Alcohols Phenols Ethers: Numerical Problems Solved Step-by-Step

Question

A 4.6 g sample of ethanol (C₂H₅OH) is completely oxidised to ethanoic acid (CH₃COOH) using acidified K₂Cr₂O₇. Calculate the mass of ethanoic acid formed and the moles of K₂Cr₂O₇ required.

Solution — Step by Step

The oxidation is $\text{C}_2\text{H}_5\text{OH} \rightarrow \text{CH}_3\text{COOH}$ , which is a 4-electron oxidation per molecule. Each $\text{Cr}_2\text{O}_7^{2-}$ accepts 6 electrons.

$M(\text{C}_2\text{H}_5\text{OH}) = 46$ g/mol, so $n = 4.6/46 = 0.1$ mol.

1 mole ethanol gives 1 mole acetic acid. So $n(\text{CH}_3\text{COOH}) = 0.1$ mol. Mass $= 0.1 \times 60 = 6$ g.

Electrons released $= 0.1 \times 4 = 0.4$ mol. Moles of $\text{Cr}_2\text{O}_7^{2-} = 0.4/6 = 0.0667$ mol.

Why This Works

Primary alcohols oxidise all the way to carboxylic acid under strong oxidisers like acidified dichromate. The trick is counting electrons: C in ethanol goes from -1 (CH₂OH carbon) to +3 (COOH carbon), a change of 4 per molecule. Balancing by electrons gives both mass and reagent quantity in one pass.

Reading the question twice before touching the pen saves more time than any shortcut. Underline what is asked — students lose marks by solving the wrong quantity, not by arithmetic errors.

Alternative Method

You can balance the full redox equation: $3\text{C}_2\text{H}_5\text{OH} + 2\text{Cr}_2\text{O}_7^{2-} + 16\text{H}^+ \rightarrow 3\text{CH}_3\text{COOH} + 4\text{Cr}^{3+} + 11\text{H}_2\text{O}$ . From the 3:2 ratio, $n(\text{dichromate}) = (2/3)(0.1) \approx 0.0667$ mol — same answer.

Common Mistake

Students often stop oxidation at acetaldehyde (a 2-electron change) and get half the dichromate. Acidified $\text{K}_2\text{Cr}_2\text{O}_7$ is a strong oxidiser — it pushes ethanol straight to the acid unless PCC or distillation is specified.

Exam Connection

This kind of question is a reliable marks-scorer in CBSE board papers and JEE Main. Examiners reuse the underlying concept with small cosmetic changes — different numbers, a renamed compound, a twist in the wording. If you understand the reasoning here, you will recognise the pattern in any future variant. That is the difference between solving one question and solving a question type.

Teachers often say, “practise till the method becomes boring.” What they mean is: work through enough variations that the decision tree inside your head becomes automatic. By the time you sit the exam, your pen should know the first three steps before your brain consciously thinks about them.

Practice Extension

Try these follow-up variations on your own and see if your method still works:

Change the starting quantity (e.g. 4.6 g becomes 9.2 g) and confirm that the final answer scales linearly.
Swap the reagent for a close analogue (e.g. $\text{KMnO}_4$ instead of $\text{K}_2\text{Cr}_2\text{O}_7$ ) and check which steps stay the same and which change.
Flip the question — given the product, work backwards to find the starting amount. This reverse drill tests whether you really understand the stoichiometry, or whether you were just plugging into a memorised chain.
Add an impurity (e.g. “the sample is only 80% pure”) and adjust the mole calculation. Purity questions are a classic twist in JEE Main.

In JEE Main, the cut-off between a 95 percentile and a 99 percentile is often just 3 to 4 questions of this exact difficulty. Practising these variations matters more than learning new topics during the last month of preparation.

Quick Recap

Identify the concept before touching numbers.
Write units on every line — they catch errors automatically.
Reach the final answer in a single bold or boxed expression so the examiner sees it instantly.
When stuck, ask: “what quantity is conserved here?” Conservation of mass, charge, and energy solve most chemistry problems.