Class 12 Chemistry · NCERT chapter 11 · 7 topics · free to read
Classification & Nomenclature
Alcohols are classified as 1° (primary), 2° (secondary), or 3° (tertiary) based on the carbon bearing the –OH group. Phenols have –OH directly attached to a benzene ring. Ethers have an oxygen bonded to two carbon groups (R–O–R'). IUPAC naming follows specific suffix and prefix rules for each class.
R–OH (alcohol) Ar–OH (phenol) R–O–R' (ether)
1°, 2°, 3° Alcohols
Classification depends on the number of carbon atoms attached to the carbon bearing the hydroxyl group. This determines reactivity in almost every reaction of alcohols.
The classification of alcohols is based on the carbon atom that carries the -OH group — not on the -OH itself. Count how many other carbon atoms are directly bonded to that C .
Primary (1°): The C bearing -OH is bonded to one other carbon. Example: CH_3CH_2OH (ethanol)
Secondary (2°): The C bearing -OH is bonded to two other carbons. Example: (CH_3)_2CHOH (propan-2-ol)
Tertiary (3°): The C bearing -OH is bonded to three other carbons. Example: (CH_3)_3COH (2-methylpropan-2-ol)
Methanol ( CH_3OH ) has ZERO carbons attached — it's a special case, often grouped with 1° alcohols
Monohydric, Dihydric & Trihydric
Alcohols are also classified by the number of hydroxyl groups present in the molecule. This affects physical properties like boiling point and viscosity dramatically.
Monohydric: One -OH group. Example: CH_3OH (methanol), C_2H_5OH (ethanol)
Trihydric (Triol): Three -OH groups. Example: HOCH_2CH(OH)CH_2OH (glycerol — used in soaps, moisturizers)
More -OH groups → more hydrogen bonding → higher boiling point and viscosity
Phenols vs Alcohols
Phenols have the –OH group directly bonded to a benzene ring carbon (sp² C), while alcohols have –OH on an sp³ carbon. This seemingly small difference makes phenols much more acidic and gives them unique reactions.
In a phenol, the -OH is directly attached to an aromatic ring carbon ( C_ sp^2 ). In an alcohol, it's attached to an aliphatic carbon ( C_ sp^3 ). This matters enormously because the lone pair on oxygen in phenol can delocalize into the benzene ring through resonance.
-OH on sp^3 carbon
-OH on sp^2 carbon (benzene ring)
Weakly acidic ( pK_a 16 – 18 )
Moderately acidic ( pK_a 10 )
Do NOT react with NaOH
React with NaOH to form sodium phenoxide
No FeCl_3 color test
Give violet/blue/green color with FeCl_3
Undergo dehydration, oxidation
Undergo electrophilic aromatic substitution
Ethers
Ethers have an oxygen atom bonded to two alkyl/aryl groups. They are classified as simple (symmetric) or mixed (asymmetric) ethers.
In ethers, oxygen is bonded to two carbon groups: R–O–R' . If both groups are the same ( R = R' ), it's a simple (symmetric) ether. If different, it's a mixed (asymmetric) ether.
Simple ether: C_2H_5–O–C_2H_5 (diethyl ether — the famous anesthetic)
Mixed ether: CH_3–O–C_2H_5 (methyl ethyl ether)
Anisole: C_6H_5–O–CH_3 (methoxybenzene — an aryl alkyl ether)
Ethers have no -OH → cannot form H-bonds with themselves → much lower boiling points than alcohols of similar molecular mass
IUPAC Nomenclature
Alcohols use the suffix '-ol', phenols use 'phenol' as the parent name, and ethers use the 'alkoxy' prefix on the longer chain. Mastering naming is essential for NEET MCQs.
The IUPAC naming rules are different for each class. The key is knowing whether the -OH or -O- group gets the suffix (higher priority) or a prefix.
Alcohols: Replace '-e' of the parent alkane with '-ol'. Number the chain to give -OH the lowest locant. Example: CH_3CH(OH)CH_3 → propan-2-ol
Phenols: Use 'phenol' as the parent name. Substituents are numbered relative to -OH at position 1. Example: 2-nitrophenol, 4-chlorophenol
Ethers: The smaller alkyl group + oxygen = alkoxy prefix; longer chain = parent alkane. Example: CH_3OC_2H_5 → methoxyethane
When -OH is a substituent (lower priority than -COOH , -CHO , C=O ), use the prefix 'hydroxy-'
Quick recall
Name: CH_3OCH_2CH_2CH_3
The shorter group is CH_3 (methyl) → prefix: methoxy. Longer chain is propane. IUPAC name: 1-methoxypropane.
