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’ (C4H11N). ‘R’ on diazotization followed by h

Question

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’ (C4H11N). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S.

(A) P=CH3–CH2–CH=CH2, Q=CH3–CH2–CH2–CH2 Cl R=CH3–CH2–CH2–NH, S=CH3–CH2–CH CH3 CH3 OH
(B) P=CH3 , Q=Cl–CH2 , R=H2N–CH2 ,S= OH
(C) P=CH3–CH=CH–CH3, Q=CH3–CH2–CH–CH3, Cl R=CH3–CH2=CH–CH3, S=CH3–CH2–CH–CH3 NH2 OH
(D) P=CH3–CH=CH–CH3,Q=CH3–CH2–CH2–CH2–Cl, R=CH3–CH2–CH2–CH2, S=CH3–CH2–CH2 NH2 CH2 OH

Prashant Kotian · Accepted Answer

Correct answer: C. CH3–CH=CH–CH3 HCl (P) CH3 – CH2 – CH – CH3 (Q) Cl NH3  CH3 – CH2 – CH – CH3 NH2 NaNO2 Aq.HCl CH3 – CH2 – CH – CH3 OH (S) (R) *

💡 Solution & Explanation

Practice 22,000+ questions like this