Consider the following sequence of reactions
`C_(2)H_(5)C -= CH overset(CH_(3)MgBr)rarr A underset((2)H_(2)O^(+))overset((1)HCHO) rarr B`
The product `(B)` is

Options

A. `CH_(3)CH_(2)OH`
B. `C_(2)H_(5)underset(CH_(3))underset(|)C-CHCH_(2)OH`
C. `C_(2)H_(5)C-=C-underset(OH)underset(|)CH-CH_(3)`
D. `C_(2)H_(5)C-=C-CH_(2)OH`

Answer

To determine the product B from the given sequence of reactions, let's analyze each step carefully.

### Step-by-Step Solution:

1. **Identify the Starting Material**: The starting material is a terminal alkyne, specifically 1-pentyne (C2H5C≡CH).

2. **Reaction with Grignard Reagent**: The terminal alkyne reacts with the Grignard reagent, CH3MgBr. In this reaction, the terminal alkyne acts as a nucleophile and attacks the electrophilic carbon of the Grignard reagent. This results in the formation of a new carbon-carbon bond. The product of this reaction (let's call it A) will be:
\[ C_2H_5C \equiv C - CH_3 \quad \text{(1-pentyne reacts with CH3MgBr)} \]

3. **Formation of an Alkyne Alcohol**: The product A (which is now a magnesium alkoxide) will then react with formaldehyde (HCHO) in the presence of an acidic workup (H3O+). The nucleophilic carbon (from the alkyne) will attack the carbonyl carbon of formaldehyde, leading to the formation of an alkoxide intermediate.

4. **Protonation**: After the nucleophilic attack, the intermediate is protonated during the acidic workup, resulting in the formation of a secondary alcohol. The structure will be:
\[ C_2H_5C(CO)H \quad \text{(where CO is the -OH group from the formaldehyde)} \]

5. **Final Product B**: The final product B can be represented as:
\[ C_2H_5C(CH_2OH)C \quad \text{(which is 3-pentanol)} \]

### Conclusion:
The product B is 3-pentanol.