Question

Identify the reagent A in the following conversion.
Alkyl halide $\rightarrow$ Alkyl nitrite

• A. $KNO_3$
• B. $NaNO_3$
• C. $AgNO_2$
• D. $KNO_2$

Hint:

Logic Tip: A classic memorization rule for ambidentate nucleophiles: Ionic salt ($KNO_2$ or $KCN$) $\rightarrow$ Attack from the more electronegative atom (Oxygen or Carbon) $\rightarrow$ Alkyl Nitrite or Alkyl Cyanide. Covalent salt ($AgNO_2$ or $AgCN$) $\rightarrow$ Attack from the less electronegative/softer atom (Nitrogen) $\rightarrow$ Nitroalkane or Isocyanide.

Answer 1

The Correct Option is D

Concept:
The nitrite ion ($NO_2^-$) is an ambidentate nucleophile, meaning it has two different atoms capable of donating electron pairs to form a new bond: the nitrogen atom and the oxygen atom. - Attack through Oxygen yields Alkyl nitrites ($R-O-N=O$). - Attack through Nitrogen yields Nitroalkanes ($R-NO_2$).
Step 1: Analyze the target product.
The desired product is an alkyl nitrite ($R-O-N=O$). This requires the nucleophilic attack to occur through the oxygen atom of the nitrite ion onto the alkyl halide ($R-X$).
Step 2: Evaluate the given reagents based on their bonding nature.

• Potassium nitrite ($KNO_2$): Potassium is a highly electropositive alkali metal. The bond between K and O is predominantly ionic ($K^+ \ ^-O-N=O$). Because the oxygen atom carries the full negative formal charge and is free in solution, it acts as the primary nucleophilic center. Thus, $KNO_2$ heavily favors the formation of alkyl nitrites.
• Silver nitrite ($AgNO_2$): The bond between Ag and O has significant covalent character. Since the oxygen is tied up in a covalent bond, the lone pair on the nitrogen atom acts as the nucleophilic center. Thus, $AgNO_2$ heavily favors the formation of nitroalkanes.
• Nitrates ($KNO_3$, $NaNO_3$): These are salts of nitric acid and contain the $NO_3^-$ ion, which does not produce alkyl nitrites in a simple nucleophilic substitution.

Since we need an alkyl nitrite, $KNO_2$ is the correct reagent.