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Unit 7: Alcohols, Phenols and Ethers

7.1 Classification and Nomenclature

  • Alcohols: Compounds containing one or more hydroxyl (-OH) groups attached to an aliphatic carbon atom. Classified as primary (\(1^\circ\)), secondary (\(2^\circ\)), or tertiary (\(3^\circ\)) depending on the carbon attached to the -OH group.
  • Phenols: Compounds containing an -OH group directly attached to a benzene ring.
  • Ethers: Compounds with an oxygen atom bonded to two alkyl or aryl groups (R-O-R’ or Ar-O-Ar’).

7.2 Alcohols

Methods of Preparation

  1. From Alkenes:
    • By acid catalyzed hydration: Alkenes react with water in the presence of acid as a catalyst to form alcohols (Markovnikov’s addition).
    • By hydroboration-oxidation: Diborane (\(B_2H_6\)) reacts with alkenes followed by oxidation with \(H_2O_2\) in aqueous sodium hydroxide. Resulting alcohol appears as anti-Markovnikov addition of water.
  2. From Carbonyl Compounds:
    • By reduction of aldehydes (to \(1^\circ\) alcohols) and ketones (to \(2^\circ\) alcohols) using \(LiAlH_4\), \(NaBH_4\), or catalytic hydrogenation.
    • By reduction of carboxylic acids and esters.
  3. From Grignard Reagents: Reaction of aldehydes and ketones with RMgX followed by hydrolysis.
    • Formaldehyde yields \(1^\circ\) alcohols.
    • Other aldehydes yield \(2^\circ\) alcohols.
    • Ketones yield \(3^\circ\) alcohols.

Identification of Primary, Secondary, and Tertiary Alcohols (Lucas Test)

The reaction of alcohols with Lucas reagent (conc. HCl + anhydrous \(ZnCl_2\)) produces alkyl chlorides, which are insoluble in water and cause turbidity.

  • \(3^\circ\) Alcohols: Produce turbidity immediately at room temperature.
  • \(2^\circ\) Alcohols: Produce turbidity after about 5 minutes.
  • \(1^\circ\) Alcohols: Do not produce turbidity at room temperature (requires heating).

Mechanism of Dehydration of Alcohols

Heating an alcohol with concentrated sulfuric acid at 443 K results in dehydration to form an alkene. Step 1: Protonation of alcohol. \[ CH_3-CH_2-\ddot{O}H + H^+ \rightleftharpoons CH_3-CH_2-\underset{+}{\ddot{O}}H_2 \quad (\text{Fast}) \] Step 2: Formation of carbocation: Cleavage of C-O bond. \[ CH_3-CH_2-\underset{+}{\ddot{O}}H_2 \rightarrow CH_3-\overset{+}{CH_2} + H_2O \quad (\text{Slow / Rate determining}) \] Step 3: Formation of ethene by elimination of a proton. \[ CH_3-\overset{+}{CH_2} \rightarrow CH_2=CH_2 + H^+ \quad (\text{Fast}) \]

7.3 Phenols

Methods of Preparation

  1. From haloarenes (Dow’s Process): Chlorobenzene is fused with NaOH at 623K and 300 atm, followed by acidification.
  2. From benzene sulphonic acid: Fused with molten NaOH followed by acidification.
  3. From diazonium salts: Warming aqueous diazonium salt solution.
  4. From cumene: Oxidation of cumene (isopropylbenzene) in air followed by acid hydrolysis gives phenol and acetone.

Acidic Nature of Phenol

Phenol is a stronger acid than aliphatic alcohols. Reason: The phenoxide ion left after the release of a proton is stabilized by resonance (delocalization of the negative charge over the benzene ring). In contrast, alkoxide ions are not resonance stabilized and instead destabilized by the +I effect of the alkyl group.

O⁻ ↔ O :- ↔ O :- ↔ O :- ↔ O⁻ Resonance structures of Phenoxide Ion

Effect of substituents on acidity:

  • Electron Withdrawing Groups (EWG) like \(-NO_2\) increase the acidity by stabilizing the phenoxide ion through resonance/inductive effect.
  • Electron Donating Groups (EDG) like an alkyl group decrease the acidity.

Electrophilic Substitution Reactions

The -OH group is a highly activating and ortho-para directing group.