Preparation of Alcohols
Alcohols can be prepared from alkenes (acid-catalyzed hydration, hydroboration-oxidation), carbonyl compounds (reduction using NaBH₄/LiAlH₄), Grignard reagents (reaction with aldehydes/ketones/esters), and by hydrolysis of esters and alkyl halides. Each method gives a specific class (1°/2°/3°) of alcohol.
R–X + NaOH_ (aq) R–OH + NaX
Acid-Catalyzed Hydration
Alkenes react with water in the presence of an acid catalyst to form alcohols. The reaction follows Markovnikov's rule — the –OH goes to the more substituted carbon.
When an alkene reacts with water in the presence of H_2SO_4 or H_3PO_4 , the elements of water ( H and OH ) add across the double bond. This is electrophilic addition and follows Markovnikov's rule: the -OH attaches to the carbon that already has fewer hydrogens (more substituted carbon).
CH_3CH=CH_2 + H_2O H_2SO_4 CH_3CH(OH)CH_3
Follows Markovnikov's rule — -OH goes to more substituted carbon
Produces 2° or 3° alcohols preferentially (not 1°)
Involves carbocation intermediate → rearrangement possible
Catalyst: dilute H_2SO_4 or H_3PO_4
Hydroboration-Oxidation
Diborane (B₂H₆) adds to alkenes in anti-Markovnikov fashion, and subsequent oxidation with H₂O₂/NaOH gives the primary alcohol. This is the go-to method when you need a 1° alcohol from an alkene.
Hydroboration-oxidation is a two-step process: first, BH_3 (from B_2H_6 ) adds across the double bond with boron going to the less substituted carbon (anti-Markovnikov). Then, oxidation with alkaline H_2O_2 replaces B with -OH .
Anti-Markovnikov addition — -OH ends up on the less substituted carbon
Gives primary alcohols from terminal alkenes
Syn addition — both H and -OH add from the same face
No carbocation intermediate → no rearrangement
Overall: anti-Markovnikov, syn addition of water
Markovnikov product
Anti-Markovnikov product
Gives 2°/3° alcohols
Gives 1° alcohols
Carbocation intermediate
No carbocation (concerted)
Rearrangement possible
No rearrangement
Anti addition (mixture)
Syn addition
Grignard Reagent Method
Grignard reagents (RMgX) react with formaldehyde to give 1° alcohols, other aldehydes to give 2° alcohols, and ketones to give 3° alcohols. This is the most versatile method for alcohol preparation.
The Grignard reagent ( RMgX , where X = halogen) is a powerful nucleophile. It attacks the electrophilic carbonyl carbon ( C=O ), forming a new C–C bond. Hydrolysis of the resulting alkoxide gives the alcohol.
RMgX + HCHO H_3O^+ 1° alcohol (one extra carbon than R)
RMgX + R'CHO H_3O^+ 2° alcohol
RMgX + R'COR'' H_3O^+ 3° alcohol
RMgX must be prepared in dry ether (anhydrous) — water destroys it instantly
Reaction with esters: RMgX + R'COOR'' excess 3° alcohol (two R groups add)
Quick recall
What alcohol do you get from CH_3MgBr + CH_3CHO ?
Methyl Grignard adds to acetaldehyde → CH_3CH(OH)CH_3 = propan-2-ol (a 2° alcohol). Aldehyde + Grignard always gives 2° alcohol.
Reduction of Aldehydes, Ketones & Acids
Carbonyl compounds can be reduced to alcohols using NaBH₄ (mild, selective) or LiAlH₄ (powerful, reduces almost everything). Catalytic hydrogenation (H₂/Pd) is another option.
Reduction of the C=O group converts aldehydes to 1° alcohols and ketones to 2° alcohols. The two main reducing agents are sodium borohydride ( NaBH_4 ) and lithium aluminium hydride ( LiAlH_4 ).
Non-selective — reduces almost all functional groups
Cheaper, easier to handle
More expensive, moisture-sensitive
Aldehyde reduce 1° alcohol
Ketone reduce 2° alcohol
Carboxylic acid/ester LiAlH_4 1° alcohol (NaBH₄ cannot do this!)
