In the formation of diazonium salt by reacting aliphatic amine with $N a N O_{2}$ and HX, the electrophile is

N_{2}^{+}

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^{+} N O_{2}

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N \equiv O^{+}

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N H_{4}^{+}

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Solution

The correct option is C $N \equiv O^{+}$

Mechanism of formation of diazonium salt:

Here, the amine attack the electrophile $N \equiv O^{+}$ . Hence, (c) is correct.

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Similar questions

Assertion (A): Both aromatic and aliphatic primary amines produce stable diazonium salts on diazotization

Reason (R): Aromatic diazonium salt exhibit resonance

Q. Diazonium salts of aliphatic amines are more stable than those of aromatic amines.
If this is true enter 1, if false enter 0.

Q. Primary amine on treating with NaNO₂ and HCl produce:

Account for the following

(i) $p K_{b}$ of aniline is more than that of methyl amine.

(ii) Ethylamine is soluble in water whereas aniline is not

(iii) Methyl amine in water reacts with ferric chloride to precipitate hydrated ferric oxide

(iv) Although amino group is o- and p- directing in aromatic electrophilic substitution, aniline on nitration gives a substantial amount of m-nitroaniline

(v) Aniline does not undergo Friedel Crafts reaction

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines

Q. What is the reason that aromatic diazonium salt is more stable than aliphatic diazonium salt?

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In the formation of diazonium salt by reacting aliphatic amine with NaNO2 and HX, the electrophile is

The correct option is C N≡O+ Mechanism of formation of diazonium salt: Here, the amine attack the electrophile N≡O+. Hence, (c) is correct.

In the formation of diazonium salt by reacting aliphatic amine with $N a N O_{2}$ and HX, the electrophile is

The correct option is C $N \equiv O^{+}$

Mechanism of formation of diazonium salt:

Here, the amine attack the electrophile $N \equiv O^{+}$ . Hence, (c) is correct.