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Step-by-Step Solution
Step 1: Identify the Functional Group from the Tests
Given that the compound forms a phenylhydrazone, it must contain a carbonyl ($C=O$) group.
A phenylhydrazone is formed by the reaction of a carbonyl compound (aldehyde or ketone) with phenylhydrazine.
Step 2: Use of Iodoform and Tollen’s Test
1. Negative Iodoform Test: This indicates that the compound does not have the $CH_3CO-$ group
(which is the requirement for a positive iodoform test). Thus, it is not a methyl ketone.
2. Negative Tollen’s Test: Tollen’s test is positive for aldehydes (due to easy oxidation
to carboxylate). The negative result means the compound is not an aldehyde.
Step 3: Analyze Effect of Reduction
The compound, on reduction, produces n-pentane ($C_5H_{12}$). Reduction of a ketone
removes the $C=O$ double bond, yielding the corresponding alkane. Thus, the core
five-carbon chain remains intact, suggesting that this carbonyl compound is a ketone
on the main carbon chain without branching.
Step 4: Determine the Location of the Carbonyl Group
The molecular formula is $C_5H_{10}O$. Since the carbonyl is not at the first carbon
(otherwise it would be an aldehyde) and is not a methyl ketone (otherwise iodoform test
would be positive), the only possibility that remains consistent with all observations is
a ketone at the third carbon, namely 3-pentanone.
Conclusion
Hence, based on these tests and the molecular formula, the compound is
3-pentanone.
