In the given sequence
`HOCH_2-CHOH-CH_2OH overset(Br_2//H_2O)(rarr)A + B overset(dil.NaOH)(rarr)(X)`,
X is

Options

A. Glyceraldehyde
B. Dihydroxyacetone
C. Fructose
D. Glucose

Answer

To solve the problem step by step, we will analyze the given sequence of reactions and identify the final product (X).

### Step 1: Identify the starting compound
The starting compound is HOCH₂-CHOH-CH₂OH, which is glycerol (a triol with three hydroxyl groups).

### Step 2: Reaction with Bromine Water
When glycerol reacts with bromine water (Br₂/H₂O), it undergoes oxidation. In this reaction:
- The primary alcohol groups (–CH₂OH) are oxidized to aldehydes (–CHO).
- The secondary alcohol group (–CHOH–) is oxidized to a ketone (–C=O).

Thus, we will have two products:
- Product A: CH₂OH (oxidized to CHO, an aldehyde)
- Product B: CHOH (oxidized to C=O, a ketone)

### Step 3: Identify Products A and B
- Product A (from the primary alcohol): CH₂OH → CHO (an aldehyde)
- Product B (from the secondary alcohol): CHOH → C=O (a ketone)

### Step 4: Reaction with Dilute NaOH
Next, the products A and B undergo a reaction with dilute NaOH. In this reaction:
- The dilute NaOH will abstract a proton from the aldehyde and ketone, forming a carbanion.
- This carbanion will then undergo an aldol condensation reaction.

### Step 5: Aldol Condensation
The aldol condensation involves the formation of a β-hydroxy aldehyde or ketone, which can further dehydrate to form an α,β-unsaturated carbonyl compound.

### Step 6: Determine Product X
After performing the aldol condensation, we will end up with a compound that has multiple hydroxyl groups and a ketone functional group. The final product (X) is a sugar known as fructose.

### Final Answer
Thus, the final product X is fructose.

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