Catalytic hydrogenation ( H_2 /Ni or Pd) also reduces C=O , but may reduce C=C too
From Alkyl Halides & Esters
Alkyl halides undergo nucleophilic substitution with aqueous NaOH/KOH to give alcohols. Esters give alcohols on acid or base hydrolysis (saponification).
Alkyl halides ( R–X ) can be converted to alcohols by nucleophilic substitution — the hydroxide ion ( OH^- ) replaces the halide. Esters ( RCOOR' ) on hydrolysis give a carboxylic acid (or its salt) and an alcohol.
The physical properties of alcohols, phenols, and ethers are dominated by hydrogen bonding. Alcohols and phenols can both donate and accept H-bonds (high b.p., water solubility), while ethers can only accept H-bonds (lower b.p., moderate solubility). Boiling points follow: phenols > alcohols > ethers for similar molecular mass.
Alcohols and phenols have an O–H bond that can form intermolecular hydrogen bonds, dramatically raising boiling points. Ethers lack O–H, so they can only accept H-bonds from other molecules.
The O–H bond in alcohols and phenols is polar — oxygen is electronegative, pulling electron density away from hydrogen. This allows the hydrogen to form a weak bond (hydrogen bond) with the oxygen of a neighboring molecule. This intermolecular H-bonding is responsible for the abnormally high boiling points of alcohols.
Alcohols & phenols → can both donate and accept H-bonds (they have O–H )
Ethers → can only accept H-bonds (lone pairs on O, but no O–H to donate)
Alkanes → no H-bonding at all (only weak London forces)
Strength of H-bonding: H–F > O–H > N–H
Boiling Point Trends
Within alcohols, boiling point increases with chain length and decreases with branching. Among isomeric alcohols, 1° > 2° > 3° in boiling point.
Ethane ( -89 °C, no H-bonding) < dimethyl ether ( -25 °C, weak dipole) < ethanol (78°C, strong H-bonding). H-bonding capability is the deciding factor.
Solubility in Water
Lower alcohols are miscible with water due to H-bonding. As chain length increases, the hydrophobic alkyl part dominates and solubility decreases. Ethers have limited water solubility.
Alcohols dissolve in water because their -OH group forms hydrogen bonds with water molecules. But the alkyl chain ( R ) is hydrophobic. As R gets bigger, the hydrophobic part wins and solubility drops.
CH_3OH , C_2H_5OH , C_3H_7OH are miscible (fully soluble) with water
From C_4 onward, solubility decreases progressively
Phenol is partially soluble in water (can form H-bonds), fully soluble above 66°C
Ethers are slightly soluble — they accept H-bonds from water but can't donate
Rule of thumb: solubility significant when -OH to C ratio 1:4
Reactions of Alcohols
Alcohols undergo a wide range of reactions: dehydration to alkenes (acid-catalyzed, Zaitsev's rule), oxidation to aldehydes/ketones/acids (PCC, KMnO₄, CrO₃), esterification with carboxylic acids, reaction with sodium to give alkoxides, and conversion to alkyl halides using HX, PCl₃, PCl₅, or SOCl₂. Reactivity order for many reactions: 3° > 2° > 1°.
R–OH conc. H_2SO_4, Alkene + H_2O
Dehydration
Acid-catalyzed dehydration removes H₂O from alcohols to give alkenes. The reaction follows Zaitsev's rule (more substituted alkene is major product). Ease of dehydration: 3° > 2° > 1°.
When heated with a strong acid catalyst (conc. H_2SO_4 or H_3PO_4 ), alcohols lose a molecule of water to form alkenes. This is an elimination reaction (E1 for 2°/3°, E2 for 1°).
1° alcohols: Need 170°C with conc. H_2SO_4 (harsh conditions)
2° alcohols: Need ~100°C with conc. H_2SO_4
3° alcohols: Need only 85°C with dilute acid (easiest to dehydrate)
Ease of dehydration: 3° > 2° > 1° (stability of carbocation intermediate)
Follows Zaitsev's rule — the more substituted alkene is the major product
Al₂O₃ at 350°C is another common dehydration condition
Oxidation
1° alcohols can be oxidized to aldehydes (mild oxidant) or carboxylic acids (strong oxidant). 2° alcohols give ketones. 3° alcohols resist oxidation under normal conditions.
Oxidation of alcohols removes hydrogen from the C–OH group. The product depends on the alcohol class and the strength of the oxidizing agent.
