Solved Examples (Dedicated Problem Set) — Class 12 Chemistry Notes

Dedicated Problem Set — JEE/NEET Level

This is a curated set of 32 fully worked, exam-level problems covering every theme of Alcohols, Phenols and Ethers — naming and classification, hydration vs hydroboration-oxidation, Grignard and reduction routes, acidity comparisons, dehydration and oxidation products, the Lucas and Victor Meyer tests, the Kolbe and Reimer-Tiemann reactions, Williamson synthesis and ether cleavage. Each solution shows the reasoning step by step, the way you should think in JEE and NEET. Work through each one yourself before reading the solution.

Naming, Classification & Acidity

Example 1. Give the IUPAC name of (CH $_3$ ) $_2$ CH-CH $_2$ -OH and classify it.

Solution: The longest chain containing -OH is propane with a methyl branch: it is 2-methylpropan-1-ol. The -OH-bearing carbon is attached to one other carbon, so it is a primary (1°) alcohol.

Example 2. Rank the acidity: ethanol, water, phenol, acetic acid.

Solution: ethanol < water < phenol < acetic acid. Alcohols are the weakest (alkyl group destabilises alkoxide); the carboxylate is most resonance-stabilised, so acetic acid is the strongest.

Hydration vs Hydroboration-Oxidation

Example 3. Propene is treated (a) with H $_2$ O/H+ and (b) with B $_2$ H $_6$ then H $_2$ O $_2$ /OH-. Give the major product of each.

Solution: (a) Acid-catalysed hydration is Markovnikov: -OH adds to the more substituted carbon → propan-2-ol (CH $_3$ CHOHCH $_3$ ). (b) Hydroboration-oxidation is anti-Markovnikov: -OH adds to the less substituted carbon → propan-1-ol (CH $_3$ CH $_2$ CH $_2$ OH).

Example 4. Which alkene and method would you choose to make butan-1-ol selectively from but-1-ene?

Solution: Use hydroboration-oxidation (B $_2$ H $_6$ , then H $_2$ O $_2$ /OH-) on but-1-ene; the anti-Markovnikov addition places -OH on the terminal carbon → butan-1-ol. (Acid hydration would give butan-2-ol.)

Example 5. Acid-catalysed hydration of 2-methylpropene gives which alcohol and why?

Solution: Markovnikov addition puts -OH on the more substituted carbon (the one giving the more stable 3° carbocation) → 2-methylpropan-2-ol (tert-butyl alcohol).

Grignard & Reduction Routes

Example 6. What alcohol forms when CH $_3$ MgBr reacts with (i) HCHO, (ii) CH $_3$ CHO, (iii) acetone, each followed by H $_3$ O+?

Solution: Grignard adds CH $_3$ to the carbonyl carbon. (i) HCHO → ethanol (1°); (ii) CH $_3$ CHO → propan-2-ol (2°); (iii) (CH $_3$ ) $_2$ CO → 2-methylpropan-2-ol (3°). Formaldehyde gives 1°, other aldehydes 2°, ketones 3°.

Example 7. How would you prepare 2-phenylpropan-2-ol using a Grignard reagent?

Solution: React CH $_3$ MgBr with acetophenone (C $_6$ H $_5$ COCH $_3$ ), then hydrolyse — the methyl adds to the carbonyl to give C $_6$ H $_5$ C(OH)(CH $_3$ ) $_2$ , i.e. 2-phenylpropan-2-ol. (Equivalently, C $_6$ H $_5$ MgBr + acetone.)

Example 8. Which reagent reduces a carboxylic acid to a primary alcohol but leaves a C=C double bond untouched?

Solution: Lithium aluminium hydride (LiAlH $_4$ ). It reduces -COOH (and esters) to -CH $_2$ OH but does not reduce isolated alkene double bonds.

Dehydration & Oxidation

Example 9. Predict the major product when butan-2-ol is heated with conc. H $_2$ SO $_4$ .

Solution: Acid dehydration gives the Saytzeff (more substituted) alkene: but-2-ene is the major product (more stable than but-1-ene).

Example 10. Arrange in order of ease of acid dehydration: butan-1-ol, butan-2-ol, 2-methylpropan-2-ol.

Solution: Dehydration goes through a carbocation, so the order follows carbocation stability: 2-methylpropan-2-ol (3°) > butan-2-ol (2°) > butan-1-ol (1°).

Example 11. Give the oxidation product (with acidified K $_2$ Cr $_2$ O $_7$ ) of (a) propan-1-ol, (b) propan-2-ol, (c) 2-methylpropan-2-ol.

Solution: (a) 1° → propanoic acid (via propanal); (b) 2° → propanone (acetone); (c) 3° → no reaction (no α-H to remove; resistant to oxidation under these conditions).

Example 12. Which mild oxidant converts a primary alcohol to an aldehyde and stops there?

