When $C_{2} H_{5} C l$ is treated with $A g C N,$ the products are:

C_{2} H_{5} C N + A g C l

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C_{2} H_{5} N C + A g C l

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C_{2} H_{5} N H C H_{3} + A g C l

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C_{2} H_{5} C O O H + A g C l

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Solution

The correct option is B $C_{2} H_{5} N C + A g C l$
When alkyl halide reacts with AgCN, isocyanide forms as product.

$C_{2} H_{5} C l + A g C N \to C_{2} H_{5} - N C + A g C l$

Hence, option B is correct.

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Similar questions

Q. The correct order of the stoichiometries of

A g C l

formed when

A g N O_{3}

in excess is treated with the complexes :

C o C l_{3} .6 N H_{3}, C o C l_{3} .5 N H_{3}, C o C l_{3} .4 N H_{3}

respectively is:

Th correct order of the stochiometrics of AgCl formed when $A g N O_{3}$ in excess is treated with the complexes: $C o C l_{3} .6 N H_{3}, C o C l_{3} .5 N H_{3}, C o C l_{3} .4 N H_{3}$ respectively is

Q. When AgCl is treated with KCN,

Q. When

A g C l

is treated with

K C N

Q. What is the structure of soluble complex formed when AgCl is treated with ethylamine?

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When C2H5Cl is treated with AgCN, the products are:

The correct option is B C2H5NC+AgClWhen alkyl halide reacts with AgCN, isocyanide forms as product.C2H5Cl+AgCN→C2H5−NC+AgClHence, option B is correct.

When $C_{2} H_{5} C l$ is treated with $A g C N,$ the products are:

The correct option is B $C_{2} H_{5} N C + A g C l$
When alkyl halide reacts with AgCN, isocyanide forms as product.

$C_{2} H_{5} C l + A g C N \to C_{2} H_{5} - N C + A g C l$

Hence, option B is correct.