Isopropylbromide reacts with silver nitrite to form major product as

The correct answer is Option (1) → 2-Nitropropane

The nitrite ion is an ambident nucleophile: it can attack through oxygen or through nitrogen. Which atom attacks depends on the nature of the nitrite salt. Potassium nitrite is ionic, so the free nitrite ion attacks through oxygen and gives an alkyl nitrite, R–O–N=O. Silver nitrite is mainly covalent (the oxygen is tied to silver), so the free lone pair on nitrogen attacks and gives a nitroalkane, \( \ce{R-NO2} \).

Isopropyl bromide with silver nitrite:

    \( \displaystyle \ce{(CH3)2CH-Br + AgNO2 -> (CH3)2CH-NO2 + AgBr v} \)

So the major product is 2-nitropropane.

Why the other options are incorrect

• Option (2) Propane-2-nitrite — Incorrect. The alkyl nitrite, \( \ce{(CH3)2CH-O-N=O} \), forms when attack is through oxygen, which is the case with ionic KNO₂.

• Option (3) Propane -2-nitrile — Incorrect. A nitrile has a –C≡N group and forms with KCN, which adds a carbon atom; nitrite contains no carbon.

• Option (4) Propane — Incorrect. This would need a reduction (for example Zn/HCl); silver nitrite only substitutes the bromine.

NCERT (2026-27 edition):
Reactions of Haloalkanes – “… Groups like cyanides and nitrites possess two nucleophilic centres and are called ambident nucleophiles.”