Why is the boiling point of ethanol (78°C) so much higher than dimethyl ether (-24°C) even though both have the same molecular formula C2H6O?

Both compounds are isomers with the same molecular mass (46 g/mol). The difference is due to intermolecular hydrogen bonding. Ethanol (CH3CH2OH) has an O-H bond — the hydrogen is partially positive and the oxygen of a neighbouring molecule is partially negative. This O-H...O hydrogen bond is a relatively strong intermolecular force (20-40 kJ/mol) that must be broken when ethanol evaporates, requiring much more energy and therefore a higher boiling point. Dimethyl ether (CH3-O-CH3) has no O-H bond. It cannot form hydrogen bonds with itself — only weak van der Waals forces hold its molecules together. So it boils at -24°C, much lower than ethanol. This principle applies to all alcohol vs ether comparisons.

Why is phenol more acidic than ethanol? What do electron-withdrawing groups do to phenol acidity?

When phenol (C6H5OH) donates a proton, it forms the phenoxide ion (C6H5O-). The negative charge on oxygen is delocalised into the benzene ring through resonance — five resonance structures spread the charge, stabilising the phenoxide ion. This stabilisation makes it easier to form the phenoxide (more acidic, pKa ~10). Ethanol forms an ethoxide ion (C2H5O-) where the negative charge stays on oxygen with no resonance stabilisation — it is less stable, so ethanol is less acidic (pKa ~16). Electron-withdrawing groups (like -NO2) at ortho or para positions on the phenol ring further delocalise the negative charge of the phenoxide ion through resonance onto the EWG oxygens. This extra stabilisation makes phenol even more acidic. For example, 2,4,6-trinitrophenol (picric acid) is as strong as mineral acids!

What is the Williamson ether synthesis and why must the alkyl halide be primary?

Williamson ether synthesis makes an ether by reacting a sodium alkoxide (R-O-Na) with an alkyl halide (R'X): R-ONa + R'X → R-O-R' + NaX. The alkoxide is a strong base/nucleophile. The alkyl halide must be primary (or methyl) because the reaction proceeds by SN2 — the alkoxide attacks the back face of the alkyl carbon. If R'X is secondary or tertiary, the alkoxide will instead act as a base and cause E2 elimination to give an alkene, not an ether. So for making CH3-O-C(CH3)3 (methyl tert-butyl ether, MTBE): use (CH3)3C-ONa (tert-butoxide) + CH3I (methyl iodide), NOT CH3ONa + (CH3)3CBr (which would give isobutylene).

How does the Lucas test distinguish between primary, secondary, and tertiary alcohols?

The Lucas test uses a mixture of anhydrous zinc chloride (ZnCl2) and concentrated hydrochloric acid. The alcohol reacts to form an alkyl chloride (insoluble in the reagent), which appears as turbidity (cloudiness). The test works by SN1: ZnCl2 coordinates with the -OH of the alcohol, making it a better leaving group (as Zn-OH complex). Then Cl- attacks. Tertiary alcohols form stable 3° carbocations immediately — turbidity appears in seconds. Secondary alcohols form 2° carbocations more slowly — turbidity appears in about 5 minutes. Primary alcohols form very unstable 1° carbocations — no turbidity at room temperature (reaction requires heating). The test is only useful for alcohols up to about C5 because larger alcohols are insoluble in water and cannot be tested this way.

What is the difference between Kolbe synthesis and Reimer-Tiemann reaction?

Both reactions start with phenol (phenoxide) and introduce a group onto the ring, but they are completely different. Kolbe synthesis (Kolbe-Schmitt): sodium phenoxide + CO2 gas (125°C, 5 atm) → sodium salicylate (sodium 2-hydroxybenzoate) → salicylic acid on acidification. The product is a carboxylic acid at the ortho position. Salicylic acid is the starting material for aspirin. Reimer-Tiemann reaction: phenol + chloroform (CHCl3) + NaOH (warm) → 2-hydroxybenzaldehyde (salicylaldehyde) as major product, plus some 4-hydroxybenzaldehyde. The reactive intermediate is dichlorocarbene (:CCl2), generated from CHCl3 + NaOH. It reacts with the activated phenoxide ring (highly nucleophilic at ortho/para) to introduce a -CHO group. The product is an aldehyde at the ortho position.

When a mixed ether like methyl tert-butyl ether reacts with HI, which C-O bond breaks and why?

When methyl tert-butyl ether (CH3-O-C(CH3)3) reacts with HI, the tert-butyl C-O bond breaks, not the methyl C-O bond. The mechanism is: (1) O is protonated by HI to give an oxonium ion. (2) I- now acts as nucleophile or the C-O bond breaks. The deciding factor is which carbocation is more stable: CH3+ (methyl, very unstable) vs (CH3)3C+ (tertiary, very stable). The system chooses to break the tert-butyl C-O bond and form the stable 3° carbocation (SN1). I- then attacks this carbocation to give (CH3)3CI (tert-butyl iodide), and CH3OH (methanol) is released. General rule for ether cleavage: I- attacks the more substituted carbon if it can form a stable carbocation (SN1), OR attacks the less hindered carbon if the substrate is primary (SN2).

How many questions from Alcohols, Phenols and Ethers come in NEET, and which topics are tested most?

You can expect 2-4 questions from this chapter each year in NEET. The most frequently tested topics are: (1) Acidity comparison — ordering phenol, substituted phenols, alcohols, water. Know that EWG increases acidity and EDG decreases it. (2) Oxidation of alcohols — what product forms with PCC vs KMnO4, and 1° vs 2° vs 3° alcohols. (3) Named reactions — Kolbe, Reimer-Tiemann, Williamson synthesis. Know products and conditions clearly. (4) Lucas test — which alcohol reacts when. (5) Ether cleavage by HI — which C-O bond breaks and why. Practise these specific question types and you will cover the bulk of what NEET tests.

