Alcohols, Phenols, and Ethers | CBSE

sources:

• text: Standard textbook reference

Alcohols contain $-\text{OH}$ bonded to an sp$^3$ carbon. Phenols have $-\text{OH}$ bonded to an aromatic ring. Ethers have the structure R-O-R’.

• Classification: primary ($1°$), secondary ($2°$), tertiary ($3°$) based on carbon bearing $-\text{OH}$
• Acidity: phenols are more acidic than alcohols (resonance stabilization of phenoxide)
• Acidity order: water < alcohols < phenols < carboxylic acids
• Dehydration: alcohols $\xrightarrow{\text{H}_2\text{SO}_4, \Delta}$ alkenes (Saytzeff’s rule)
• Oxidation: primary alcohols $\rightarrow$ aldehydes $\rightarrow$ carboxylic acids
• Oxidation: secondary alcohols $\rightarrow$ ketones
• Tertiary alcohols resist oxidation (no H on the carbon bearing $-\text{OH}$)
• Williamson synthesis: RONa + R’X $\rightarrow$ ROR’ (for ethers)
• Phenol reactions: electrophilic aromatic substitution (bromination, nitration)

Problem: Arrange ethanol, phenol, and water in decreasing order of acidity.

Solution:

Acidity is determined by the stability of the conjugate base.

• Phenol: Phenoxide ion is stabilized by resonance (negative charge delocalized into the ring). $pK_a = 10$

• Water: Hydroxide ion has no resonance stabilization. $pK_a = 15.7$

• Ethanol: Ethoxide ion is destabilized by the electron-donating ethyl group. $pK_a = 16$

Decreasing acidity: phenol > water > ethanol

Common mistake: Assuming alcohols are more acidic than water. In reality, water is slightly more acidic than most simple alcohols.

Problem: Write the products of dehydration of 2-methylpropan-2-ol with concentrated H$_2$SO$_4$.

Solution:

2-methylpropan-2-ol is a tertiary alcohol. Dehydration proceeds via an E1 mechanism:

$\text{(CH}_3)_3\text{COH} \xrightarrow{\text{H}_2\text{SO}_4, \Delta} \text{(CH}_3)_2\text{C}=\text{CH}_2 + \text{H}_2\text{O}$

Product: 2-methylpropene (only one product possible, Saytzeff’s rule gives the same result).

Tertiary alcohols dehydrate most efficiently (most stable carbocation intermediate).

Common mistake: Using SN2 conditions for tertiary alcohols. Tertiary alcohols undergo E1/SN1, not E2/SN2.

Problem: How would you prepare diethyl ether using Williamson synthesis?

Solution:

Williamson synthesis: alkoxide + primary alkyl halide $\rightarrow$ ether

$\text{CH}_3\text{CH}_2\text{O}^-\text{Na}^+ + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 + \text{NaBr}$

Sodium ethoxide + bromoethane $\rightarrow$ diethyl ether + NaBr

The alkyl halide must be primary (or methyl) to avoid elimination. Using a tertiary halide gives elimination instead of substitution.

Common mistake: Using a tertiary alkyl halide with a strong base like alkoxide. This gives elimination (E2) rather than substitution (SN2).

1. Write the reaction of phenol with bromine water.
2. How would you convert ethanol to ethoxyethane?
3. Explain why phenol is more acidic than cyclohexanol.

Alcohols are solvents (ethanol, methanol), fuels (methanol), and precursors to many chemicals. Phenols are used in disinfectants, plastics (BPA), and pharmaceuticals (aspirin).

• Acidity: phenol > water > alcohol (as a rule)
• Dehydration: follows Saytzeff’s rule, tertiary > secondary > primary
• Oxidation: primary $\rightarrow$ aldehyde $\rightarrow$ acid; secondary $\rightarrow$ ketone
• Williamson synthesis: primary halide + alkoxide (avoid elimination)
• Phenol: activates ring for electrophilic substitution (ortho/para directing)

Problem: What are the products when (a) propan-1-ol and (b) propan-2-ol are heated with acidified potassium dichromate?

Solution:

(a) Propan-1-ol is a primary alcohol. Oxidation proceeds in two steps:

First oxidation: $3\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 \rightarrow 3\text{CH}_3\text{CH}_2\text{CHO} + \text{Cr}_2(\text{SO}_4)_3 + \text{K}_2\text{SO}_4 + 7\text{H}_2\text{O}$

Product: propanal (an aldehyde)

Further oxidation: $\text{CH}_3\text{CH}_2\text{CHO} \xrightarrow{[\text{O}]} \text{CH}_3\text{CH}_2\text{COOH}$

Product: propanoic acid (a carboxylic acid)

(b) Propan-2-ol is a secondary alcohol. Oxidation gives a ketone:

$\text{CH}_3\text{CH(OH)CH}_3 \xrightarrow{[\text{O}]} \text{CH}_3\text{COCH}_3$

Product: propanone (acetone, a ketone)

Common mistake: Forgetting that primary alcohols can be oxidised further to carboxylic acids. To stop at the aldehyde stage, use PCC (pyridinium chlorochromate) as the oxidising agent.

