Alcohols, Phenols and Ethers • Topic 1 of 3

Alcohols

An alcohol contains one or more hydroxyl (–OH) groups bonded to an sp3 carbon of an alkyl or substituted alkyl group. Depending on the number of –OH groups they are mono-, di- or tri-hydric; depending on the carbon bearing the –OH they are primary (1°), secondary (2°) or tertiary (3°). In ethanol CH3CH2OH the carbon carries one other carbon, so it is 1°; in propan-2-ol (CH3)2CHOH it is 2°; in 2-methylpropan-2-ol (CH3)3COH it is 3°.

IUPAC nomenclature

Replace the –e of the parent alkane with –ol and number the chain so the carbon bearing –OH gets the lowest locant: CH3CH2CH2OH is propan-1-ol, CH3CH(OH)CH3 is propan-2-ol, and HOCH2CH2OH is ethane-1,2-diol (glycol).

Preparation

From alkenes. Acid-catalysed hydration adds water by Markovnikov's rule, giving the more substituted alcohol via a carbocation (so rearrangement is possible). Oxymercuration–demercuration (Hg(OAc)2/H2O then NaBH4) gives the Markovnikov alcohol with no rearrangement. Hydroboration–oxidation (B2H6 then H2O2/OH–) adds water with anti-Markovnikov regiochemistry, placing –OH on the less substituted carbon.

From carbonyl compounds. Aldehydes reduce to 1° alcohols and ketones to 2° alcohols (H2/Ni, or NaBH4/LiAlH4). Carboxylic acids and esters need the stronger LiAlH4 to give 1° alcohols.

From Grignard reagents. RMgX adds to a carbonyl and aqueous work-up gives the alcohol: methanal → 1°, other aldehydes → 2°, ketones → 3° alcohol.

Physical properties

The polar –OH group lets alcohols form intermolecular hydrogen bonds, so their boiling points are far higher than those of comparable hydrocarbons or ethers (ethanol b.p. 78°C vs dimethyl ether –24°C). Lower alcohols are miscible with water because they H-bond to it; solubility falls as the alkyl chain grows.

Chemical reactions

Acidity. The O–H bond is weakly acidic; alcohols react with active metals (Na, K) to release H2 and form alkoxides. Acidity order is 1° > 2° > 3° (electron-releasing alkyl groups destabilise the alkoxide).

With HX / Lucas test. Alcohols give alkyl halides with HX; reactivity 3° > 2° > 1°. The Lucas reagent (conc. HCl + anhyd. ZnCl2) distinguishes them: 3° gives immediate turbidity, 2° in about 5 min, 1° no turbidity at room temperature. Esterification with carboxylic acids gives esters; dehydration with conc. H2SO4 gives alkenes (ease 3° > 2° > 1°); oxidation converts 1° → aldehyde → acid and 2° → ketone, while 3° resists oxidation.

Distinguishing 1°, 2° and 3° alcohols
Test / propertyPrimary (1°)Secondary (2°)Tertiary (3°)
Lucas reagent (HCl + ZnCl2)No turbidity at room temp.Turbidity in ~5 minImmediate turbidity
Mild oxidation (Cu/573 K or KMnO4)Aldehyde → carboxylic acidKetone (no further C–C break)No reaction (resists)
Victor Meyer / structure–CH2OH (one C on C–OH)>CHOH (two C)–C(OH) with three C
Acidity of O–HHighestIntermediateLowest
Solved Examples
1
Worked Example
Give the IUPAC name and class (1°/2°/3°) of (CH3)3COH.
Solution
  1. Longest chain through the C–OH carbon is propane (3 C) with a methyl substituent.
  2. The carbon bearing –OH is C-2 and carries three carbon groups, so it is tertiary.
  3. Name: 2-methylpropan-2-ol.

Answer: 2-methylpropan-2-ol; tertiary (3°) alcohol.

2
Worked Example
Which alcohol forms when propene undergoes (a) acid hydration and (b) hydroboration–oxidation?
Solution
  1. Acid hydration follows Markovnikov: H+ adds to give the more stable 2° carbocation, –OH lands on C-2.
  2. Product (a): propan-2-ol, CH3CH(OH)CH3.
  3. Hydroboration–oxidation is anti-Markovnikov: –OH goes to the terminal (less substituted) carbon.
  4. Product (b): propan-1-ol, CH3CH2CH2OH.