  • Nitration: Dilute \(HNO_3\) yields a mixture of ortho and para nitrophenols. Concentrated \(HNO_3\) with conc. \(H_2SO_4\) yields 2,4,6-trinitrophenol (picric acid).
  • Halogenation: With aqueous bromine, it gives a white precipitate of 2,4,6-tribromophenol. With bromine in \(CS_2\) at low temperature, it gives ortho and para bromophenols.
  • Kolbe’s Reaction: Phenoxide ion treated with \(CO_2\) followed by acidification yields salicylic acid (2-hydroxybenzoic acid).
  • Reimer-Tiemann Reaction: Treating phenol with chloroform in the presence of aq. NaOH yields salicylaldehyde.

7.4 Ethers

  • Preparation: Dehydration of alcohols (at 413 K) or Williamson’s synthesis (reaction of an alkyl halide with sodium alkoxide).
  • Cleavage: Ethers are cleaved by strong acids like HI to give alcohols and alkyl iodides. Cleavage of the C-O bond in mixed ethers generally happens such that the halogen attaches to the smaller/less sterically hindered alkyl group for primary/secondary groups.

Competency-Based Questions (CBQs)

Q1. (CBSE 2021) How will you distinguish between propan-1-ol and propan-2-ol?


Answer: Propan-1-ol is a primary (\(1^\circ\)) alcohol, whereas propan-2-ol is a secondary (\(2^\circ\)) alcohol. They can be distinguished using the Lucas Test.

  • Add a few drops of Lucas reagent (anhydrous \(ZnCl_2\) in concentrated HCl) to the unknown alcohol.
  • If turbidity (cloudiness) appears after about 5 minutes, it is propan-2-ol. \[ CH_3-CH(OH)-CH_3 + HCl \xrightarrow{ZnCl_2} CH_3-CH(Cl)-CH_3 \downarrow (\text{turbidity in 5 min}) + H_2O \]
  • If no turbidity appears at room temperature (even after 10-15 minutes), it is propan-1-ol. \[ CH_3-CH_2-CH_2-OH + HCl \xrightarrow{ZnCl_2} \text{No reaction at room temp.} \]

Q2. (Sample Paper 2024) Arrange the following compounds in decreasing order of their acid strength: Phenol, 4-nitrophenol, 3-nitrophenol, and 4-methylphenol. Give reason.


Answer: Decreasing order of acid strength: 4-nitrophenol > 3-nitrophenol > Phenol > 4-methylphenol

Reason:

  • 4-nitrophenoI: The \(-NO_2\) group is a strong electron-withdrawing group (-R and -I effects). At the para position, it can stabilize the phenoxide ion exceptionally well through extended conjugation (resonance). Thus, it is the strongest acid.
  • 3-nitrophenol: The \(-NO_2\) group at the meta position also exerts a localized electron-withdrawing (-I) effect (but no resonance or -R effect onto the phenoxide oxygen). Hence, it increases acidity compared to phenol, but not as much as in 4-nitrophenol.
  • Phenol: The reference compound.
  • 4-methylphenol (p-Cresol): The methyl group (\(-CH_3\)) is an electron-donating group (+I and hyperconjugation). It intensifies the negative charge on the phenoxide ion, destabilizing it and decreasing the acidic strength below that of phenol.

Q3. (CBSE 2019) Anisole (methoxybenzene) on reaction with HI gives phenol and methyl iodide, and not iodobenzene and methanol. Explain why.


Answer: Anisole has the structure \(C_6H_5-O-CH_3\). When it reacts with HI, it first undergoes protonation to form a methylphenyl oxonium ion: \[ C_6H_5-\ddot{O}-CH_3 + HI \rightarrow C_6H_5-\underset{+}{\ddot{O}}(H)-CH_3 + I^- \]

The bond between the oxygen atom and the phenyl ring (\(C_6H_5-O\)) is much stronger and harder to break because of the partial double bond character due to resonance and the \(sp^2\) hybridized nature of the carbon of the benzene ring. In contrast, the bond between oxygen and the methyl group (\(O-CH_3\)) is a weaker single bond (between \(A \) sp³ carbon and Oxygen). Therefore, the nucleophile (\(I^-\) ion) attacks the less sterically hindered \(CH_3\) group (via \(S_N2\) mechanism), cleaving the \(O-CH_3\) bond to give methyl iodide (\(CH_3I\)) and phenol (\(C_6H_5OH\)).