A hydrocarbon 'P' \(\left(\mathrm{C}_{4} \mathrm{H}_{8}\right)\) on reaction with HCl gives an optically active compound 'Q' \(\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\right)\) which on reaction with one mole of ammonia gives compound 'R' \(\left(\mathrm{C}_{4} \mathrm{H}_{11} \mathrm{~N}\right)\). 'R' on diazotization followed by hydrolysis gives 'S'. Identify \(\mathrm{P}, \mathrm{Q}, \mathrm{R}\) and S.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Compounds Containing Nitrogen

A hydrocarbon 'P'

(C_{4} H_{8})

on reaction with HCl gives an optically active compound 'Q'

(C_{4} H_{9} Cl)

which on reaction with one mole of ammonia gives compound 'R'

(C_{4} H_{11} N)

. 'R' on diazotization followed by hydrolysis gives 'S'. Identify

P, Q, R

and S.

(A)(see structures)
A
B
(B)(see structures)
A
B
(C)(see structures)
A
B
✓ Correct
(D)(see structures)
A
B

Answer: (C)

Solution

Because Q is optically active, addition of HCl to the C

_{4}

_{8}

hydrocarbon must build a stereocentre. But-2-ene, CH

_{3}

CH=CHCH

_{3}

, adds HCl to give 2-chlorobutane, CH

_{3}

CHClCH

_{2}

_{3}

, whose C-2 carries four different groups and is chiral. Treating 2-chlorobutane with one mole of ammonia gives butan-2-amine (R), and diazotisation of this primary amine followed by hydrolysis replaces

-

_{2}

-

OH to give butan-2-ol (S). The set but-2-ene, 2-chlorobutane, butan-2-amine and butan-2-ol identifies P, Q, R and S.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Compounds Containing Nitrogen

Solution

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