JEE Mains · Chemistry · STD 12 - 6. Haloalkanes and Haloarenes
Match List \(I\) with List \(II\) \(1-\) Bromopropane is reacted with reagents in List \(I\) to give product in List \(II\)
| List\(I-\) Reagent | List\(II-\)Product |
| \(A\) \(KOH (\text { alc) }\) | \(I\) Nitrile |
| \(B\) \(KCN \text { (alc) }\) | \(II\) Ester |
| \(C\) \(AgNO _2\) | \(III\) Alkene |
| \(D\) \(H _3 CCOOAg\) | \(IV\) Nitroalkane |
- A \(A-IV,B-III,C-II,D-I\)
- B \(A-III,B-I,C-IV,D-II\)
- C \(A-I,B-II,C-III,D-IV\)
- D \(A-I,B-III,C-IV,D-II\)
Answer & Solution
Correct Answer
(B) \(A-III,B-I,C-IV,D-II\)
Step-by-step Solution
Detailed explanation
\(CH _3- CH _2- CH _2- Br + KOH ( Alc ) \rightarrow CH _3-\underset{\text { (Alkene) }}{ CH }= CH _2\) \(CH _3- CH _2- CH _2- Br + KCN ( Alc ) \rightarrow CH _3- CH _2- CH _2- CN\) \(CH _3- CH _2- CH _2- Br + AgNO _2 \rightarrow CH _3- CH _2- CH _2- COO _2+ AgBr \downarrow\)…
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