Alcohols, Phenols & Ethers: Prep & Basics
Module 1 | CBSE Class 12 Chemistry | Organic Chemistry
1. Introduction & Classification
Alcohols and phenols are formed when a hydrogen atom in a hydrocarbon (aliphatic and aromatic respectively) is replaced by an -OH (hydroxyl) group. The substitution of a hydrogen atom in a hydrocarbon by an alkoxy or aryloxy group (R-O/Ar-O) yields ethers.
Classification Based on Number of -OH Groups
Alcohols and phenols may be classified as mono-, di-, tri- or polyhydric depending on whether they contain one, two, three, or many hydroxyl groups respectively.
Classification Based on Hybridization of C-OH Bond
- Compounds containing sp3 C-OH bond:
- Primary (1°), Secondary (2°), Tertiary (3°) Alcohols: The -OH group is attached to primary, secondary, and tertiary carbon atoms, respectively.
- Allylic Alcohols: The -OH group is attached to an sp3 hybridized carbon adjacent to a carbon-carbon double bond (CH2=CH-CH2-OH).
- Benzylic Alcohols: The -OH group is attached to an sp3 hybridized carbon next to an aromatic ring (C6H5-CH2-OH).
- Compounds containing sp2 C-OH bond:
- Vinylic Alcohols: The -OH group is attached to a carbon-carbon double bond (CH2=CH-OH).
- Phenols: The -OH group is attached directly to the sp2 carbon of a benzene ring.
2. Nomenclature (Alcohols, Phenols & Ethers)
Alcohols: The IUPAC name is derived from the name of the corresponding alkane by substituting the suffix 'e' with 'ol'. The longest carbon chain containing the -OH group is selected, and numbering starts from the end nearest to the -OH group.
Phenols: The simplest hydroxy derivative of benzene is phenol. It is both a common and IUPAC name. Substituted phenols are named as derivatives of phenol (ortho, meta, para or 1,2 / 1,3 / 1,4).
Ethers: Common names are derived by naming the two alkyl/aryl groups alphabetically and adding the word 'ether'. IUPAC names treat ethers as hydrocarbon derivatives where the larger group is the parent alkane, and the smaller group along with the oxygen forms an alkoxy substituent (e.g., Methoxyethane).
| Structure | Common Name | IUPAC Name |
|---|---|---|
| CH3-CH(OH)-CH3 | Isopropyl alcohol | Propan-2-ol |
| HO-CH2-CH2-OH | Ethylene glycol | Ethane-1,2-diol |
| C6H5-OCH3 | Anisole | Methoxybenzene |
| o-CH3-C6H4-OH | o-Cresol | 2-Methylphenol |
3. Structure of Functional Groups (Highly Tested)
The structural parameters (bond angles and bond lengths) of alcohols, phenols, and ethers are critical due to differences in hybridization and resonance.
- Methanol (Alcohol): The C-O-H bond angle is slightly less than the regular tetrahedral angle (109°28'). It is 108.9°. This is due to the repulsion between the two unshared electron pairs (lone pairs) on the oxygen atom.
- Methoxymethane (Ether): The C-O-C bond angle is 111.7°. This angle is slightly greater than the tetrahedral angle due to the repulsive interaction (steric hindrance) between the two bulky alkyl groups attached to oxygen.
The C-O bond length in phenol (136 pm) is noticeably shorter than that in methanol (142 pm). This is due to:
(i) The partial double bond character of the C-O bond resulting from the conjugation of the unshared electron pair of oxygen with the aromatic ring.
(ii) The oxygen atom is attached to an sp2 hybridized carbon in phenol (which is more electronegative and smaller), whereas in methanol, it is attached to an sp3 hybridized carbon.
4. Preparation of Alcohols
4.1 From Alkenes
A. Acid Catalyzed Hydration
Alkenes react with water in the presence of an acid catalyst (like conc. H2SO4) to form alcohols. In unsymmetrical alkenes, the addition reaction takes place in accordance with Markovnikov’s rule.
Mechanism (3 Steps):
Step 1: Protonation of alkene to form carbocation by electrophilic attack of H3O+.
Step 2: Nucleophilic attack of water on carbocation.
Step 3: Deprotonation to form an alcohol.
B. Hydroboration-Oxidation (Very Important)
Diborane (B2H6) reacts with alkenes to give trialkyl boranes as an addition product. This is then oxidized to alcohol by hydrogen peroxide (H2O2) in the presence of aqueous sodium hydroxide.
(CH3-CH2-CH2)3B + 3H2O2 →(OH-/H2O) 3CH3-CH2-CH2-OH + B(OH)3
4.2 From Carbonyl Compounds
A. Reduction of Aldehydes and Ketones
Aldehydes are reduced to primary alcohols, and ketones are reduced to secondary alcohols. Reducing agents include finely divided metals (catalytic hydrogenation with Pt/Pd/Ni) or chemical reagents like Sodium borohydride (NaBH4) or Lithium aluminium hydride (LiAlH4).
