In the reaction series: $C_{6} H_{5} C l N H_{3} C u_{2} O, 20 0^{o} C X; X H N O_{2} 0 - 5^{0} C Z; X + Z \to A$ , the number of $σ$ and $π$ bonds in $A$ are:

Question

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Answer 1

Nucleophilic substitution of chloro-benzene with ammonia gives aniline.

$C_{6} H_{5} C l N H_{3} C u_{2} O, 200^{0} C X C_{6} H_{5} N H_{2} + H C l$

Diazotization of aniline gives benzene diazonium chloride.

$X C_{6} H_{5} N H_{2} H N O_{2} 0 - 5^{0} C Z C_{6} H_{5} N_{2} Cl$

Azo coupling of aniline with benzenediazonium chloride gives p-aminoazobenzene.

$X C_{6} H_{5} N H_{2} + Z C_{6} H_{5} N_{2} Cl \to A C_{6} H_{5} - N = N - C_{6} H_{4} - N H_{2}$

The number of $σ$ and $π$ bonds in $C_{6} H_{5} - N = N - C_{6} H_{4} - N H_{2}$ are $27$ $σ$ and $7$ $π$ .

Hence, the correct option is C.

Answer 2

Nucleophilic substitution of chloro-benzene with ammonia gives aniline.

C_{6} H_{5} C l N H_{3} C u_{2} O, 200^{0} C X C_{6} H_{5} N H_{2} + H C l

Diazotization of aniline gives benzene diazonium chloride.

X C_{6} H_{5} N H_{2} H N O_{2} 0 - 5^{0} C Z C_{6} H_{5} N_{2} Cl

Azo coupling of aniline with benzenediazonium chloride gives p-aminoazobenzene.

X C_{6} H_{5} N H_{2} + Z C_{6} H_{5} N_{2} Cl \to A C_{6} H_{5} - N = N - C_{6} H_{4} - N H_{2}

The number of

σ

and

π

bonds in

C_{6} H_{5} - N = N - C_{6} H_{4} - N H_{2}

are

27

σ

and

7

π

.

Hence, the correct option is C.