CUET · CHEMISTRY · PYQ PAPER 2023
2-Methyl propene can be prepared by the reaction of:
- A \((\text{CH}_3)_3\text{C} - \text{Br} + \text{CH}_3\text{ONa}\)
- B \((\text{CH}_3)_3\text{C} - \text{ONa} + \text{CH}_3\text{Br}\)
- C \((\text{CH}_3)_3\text{C} - \text{ONa} + \text{CH}_3 – \text{CH}_2 - \text{Br}\)
- D \((\text{CH}_3)_3\text{C} - \text{ONa} + \text{H}_3\text{C} – \text{CH}(\text{Br}) – \text{CH}_3\)
Answer & Solution
Correct Answer
(A) \((\text{CH}_3)_3\text{C} - \text{Br} + \text{CH}_3\text{ONa}\)
Step-by-step Solution
Detailed explanation
\((\text{CH}_3)_3\text{C} - \text{Br} + \text{CH}_3\text{ONa} \rightarrow \text{CH}_2=\text{C}(\text{CH}_3)_2 + \text{CH}_3\text{OH} + \text{NaBr}\)
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\(KMnO _4\) is prepared by the fusion of \(MnO _2\) with an alkali metal hydroxide and an oxidizing agent like \(KNO _3\) to give a dark-green manganate ion which disproportionate to give permanganate as follows.
\(\begin{array}{l}2 MnO _2+4 KOH + O _2 \rightarrow 2 K_2 MnO _4+2 H _2 O \\ 3 K_2 MnO _4+4 H ^{+} \rightarrow 2 KMnO _4+ MnO _2+2 H _2 O \end{array}\)
On heating \(K _2 MnO _4\) decomposes at 513 K to give \(K _2 MnO _4\). Permanganate ion is tetrahedral and diamagnetic
Acidified \(KMnO _4\) acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
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