Alcohols Phenols and Ethers Class 12 Notes

This chapter is split into three functional group families — alcohols, phenols and ethers. This page gives you the complete alcohols phenols and ethers class 12 notes: how to name each class, the ways they are prepared, their physical properties, and the reactions you must know for exams — acidity, Lucas test, dehydration, oxidation, electrophilic substitution and ether cleavage.

By the end, you’ll be able to compare acid strength, apply the Lucas test, carry out Williamson synthesis, and write stepwise mechanisms (NCERT, p. 1–3).

Types of Alcohols, Phenols and Ethers

Classification makes the chemistry manageable. Alcohols and phenols are grouped by the number of –OH groups (mono-, di-, tri-, polyhydric), while alcohols are further split by the hybridisation of the carbon carrying the –OH group (NCERT, p. 2).

When the –OH is on an sp³ carbon, the alcohol is primary (1°), secondary (2°) or tertiary (3°) depending on how many carbon atoms that carbon is bonded to. The figures below show the general structures.

Special classes you must recognise (NCERT, p. 2–3):

• Allylic alcohol — –OH on a carbon next to a C=C bond, e.g. \( CH_2=CH-CH_2OH \) (prop-2-en-1-ol).
• Benzylic alcohol — –OH on a carbon next to a benzene ring, e.g. \( C_6H_5-CH_2OH \).
• Vinylic alcohol — –OH directly on a C=C carbon, e.g. \( CH_2=CH-OH \).

Ethers are classified by the groups on oxygen: simple/symmetrical (same groups, e.g. \( C_2H_5OC_2H_5 \)) or mixed/unsymmetrical (different groups, e.g. \( C_2H_5OCH_3 \)) (NCERT, p. 3).

The geometry of the functional groups

The structure of the –OH and –O– groups is a favourite one-mark question (NCERT, p. 6–7).

• Alcohols: oxygen is sp³ hybridised, bonded to carbon by a sigma bond. The C–O–H bond angle is slightly less than the tetrahedral angle (109°-28′) because lone pairs on oxygen repel the bond pairs.
• Phenols: –OH is attached to an sp² hybridised ring carbon. The C–O bond length is 136 pm, shorter than in methanol, because the oxygen lone pair is partly conjugated with the ring (partial double-bond character).
• Ethers: the four electron pairs around oxygen (two bonds, two lone pairs) are roughly tetrahedral, but the C–O–C angle is a little larger than the tetrahedral angle due to repulsion between the two bulky R groups. The C–O bond length is 141 pm, close to that in alcohols.

IUPAC Nomenclature – Quick Tables

Naming is guaranteed marks if you follow a fixed order of operations: pick the longest chain, number from the end nearest the –OH, and give the position of every substituent and the –OH itself (NCERT, p. 3–4).

Common names still appear in questions: CH₃OH is methyl alcohol, C₂H₅OH is ethyl alcohol, C₆H₅OCH₃ is anisole, C₆H₅OC₂H₅ is phenetole (NCERT, p. 3–5).

Cyclic alcohols take the prefix cyclo and the –OH is assumed to be on C-1 (NCERT, p. 4).

Preparation Methods at a Glance

You need each method’s start material, reagent, condition and example. These three tables cover the whole chapter.

Preparation of alcohols

The Grignard logic decides the product class: with methanal you always get a primary alcohol, with other aldehydes a secondary alcohol, and with ketones a tertiary alcohol (NCERT, p. 9).

Preparation of phenols

Preparation of ethers

At 443 K with \( H_2SO_4 \), ethanol dehydrates to ethene; at 413 K the main product is ethoxyethane (NCERT, p. 23). The temperature decides the product.

Physical Properties Comparison

The single most tested fact: alcohols and phenols boil much higher than ethers, haloalkanes and hydrocarbons of comparable molecular mass. The reason is intermolecular hydrogen bonding (NCERT, p. 11).

Note that ethanol and methoxymethane have the same mass, yet differ by ~100 K in boiling point.

Solubility: alcohols and phenols dissolve in water because the –OH can hydrogen-bond with water. Solubility decreases as the hydrophobic alkyl/aryl part grows (NCERT, p. 12). Ethers are comparably soluble to alcohols of the same mass because the ether oxygen also forms hydrogen bonds with water.

Key Chemical Reactions – Logic and Mechanisms

Organise every reaction by which bond breaks(NCERT, p. 12): the O–H bond breaks when the alcohol acts as a nucleophile; the C–O bond breaks when the protonated alcohol acts as an electrophile.

1. O–H bond cleavage: acidity

Reaction with metals: alcohols and phenols give alkoxides/phenoxides and hydrogen, e.g. \( 2ROH + 2Na \rightarrow 2RONa + H_2 \). Phenols also dissolve in aqueous NaOH, which alcohols do not — this is the classic distinction (NCERT, p. 13).

