The major products formed in the reaction
`C_(6)H_(5)CHO+CH_(3)NO_(2) overset(NaOH)underset("Heat")to(X)`

Explanation

To solve the reaction between benzaldehyde (C₆H₅CHO) and nitromethane (CH₃NO₂) in the presence of sodium hydroxide (NaOH) and heat, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Benzaldehyde (C₆H₅CHO)
- Nitromethane (CH₃NO₂)

### Step 2: Understand the Role of the Base
Sodium hydroxide (NaOH) acts as a base in this reaction. It will deprotonate the nitromethane, leading to the formation of a nucleophile. The hydrogen atom from the methyl group (CH₃) of nitromethane is abstracted by the hydroxide ion (OH⁻).

### Step 3: Formation of the Nucleophile
When NaOH abstracts a proton from nitromethane, it forms a carbanion:
- CH₃NO₂ → CH₂NO₂⁻ (carbanion)

### Step 4: Nucleophilic Attack
The carbanion (CH₂NO₂⁻) acts as a nucleophile and attacks the electrophilic carbon of the carbonyl group in benzaldehyde (C₆H₅CHO). The carbonyl carbon is electrophilic due to the electron-withdrawing nature of the oxygen atom.

### Step 5: Formation of the Intermediate
The nucleophilic attack results in the formation of an intermediate:
- C₆H₅C(OH)(CH₂NO₂) (this is a tetrahedral intermediate)

### Step 6: Elimination of Water
Upon heating, the intermediate undergoes dehydration (loss of water). The hydroxyl group (OH) from the intermediate and a hydrogen from the methylene group (CH₂) are eliminated, leading to the formation of an alkene.

### Step 7: Final Product
The final product of the reaction is:
- C₆H₅CH=CHNO₂ (Benzene with a double bond to a carbon that is attached to a nitro group)

### Conclusion
The major product formed in the reaction of benzaldehyde with nitromethane in the presence of NaOH and heat is:
- C₆H₅CH=CHNO₂