1° alcohol PCC aldehyde (stops here — PCC is mild)
1° alcohol KMnO_4 or CrO_3 carboxylic acid (strong oxidant, goes all the way)
2° alcohol any oxidant ketone (cannot be oxidized further without breaking C–C bond)
3° alcohol normal conditions NO reaction (no H on C–OH carbon to remove)
PCC = pyridinium chlorochromate ( C_5H_5NH^+ CrO_3Cl^- ) — the go-to mild oxidant for stopping at aldehyde
Quick recall
What happens when 2-methylpropan-2-ol is treated with acidified KMnO_4 ?
Nothing significant under normal conditions — it's a 3° alcohol with no hydrogen on the C–OH carbon. 3° alcohols resist oxidation. (Under extreme conditions, C–C bond cleavage can occur, but that's not tested in NEET.)
Esterification
Alcohols react with carboxylic acids in the presence of an acid catalyst (Fischer esterification) to form esters. This reversible reaction is driven forward by removing water.
R–OH + R'–COOH H^+ leftharpoons R'–COOR + H_2O
Catalyst: concentrated H_2SO_4 (protonates the carboxylic acid, makes it more electrophilic)
Reversible reaction — use excess alcohol or remove water to shift equilibrium forward
Mechanism involves nucleophilic addition-elimination at the carbonyl carbon
Alcohols react with sodium metal to produce hydrogen gas and sodium alkoxide. The rate of reaction decreases from 1° to 3° due to steric hindrance. This reaction is used to detect the presence of an active hydrogen.
2R–OH + 2Na 2R–ONa + H_2
All alcohols react with Na — confirms presence of active (acidic) hydrogen
Phenol also reacts with Na, more vigorously than alcohols (phenol is more acidic)
Ethers do NOT react with Na (no active hydrogen)
This is different from acidity order — acidity in aqueous solution is 3° < 2° < 1° in the gas phase, but in water the order can reverse due to solvation effects
Conversion to Alkyl Halides
Alcohols can be converted to alkyl halides using various reagents. The choice of reagent determines the mechanism, stereochemistry, and byproducts.
Replacing the -OH group with a halide ( -X ) is one of the most important reactions of alcohols. Multiple reagents can do this, each with different advantages.
PCl₅: R–OH + PCl_5 R–Cl + POCl_3 + HCl (fumes of HCl — used as a test for -OH )
SOCl₂ (thionyl chloride): R–OH + SOCl_2 pyridine R–Cl + SO_2 + HCl — BEST method (gaseous byproducts escape, pure product)
Lucas reagent ( ZnCl_2 + conc. HCl ): used to distinguish 1°/2°/3° alcohols
Phenols
Phenols are prepared industrially from cumene and in the lab from diazonium salts. Phenols are much more acidic than alcohols because the phenoxide ion is stabilized by resonance with the benzene ring. Key reactions include Kolbe's reaction (salicylic acid), Reimer-Tiemann (salicylaldehyde), Kolbe-Schmitt, azo coupling, and electrophilic aromatic substitution (–OH is a powerful activating, ortho/para-directing group).
C_6H_5OH + NaOH C_6H_5ONa + H_2O
Preparation of Phenols
The cumene process is the dominant industrial method (from benzene via cumene hydroperoxide). Lab methods include hydrolysis of diazonium salts, chlorobenzene (Dow process), and sodium benzenesulphonate.
The cumene process is the industrial method used worldwide. Benzene is alkylated with propene to form cumene (isopropylbenzene), which is then oxidized by air to cumene hydroperoxide, and finally treated with dilute acid to give phenol + acetone (both valuable products).
C_6H_6 CH_3CH=CH_2, H_3PO_4 C_6H_5CH(CH_3)_2 O_2, then H^+ C_6H_5OH + CH_3COCH_3
From sulphonate: C_6H_5SO_3Na NaOH, fuse C_6H_5ONa H^+ C_6H_5OH
Acidity of Phenols
Phenol is more acidic than alcohols because the phenoxide ion (C₆H₅O⁻) is stabilized by resonance — the negative charge delocalizes into the benzene ring. Electron-withdrawing groups increase acidity; electron-donating groups decrease it.
Phenol ( pK_a 10 ) is about 10⁶ times more acidic than ethanol ( pK_a 16 ). The reason is entirely about the stability of the conjugate base. When phenol loses H^+ , the resulting phenoxide ion ( C_6H_5O^- ) is stabilized by delocalization of the negative charge across the benzene ring (5 resonance structures).
