Which one of the following reaction would produce secondary alcohol?

Options

A. `C_(6)H_(5)-overset(O)overset(||)(C)-CH_(3)underset(2 . H^(+))overset(1.CH_(3) MgBr)to`
B. `C_(6)H_(5)-overset(O)overset(||)(C)-CH_(3)underset(2 . H^(+))overset(1.1.LiAIH_(4))to`
C. `CH_(3) CHO underset(2 . H^(+))overset(1.LiAIH_(4))to`
D. `CH_(3) - overset(O)overset(||)(C) - CH_(3) underset(2 . Br_(2)) overset(1.^(-)OH)to`

Answer

To determine which reaction produces a secondary alcohol, we need to analyze the given reactions step by step.

### Step 1: Understand the structure of secondary alcohols
A secondary alcohol has the general structure where the carbon atom attached to the hydroxyl (OH) group is connected to one hydrogen atom and two alkyl groups (R1 and R2).

### Step 2: Analyze the first reaction
The first reaction involves a Grignard reagent (C6H5C(=O)CH3 + MgBr). The Grignard reagent generates a nucleophile (CH3-) that attacks the carbonyl carbon. This leads to the formation of a tertiary alcohol because the carbon attached to the OH group will have no hydrogen atoms (it will be connected to three other carbons). Therefore, this reaction does not produce a secondary alcohol.

### Step 3: Analyze the second reaction
The second reaction involves the reduction of a ketone (C6H5C(=O)CH3) using LiAlH4. Here, the hydride ion from LiAlH4 attacks the electron-deficient carbonyl carbon. The resulting structure has the OH group attached to a carbon that is connected to one hydrogen and two alkyl groups (C6H5 and CH3). This confirms that the product is a secondary alcohol.

### Step 4: Analyze the third reaction
The third reaction involves the reduction of an aldehyde (CH3CHO) with LiAlH4. The hydride ion attacks the carbonyl carbon, resulting in a primary alcohol (since the carbon with the OH group is attached to two hydrogens and one alkyl group). Thus, this reaction does not yield a secondary alcohol.

### Step 5: Analyze the fourth reaction
The fourth reaction involves a ketone (CH3C(=O)H) reacting with a base (OH-). The base abstracts a proton, forming a carbanion, which then attacks a bromine electrophile. This reaction does not lead to the formation of an alcohol; instead, it produces a bromo compound, which is not an alcohol.

### Conclusion
Based on the analysis, the second reaction (C6H5C(=O)CH3 with LiAlH4) is the only one that produces a secondary alcohol.

### Final Answer
The reaction that produces a secondary alcohol is the second option: C6H5C(=O)CH3 + LiAlH4.

---