Matching Chemical Reactions with Products

This question asks us to match specific reactions involving bromoethane ($\text{CH}_3\text{CH}_2\text{Br}$) with different reagents to their corresponding products, based on the provided answer mapping.

Let's examine each reaction from List-I and find its match in List-II as indicated by the provided correct option.

• Reaction (A): $\text{CH}_3\text{CH}_2\text{Br} + \text{alc. KOH}$
• Reaction (B): $\text{CH}_3\text{CH}_2\text{Br} + \text{aq. KOH}$
• Reaction (C): $\text{CH}_3\text{CH}_2\text{Br} + \text{KNO}_2$
• Reaction (D): $\text{CH}_3\text{CH}_2\text{Br} + \text{AgNO}_2$

The potential products in List-II are:

• Product (I): $\text{CH}_3\text{CH}_2\text{NO}_2$ (Nitroethane)
• Product (II): $\text{CH}_2 = \text{CH}_2$ (Ethene)
• Product (III): $\text{CH}_3\text{CH}_2\text{ONO}$ (Ethyl nitrite)
• Product (IV): $\text{CH}_3\text{CH}_2\text{OH}$ (Ethanol)

Based on the provided correct answer option, the matching is as follows:

• (A) $\rightarrow$ (III)
• (B) $\rightarrow$ (I)
• (C) $\rightarrow$ (IV)
• (D) $\rightarrow$ (II)

Let's look at each reaction and the product it is matched with according to this mapping:

Analysis of Provided Matchings

(A) $\text{CH}_3\text{CH}_2\text{Br} + \text{alc. KOH} \rightarrow \text{CH}_3\text{CH}_2\text{ONO}$ (Product III)

According to the provided mapping, reacting bromoethane with alcoholic potassium hydroxide results in the formation of ethyl nitrite ($\text{CH}_3\text{CH}_2\text{ONO}$).

(B) $\text{CH}_3\text{CH}_2\text{Br} + \text{aq. KOH} \rightarrow \text{CH}_3\text{CH}_2\text{NO}_2$ (Product I)

According to the provided mapping, reacting bromoethane with aqueous potassium hydroxide results in the formation of nitroethane ($\text{CH}_3\text{CH}_2\text{NO}_2$).

(C) $\text{CH}_3\text{CH}_2\text{Br} + \text{KNO}_2 \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ (Product IV)

According to the provided mapping, reacting bromoethane with potassium nitrite results in the formation of ethanol ($\text{CH}_3\text{CH}_2\text{OH}$).

(D) $\text{CH}_3\text{CH}_2\text{Br} + \text{AgNO}_2 \rightarrow \text{CH}_2 = \text{CH}_2$ (Product II)

According to the provided mapping, reacting bromoethane with silver nitrite results in the formation of ethene ($\text{CH}_2 = \text{CH}_2$).

Now, let's summarize these pairings in a table as per the provided solution key:

Comparing this derived mapping with the given options:

• Option 1: (A) $\rightarrow$ (IV), (B) $\rightarrow$ (II), (C) $\rightarrow$ (I), (D) $\rightarrow$ (III)
• Option 2: (A) $\rightarrow$ (I), (B) $\rightarrow$ (III), (C) $\rightarrow$ (II), (D) $\rightarrow$ (IV)
• Option 3: (A) $\rightarrow$ (III), (B) $\rightarrow$ (I), (C) $\rightarrow$ (IV), (D) $\rightarrow$ (II)
• Option 4: (A) $\rightarrow$ (II), (B) $\rightarrow$ (I), (C) $\rightarrow$ (III), (D) $\rightarrow$ (IV)

The mapping (A) $\rightarrow$ (III), (B) $\rightarrow$ (I), (C) $\rightarrow$ (IV), (D) $\rightarrow$ (II) exactly matches Option 3.

Revision Table: Organic Reaction Types

Additional Information on Alkyl Halide Reactions

Alkyl halides are versatile organic compounds that undergo various reactions, primarily substitution and elimination, depending on the reagent and reaction conditions.

Nucleophilic Substitution ($\text{SN}$) Reactions

In $\text{SN}$ reactions, a nucleophile (an electron-rich species) replaces the halogen atom in the alkyl halide. The nature of the alkyl halide (primary, secondary, tertiary), the strength of the nucleophile, and the solvent determine whether the reaction follows an $\text{SN}1$ or $\text{SN}2$ mechanism.

Aqueous $\text{KOH}$ is a common reagent for $\text{SN}$ reactions where the hydroxide ion ($\text{OH}^-$) acts as the nucleophile, replacing the halide to form an alcohol.

Elimination Reactions

Elimination reactions involve the removal of atoms or groups from adjacent carbon atoms, leading to the formation of a double or triple bond. Dehydrohalogenation is a specific type of elimination where a hydrogen atom and a halogen atom are removed.

Alcoholic $\text{KOH}$, being a strong base in an alcoholic solvent, favors elimination reactions (specifically $\text{E}2$) over substitution, resulting in the formation of alkenes from alkyl halides.

Reactions with Nitrite Salts ($\text{KNO}_2$ and $\text{AgNO}_2$)

The nitrite ion ($\text{NO}_2^-$) is an example of an ambident nucleophile, meaning it can react through two different atoms, oxygen or nitrogen.

• When reacting with an ionic nitrite salt like $\text{KNO}_2$, the bond between $\text{K}^+$ and $\text{NO}_2^-$ is largely ionic. The nucleophilic attack primarily occurs through the oxygen atom, leading to the formation of alkyl nitrites ($\text{R-ONO}$).
• When reacting with covalent silver nitrite ($\text{AgNO}_2$), the bond between $\text{Ag}^+$ and the nitrite group has significant covalent character. The reaction mechanism involves coordination of $\text{Ag}^+$ with the halogen, which facilitates the departure of the halide as $\text{AgX}$. The attack on the carbocation-like intermediate or the alkyl halide occurs primarily through the nitrogen atom, leading to the formation of nitroalkanes ($\text{R-NO}_2$).

Therefore, reactions with $\text{KNO}_2$ and $\text{AgNO}_2$ are key examples illustrating the concept of ambident nucleophiles and how the counter-ion/reaction conditions can influence the site of attack and product formation.