Intramolecular aldol condensation :
The aldol condensation also offer a convenient way to synthesize molecules with five and six membered ring.This can be done by an intramolecular aldol condensation using a dialdehyde, a keto aldehyde or a diketone as the substrate.The major product is formed by the attack of the enolate from the ketone side of the molecule that adds to the aldehyde group.The reason the aldehyde group undergoes addition preferentially may arise from the greater reactivity of aldehyde towards nucleophilic addition.In reaction of this type five membered rings from far more readily than seven membered rings and six membered rings are more favourable than four or eight membered rings when possible.
The true statement about the major product of `CH_3-oversetOoverset(||)C-CH_2-CH_2-CH_2-CH_2-oversetoverset(O)(||)C-H` in reaction with aq.NaOH followed by heating is.

Options

A. It gives yellow ppt with `I_2//overset(Theta)OH`
B. It gives silver mirror with Tollen's reagent
C. It gives fehling's solution test .
D. It does not give yellow ppt with 2,4-DNP

Explanation

To solve the problem regarding the intramolecular aldol condensation of the compound `CH3-C(=O)-CH2-CH2-CH2-CH2-C(=O)-H` in the presence of aqueous NaOH followed by heating, we will follow these steps:

### Step 1: Identify the Structure
The given compound can be represented as follows:
- It has two carbonyl groups: one is an aldehyde (-CHO) and the other is a ketone (C=O).
- The structure can be drawn as:

```
CH3-C(=O)-CH2-CH2-CH2-CH2-CHO
```

### Step 2: Formation of Enolate Ion
In the presence of a base (NaOH), the alpha hydrogen (the hydrogen adjacent to the carbonyl group) on the ketone can be removed, forming an enolate ion.

- The enolate ion is formed from the ketone side:

```
CH3-C(=O)-CH2-CH2-CH2-CH2-CH2-O^-
```

### Step 3: Nucleophilic Attack
The enolate ion will then attack the carbonyl carbon of the aldehyde group. The reaction can be represented as:

- The nucleophilic attack of the enolate on the aldehyde carbon results in the formation of a five-membered ring:

```
CH3-C(=O)-CH2-CH2-CH2-CH2-CHO
```

### Step 4: Ring Closure
The nucleophilic attack leads to the formation of a five-membered ring. The structure can be represented as:

```
O
/ \
CH2 C
| |
CH2 CH3
```

### Step 5: Elimination of Water
Upon heating, the compound undergoes dehydration (loss of water) to form a double bond, resulting in a more stable alkene product.

### Step 6: Identify the Major Product
The major product is stabilized by resonance. The double bond formed can be represented as:

```
CH3-C(=O)-CH=CH-CH2-CH2
```

### Step 7: Check for Functional Group Tests
1. **Iodoform Test**: The product contains a methyl ketone (C=O with a CH3 group), so it will give a yellow precipitate with I2 and NaOH.
2. **Tollens' Test**: Ketones do not give a silver mirror with Tollens' reagent.
3. **Fehling's Test**: Ketones do not react with Fehling's solution.
4. **2,4-DNP Test**: Ketones do react with 2,4-DNP, forming a yellow precipitate.

### Conclusion
The only true statement about the major product is that it gives a yellow precipitate with I2 and NaOH (iodoform test).

### Final Answer
- The true statement about the major product is: **It gives yellow PPT with I2 and OH-.**