Match List I with List II

Choose the correct answer from the options given below:

A-II, B-IV, C-III, D-I

The correct answer is option 3. A-II, B-IV, C-III, D-I.

Let us break down each of the reactions listed in the question to understand why the correct answer is option 3 (A-II, B-IV, C-III, D-I):

A. \(CH_3CH_2Br\, \ +\, \  alc. KOH\): II. \(CH_2=CH_2\)

Reaction Type: Elimination (Dehydrohalogenation)

When ethyl bromide \((CH_3CH_2Br)\) reacts with alcoholic KOH (potassium hydroxide), an elimination reaction occurs. The bromine atom is removed along with a hydrogen atom from the adjacent carbon atom, forming ethene \((CH_2=CH_2)\). Alcoholic KOH promotes this type of reaction because it acts as a strong base.

B. \(CH_3CH_2Br\, \ +\, \  aq. KOH\): IV. \(CH_3CH_2OH\)

Reaction Type: Nucleophilic substitution (S\(_\text{N}\)2)

When ethyl bromide reacts with aqueous KOH, a nucleophilic substitution reaction occurs. The hydroxide ion \((OH^-)\) from aqueous KOH replaces the bromine atom, resulting in the formation of ethanol \((CH_3CH_2OH)\).

C. \(CH_3CH_2Br\, \ +\, \  KNO_2\) : III. \(CH_3CH_2ONO\)

Reaction Type: Nucleophilic substitution

When ethyl bromide reacts with potassium nitrite \((KNO_2)\), there are two possible products depending on the reaction conditions. Here, potassium nitrite acts as a nucleophile, leading to the formation of ethyl nitrite \((CH_3CH_2ONO)\), where the nitrite group \((ONO^-)\) attaches to the ethyl group.

D. \(CH_3CH_2Br\, \ +\, \  AgNO_2\) : I. \(CH_3CH_2NO_2\)

Reaction Type: Nucleophilic substitution (S\(_\text{N}\)2)

When ethyl bromide reacts with silver nitrite \((AgNO_2)\), silver bromide precipitates out, and the nitrite ion \((NO_2^-)\) from silver nitrite acts as a nucleophile. This results in the formation of nitroethane \((CH_3CH_2NO_2)\).