Q.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.

Concept: Cannizzaro Reaction — but here the key is identifying an aldehyde/ketone that gives iodoform and cleaves into two acids on oxidation.

Step 1: Find the molecular formula.

% oxygen = $100 - (69.77 + 11.63) = 18.60\%$.

Moles per 100 g: C = $\frac{69.77}{12} \approx 5.814$, H = $\frac{11.63}{1} = 11.63$, O = $\frac{18.60}{16} \approx 1.1625$.

Divide by smallest (1.1625): C ≈ 5, H ≈ 10, O ≈ 1 → empirical formula = $\text{C}_5\text{H}_{10}\text{O}$.

Molecular mass = 86, empirical mass = 86 → molecular formula = $\text{C}_5\text{H}_{10}\text{O}$.

Step 2: Interpret the chemical tests.

• Does not reduce Tollens’ reagent → not an aldehyde (so it’s a ketone).
• Forms addition compound with NaHSO₃ → confirms a carbonyl group (ketone).
• Positive iodoform test → must have a $\text{CH}_3\text{CO}-$ group (methyl ketone).
• Vigorous oxidation gives ethanoic acid ($\text{CH}_3\text{COOH}$) and propanoic acid ($\text{C}_2\text{H}_5\text{COOH}$) → the carbon skeleton breaks at the carbonyl, giving a 2‑carbon and a 3‑carbon acid. …

1. Molecular formula from composition

$O\% = 100 - (69.77 + 11.63) = 18.60\%$. Taking 100 g:

Dividing by the smallest (1.16): $C:H:O = 5:10:1$, so the empirical formula is $C_5H_{10}O$ (mass $=86$). Since the molecular mass is also 86, the molecular formula is $C_5H_{10}O$.

2. Interpreting the tests

• Does not reduce Tollens/Fehling -> not an aldehyde, so it is a ketone.
• Forms an addition compound with $NaHSO_3$ -> confirms a $>C=O$ group.
• Positive iodoform test -> contains a $CH_3CO-$ group.

So (A) is a methyl ketone of formula $C_5H_{10}O$: either pentan-2-one or 3-methylbutan-2-one.

3. Using the oxidation products …

Method: Retrospective Analysis from Chemical Tests & Combustion Data

This method works backwards from the given data — first determine the molecular formula, then use chemical tests to narrow down the functional groups, and finally deduce the structure from the oxidation products.

Step 1: Find the molecular formula from percentage composition

• Carbon: $\frac{69.77}{12} = 5.814$
• Hydrogen: $\frac{11.63}{1} = 11.63$
• Oxygen (by difference): $100 - (69.77 + 11.63) = 18.6\%$ $\frac{18.6}{16} = 1.1625$

Divide by the smallest (1.1625):

• C: $\frac{5.814}{1.1625} \approx 5$
• H: $\frac{11.63}{1.1625} \approx 10$
• O: $\frac{1.1625}{1.1625} = 1$

Empirical formula: $\text{C}_5\text{H}_{10}\text{O}$

Empirical mass: $5 \times 12 + 10 \times 1 + 16 = 86$

Given molecular mass = 86, so molecular formula = $\text{C}_5\text{H}_{10}\text{O}$

Step 2: Interpret the chemical tests

Since it’s not an aldehyde (no Tollens reduction) but has a carbonyl (NaHSO$_3$ test) and gives iodoform — it must be a methyl ketone ($\text{CH}_3\text{CO}-$).

Step 3: Use oxidation products to find the carbon skeleton

• Vigorous oxidation gives ethanoic acid ($\text{CH}_3\text{COOH}$) and propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$).
• This means the original molecule had a carbon chain that breaks at the carbonyl group during oxidation. …

Here are the common mistakes students make when solving this exact type of problem (Cannizzaro + iodoform + oxidation), and how to avoid each.

1. Mistake: Calculating the wrong empirical formula

The error:

Students often round off percentages incorrectly or forget that oxygen is the remainder.

• Given: C = 69.77%, H = 11.63%
• Rest oxygen = $100 - (69.77 + 11.63) = 18.60\%$
• Moles:
  • C: $\frac{69.77}{12} \approx 5.814$
  • H: $\frac{11.63}{1} \approx 11.63$
  • O: $\frac{18.60}{16} \approx 1.1625$

Dividing by smallest (1.1625):

• C: $5.814 / 1.1625 \approx 5$
• H: $11.63 / 1.1625 \approx 10$
• O: $1.1625 / 1.1625 = 1$

Empirical formula: $\mathbf{C_5H_{10}O}$

Empirical mass: $5(12) + 10(1) + 16 = 86$

Since molecular mass is also 86, molecular formula = $\mathbf{C_5H_{10}O}$.

How to avoid:

• Always calculate oxygen by subtraction.
• Divide each mole value by the smallest mole value.
• Check if empirical mass matches given molecular mass — if yes, they are the same.

2. Mistake: Ignoring the "does not reduce Tollens' reagent" clue

The error:

Students assume the compound is an aldehyde because it forms an addition compound with $\ce{NaHSO3}$.

Why it's wrong:

• Tollens' reagent is reduced only by aldehydes (and $\alpha$-hydroxy ketones).
• A negative Tollens' test means no aldehyde group is present.
• But it still forms a bisulfite addition compound — this is possible with ketones (especially methyl ketones) and some cyclic ketones.

How to avoid:

• Remember: Both aldehydes and ketones form bisulfite addition products.
• Negative Tollens' → not an aldehyde → must be a ketone.

3. Mistake: Misinterpreting the positive iodoform test

The error:

Students think any ketone gives a positive iodoform test.

Correction:

Iodoform test is positive only for:

• Methyl ketones ($\ce{R-CO-CH3}$)
• Ethanol and secondary alcohols with $\ce{CH3CH(OH)-}$ group
• Acetaldehyde ($\ce{CH3CHO}$)

Since the compound is a ketone (from clue 2), it must have the $\ce{CH3CO-}$ (methyl carbonyl) group.

How to avoid:

• Memorise the exact structural requirement: $\ce{CH3CO-}$ or $\ce{CH3CH(OH)-}$.
• For a $\ce{C5H10O}$ ketone, the only way to have a methyl carbonyl is: $\ce{CH3CO-CH2CH2CH3}$ or $\ce{CH3CO-CH(CH3)2}$.

4. Mistake: Forgetting the oxidation product clue

The error:

Students stop after identifying the methyl ketone and don't check the oxidation products.

Given: Vigorous oxidation gives ethanoic acid ($\ce{CH3COOH}$) and propanoic acid ($\ce{CH3CH2COOH}$).

What this means:

• Vigorous oxidation of a ketone cleaves the carbon chain at the carbonyl group.
• The two fragments become carboxylic acids.
• If we get $\ce{CH3COOH}$ and $\ce{CH3CH2COOH}$, the original ketone must be: $\ce{CH3COCH2CH2CH3}$ (pentan-2-one)

How to avoid:

• Draw the oxidation cleavage: $\ce{R-CO-R'}$ → $\ce{R-COOH}$ + $\ce{R'-COOH}$ …