Phenoxide ion has 5 resonance structures — charge spread over O and ortho/para carbons of the ring
Alkoxide ion ( RO^- ) has NO resonance stabilization — charge stuck on oxygen alone
Phenol reacts with NaOH (soluble in base); alcohols do NOT react with NaOH
But phenol is still a weak acid — does NOT react with NaHCO_3 (unlike carboxylic acids)
Acidity order: RCOOH > C_6H_5OH > H_2O > ROH
Effect of substituents on phenol acidity:
Electron-withdrawing groups ( -NO_2 , -CN , -X ) at ortho/para positions increase acidity (stabilize phenoxide further)
2,4,6-Trinitrophenol (picric acid) is a strong acid ( pK_a 0.4 ) — three -NO_2 groups stabilize the phenoxide massively
Reactions of Phenols
Phenols undergo several named reactions important for NEET: Kolbe's reaction (→ salicylic acid), Reimer-Tiemann (→ salicylaldehyde), coupling with diazonium salts, and facile electrophilic aromatic substitution due to the activating –OH group.
The -OH group on the benzene ring is a strong activating group (donates electron density via resonance). This makes phenols extremely reactive toward electrophilic aromatic substitution — much more reactive than benzene itself.
Ethers are prepared primarily by Williamson synthesis (sodium alkoxide + alkyl halide). They are cleaved by HI to give alcohols and alkyl iodides. Anisole undergoes electrophilic aromatic substitution where –OCH₃ is an activating, ortho/para-directing group. Ethers are relatively unreactive — making them excellent solvents.
R–ONa + R'–X R–O–R' + NaX
Williamson Synthesis
Sodium alkoxide (strong nucleophile) reacts with a primary alkyl halide via Sₙ2 to form an ether. This is the most versatile and important method for ether synthesis.
In Williamson synthesis, a sodium alkoxide ( R–ONa ) acts as the nucleophile and attacks a primary alkyl halide ( R'–X ) via S _N 2 mechanism to form the ether.
R–ONa + R'–X S_N2 R–O–R' + NaX
Alkyl halide must be primary (or methyl) — S _N 2 works best with unhindered substrates
If 2° or 3° halide is used → elimination (alkene) dominates instead of substitution
To make a mixed ether R–O–R' : use the bulkier group as alkoxide and the smaller group as halide
Example: To make tert-butyl methyl ether → use (CH_3)_3CONa + CH_3I (NOT CH_3ONa + (CH_3)_3CCl )
Intramolecular Williamson synthesis can form cyclic ethers (epoxides, THF)
Quick recall
How would you prepare anisole ( C_6H_5OCH_3 ) by Williamson synthesis?
C_6H_5ONa (sodium phenoxide) + CH_3I (methyl iodide) → C_6H_5OCH_3 + NaI . Phenoxide is the nucleophile, methyl halide is the substrate.
Cleavage of Ethers by HI
Ethers are generally unreactive, but concentrated HI (or HBr) can cleave the C–O bond. The products depend on whether the ether is symmetric, asymmetric, or aryl.
Ethers are resistant to most reagents (that's why they're great solvents), but concentrated HI can cleave the C–O bond. HI is used because I^- is an excellent nucleophile.
R–O–R' + HI R–OH + R'–I (or both as R–I with excess HI)
Simple (symmetric) ether + excess HI → two molecules of alkyl iodide + H_2O
Mixed ether with one 3° group: I^- attacks the 3° carbon (S _N 1) → 3° alkyl iodide + alcohol
Mixed ether with 1° and 1° groups: I^- attacks the smaller/less hindered group (S _N 2)
Aryl alkyl ether (anisole): C_6H_5OCH_3 + HI C_6H_5OH + CH_3I (phenol + methyl iodide, NOT iodobenzene!)
Reactivity of HX for cleavage: HI > HBr >> HCl (nucleophilicity of halide ion)
Electrophilic Substitution of Anisole
The –OCH₃ group in anisole is an activating group (donates electrons via resonance). It directs incoming electrophiles to ortho and para positions and makes the ring more reactive than benzene.
Anisole ( C_6H_5OCH_3 ) undergoes electrophilic aromatic substitution (EAS) more readily than benzene. The -OCH_3 group donates electron density into the ring through resonance (lone pair on O conjugates with the ring), activating it toward electrophiles.
