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Q8.19:
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

Solution :

Step 1: Determining the Empirical Formula

First, calculate the percentage of oxygen: $100 - (69.77 + 11.63) = 18.60\%$. [Per the Law of Conservation of Mass].

ElementPercentageAtomic MassRelative MolesSimplest Ratio
C69.7712$69.77/12 = 5.814$$5.814/1.1625 \approx 5$
H11.631$11.63/1 = 11.63$$11.63/1.1625 \approx 10$
O18.6016$18.60/16 = 1.1625$$1.1625/1.1625 = 1$

Empirical formula: $C_5H_{10}O$. Empirical formula mass: $(5 \times 12) + (10 \times 1) + (1 \times 16) = 86$ g/mol. Since the given molecular mass is 86, the Molecular Formula is $C_5H_{10}O$.

Step 2: Analyzing Functional Group Properties

  • Negative Tollens' Test: The compound is a ketone, not an aldehyde. [Justification: Tollens' reagent reacts only with aldehydes].
  • Sodium Hydrogensulphite Addition: Confirms the presence of a carbonyl group ($>C=O$). [Justification: Characteristic reaction of ketones with low steric hindrance].
  • Positive Iodoform Test: Confirms the presence of a methyl ketone group ($CH_3-CO-$). [Justification: The $CH_3-CO-$ group is oxidized to $CHI_3$ (iodoform) and a carboxylate ion by $I_2/OH^-$].

Step 3: Determining Structural Isomers via Oxidation

Vigorous oxidation of a ketone involves the cleavage of the $C-C$ bond adjacent to the carbonyl group. Oxidation of the unknown compound yields ethanoic acid ($CH_3COOH$) and propanoic acid ($CH_3CH_2COOH$).

Given the formula $C_5H_{10}O$ and the methyl ketone requirement, the structure must be $CH_3-CO-CH_2-CH_2-CH_3$ (Pentan-2-one).

CH₃ C=O CH₂ CH₂ CH₃

Step 4: Verification of Oxidation Products

Cleavage of Pentan-2-one during vigorous oxidation:

$CH_3-CO-CH_2-CH_2-CH_3 \xrightarrow{[O]} CH_3COOH + CH_3CH_2COOH$

The cleavage at the $C_2-C_3$ bond produces two-carbon (ethanoic) and three-carbon (propanoic) fragments, matching the empirical evidence provided. [Popoff’s Rule: Oxidation of unsymmetrical ketones occurs by cleavage of the $C-C$ bond on either side of the carbonyl to produce a mixture of carboxylic acids].

Final Solution: The compound is Pentan-2-one, with the structure $CH_3COCH_2CH_2CH_3$.


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