question_answer 98)

  A hydrocarbon 'A',  on reaction with HCl gives a compound 'B', , which on reaction with 1 mol of  gives compound 'C', . On reacting with  and HCl followed by treatment with water, compound 'C' yields an optically active alcohol, 'D'. Ozonolysis of 'A' gives 2 mols of acetaldehyde. Identify compounds 'A' to 'D?. Explain the reactions involved.

Answer:

                  (i) M.F. of compound A  corresponds to the general formula where n = 4, therefore, A is an alkene. Since ozonolysis of compound A  gives two moles of acetaldehyde, therefore, alkene (A) must be symmetrical, i.e., but-2-ene (ii) Since compound (A), i.e., but-2-ene reacts with HCl to form compound (B), therefore, (B) must be 2-chlorobutane.   (iii) Since compound (B), i.e., 2-chlorobutane reacts with one mole of to give compound (C), therefore, compound (C) must be a 1° amine, i.e., butan-2-amine   (iv) Since compound (C) reacts with  to give an alcohol (D), therefore, compound (D) must be butan-2-ol. Please note that 2-chlorobutane (B), butan-2-amine (C) and butan-2-ol (D) contain a chiral carbon, therefore, all these compounds are optically active.