Question

HCl is added to a hydrocarbon 'A'
(CH) to give a compound 'B' which
on hydrolysis with aqueous alkali
forms tertiary alcohol 'C' (CHO).
Identify 'A', 'B' and 'C'.​

Answer 1

[N.B. - the formulae are ($C_4H_8$) and not (CH), and ($C_4H_1_0O$) instead of (CHO)]

As per the given formula, it can be inferred that the organic compound $C_4H_8$ is 2-methylpropane, or Butene. When Butene reacts with hydrochloric acid, it results in the formation of

2-methyl-2-chloropropane.

The structural diagram has been attached below.

So, in the first case, we get 2-methyl-2-chloropropane after the completion of the reaction. In the next step, on hydrolysis with aqueous alkali, we get $C_4H_1_0O$, or 2-methyl-2-propanol.

Ans) A = 2-methylpropane

B = 2-methyl-2-chloropropane

C = 2-methyl-2-propanol

Answer 2

Answer:

N.B. - the formulae are (C_4H_8C4H8 ) and not (CH), and (C_4H_1_0O ) instead of (CHO)]

As per the given formula, it can be inferred that the organic compound C_4H_8C4H8 is 2-methylpropane, or Butene. When Butene reacts with hydrochloric acid, it results in the formation of

2-methyl-2-chloropropane.

The structural diagram has been attached below.

So, in the first case, we get 2-methyl-2-chloropropane after the completion of the reaction. In the next step,n hydrolysis with aqueous alkali, we get C_4H_1_0O , or 2-methyl-2-propanol.

Ans) A = 2-methylpropane

B = 2-methyl-2-chloropropane

C = 2-methyl-2-propanol