Which of the following compounds will form the precipitate with aq. AgNO3 solution most readily?

Step 1: Understand the $\ce{AgNO3}$ test

Silver nitrate precipitates AgX from reactive halides:

$\ce{R-X + AgNO3 -> AgX v + R-ONO2}$

Reactivity depends on:

• Ease of C–X bond breaking
• Carbocation stability (if $S_N1$)
• Accessibility (if $S_N2$)

Step 2: Analyze each option

(a) Allylic bromide:

• $\ce{CH2=CH-CH2Br}$
• Allylic position (resonance-stabilized carbocation)
• Very reactive
• Forms precipitate readily ✓✓

(b) Benzyl bromide:

• $\ce{C6H5-CH2Br}$
• Benzylic position (resonance with benzene ring)
• Extremely reactive (most stable carbocation)
• Forms precipitate most readily ✓✓✓

(c) N-containing bromide:

• If it's an amine derivative
• Depends on structure
• Generally less reactive than allylic/benzylic

(d) Aryl bromide:

• $\ce{C6H5Br}$ or derivative
• Does not react with $\ce{AgNO3}$ under normal conditions
• Aryl halides are unreactive ✗

Step 3: Rank reactivity

Benzylic > Allylic > Primary > Secondary > Aryl (no reaction)

Benzylic bromide is most reactive because:

• Benzylic carbocation is highly stabilized
• Resonance with benzene ring
• Multiple resonance structures

Answer: (b) Benzyl bromide

Key Points:

Reactivity toward $\ce{AgNO3}$:

| Type | Reactivity | Reason | |------|------------|--------| | Benzylic | Highest | Resonance with benzene | | Allylic | High | Resonance with C=C | | 3° | High | Carbocation stability | | 2° | Moderate | Less stable carbocation | | 1° | Low | Unstable carbocation | | Aryl/Vinyl | None | Cannot form carbocation |

Benzylic carbocation:

[C6H5-CH2]+ ↔ Multiple resonance structures with + on ring

Highly stabilized by delocalization into aromatic ring.