In the given reaction
`CH_3-CH_2-C-=C-Hunderset((ii)H_2O_2//Ooverset(Theta)H)overset((i)BH_3)rarr(X)`, (X)will be

Options

A. Butanal
B. Butanone
C. 2 - butanol
D. 1 - butanol

Answer

To solve the given reaction step by step, we will analyze the addition of BH3 and H2O2 to the alkyne, following the anti-Markovnikov rule.

### Step-by-Step Solution:

1. **Identify the Reactant:**
The reactant is 1-butyne (CH3-CH2-C≡C-H). It is an alkyne with a triple bond between the second and third carbon atoms.

2. **Addition of BH3:**
When BH3 is added to the alkyne, it undergoes hydroboration. According to the anti-Markovnikov rule, the boron atom will add to the carbon with more hydrogen atoms. Therefore, the boron will attach to the terminal carbon (the one with the hydrogen), and the hydrogen will add to the carbon with the triple bond.

The product after this step will be:
\[ CH3-CH2-C(BH2)-CH3 \]
(This is trialkyl boron, where the boron is bonded to the carbon that was part of the triple bond.)

3. **Oxidation with H2O2 and OH-:**
The next step involves treating the boron compound with hydrogen peroxide (H2O2) in the presence of a base (OH-). This leads to the replacement of the boron with a hydroxyl group (OH).

The reaction will convert the trialkyl boron compound into an alcohol:
\[ CH3-CH2-CH(BH2)-CH3 \xrightarrow{H2O2, OH-} CH3-CH2-CH(OH)-CH3 \]
This product is an enol form.

4. **Tautomerization to Keto Form:**
The enol can undergo tautomerization to form a ketone. In this case, the double bond shifts, and a hydrogen atom is transferred to the oxygen, resulting in the formation of a carbonyl compound (ketone).

The final product after tautomerization is:
\[ CH3-CH2-CO-CH3 \]
This compound is butan-2-one (or methyl ethyl ketone).

5. **Conclusion:**
The product (X) formed from the reaction is butan-2-one.

### Final Answer:
(X) will be butan-2-one.