Question

The products A and B in the following reactions, respectively, are:

• A. CH$_3$-CH$_2$-CH$_2$-NO$_2$, CH$_3$-CH$_2$-CH$_2$-NC
• B. CH$_3$-CH$_2$-CH$_2$-ONO, CH$_3$-CH$_2$-CH$_2$-CN
• C. CH$_3$-CH$_2$-CH$_2$-ONO, CH$_3$-CH$_2$-CH$_2$-NC
• D. CH$_3$-CH$_2$-CH$_2$-NO$_2$, CH$_3$-CH$_2$-CH$_2$-CN

Hint:

AgX reagents (where X can be NO$_2$ or CN) are often used for nucleophilic substitutions in organic synthesis to introduce nitro or cyano groups respectively.

Answer 1

The Correct Option is C

When an alkyl halide reacts with $ AgNO_2 $, the product is a nitroalkane ($ R-NO_2 $).
This is because silver nitrite is a covalent compound, and the nitrogen atom donates the electron pair to form a covalent bond with the alkyl group.

When an alkyl halide reacts with $ AgCN $, the product is an alkyl isocyanide ($ R-NC $).
This happens because silver cyanide is predominantly covalent. The carbon atom is more electronegative than silver, so it prefers to form a bond with silver. The nitrogen atom has a lone pair of electrons, which attacks the carbon atom of the alkyl halide, leading to the formation of an isocyanide.

In the given reactions:

$$ CH_3-CH_2-CH_2-Br \xrightarrow{AgNO_2} CH_3-CH_2-CH_2-NO_2 \quad (A) $$ $$ CH_3-CH_2-CH_2-Br \xrightarrow{AgCN} CH_3-CH_2-CH_2-NC \quad (B) $$

Therefore:

• A is $ CH_3-CH_2-CH_2-NO_2 $
• B is $ CH_3-CH_2-CH_2-NC $

Final Answer:
The final answer is $ CH_3 - CH_2 - CH_2 - NO_2,\ CH_3 - CH_2 - CH_2 - NC $.

Answer 2

Step 1: Understand the given reaction sequence.
We are given two reactions involving 1-bromopropane (CH₃-CH₂-CH₂-Br):
1. Reaction with AgNO₂
2. Reaction with AgCN

We are asked to find the products A and B respectively.

Step 2: Reaction with AgNO₂.
When an alkyl halide reacts with silver nitrite (AgNO₂), the product formed is an alkyl nitrite (R–ONO) instead of a nitro compound (R–NO₂).
This happens because the bonding in AgNO₂ occurs through the oxygen atom, allowing nucleophilic substitution to yield an alkyl nitrite.

Therefore, the reaction is: \[ \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{Br} \xrightarrow{AgNO_2} \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{ONO} \] Hence, the product A = CH₃-CH₂-CH₂-ONO.

Step 3: Reaction with AgCN.
When an alkyl halide reacts with silver cyanide (AgCN), the product formed is an alkyl isocyanide (R–NC) rather than an alkyl cyanide (R–CN).
This happens because AgCN has covalent bonding between silver and carbon, so the nucleophilic attack occurs through the nitrogen atom.

Therefore, the reaction is: \[ \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{Br} \xrightarrow{AgCN} \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{NC} \] Hence, the product B = CH₃-CH₂-CH₂-NC.

Step 4: Conclusion.
- Product A: CH₃-CH₂-CH₂-ONO
- Product B: CH₃-CH₂-CH₂-NC

Final Answer:
\[ \boxed{\text{A = CH}_3\text{-CH}_2\text{-CH}_2\text{-ONO,} \quad \text{B = CH}_3\text{-CH}_2\text{-CH}_2\text{-NC}} \]