Compound (X) containing chlorine on treatment with strong ammonia gives a solid Y which is free from chlorine. (Y) analysed as: (C) = 49.31 %, H = 9.59%, and N = 19.18% and reacts with $B r_{2}$ and caustic soda to give a basic compound (Z). (Z) reacts with ${H N O}_{2}$ to give ethanol. Suggest structures for (X), (Y), and (Z).

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Solution

In compound (Y) the percentage of $C = 49.31$

The percentage of $H = 9.59$

The percentage of $N = 19.18$

The percentage of $O = 100 - % C - % H - % N$

$∴$ $= 100 - 78.08 = 21.92 %$

Empirical formula of compound (Y):

Element Percentage At.wt Relative number Ratio
C 49.31 12 49.31/12=4.10 4.10/1.37=3
H 9.59 1 9.59/1=9.59 9.59/1.37=7
N 19.18 14 19.18/14=1.37 1.37/1.37=1
O 21.92 16 21.92/16=1.37 1.37/1.37=1

Hence, empirical formula of compound $Y$ is $C_{3} H_{7} - N O$ . On the basis of the empirical formula, its molecular formula may be equal to its empirical formula.

Hence, compound $Y$ is $C H_{3} C H_{2} C O N H_{2}$ because it gives Hoffmann's bromamide reaction with $B r_{2}$ and caustic soda and gives a basic compound $Z$ . $Z$ reacts with $H N O_{2}$ to give ethanol.

$C H_{3} C H_{2} C O N H_{2} + B r_{2} + 4 K O H \to C H_{2} C H_{2} N H_{2} + 2 K B r + K_{2} C O_{3} + 2 H_{2} O$
$(Y)$ $(Z)$

$C H_{3} C H_{2} N H_{2} + H N O_{2} \to C H_{3} C H_{2} O H + N_{2} + H_{2} O$
$(Z)$ Ethanol

Compound (Y) is obtained with compound (X) (which has chlorine) on treatment with strong ammonia, So, compound (X) $C H_{3} C H_{2} C O C l$

$C H_{2} C H_{2} C O C l + N H_{3} \to C H_{3} C H_{2} C O N H_{2}$
Propionyl chlorine Compound (Y)
Compound (X)

Hence, compound are:

$(X) = C H_{3} - C H_{2} - C | | O - C l$

$(Y) = C H_{3} - C H_{2} - C | | O - N H_{2}$

$(Z) = C H_{3} - C H_{2} - N H_{2}$

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