Solution: PCC (pyridinium chlorochromate) in dichloromethane oxidises a 1° alcohol to the aldehyde without over-oxidation to the acid.

Lucas & Victor Meyer Tests

Example 13. Three bottles contain butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. How does the Lucas test identify each?

Solution: Add Lucas reagent (conc. HCl + ZnCl $_2$ ). 2-methylpropan-2-ol (3°) → immediate turbidity; butan-2-ol (2°) → turbidity in about 5 min; butan-1-ol (1°) → no turbidity at room temperature.

Example 14. A colourless alcohol gives a blue colour in the Victor Meyer test. What is its class?

Solution: Blue colour ⇒ secondary (2°) alcohol (1° gives blood-red, 3° gives colourless).

Example 15. Why does a tertiary alcohol give no colour in the Victor Meyer test?

Solution: A 3° alcohol cannot be converted into the nitro compound needed for the test to proceed, so it does not form the coloured nitrolic acid/ pseudonitrol derivative; hence no colour develops and the solution stays colourless.

Phenol — Named Reactions & EAS

Example 16. Phenol reacts with bromine water. Give the product and explain the conditions.

Solution: In bromine water (polar, no catalyst), the highly activated ring undergoes trisubstitution to give a white precipitate of 2,4,6-tribromophenol. (In CS $_2$ at low temperature, monobromination gives mainly p-bromophenol.)

Example 17. Name and give the product of: sodium phenoxide + CO $_2$ (400 K, 4-7 atm) then H+.

Solution: This is Kolbe's reaction; the product is salicylic acid (2-hydroxybenzoic acid).

Example 18. Phenol + CHCl $_3$ + aqueous NaOH, then H+ gives which product, and what is the reaction called?

Solution: This is the Reimer-Tiemann reaction, giving salicylaldehyde (2-hydroxybenzaldehyde) (the -CHO enters mainly ortho).

Example 19. What is observed when neutral FeCl $_3$ is added to phenol, and why is it diagnostic?

Solution: A violet/purple colour appears due to a coloured iron(III)-phenoxide complex. Alcohols give no colour, so this distinguishes a phenol from an alcohol.

Williamson Synthesis & Ether Cleavage

Example 20. Design a Williamson synthesis of 2-methoxy-2-methylpropane (tert-butyl methyl ether).

Solution: Use the primary halide: sodium tert-butoxide ((CH $_3$ ) $_3$ CO^- Na+) + CH $_3$ I. The methyl iodide undergoes SN2; the bulky alkoxide supplies the oxygen. (The reverse pair — CH $_3$ O^- Na+ + (CH $_3$ ) $_3$ CBr — would give an alkene by elimination.)

Example 21. Give the products of CH $_3$ -O-CH $_2$ CH $_3$ with excess HI.

Solution: SN2 cleavage attacks the smaller methyl group first → CH $_3$ I + C $_2$ H $_5$ OH; with excess HI the ethanol also converts → CH $_3$ I + C $_2$ H $_5$ I.

Example 22. What are the products of anisole (C $_6$ H $_5$ OCH $_3$ ) + HI, and why?

Solution: The aryl-O bond does not cleave (partial double-bond character), so the alkyl-O bond breaks → phenol + CH $_3$ I. Phenol is not converted further.

Example 23. (CH $_3$ ) $_3$ C-O-CH $_3$ + HI gives which products? Explain the regiochemistry.

Solution: The stable 3° carbocation forms (SN1), so the tert-butyl group becomes the iodide: products are (CH $_3$ ) $_3$ CI + CH $_3$ OH.

Mixed JEE/NEET Challenge

Example 24. An organic compound A (C $_3$ H $_8$ O) does not react with Lucas reagent at room temperature and on oxidation gives a compound that answers the iodoform test negatively but gives Tollens' test. Identify A.

Solution: No quick Lucas reaction ⇒ primary alcohol. C $_3$ H $_8$ O primary alcohol is propan-1-ol; oxidation gives propanal (positive Tollens, negative iodoform — it has no CH $_3$ CO or CH $_3$ CH(OH) group). A = propan-1-ol.

Example 25. Compound B (C $_3$ H $_8$ O) gives an immediate yellow precipitate in the iodoform test and turbidity with Lucas reagent in 5 minutes. Identify B.

Solution: Positive iodoform ⇒ a CH $_3$ -CH(OH)- group; Lucas in about 5 min ⇒ secondary alcohol. B = propan-2-ol.

Example 26. Why is the boiling point order: ethanol > dimethyl ether, but methanol > water is false?

Solution: Ethanol H-bonds (high b.p.), dimethyl ether does not — so ethanol > dimethyl ether is correct. Water has two O-H bonds and forms a 3-D H-bond network, so water (100 °C) > methanol (65 °C) — methanol > water is false.