Problem: Write the equation for the reaction between ethanol and ethanoic acid in the presence of concentrated sulfuric acid.

Solution:

This is a Fischer esterification — a condensation reaction between a carboxylic acid and an alcohol:

$\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \xrightleftharpoons[\Delta]{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$

The product is ethyl ethanoate (an ester) with a fruity smell.

The reaction is reversible. Concentrated $\text{H}_2\text{SO}_4$ acts as both a catalyst and a dehydrating agent, shifting the equilibrium towards the ester.

Common mistake: Forgetting that the reaction is reversible. Using excess alcohol or removing water drives the equilibrium towards ester formation (Le Chatelier’s principle).

Problem: Write equations showing the reaction of phenol with (a) NaOH and (b) Na. Explain why phenol does not react with $\text{NaHCO}_3$.

Solution:

(a) Phenol reacts with NaOH to form sodium phenoxide (a salt): $\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{H}_2\text{O}$

(b) Phenol reacts with sodium metal: $2\text{C}_6\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_6\text{H}_5\text{ONa} + \text{H}_2$

(c) Phenol does not react with $\text{NaHCO}_3$ because phenol is a weaker acid than carbonic acid ($\text{H}_2\text{CO}_3$). The $\text{p}K_a$ of phenol is 10, while $\text{p}K_a$ of $\text{H}_2\text{CO}_3$ is 6.4. A weaker acid cannot displace a stronger acid from its salt.

Common mistake: Assuming that because phenol is more acidic than alcohols, it behaves like a carboxylic acid. Phenol is still a very weak acid ($pK_a = 10$) compared to carboxylic acids ($pK_a \approx 4$-$5$).

1. For acidity questions, always compare the stability of the conjugate base (phenoxide vs. alkoxide vs. hydroxide).
2. In Williamson synthesis, always use a primary alkyl halide to minimise elimination.
3. Oxidation of primary alcohols can stop at the aldehyde stage using PCC; without it, the carboxylic acid forms.
4. Phenol undergoes electrophilic aromatic substitution at the ortho and para positions due to activation by the $-\text{OH}$ group.
5. Esterification is reversible; use Le Chatelier’s principle to predict the effect of changing conditions.
6. Dehydration follows Saytzeff’s rule: the more substituted alkene is the major product.
7. Tertiary alcohols dehydrate most efficiently (most stable carbocation); primary alcohols require harsher conditions.
8. Remember the Lucas test: tertiary alcohols react immediately, secondary within 5 minutes, primary do not react at room temperature.

The -OH group is a double-edged sword: Think of the hydroxyl group as a magnetic pull on electrons. In alcohols, it pulls electron density toward itself, making the O-H bond polar and the carbon slightly positive. This polarity drives most of alcohol chemistry — nucleophiles attack the carbon, acids efficiently donate the proton, and the oxygen can participate in hydrogen bonding.

Why it matters: Alcohols are the bridge between simple organic molecules and complex pharmaceuticals. The ability to oxidize, dehydrate, and form ethers makes them essential intermediates in synthesis. Understanding acidity trends helps predict which reactions will work.

The key insight: Acidity is really about conjugate base stability — the more stable the anion after losing a proton, the stronger the acid.

Confusing the acidity order of alcohols, water, and phenols. Many students assume alcohols are more acidic than water because the -OH group is present in both. In reality, water (pKa = 15.7) is slightly more acidic than most simple alcohols (pKa = 16-18) because the ethoxide ion is destabilised by the electron-donating alkyl group, making the conjugate base less stable.

Using a tertiary alkyl halide in Williamson synthesis. Williamson synthesis requires a primary (or methyl) alkyl halide to favour SN2 substitution over E2 elimination. When a tertiary halide reacts with a strong base like alkoxide, elimination dominates and no ether is formed. Always pair a primary halide with the alkoxide for successful ether synthesis.

Assuming phenol reacts with sodium bicarbonate. Phenol (pKa = 10) is a weaker acid than carbonic acid (pKa = 6.4), so it cannot displace CO2 from NaHCO3. Only carboxylic acids (pKa = 4-5) are acidic enough to react with sodium bicarbonate. This is a reliable test to distinguish phenols from carboxylic acids.

• Organic Chemistry Fundamentals — Alcohols are synthesised from haloalkanes via nucleophilic substitution, connecting the two functional group families.
• Carboxylic Acids — Oxidation of alcohols produces carboxylic acids, linking alcohol chemistry to the broader oxidation-reduction framework.
• Chemical Kinetics — Reaction rates of alcohol substitution and elimination depend on concentration and temperature, connecting organic reactions to kinetics.