl) which on reaction with one mole of ammonia gives compound ' R^'(C₄H₁₁ N). ' R^' on diazotization followed by hydrolysis gives ' S '. Identify P,Q,R and S .
- (a) P=CH₃-CH=CH-CH₃, Q=CH₃-CH₂-CH₂-CH₂-Cl, R=CH₃-CH₂-CH₂-CH₂, S=CH₃-CH₂-CH₂
- (b) P = CH₃-CH=CH-CH₃, Q = CH₃-CH₂-CH-CH₃, R = CH₃-CH₂-CH-CH₃, S = CH₃-CH₂-CH-CH₃
- (c)
- (d) P=CH₃-CH₂-CH=CH₂, Q=CH₃-CH₂-CH₂-CH₂, R=CH₃-CH₂-CH₂-NH, S=CH₃-CH₂-CH(OH |)
Answer and solution
Answer (b) P = CH₃-CH=CH-CH₃, Q = CH₃-CH₂-CH-CH₃, R = CH₃-CH₂-CH-CH₃, S = CH₃-CH₂-CH-CH₃
Solution
Chemistry ma’am · 4 steps
Start from the words 'optically active': Q has to be 2-chlorobutane, and the whole chain follows from it.
- Pin down Q. C₄H₉Cl is optically active only if the Cl sits on a carbon with four different groups. In 2-chlorobutane, C2 carries H, Cl, CH₃ and C₂H₅, so it is chiral. Every other C₄H₉Cl isomer is achiral, and 1-chlorobutane is one of them.
- Work back to P. C₄H₈ with HCl is an alkene addition. HCl adds across the double bond of but-2-ene, CH₃CH=CHCH₃, and puts Cl on C2, which gives 2-chlorobutane.
CH₃CH=CHCH₃ →(HCl) CH₃CH₂CHClCH₃ →(NH₃) CH₃CH₂CH(NH₂)CH₃ →(HNO₂, H₂O) CH₃CH₂CH(OH)CH₃
- R and S. One mole of NH₃ replaces Cl by NH₂, giving butan-2-amine, a primary amine C₄H₁₁N. Nitrous acid turns it into an alkyl diazonium ion, which loses N₂. Water attacks the carbocation left behind, giving butan-2-ol.
- So. P is but-2-ene, Q is 2-chlorobutane, R is butan-2-amine and S is butan-2-ol. That is option (b).
The idea to keep. In a reaction chain, begin at the strongest clue, such as optical activity that needs a chiral carbon, and work outwards. Then check that the carbon count stays the same at every step.
Answer from the NTA final answer key. Solution by PrepWiser.