An organic compound 'X' having molecular formula C₅H₁₀O yields phenyl hydrazone and gives negative response to the Iodoform test and Tollen's test. It produces n-pentane on reduction. What is ‘X’?

3-Pentanone 

The correct answer is option 1. 3-Pentanone.

Let us break down the problem step by step to identify the compound 'X':

The given molecular formula is \( \text{C}_5\text{H}_{10}\text{O} \). This suggests that the compound is either a ketone, aldehyde, or alcohol. Phenyl hydrazone formation occurs when a compound reacts with phenylhydrazine (\( \text{C}_6\text{H}_5\text{NHNH}_2 \)), indicating the presence of a carbonyl group (C=O). This means the compound is either a ketone or an aldehyde.

The iodoform test is used to identify methyl ketones (\( \text{R-CO-CH}_3 \)) and secondary alcohols with a methyl group adjacent to the hydroxyl group (\( \text{R-CH(OH)-CH}_3 \)). A negative iodoform test means the compound does not have the \( \text{R-CO-CH}_3 \) or \( \text{R-CH(OH)-CH}_3 \) structure.

Tollen's test is used to identify aldehydes. A negative Tollen's test indicates the compound is not an aldehyde.

Reduction of the compound to n-pentane suggests that the carbonyl group is not at the end of the carbon chain, as this would yield a primary alcohol or an aldehyde upon reduction, which would then yield a different product. Instead, a ketone in the middle of the chain would be reduced to a straight-chain alkane.

Conclusion:

3-Pentanone: It has the molecular formula \( \text{C}_5\text{H}_{10}\text{O} \).

It forms a phenyl hydrazone, indicating the presence of a carbonyl group. It gives a negative iodoform test because it does not have the structure \( \text{R-CO-CH}_3 \). It gives a negative Tollen's test, confirming it is not an aldehyde. On reduction, 3-pentanone yields n-pentane, as the ketone in the middle of the chain is reduced to the corresponding alkane. Therefore, the compound 'X' is 3-Pentanone.