Wolff-kishner reduction (hydrazine, koh, ethylene glycol, 130°c) of the compound shown gave compound
1 answer:
The scheme is shown below, the steps involved are as follow,
Step one: Reduction:
The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation:
The double bond present in starting compound when treated with m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction:
The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.
Step four: Oxidation:
The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H</span>₂CrO₄).
Step one: Reduction:
The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation:
The double bond present in starting compound when treated with m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction:
The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.
Step four: Oxidation:
The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H</span>₂CrO₄).
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Answer: The correct answer is option B i.e., 2.78 mol
Explanation:
Aluminium reacts with iodine to form Aluminium iodide
From the equation, it is clear that 3 moles of iodine reacts with 2 moles of Aluminium to form Aluminium iodide
We know
For 2 moles of Aluminium,
3 moles of Iodine reacts with 54 g of Aluminium
? moles of iodine react with 50 g of Aluminium