Alcohols Phenols Ethers: Previous Year Questions with Solutions

Question

(JEE Main 2023) Which of the following compounds gives a yellow precipitate with iodine in aqueous NaOH (iodoform test): (a) methanol, (b) ethanol, (c) propan-1-ol, (d) tert-butanol?

Solution — Step by Step

Iodoform test is positive only for $\text{CH}_3\text{CO-}$ or $\text{CH}_3\text{CH(OH)-}$ groups — methyl ketones and methyl carbinols.

Methanol ( $\text{CH}_3\text{OH}$ ): no CH(OH) adjacent to methyl — negative. Ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ ): has $\text{CH}_3\text{CH(OH)H}$ — positive. Propan-1-ol: the OH is on the terminal carbon, not on a carbon bearing a methyl — negative. Tert-butanol: no $\alpha$ -H on the carbon bearing OH — negative.

Ethanol (b) gives the positive iodoform test.

Why This Works

The iodoform test oxidises a secondary alcohol of the form $\text{CH}_3\text{CH(OH)R}$ to a methyl ketone $\text{CH}_3\text{COR}$ , then successively replaces the three $\alpha$ -H atoms with iodine, and finally cleaves to give $\text{CHI}_3$ (yellow) and $\text{RCOO}^-$ . Ethanol is the smallest alcohol that fits.

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

A quick shortcut: any alcohol of the form $\text{CH}_3\text{CH(OH)-}$ or $\text{CH}_3\text{CO-}$ passes. Scan mentally for a methyl attached to a CH(OH) or C=O.

Common Mistake

Some students mark methanol as positive, thinking ‘methyl group is present’. Methanol has no $\alpha$ -carbon at all — it only has the OH carbon. The rule needs a methyl adjacent to the CHOH or CO, not the CHOH itself being methyl.

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.