Question 25 of 26advanced🔍 AnalyzeLong Answer5 marks

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.

Correct Answer

Step 1: Determination of Molecular Formula

Given: Carbon = 69.77%, Hydrogen = 11.63%, Oxygen = 100 - (69.77 + 11.63) = 18.60%

Moles calculation:

  • Moles of C = 69.77/12 = 5.81
  • Moles of H = 11.63/1 = 11.63
  • Moles of O = 18.60/16 = 1.16

Simplest ratio:

  • C : H : O = 5.81/1.16 : 11.63/1.16 : 1.16/1.16 = 5 : 10 : 1

Empirical formula = C₅H₁₀O Empirical formula mass = (5×125 \times 12) + (10×110 \times 1) + 16 = 86 Since molecular mass = 86, Molecular formula = C₅H₁₀O

Step 2: Analysis of Functional Groups

The compound does not reduce Tollens' reagent, indicating it is not an aldehyde but a ketone.

The compound forms addition compound with sodium hydrogensulphite, which is characteristic of methyl ketones.

The compound gives positive iodoform test, confirming the presence of CH₃CO- group (methyl ketone).

Step 3: Determination of Structure from Oxidation Products

On vigorous oxidation, the compound gives ethanoic acid (CH₃COOH) and propanoic acid (CH₃CH₂COOH). This indicates the ketone has:

  • Methyl group (CH₃-) on one side of carbonyl
  • Propyl group (CH₃CH₂CH₂-) on the other side

Structure: CH₃-CO-CH₂-CH₂-CH₃

Name: Pentan-2-one

Exercise: EXERCISES | Q: 8.19 | (Page 31)
For More Understanding

Explanation

This question tests multiple concepts: empirical formula calculation, distinction between aldehydes and ketones using Tollens' test, identification of methyl ketones through iodoform test and NaHSO₃ addition, and structure elucidation from oxidation products. The key insight is that vigorous oxidation of ketones cleaves the C-C bond adjacent to carbonyl, producing carboxylic acids from both sides. Since ethanoic acid (from CH₃CO-) and propanoic acid (from CH₃CH₂CH₂-) are formed, the ketone must be pentan-2-one.

Solution Steps

  1. Step 1: Calculate empirical formula from percentage composition - C₅H₁₀O

  2. Step 2: Confirm molecular formula equals empirical formula (mass = 86)

  3. Step 3: Identify functional group as ketone (negative Tollens' test)

  4. Step 4: Confirm methyl ketone from positive iodoform test and NaHSO₃ addition

  5. Step 5: Deduce structure from oxidation products - pentan-2-one

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