Match List I with List II
| List I | List II |
| A. \(CH_3CH_2Br\, \ +\, \ alc. KOH\) | I. \(CH_3CH_2NO_2\) |
| B. \(CH_3CH_2Br\, \ +\, \ aq. KOH\) | II. \(CH_2=CH_2\) |
| C. \(CH_3CH_2Br\, \ +\, \ KNO_2\) | III. \(CH_3CH_2ONO\) |
| D. \(CH_3CH_2Br\, \ +\, \ AgNO_2\) | IV. \(CH_3CH_2OH\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. A-II, B-IV, C-III, D-I.
| List I | List II |
| A. \(CH_3CH_2Br\, \ +\, \ alc. KOH\) | II. \(CH_2=CH_2\) |
| B. \(CH_3CH_2Br\, \ +\, \ aq. KOH\) | IV. \(CH_3CH_2OH\) |
| C. \(CH_3CH_2Br\, \ +\, \ KNO_2\) | III. \(CH_3CH_2ONO\) |
| D. \(CH_3CH_2Br\, \ +\, \ AgNO_2\) | I. \(CH_3CH_2NO_2\) |
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 \((CH_3CH_2NO_2)\).
