`CH_(3)-C-=CHunderset("catalyst")overset("Linddar's")rarr(A)underset(H_(2)O)overset("Conc"H_(2)SO_(4))rarr(B)`
What are the products (A) and (B) in the given reaction?

Options

A. `CH_(3)-CH=CH_(2), CH_(3)-underset(OH)underset(|)(CH)-CH_(3)`
B. `CH_(2)=CH-CH_(3),CH_(3)-underset(OH)underset(|)(CH)-CH_(3)`
C. `CH_(2)=CH-CH_(3).CH_(3)-C-=CH`
D. `CH_(3)-CH=CH_(2).CH_(3)-underset(OSO_(2)OH)underset(|)(CH)-CH_(3)`

Answer

To solve the problem step by step, we will analyze the given reaction involving propyne (CH₃-C≡CH) and its subsequent reactions.

### Step 1: Identify the starting compound
The starting compound is propyne, which has the structure:
\[ \text{CH}_3-\text{C} \equiv \text{C}\text{H} \]

### Step 2: Reaction with Lindlar's Catalyst
Propyne reacts with Lindlar's catalyst. Lindlar's catalyst is a palladium catalyst that is poisoned to selectively reduce alkynes to cis-alkenes.

The reaction can be represented as:
\[ \text{CH}_3-\text{C} \equiv \text{C}\text{H} \xrightarrow{\text{Lindlar's catalyst}} \text{CH}_3-\text{C}=\text{CH}_2 \]

The product (A) formed is **propene (CH₃-CH=CH₂)**.

### Step 3: Reaction with Concentrated H₂SO₄ and Water
Next, the product propene (A) undergoes hydrolysis in the presence of concentrated sulfuric acid (H₂SO₄) and water (H₂O).

The mechanism involves the formation of a carbocation intermediate:
1. The concentrated H₂SO₄ donates a proton (H⁺) to propene, forming a carbocation:
\[ \text{CH}_3-\text{C}=\text{CH}_2 + \text{H}^+ \rightarrow \text{CH}_3-\text{C}^+(\text{H})-\text{CH}_2 \]

2. The carbocation is then attacked by water (H₂O), leading to the formation of an alcohol:
\[ \text{CH}_3-\text{C}^+(\text{H})-\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3-\text{C}(\text{OH})(\text{H})-\text{CH}_3 \]

The product (B) formed is **2-propanol (CH₃-CHOH-CH₃)**.

### Summary of Products
- Product (A) is **propene (CH₃-CH=CH₂)**.
- Product (B) is **2-propanol (CH₃-CHOH-CH₃)**.

### Final Answer
- A: Propene (CH₃-CH=CH₂)
- B: 2-Propanol (CH₃-CHOH-CH₃)

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