Why are phenols more acidic than alcohols? The phenoxide ion is stabilised by delocalisation of the negative charge over the ring, while the alkoxide has a localised charge on oxygen. A more stable conjugate base means a stronger acid.

The sp² ring carbon is also more electronegative than an sp³ carbon, which pulls electron density away from O–H (NCERT, p. 13–14).

Effect of substituents on phenol acidity (NCERT, p. 14):

• Electron-withdrawing groups (EWG) like –NO₂ increase acidity, especially at ortho and para positions, because they stabilise the phenoxide. More EWG = stronger acid.
• Electron-releasing groups (ERG) like –CH₃ decrease acidity. Cresols are weaker acids than phenol.

Smaller pKa = stronger acid. Phenol is about a million times more acidic than ethanol (NCERT, p. 14).

2. Esterification

Alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters. The reaction with acid chloride needs pyridine to neutralise the HCl produced and drive the equilibrium right. Acetylation of salicylic acid forms aspirin (NCERT, p. 15).

3. C–O bond cleavage in alcohols

Reaction with HX: \( ROH + HX \rightarrow RX + H_2O \). This is the basis of the Lucas test (conc. HCl + anhydrous ZnCl₂):

• Tertiary alcohol: immediate turbidity (halide forms fast).
• Secondary alcohol: cloudy within ~5 minutes.
• Primary alcohol: no turbidity at room temperature.

Memory device: “Lucas test – primary: no cloud at RT; secondary: cloudy in 5 mins; tertiary: immediate.” The turbidity appears because the alkyl halide is immiscible with the aqueous reagent (NCERT, p. 16).

Dehydration: loss of water to give an alkene. Ethanol needs conc. \( H_2SO_4 \) at 443 K; secondary and tertiary alcohols dehydrate under milder conditions. Ease of dehydration: Tertiary > Secondary > Primary, because the carbocation intermediate is more stable when more substituted (NCERT, p. 16–17).

Oxidation: the O–H and a C–H bond break to form a C=O.

• Primary alcohol → aldehyde (with PCC or \( CrO_3 \) in anhydrous medium) → carboxylic acid (with strong oxidiser like acidified \( KMnO_4 \)).
• Secondary alcohol → ketone (with \( CrO_3 \)).
• Tertiary alcohol — resistant to oxidation under normal conditions; strong oxidisers cleave C–C bonds to give a mixture of smaller acids.

Dehydrogenation over hot Cu at 573 K gives aldehydes (from 1°) and ketones (from 2°), while 3° alcohols dehydrate to alkenes (NCERT, p. 18).

4. Reactions of phenols – electrophilic substitution

The –OH group activates the ring and directs incoming groups to ortho and para positions (NCERT, p. 19).

Steam volatility of nitrophenols: o-nitrophenol shows intramolecular hydrogen bonding, so molecules do not associate and it is steam volatile. p-nitrophenol shows intermolecular H-bonding, causing association, so it is less volatile and stays behind (NCERT, p. 19).

5. Reactions of ethers

Cleavage of C–O bond: ethers are the least reactive functional group and need drastic conditions, with excess HX. Order of reactivity: HI > HBr > HCl (NCERT, p. 25–26).

• Dialkyl ethers give two alkyl halides: \( ROR + HI \rightarrow RI + ROH \), then \( ROH + HI \rightarrow RI + H_2O \).
• Alkyl aryl ethers cleave at the alkyl–oxygen bond (it is weaker than the aryl–O bond, which has partial double-bond character). Example: ethoxybenzene + HI → phenol + ethyl iodide (the aryl–O bond survives).
• Mixed ethers with two alkyl groups: the smaller (less substituted) alkyl group leaves as the halide by an \( S_N2 \) path when both are primary/secondary. But if one group is tertiary, a stable tertiary carbocation forms and the mechanism is \( S_N1 \), so the tertiary halide is produced.

Electrophilic substitution on anisole: the –OCH₃ group activates the ring and directs ortho/para, just like –OH. Anisole brominates in ethanoic acid even without a Lewis acid catalyst, and the para isomer is obtained in 90% yield (NCERT, p. 27).