-OCH_3 is an activating, ortho/para-directing group
Bromination: Anisole + Br_2/CH_3COOH → mixture of ortho- and para-bromoanisole (para is major)
Friedel-Crafts alkylation: Anisole + CH_3Cl/AlCl_3 → ortho- and para-methoxytoluene
Nitration: Anisole + dilute HNO_3 → mixture of ortho- and para-nitroanisole
Para product usually dominates due to less steric hindrance than ortho
Activating — reacts faster
Reference compound
Ortho/para-directing
No directing preference
Electron-rich ring (resonance donation)
Neutral electron density
Reacts with weak electrophiles
Needs strong electrophiles + catalyst
Identification Tests
Several qualitative tests help identify and distinguish alcohols, phenols, and ethers. The Lucas test differentiates 1°/2°/3° alcohols by turbidity timing. Victor Meyer test gives different colored products. The iodoform test identifies methyl carbinols (–CHOH–CH₃). The FeCl₃ test detects phenols (violet/blue/green color). The 2,4-DNP test identifies carbonyl compounds (relevant when alcohols are oxidized).
Lucas test: ROH + HCl/ZnCl_2 RCl (turbidity)
Lucas Test
The Lucas reagent (conc. HCl + anhydrous ZnCl₂) converts alcohols to alkyl chlorides. Since alkyl chlorides are insoluble, the solution turns turbid. The speed of turbidity reveals the alcohol class.
The Lucas test exploits the fact that 3° alcohols react fastest with HCl/ZnCl_2 (S _N 1, stable carbocation forms instantly), 2° react slower, and 1° barely react at room temperature. The product alkyl chloride ( R–Cl ) is insoluble in the aqueous reagent, causing turbidity (cloudiness).
Only works for low molecular weight alcohols ( 6 carbons) that are soluble in the reagent
Victor Meyer Test
The Victor Meyer test converts alcohols to nitro compounds via iodoalkanes, then treats with HNO₂. The resulting product's behavior with NaOH and HNO₂ gives different colors for 1°, 2°, and 3° alcohols.
In the Victor Meyer test, the alcohol is first converted to an alkyl iodide, then to a nitroalkane using AgNO_2 . The nitroalkane is then treated with HNO_2 (nitrous acid) and the product's behavior with NaOH reveals the alcohol class.
1° alcohol → 1° nitroalkane: Reacts with HNO_2 to form nitrolic acid → dissolves in NaOH giving blood RED color
2° alcohol → 2° nitroalkane: Reacts with HNO_2 to form pseudo-nitrole → BLUE color
3° alcohol → 3° nitroalkane: Does NOT react with HNO_2 (no -hydrogen) → NO COLOR (colorless)
Steps: R–OH → R–I → R–NO₂ → test with HNO_2 + NaOH
Iodoform Test
Alcohols with the structure CH₃–CH(OH)– (methyl carbinols, i.e., secondary alcohols with a methyl group on the carbinol carbon) give a positive iodoform test — producing yellow crystals of CHI₃ with I₂/NaOH.
The iodoform test uses I_2/NaOH (or NaOI ). Alcohols with the CH_3CH(OH)– structure are first oxidized to methyl ketones ( CH_3CO– ), which then undergo haloform reaction to give iodoform ( CHI_3 ) — a yellow precipitate with a characteristic smell.
Positive test: Yellow precipitate of CHI_3 (iodoform) — confirms CH_3CH(OH)– or CH_3CO– group
Ethanol ( CH_3CH_2OH ) gives a positive test — only 1° alcohol to do so!
Propan-2-ol ( CH_3CHOHCH_3 ) gives positive test (has CH_3CHOH– )
Propan-1-ol ( CH_3CH_2CH_2OH ) gives negative test (no CH_3CHOH– group)
Methanol gives negative test (it's CH_3OH , not CH_3CHOH– )
FeCl₃ Test
Phenols give a characteristic violet, blue, or green coloration with neutral FeCl₃ solution. This is due to the formation of a colored iron-phenol complex. Most alcohols and ethers do not give this test.
When a phenol is treated with neutral FeCl_3 solution, a colored complex is formed. The most common color is violet/purple for phenol itself, but substituted phenols can give blue, green, or red colors depending on the substituent.
Phenol → violet/purple color with FeCl_3
Resorcinol (1,3-benzenediol) → dark violet
Catechol (1,2-benzenediol) → green color
Hydroquinone (1,4-benzenediol) → blue color (transient, then turns brown)
Alcohols → NO color change with FeCl_3 (negative test)
Enols (like acetoacetic ester) also give positive FeCl_3 test — not unique to phenols
Distinguishing Tests Summary
A consolidated comparison of all key tests used to distinguish between different alcohols, phenols, and ethers in NEET examinations.