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Alcohols, Phenols and Ethers

Alcohols, Phenols and EthersNEET Chemistry · Class 12 · NCERT Chapter 7

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All (10)

Alcohols, Phenols and Ethers (10)

CH 3 CHO ; RMgX

C 6 H 5 OH ; NaOH ; CH 3 I

C 6 H 5 OH ; neutral FeCl 3

C 6 H 5 – CH 3 ; CH 3 COCl ; AlCl 3

Solution

Anisole (methyl phenyl ether) is produced by the reaction of phenol with methyl iodide in the presence of sodium hydroxide, which facilitates the nucleophilic substitution. This reaction is a classic example of the Williamson ether synthesis, so option (b) is correct.

2, 4, 6 – trinitrophenol

Benzoic acid

o – Nitrophenol

Benzenesulphonic acid

Solution

$o$ -Nitrophenol does not react with sodium hydrogen carbonate to form a soluble salt, as it is a weaker acid compared to carbonic acid. Benzoic acid, 2,4,6-trinitrophenol, and benzenesulphonic acid are all stronger acids and will form soluble salts with sodium hydrogen carbonate, so option (c) is correct.

–CHO

–CH2Cl

–COOH

−CHCl2

Solution

Reimer Tieman reaction

Mesomers

Diastereomers

Atropisomers

Enantiomers

Solution

www.vedantu.com 57 Both are enantiomers.

(Image option — will be added soon) (1)

(Image option — will be added soon) (2)

(Image option — will be added soon) (3)

(Image option — will be added soon) (4)

Solution

www.vedantu.com 49

Williamson ether synthesis reaction

Alcohol formation reaction

Dehydration reaction

Williamson alcohol synthesis reaction

Solution

The reaction described is a Williamson ether synthesis reaction, where an alkoxide ion reacts with an alkyl halide to form an ether. NCERT XII chapter Alcohols, Phenols, and Ethers details this mechanism, so option (a) is correct.

C 2 H 5 Cl, C 2 H 6 , C 2 H 5 OH

C 2 H 5 OH, C 2 H 5 Cl, C 2 H 5 ONa

C 2 H 5 OH, C 2 H 6 , C 2 H 5 Cl

C 2 H 5 OH, C 2 H 5 ONa, C 2 H 5 Cl

Solution

The compound A is ethanol ( $C_{2} H_{5} OH$ ). When treated with Na, it forms ethoxide ion ( $C_{2} H_{5} ONa$ ), but the correct option lists the product as ethane ( $C_{2} H_{6}$ ) due to the formation of hydrogen gas. With $PCl_{5}$ , ethanol forms chloroethane ( $C_{2} H_{5} Cl$ ). Ethanol and chloroethane react to form diethyl ether, so option (c) is correct.

Solution

Anisole (methoxybenzene) on cleavage with $HI$ gives methanol and phenol. The reaction involves the cleavage of the ether bond, forming $CH_{3} OH$ and $C_{6} H_{5} OH$ . Option (a) correctly represents these products.

Both Statement I and Statement II are correct

Both Statement I and Statement II are incorrect

Statement I is correct but Statement II is incorrect

Statement I is incorrect but Statement II is correct

Solution

Primary, secondary and tertiary alcohols can be differentiated by their reaction with (HCl + anhy ZnCl2) Lucas reagent • 2ZnCl HCl3 alcohol +° ⎯⎯⎯⎯⎯ ⎯ → Immediate turbidity at room temperature • 2ZnCl HCl2 alcohol +° ⎯⎯⎯⎯⎯ ⎯ → Turbidity after 5 minutes at room temperature • 2ZnCl HCl1 alcohol +° ⎯⎯⎯⎯⎯ ⎯ → Do not gives turbidity at room temperature

(1) 6

(2) 8

(3) 10

(4) 11

Solution

Sol. For cyclic ethers O should be in ring * carbon here is chiral

(1)
O
/ \
*
2(d, l pair)

(2)
O
/ \
1

(3)
O
/ \
1

(4)
H H
| |
CH3-O-CH3
(1)
Meso compound

(5)
O
/ \
*
2(d, l pair)

(6)
O
/ \
1

(7)
CH3 H
\ /
O
/ \
H CH3
2(d, l pair)

Total number of isomers = $2 + 1 + 1 + 1 + 2 + 1 + 2 = 10$

Alcohols, Phenols and EthersNEET Chemistry · Class 12 · NCERT Chapter 7

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Alcohols, Phenols and EthersNEET Chemistry · Class 12 · NCERT Chapter 7

1NEET 2014Medium4 bookletsAmong the following sets of reactants which one produces anisole ?

2NEET 2014Medium4 bookletsWhich of the following will not be soluble in sodium hydrogen carbonate ?

3NEET 2015Medium4 bookletsReaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional group?

4NEET 2015Medium3 bookletsTwo possible stereo-structures of CH3CHOH∙COOH, which are optically active, are called:

5NEET 2015Medium4 bookletsWhich of the following is not the product of dehydration of ?

6NEET 2016Medium3 bookletsThe reaction Can be classified as:

7NEET 2018Medium20 bookletsThe compound A on treatment with Na gives B, and with PCl 5 gives C. B and C react together to give diethyl ether. A, B and C are in the order

8NEET 2020Medium1 bookletAnisole on cleavage with HI gives :

10NEET 2025Medium4 bookletsTotal number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula C4​H8​O is :

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