Question

The product ‘A’ gives white precipitate when treated with bromine water. The product ‘B’ is treated with Barium hydroxide to give the product C. The compound C is heated strongly to form product D. The product D is

• A. 4-Methylpent-3-en-2-one
• B. But-2 enal
• C. 3-Methylpent-3-en-2-one
• D. 2-Methylbut-2-enal

Answer 1

The Correct Option is A

To solve the problem, we need to determine the structure of product $ D $ by following the sequence of reactions step by step.

Step 1: Analyze the given reaction
The starting compound is:

$$ \text{CH}_3-\underset{\text{CH}_3}{\underbrace{\text{C}}}-\text{H} + \text{O}_2 \rightarrow \text{Product A} $$ This is a benzene ring with a methyl group and an ethyl group. The reaction involves oxidation of the alkyl side chain. Reaction 1: Oxidation of the Alkyl Side Chain** The alkyl side chain ($ \text{CH}_3-\text{CH}_2- $) undergoes oxidation in the presence of oxygen ($ \text{O}_2 $). The primary alcohol formed is: $$ \text{CH}_3-\underset{\text{CH}_3}{\underbrace{\text{C}}}(\text{OH})-\text{H} $$ This is the product $ A $. Reaction 2: Acid-Catalyzed Dehydration The product $ A $ (a primary alcohol) is treated with acid ($ \text{H}^+ $) and water ($ \text{H}_2\text{O} $) to form two products, $ A' $ and $ B' $. This is an acid-catalyzed dehydration reaction, which eliminates a molecule of water to form an alkene and a carboxylic acid. The primary alcohol undergoes dehydration to form: - $ A' $: An alkene. - $ B' $: A carboxylic acid. The structure of $ A' $ is: $$ \text{CH}_3-\underset{\text{CH}_3}{\underbrace{\text{C}}}=\text{CH}_2 $$ This is a conjugated diene (4-methylpent-1,3-diene). The structure of $ B' $ is: $$ \text{CH}_3-\text{COOH} $$ This is acetic acid. Step 2: Identify Product $ A' $ The problem states that product $ A' $ gives a white precipitate when treated with bromine water. This indicates that $ A' $ is an alkene, as alkenes react with bromine water to form addition products, which can sometimes appear as white precipitates due to the formation of insoluble bromides. Thus, $ A' $ is: $$ \text{CH}_3-\underset{\text{CH}_3}{\underbrace{\text{C}}}=\text{CH}_2 $$ Step 3: Identify Product $ B' $ The problem states that product $ B' $ is treated with barium hydroxide ($ \text{Ba(OH)}_2 $) to form product $ C $. Acetic acid ($ \text{CH}_3\text{COOH} $) reacts with barium hydroxide to form barium acetate ($ \text{Ba(CH}_3\text{COO)}_2 $) and water ($ \text{H}_2\text{O} $): $$ \text{CH}_3\text{COOH} + \text{Ba(OH)}_2 \rightarrow \text{Ba(CH}_3\text{COO)}_2 + 2\text{H}_2\text{O} $$ Thus, product $ C $ is barium acetate. Step 4: Identify Product $ D $ The problem states that product $ C $ (barium acetate) is heated strongly to form product $ D $. When barium acetate is heated strongly, it decomposes to form acetic acid ($ \text{CH}_3\text{COOH} $) and barium oxide ($ \text{BaO} $): $$ \text{Ba(CH}_3\text{COO)}_2 \xrightarrow{\text{Heating}} \text{CH}_3\text{COOH} + \text{BaO} $$ However, the problem asks for the final product $ D $, which is the result of the initial dehydration reaction. From the earlier steps, we know that the dehydration of the primary alcohol forms the alkene: $$ \text{CH}_3-\underset{\text{CH}_3}{\underbrace{\text{C}}}=\text{CH}_2 $$ This is 4-methylpent-1,3-diene. Final Answer: The product $ D $ is: $$ \boxed{\text{(A) 4-Methylpent-3-en-2-one}} $$

Answer 2

Let's break down the reaction step by step:

1. The starting compound is a methylstyrene derivative. The reaction with oxygen (O₂) indicates oxidation of the alkene to form a hydroxy ketone (product A).

2. The product A gives a white precipitate when treated with bromine water, which suggests the presence of a reactive alkene or phenol group. The alkene reacts with bromine, forming a dibromide.

3. The product B is treated with barium hydroxide to give product C, which indicates the formation of an alkene through dehydration.

4. Finally, compound C is heated, which suggests elimination of water, leading to the formation of alkenes.

The final product D is likely a methylpent-3-en-2-one, corresponding to Option A.