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.
- P=CH3-CH=CH-CH3, Q=CH3-CH2-CH2-CH2-Cl, R=CH3-CH2-CH2- NH2 |CH2, S=CH3-CH2- OH |CH2 |CH2
- P=CH3-CH=CH-CH3, Q=CH3-CH2- Cl |CH-CH3, R=CH3-CH2- NH2 |CH-CH3, S=CH3-CH2- OH |CH-CH3
- P = [figure] , Q = [figure] , R = [figure] , S = [figure]
- P=CH3-CH2-CH=CH2, Q=CH3-CH2-CH2- Cl |CH2, R=CH3-CH2-CH2- CH3 |NH, S=CH3-CH2- OH | CH3 |CH
Answer
(2) P=CH3-CH=CH-CH3, Q=CH3-CH2- Cl |CH-CH3, R=CH3-CH2- NH2 |CH-CH3, S=CH3-CH2- OH |CH-CH3
Solution
Q (C4H9Cl) is optically active it has a chiral carbon Q = CH3-CH2- *CHCl-CH3 (2-chlorobutane).
(1-Chlorobutane, (chloromethyl)cyclopropane and 1-chloro-2-methylpropane have no chiral carbon.)
P (C4H8) gives 2-chlorobutane with HCl P = CH3-CH=CH-CH3 (but-2-ene):
CH3-CH=CH-CH3 + HCl CH3-CH2-CHCl-CH3
Ammonolysis with one mole of NH3 gives the primary amine:
CH3-CH2-CHCl-CH3 + NH3 CH3-CH2-CH(NH2)-CH3 + HCl
R = butan-2-amine ( C4H11N)
A primary aliphatic amine with nitrous acid (NaNO2 + HCl) forms an unstable diazonium salt, which is hydrolysed to the alcohol with evolution of N2:
CH3-CH2-CH(NH2)-CH3 NaNO2 + HCl CH3-CH2-CH(OH)-CH3 + N2
S = butan-2-ol.
Answer: (2)