R-CHO + H2 →(Pd) R-CH2-OH (1° Alcohol)
R-CO-R' + NaBH4 → R-CH(OH)-R' (2° Alcohol)
B. Reduction of Carboxylic Acids and Esters
Carboxylic acids are reduced to primary alcohols by the powerful reducing agent LiAlH4. However, LiAlH4 is an expensive reagent. Commercially, acids are first converted to esters, followed by catalytic hydrogenation.
R-COOH →(LiAlH4, then H2O) R-CH2OH
4.3 From Grignard Reagents
Alcohols are produced by the reaction of Grignard reagents (R-Mg-X) with aldehydes and ketones. The first step is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct. The adduct is then hydrolyzed to yield an alcohol.
1. Formaldehyde (Methanal, HCHO) + Grignard Reagent → Primary (1°) Alcohol.
2. Any other Aldehyde (R-CHO) + Grignard Reagent → Secondary (2°) Alcohol.
3. Ketones (R-CO-R') + Grignard Reagent → Tertiary (3°) Alcohol.
Example: HCHO + CH3MgBr → HCH(OMgBr)CH3 →(H2O) CH3CH2OH + Mg(OH)Br
5. Preparation of Phenols
Phenol (Carbolic acid) was first isolated from coal tar. Today, it is synthetically prepared from benzene derivatives.
5.1 From Haloarenes (Dow's Process)
Chlorobenzene is fused with NaOH at 623 K and 300 atmospheric pressure. Phenol is obtained by acidification of the sodium phenoxide produced.
5.2 From Diazonium Salts
Aniline is treated with nitrous acid (NaNO2 + HCl) at 273-278 K to form benzene diazonium chloride. This diazonium salt is extremely reactive and is converted to phenol simply by warming with water or treating with dilute acids.
C6H5N2+Cl- + H2O →(Warm) C6H5OH + N2 + HCl
5.3 From Cumene (Commercial Method - Board Favorite)
Most of the worldwide production of phenol is from cumene (isopropylbenzene). Cumene is oxidized in the presence of air to form cumene hydroperoxide. This intermediate is treated with dilute acid to yield phenol and acetone.
A large quantity of Acetone, which is a highly valuable solvent and by-product, is produced in this reaction. Thus, this process yields two highly important commercial chemicals simultaneously.
Cumene Hydroperoxide →(H+, H2O) C6H5OH (Phenol) + CH3COCH3 (Acetone)
6. NCERT Solved Examples (Step-by-Step)
NCERT Example 11.2: Give the structures and IUPAC names of the products expected from the following reactions:
(a) Catalytic reduction of butanal.
(b) Hydration of propene in the presence of dilute sulphuric acid.
(c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis.
(a) Butanal (CH3CH2CH2CHO) on catalytic reduction (H2/Pd) gives a primary alcohol.
Product: CH3CH2CH2CH2OH (Butan-1-ol).
(b) Hydration of propene (CH3CH=CH2) with dilute acid follows Markovnikov's rule. The -OH group attaches to the middle carbon.
Product: CH3CH(OH)CH3 (Propan-2-ol).
(c) Propanone (Acetone, CH3COCH3) is a ketone. Reaction with a Grignard reagent (CH3MgBr) yields a tertiary alcohol.
Product: (CH3)3C-OH (2-Methylpropan-2-ol).
7. Previous Year Questions (PYQs) & Exhaustive Question Bank
Part A: Conceptual (1-2 Marks)
Q1. Why is the C-O-C bond angle in ethers slightly greater than the tetrahedral angle (109.5°)?
Q2. Why is the C-O bond length in phenol shorter than that in methanol?
1. Resonance: The lone pair of electrons on the oxygen atom is in conjugation with the aromatic ring, giving the C-O bond a partial double bond character.
2. Hybridization: The carbon atom in phenol is sp2 hybridized (more s-character, more electronegative, smaller size), while in methanol it is sp3 hybridized.
Part B: Assertion-Reason Type (1 Mark)
Q3. Assertion (A): Addition of water to an alkene via hydroboration-oxidation yields anti-Markovnikov alcohol.
Reason (R): Diborane adds to the alkene such that the boron atom attaches to the more substituted carbon atom.
Hydroboration-oxidation indeed yields the anti-Markovnikov product. However, the reason is that Boron (being less electronegative than Hydrogen) acts as the electrophile and attaches to the less substituted carbon atom (the one with more hydrogens) due to less steric hindrance, while hydrogen attaches to the more substituted carbon.
Part C: Synthesis and Reactions (3 Marks)
Q4. How are the following conversions carried out?
(i) Propene to Propan-1-ol
(ii) Aniline to Phenol
(iii) Cumene to Phenol
(i) Propene to Propan-1-ol: This requires Anti-Markovnikov addition of water. It is done via Hydroboration-Oxidation. React propene with diborane (B2H6) followed by oxidation with H2O2 in an alkaline medium (NaOH).
(ii) Aniline to Phenol: First, diazotize aniline by treating it with NaNO2 + HCl at 0-5°C to form Benzene diazonium chloride. Then, warm it with water to yield phenol and nitrogen gas.
(iii) Cumene to Phenol: Oxidize cumene with O2 (air) to form Cumene hydroperoxide. Treat this intermediate with dilute acid (H+/H2O) to yield phenol and acetone as a by-product.
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