Worked Examples (Step-by-Step)

Common Mistakes and Corrections

Exam Tips and Scoring Patterns

From the chapter’s end-of-chapter exercises, the recurring asked skills are (NCERT, p. 30–33):

• Draw structure from IUPAC name — practised in Q. 7.2 (e.g. 2-methylbutan-2-ol, 1-ethoxypropane, cyclohexylmethanol). Number the chain from the –OH end and check every substituent.
• Name a compound — Q. 7.1 and 7.23 cover alcohols and ethers. For ethers, the larger group is the parent; the smaller becomes an alkoxy prefix.
• Compare acid strength — Q. 7.14 and 7.15 ask you to justify phenol vs ethanol and o-nitrophenol vs o-methoxyphenol. A full-marks answer cites resonance stabilisation of the phenoxide and the position of the –NO₂.
• Write a mechanism — Q. 7.11 (hydration of ethene), 7.19 (acid dehydration of ethanol), 7.30 (HI with methoxymethane). Mechanisms carry 3–5 marks: write every step, label the slow step, show the protonation and deprotonation.
• Give the reason for a physical property — Q. 7.4 (propanol vs butane), 7.8 (o-nitrophenol steam volatile), 7.22 (ethanol vs methoxymethane). Always name hydrogen bonding as the cause.
• Devise a synthesis — Q. 7.13 and 7.20 test Grignard, hydration, reduction and Williamson routes. State the reagent and the condition, e.g. “propene + H₂O/H⁺ gives propan-2-ol.”

For a 3–4 mark “give the mechanism” question, do not skip steps — the protonation of the alkene/ether, the carbocation formation (rate-determining), and the final deprotonation are all needed for full credit.

Quick Revision Recap

Reagent quick list

• Lucas test: conc. HCl + anhydrous ZnCl₂.
• Mild oxidation of 1° alcohol to aldehyde: PCC or \( CrO_3 \) (anhydrous).
• Strong oxidation to acid: acidified \( KMnO_4 \).
• Hydration of alkene: \( H_2O \), catalytic \( H^+ \).
• Reduction of ester/acid: \( LiAlH_4 \).
• Ether formation: Williamson — alkoxide + primary alkyl halide.

Acidity order memory aid

“Nitro Increases, Methyl Decreases” (NIMD): EWG (like –NO₂) raises acidity, ERG (like –CH₃) lowers it. General order: phenol < nitrophenol < dinitrophenol < trinitrophenol, while cresols < phenol.

Ether cleavage rule

HI > HBr > HCl, and the alkyl–O bond breaks in aryl alkyl ethers. If one alkyl group is tertiary, \( S_N1 \) gives the tertiary halide; otherwise the less substituted alkyl group becomes the halide by \( S_N2 \).

Frequently Asked Questions

Why is phenol more acidic than ethanol? Explain with resonance structures.

The direct answer: the phenoxide ion is stabilised by resonance delocalisation of the negative charge over the benzene ring, while the ethoxide ion has a localised charge on oxygen. A more stable conjugate base means a stronger acid, so phenol (pKa 10.0) is about a million times more acidic than ethanol (pKa 15.9)

How does a nitro group affect acidity? Why is para-nitrophenol a stronger acid than phenol?

The –NO₂ group is electron-withdrawing. It stabilises the phenoxide ion by delocalising the negative charge, and this effect is strongest at the ortho and para positions. p-Nitrophenol (pKa 7.1) is therefore a much stronger acid than phenol (pKa 10.0). Ortho and para isomers are stronger than meta because the meta position cannot delocalise the negative charge effectively.

Why can’t Williamson ether synthesis be used to prepare t-butyl ethyl ether from t-butyl chloride and sodium ethoxide?

Because sodium ethoxide is a strong base as well as a nucleophile. With a tertiary halide like t-butyl chloride, elimination dominates over substitution, giving 2-methylpropene (an alkene) instead of the ether. The correct route is to use sodium t-butoxide with a primary halide, e.g. sodium t-butoxide + ethyl bromide gives t-butyl ethyl ether (NCERT, p. 24).

What is the Lucas test? When does a primary alcohol give turbidity?

The Lucas test distinguishes 1°, 2° and 3° alcohols by their reaction with a mixture of conc. HCl and anhydrous ZnCl₂. A primary alcohol does not give turbidity at room temperature; turbidity appears only on heating, because the primary carbocation is too unstable to form quickly. Tertiary alcohols give immediate cloudiness; secondary give turbidity within about five minutes.

Why is o-nitrophenol steam volatile and p-nitrophenol not?

o-Nitrophenol forms an intramolecular hydrogen bond, so its molecules do not associate and it vaporises easily with steam. p-Nitrophenol forms intermolecular hydrogen bonds, causing molecules to associate into larger clusters that are less volatile, so it stays behind during steam distillation (NCERT, p. 19).

Why do ethers have lower boiling points than alcohols?

Ethers cannot form intermolecular hydrogen bonds between their own molecules (no –OH). Alcohols do, so they need more energy to separate molecules into the vapour phase. Ethoxyethane boils at 308 K, while butan-1-ol of similar mass boils at 390 K (NCERT, p. 25).

For the official NCERT textbook used for this chapter, see the NCERT Class 12 Chemistry Part II PDF on ncert.nic.in.

Reference: NCERT Class 12 Chemistry textbook, chapter Alcohols, Phenols and Ethers.