The end product of the following reaction
`CH_(3)MgBr+CH_(2)=CH-CHOoverset(H_(3)O^(+))rarr` is :

Options

A. `CH_(2)=CH-overset(OH)overset(|)underset(CH_(3))underset(|)C-H`
B. `underset(OH)underset(|)CH_(2)CH=CHCH_(3)`
C. `CH_(3)CH_(2)CH_(2)CHO`
D. None of these is correct.

Explanation

To solve the question regarding the end product of the reaction between CH₃MgBr and CH₂=CH-CHO in the presence of H₃O⁺, we can follow these steps:

### Step 1: Identify the Reactants
We have two reactants:
1. CH₃MgBr (a Grignard reagent)
2. CH₂=CH-CHO (an aldehyde)

### Step 2: Understand the Reaction Mechanism
Grignard reagents react with carbonyl compounds (like aldehydes and ketones) to form alcohols. In this case, CH₂=CH-CHO contains an aldehyde functional group.

### Step 3: Nucleophilic Attack
The nucleophilic carbon (from CH₃MgBr) will attack the electrophilic carbon of the carbonyl group (C=O) in the aldehyde:
- The carbonyl carbon (C=O) is attacked by the nucleophilic carbon from CH₃MgBr.
- This results in the formation of a tetrahedral intermediate.

### Step 4: Formation of the Alkoxide Intermediate
After the nucleophilic attack, the intermediate will be:
- CH₃-CH₂-CH=CH-O⁻ (alkoxide ion)

### Step 5: Protonation of the Alkoxide
When H₃O⁺ is added, it protonates the alkoxide ion to form the corresponding alcohol:
- CH₃-CH₂-CH=CH-OH (an alcohol)

### Step 6: Identify the Final Product
The product formed is an alcohol, specifically an allylic alcohol. The structure can be represented as:
- CH₃-CH₂-CH=CH-OH

### Step 7: Keto-Enol Tautomerism
The alcohol can exist in two forms: the enol form and the keto form. The keto form can be represented as:
- CH₃-CH₂-CHO (which is a ketone)

### Final Answer
The end product of the reaction is CH₃-CH₂-CHO (the keto form).