Isopropylbromide reacts with silver nitrite to form major product as

2-Nitropropane

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

Core Concept:

Nitrite ion (NO₂⁻) is an ambident nucleophile and can attack either through nitrogen or oxygen, but the nature of metal ion determines the mode of attack.

Explanation:

Nitrite ion (NO₂⁻) has two possible points of attachment:

• Nitrogen → forms nitro compound (R–NO₂)
• Oxygen → forms alkyl nitrite (R–ONO)

When sodium nitrite (NaNO₂) is used, the bond is ionic and oxygen attack is favoured, producing alkyl nitrites.

However, in silver nitrite (AgNO₂), the bond is more covalent. Silver coordinates with oxygen, making nitrogen the preferred site for nucleophilic attack. This leads to substitution via nitrogen and formation of nitroalkane.

Thus:

Isopropyl bromide reacts with AgNO₂ to form 2-nitropropane.

$(CH_3)_2CHBr+AgNO_2→(CH_3)_2CHNO_2+AgBr$

Option 1: 2-Nitropropane

Correct because AgNO₂ promotes nitrogen attack, forming nitroalkane.

Option 2: Propane-2-nitrite

Incorrect because this product forms when NaNO₂ is used, where oxygen attack occurs.

Option 3: Propane-2-nitrile

Incorrect because nitriles form with cyanide ion (CN⁻), not nitrite.

Option 4: Propane

Incorrect because substitution occurs instead of reduction.