Answer: (a) propan-2-ol; (b) propan-1-ol.

3
Worked Example
Predict the product when ethanal (CH3CHO) reacts with CH3MgBr followed by aqueous acid.
Solution
  1. The Grignard carbanion CH3– adds to the carbonyl carbon of ethanal.
  2. This gives an alkoxide CH3CH(OMgBr)CH3.
  3. Aqueous acid protonates the alkoxide to the alcohol.
  4. An aldehyde + Grignard gives a 2° alcohol.

Answer: propan-2-ol, CH3CH(OH)CH3 (a secondary alcohol).

4
Worked Example
Arrange ethanol, propan-2-ol and 2-methylpropan-2-ol in order of reactivity toward Lucas reagent and explain.
Solution
  1. Lucas reaction proceeds via a carbocation (SN1); rate depends on cation stability.
  2. 3° cation is most stable, 1° least.
  3. Hence 2-methylpropan-2-ol (3°) reacts instantly, propan-2-ol (2°) in minutes, ethanol (1°) not at room temperature.

Answer: 2-methylpropan-2-ol > propan-2-ol > ethanol.

5
Worked Example
Why does ethanol (b.p. 78°C) boil much higher than its isomeric ether dimethyl ether (b.p. –24°C)?
Solution
  1. Both have formula C2H6O and similar molar mass.
  2. Ethanol has an O–H group and forms intermolecular hydrogen bonds.
  3. Dimethyl ether has no O–H, so it cannot H-bond (only weak dipole–dipole forces).
  4. Breaking the H-bond network in ethanol needs more energy, raising its boiling point.

Answer: Extensive intermolecular hydrogen bonding in ethanol, absent in the ether, gives it the much higher boiling point.

6
Worked Example
Identify the products of oxidation of (a) propan-1-ol and (b) propan-2-ol with acidified KMnO4.
Solution
  1. Propan-1-ol is 1°: oxidises first to propanal, then to propanoic acid.
  2. With strong oxidant (acidified KMnO4) it goes all the way to the acid.
  3. Propan-2-ol is 2°: oxidises to the ketone propan-2-one (acetone); it cannot oxidise further without C–C cleavage.

Answer: (a) propanoic acid CH3CH2COOH; (b) propan-2-one (acetone) CH3COCH3.

Key Points

  • Alcohols have –OH on sp3 carbon; classed as 1°/2°/3° by the number of carbons on the C–OH carbon; IUPAC = alkan-n-ol.
  • Preparations: alkene hydration (Markovnikov), oxymercuration (Markovnikov, no rearrangement), hydroboration–oxidation (anti-Markovnikov), carbonyl reduction, and Grignard addition (HCHO→1°, RCHO→2°, ketone→3°).
  • Hydrogen bonding gives high boiling points and water-miscibility of lower members; both fall as the alkyl chain lengthens.
  • Reactions: acidic O–H (Na → alkoxide + H2), with HX (3°>2°>1°), esterification, dehydration to alkenes (3°>2°>1°), oxidation (1°→aldehyde→acid, 2°→ketone, 3° resists).
  • Lucas test distinguishes alcohols: 3° immediate turbidity, 2° in ~5 min, 1° none at room temperature; acidity order 1°>2°>3°.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Hydroboration–oxidation of but-1-ene gives mainly:
Explanation: Hydroboration–oxidation is anti-Markovnikov, so –OH adds to the terminal carbon giving butan-1-ol.
Q2.The correct order of reactivity toward Lucas reagent is:
Explanation: The Lucas (SN1) reaction goes through a carbocation, most stable for 3°, so 3° > 2° > 1°.
Q3.A Grignard reagent reacting with a ketone, after work-up, gives a:
Explanation: Addition of RMgX to a ketone places three carbon groups on the new C–OH carbon, giving a tertiary alcohol.
Q4.Which alcohol does NOT give a ketone or aldehyde on oxidation?
Explanation: 2-Methylpropan-2-ol is tertiary; it has no H on the C–OH carbon and resists oxidation under normal conditions.
Practice more MCQs → 📝 Assignment · 30 marks