Alcohols Phenols Ethers: Tricky Problems from JEE/NEET

Question

(JEE Advanced style) A phenol X on treatment with $\text{CHCl}_3/\text{NaOH}$ followed by acidification gives a compound Y. Y reacts with Tollens’ reagent to give a silver mirror. Identify the reaction and predict the major product if X is $p$ -cresol.

Solution — Step by Step

$\text{CHCl}_3 + \text{NaOH}$ with a phenol is the Reimer-Tiemann reaction. It introduces a $-\text{CHO}$ group ortho to the $-\text{OH}$ .

$p$ -cresol has $-\text{OH}$ at position 1 and $-\text{CH}_3$ at position 4. The ortho positions (2 and 6) are both available. The $-\text{CHO}$ goes to one of them — position 2.

Y is 2-hydroxy-5-methylbenzaldehyde (a salicylaldehyde-type compound with a methyl at the 5-position).

Y has a free $-\text{CHO}$ group. Tollens’ reagent (ammoniacal $\text{AgNO}_3$ ) oxidises aldehydes to carboxylates, depositing a silver mirror. Confirmed.

Why This Works

Reimer-Tiemann generates a dichlorocarbene ( $:\text{CCl}_2$ ) in situ from $\text{CHCl}_3$ and $\text{OH}^-$ . This electrophile attacks the ortho position of the phenoxide (the ring is activated and ortho/para-directing through the $-\text{O}^-$ ). Hydrolysis of the intermediate gives the ortho-hydroxyaldehyde.

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

If you use $\text{CCl}_4$ instead of $\text{CHCl}_3$ , the same mechanism gives a carboxylic acid — the Reimer-Tiemann ‘acid variant’. Knowing both lets you predict products from the reagent alone.

Common Mistake

A common trap is to place the $-\text{CHO}$ para to the $-\text{OH}$ . In $p$ -cresol, para is blocked by $-\text{CH}_3$ , so para attack cannot happen. Even in unblocked phenols, Reimer-Tiemann is strongly ortho-selective because the transition state is cyclic and geometrically favours ortho attack.

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.