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.
D.
For Q (C_4H_9Cl) to be optically active, it must have a chiral center. P (C_4H_8) = but-2-ene (CH_3-CH=CH-CH_3) P + HCl → Q: CH_3-CH=CH-CH_3 + HCl CH_3-CH(Cl)-CH_2-CH_3 (2-chlorobutane, chiral at C-2) Q + NH_3 → R: 2-chlorobutane + NH_3 CH_3-CH(NH_2)-CH_2-CH_3 (2-aminobutane) R → diazotization → hydrolysis → S: 2-aminobutane NaNO_2/HCl diazonium salt H_2O CH_3-CH(OH)-CH_2-CH_3 (butan-2-ol)
Related: Chemistry — Amines · All PYQ Banks