Example 27. Predict the major monobromination product of anisole and justify.

Solution: -OCH $_3$ is activating, o/p-directing; the ortho position is sterically hindered, so the major product is p-bromoanisole.

Example 28. An alcohol C $_4$ H $_{10}$ O is optically active and gives a 2° classification by the Lucas test (turbidity in about 5 min). Identify it.

Solution: A chiral 2° C $_4$ H $_{10}$ O alcohol is butan-2-ol (CH $_3$ CH(OH)CH $_2$ CH $_3$ ) — its C-2 is a stereocentre, and it is secondary.

Example 29. Convert phenol to salicylaldehyde in one named reaction; name the reagent.

Solution: Reimer-Tiemann reaction: phenol + CHCl $_3$ + aqueous NaOH, then acidification → salicylaldehyde (2-hydroxybenzaldehyde).

Example 30. Which gives a faster SN1 dehydration: cyclohexanol or 1-methylcyclohexanol? Why?

Solution: 1-Methylcyclohexan-1-ol (a 3° alcohol) dehydrates faster because it forms a more stable tertiary carbocation than the secondary cation from cyclohexanol.

JEE Main & Advanced Level Solved Examples

These problems demand substituent-effect reasoning, carbocation mechanisms, regiochemistry and structure elucidation — the level that separates top scorers. Decide the mechanism (SN1/SN2/E1) and the dominant electronic effect before you answer.

Example 31: Order the acidity of substituted phenols [JEE]

Arrange in increasing acidity: phenol, p-cresol (4-methylphenol), p-chlorophenol and p-nitrophenol (approximate pK $_a$ values 10.0, 10.2, 9.4, 7.1).

Solution: Acidity rises when the substituent stabilises the phenoxide ion (electron-withdrawing) and falls when it destabilises it (electron-donating).

-NO $_2$ (strong -R and -I): delocalises the negative charge onto its oxygens through para resonance → biggest increase.
-Cl (-I dominant, weak +R): mild increase over phenol.
-CH $_3$ (+I and hyperconjugation, electron-donating): destabilises the phenoxide → less acidic than phenol.

Increasing acidity: p-cresol < phenol < p-chlorophenol < p-nitrophenol (matching pK $_a$ 10.2 > 10.0 > 9.4 > 7.1).

Example 32: o-nitrophenol vs p-nitrophenol [JEE]

Why is o-nitrophenol steam-volatile and lower-boiling than p-nitrophenol, so that the two can be separated by steam distillation?

Solution: In o-nitrophenol the -OH and the neighbouring -NO $_2$ are close enough to form an intramolecular hydrogen bond (chelation), so its molecules do not associate strongly with one another — it is volatile in steam and has a lower boiling point. In p-nitrophenol the groups are too far apart for intramolecular bonding, so molecules link through intermolecular hydrogen bonds, giving strong association, higher boiling point and non-volatility. Steam distillation therefore carries over only the ortho isomer.

JEE/Advanced — Dehydration, Rearrangement & Reduction

Example 33: Dehydration with carbocation rearrangement [JEE Advanced]

Predict the major alkene from the acid-catalysed dehydration of 3,3-dimethylbutan-2-ol, CH $_3$ -CH(OH)-C(CH $_3$ ) $_3$ .

Solution: Protonation and loss of water give a secondary carbocation at C2. Adjacent C3 is a quaternary carbon, so a 1,2-methyl shift moves a methyl from C3 to C2, generating the more stable tertiary cation at C3. Loss of a proton then gives the highly substituted alkene 2,3-dimethylbut-2-ene [(CH $_3$ ) $_2$ C=C(CH $_3$ ) $_2$ , tetramethylethylene] as the major product — not the un-rearranged 3,3-dimethylbut-1-ene. Skeletal rearrangement is the giveaway of a carbocation intermediate.

Example 34: Chemoselective reduction — NaBH $_4$ vs LiAlH $_4$ [JEE]

A molecule contains both a ketone and a carboxylic acid group. Which reagent reduces only the ketone, and which reduces both groups to alcohols?

Solution: NaBH $_4$ is a mild hydride donor: it reduces aldehydes and ketones to alcohols but leaves carboxylic acids (and esters) untouched — so it converts only the ketone to a secondary alcohol. LiAlH $_4$ is far more powerful and reduces the carboxylic acid as well, giving a primary alcohol, in addition to reducing the ketone. Use NaBH $_4$ for chemoselective ketone reduction; use LiAlH $_4$ when the acid must also be reduced.

JEE/Advanced — Ethers: Williamson Synthesis & Cleavage

Example 35: Choosing the right Williamson pair [JEE Main]

Williamson synthesis makes ethers by an SN2 reaction of an alkoxide with an alkyl halide. Which pair correctly prepares tert-butyl methyl ether, (CH $_3$ ) $_3$ C-O-CH $_3$ ?

