The major product 'H' of the given reaction sequence is
`CH_(3)-CH_(2)-CO-CH_(3) overset(.^(Theta)CN)to G underset("Heat")overset(95%H_(2)SO_(4))toH`

Options

A. `CH_(3)-CH=underset(CH_(3))underset(|)(C)-COOH`
B. `CH_(3)-CH=underset(CH_(3))underset(|)(C)-CN`
C. `CH_(3)CH_(2)-underset(CH_(3))underset(|)overset(OH)overset(|)(C)-COOH`
D. `CH_(3)CH=underset(CH_(3))underset(|)(C)-CONH_(2)`

Explanation

To solve the problem, we will analyze the reaction sequence step by step.

### Step 1: Identify the starting compound
The starting compound is **CH₃-CH₂-CO-CH₃**, which is **butan-2-one** (a ketone).

### Step 2: Reaction with CN⁻
When butan-2-one reacts with **CN⁻** (cyanide ion), it undergoes a nucleophilic addition reaction. The cyanide ion attacks the carbonyl carbon (C=O), resulting in the formation of a cyanohydrin.

- The reaction can be represented as:
\[ CH₃-CH₂-CO-CH₃ + CN⁻ \rightarrow G \]
- Here, **G** is the cyanohydrin formed from the addition of CN⁻ to the carbonyl group.

### Step 3: Treatment with concentrated H₂SO₄ and heating
Next, the cyanohydrin **G** is treated with **95% H₂SO₄** and heated. The concentrated sulfuric acid will protonate the hydroxyl group (-OH) of the cyanohydrin, making it a better leaving group.

- The reaction can be represented as:
\[ G \overset{95\% H₂SO₄}{\rightarrow} \text{Carbocation} \]
- The protonated hydroxyl group (H₂O) will leave, forming a carbocation.

### Step 4: Dehydration to form an alkene
The carbocation formed can undergo a dehydration reaction where a beta hydrogen (H) is eliminated. This results in the formation of a double bond, leading to the formation of an alkene.

- The reaction can be represented as:
\[ \text{Carbocation} \rightarrow \text{Alkene (H)} \]
- The major product **H** is thus **CH₃-CH=CH-CH₃**, which is **but-2-ene**.

### Final Answer
The major product **H** of the given reaction sequence is **but-2-ene (CH₃-CH=CH-CH₃)**.

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