A hydrocarbon 'P' () on reaction with HCl gives an optically active compound 'Q' () which on reaction with one mole of ammonia gives compound 'R' (). 'R' on diazotization followed by hydrolysis gives 'S'. Identify P, Q, R and S.
- A
P=CH₃-CH₂-CH=CH₂, Q=CH₃-CH₂-CH₂-CH₂Cl, R=CH₃-CH₂-CH₂-NH₂, S=CH₃-CH₂-CH(OH)CH₃
- B
P=CH₃-CH=CH-CH₃, Q=CH₃-CH₂-CH₂-CH₂-Cl, R=CH₃-CH₂-CH(NH₂)CH₃, S=CH₃-CH₂-CH₂-CH₂OH
- C
P=CH₃-CH=CH-CH₃, Q=CH₃-CH₂-CH(Cl)CH₃, R=CH₃-CH₂-CH(NH₂)CH₃, S=CH₃-CH₂-CH(OH)CH₃
- D
P=Cyclobutane, Q=Chlorocyclobutane, R=Cyclobutylamine, S=Cyclobutanol