Solution: Use sodium tert-butoxide + iodomethane: (CH $_3$ ) $_3$ C-O $^-$ + CH $_3$ I → (CH $_3$ ) $_3$ C-O-CH $_3$ . The SN2 step occurs at the unhindered methyl carbon, so it works cleanly.

The reverse pair — tert-butyl halide + sodium methoxide — fails: methoxide is a strong base and the tertiary halide has no accessible backside, so E2 elimination dominates, giving isobutylene instead of the ether. Rule: in Williamson synthesis put the SN2 leaving group on the less hindered carbon (methyl/primary).

Example 36: Regiochemistry of ether cleavage by HI [JEE Advanced]

Give the products when (a) anisole (C $_6$ H $_5$ -O-CH $_3$ ) and (b) tert-butyl methyl ether ((CH $_3$ ) $_3$ C-O-CH $_3$ ) are heated with HI, and explain the difference.

Solution:

(a) Anisole → phenol + CH $_3$ I. I $^-$ can only attack the sp $^3$ methyl carbon (SN2); the aryl-O bond cannot break (the aryl carbon is sp $^2$ with partial double-bond character). Even with excess HI the phenol is not converted to iodobenzene.
(b) tert-Butyl methyl ether → (CH $_3$ ) $_3$ C-I + CH $_3$ OH. Here the tert-butyl group forms a stable tertiary carbocation (SN1), so I $^-$ bonds to the tert-butyl carbon and the methyl leaves as methanol.

Takeaway: with a 1°/methyl ether, I $^-$ attacks the less hindered carbon (SN2); with a 3° ether, the halide goes to the carbon that gives the stable carbocation (SN1).

JEE/Advanced — Mechanisms, Tests & Structure ID

Example 37: The $^{18}$ O esterification experiment [JEE Advanced]

When a carboxylic acid is esterified with an alcohol whose oxygen is labelled with $^{18}$ O, where does the label end up — in the ester or in the water? What does this prove?

Solution: The label is found in the ester, RCO- $^{18}$ O-R', and the water produced is ordinary H $_2$ O. This shows that the bond broken in the acid is the acyl-oxygen (C-OH) bond, while the alcohol keeps its oxygen. In other words, the -OH lost as water comes from the carboxylic acid, not from the alcohol — the mechanism is acyl-oxygen fission, not alkyl-oxygen fission.

Example 38: Which alcohols give the iodoform test [JEE/NEET]

Of methanol, ethanol, propan-1-ol, propan-2-ol and 2-methylpropan-2-ol, which give a positive iodoform test, and why?

Solution: The iodoform (CHI $_3$ ) test is positive only for alcohols with a CH $_3$ -CH(OH)- unit, which I $_2$ /NaOH oxidises to a CH $_3$ -CO- (methyl carbonyl) group. So:

Positive: ethanol (CH $_3$ CH $_2$ OH → CH $_3$ CHO pattern) and propan-2-ol (CH $_3$ -CH(OH)-CH $_3$ ).
Negative: methanol (no such carbon), propan-1-ol (CH $_3$ CH $_2$ CH $_2$ OH — the carbinol carbon carries no methyl) and 2-methylpropan-2-ol (tertiary, no -CH(OH)- hydrogen).

Example 39: Identify the ether by structure elucidation [JEE Advanced]

A neutral compound A (molecular formula C $_4$ H $_{10}$ O) gives no hydrogen with sodium metal, and on heating with excess HI yields a single alkyl iodide. Identify A and explain.

Solution: No reaction with Na means no -OH group, so A is an ether (not an alcohol). Cleavage by HI giving only one alkyl iodide means the two alkyl groups on oxygen are identical, i.e. A is a symmetrical ether. The only symmetrical C $_4$ H $_{10}$ O ether is diethyl ether, C $_2$ H $_5$ -O-C $_2$ H $_5$ , which with excess HI gives 2 C $_2$ H $_5$ I (via C $_2$ H $_5$ OH formed first). So A = diethyl ether.

Example 40: Reimer-Tiemann mechanism [JEE]

Phenol heated with CHCl $_3$ and aqueous NaOH (~340 K), followed by hydrolysis, gives salicylaldehyde. What is the attacking electrophile, and why is the product mainly the ortho isomer?

Solution: Hydroxide first deprotonates CHCl $_3$ and eliminates Cl $^-$ to generate dichlorocarbene (:CCl $_2$ ), the electrophile. It attacks the highly nucleophilic phenoxide ring, preferentially at the ortho position (favoured by coordination/H-bonding of the phenoxide oxygen with the reagent), giving an o-(dichloromethyl) intermediate; alkaline hydrolysis of the -CHCl $_2$ group then yields the -CHO of 2-hydroxybenzaldehyde (salicylaldehyde).