Alcohols — Concepts, Formulas & Examples

Alcohols are organic compounds with an OH group attached to a carbon atom. They are the gateway to much of organic chemistry. CBSE Class 12 and NEET test preparation, reactions and differentiation between primary, secondary and tertiary alcohols.

Core Concepts

Classification

Primary (OH on carbon with one other carbon), secondary (OH on carbon with two other carbons), tertiary (OH on carbon with three other carbons). Each class reacts differently.

Methanol (CH $_3$ OH) is a special case — the OH is on a carbon bonded to no other carbons, but it is still classified as primary. The classification matters because the reactivity pattern changes dramatically with each class — oxidation products differ, substitution mechanisms differ, and even the speed of dehydration differs.

Preparation

From alkenes by hydration. From haloalkanes by SN reaction with OH $^-$ . From aldehydes by reduction with NaBH $_4$ or LiAlH $_4$ — gives primary. From ketones by reduction — gives secondary.

Key preparation methods in detail:

Method	Reagent	Product	Notes
Acid-catalysed hydration	H $_2$ O + H $_2$ SO $_4$	Markovnikov alcohol	Follows Markovnikov’s rule
Hydroboration-oxidation	BH $_3$ , then H $_2$ O $_2$ /NaOH	Anti-Markovnikov alcohol	Gives primary from terminal alkenes
Grignard reaction	RMgBr + carbonyl, then H $_3$ O $^+$	Primary/secondary/tertiary	Most versatile method
Reduction of aldehyde	NaBH $_4$ or LiAlH $_4$	Primary alcohol	Mild, selective
Reduction of ketone	NaBH $_4$ or LiAlH $_4$	Secondary alcohol	Same reagents as aldehyde
Reduction of ester	LiAlH $_4$ (not NaBH $_4$ )	Primary alcohol	NaBH $_4$ is too mild for esters

$\text{R-MgBr} + \text{HCHO} \xrightarrow{\text{H}_3\text{O}^+} \text{R-CH}_2\text{OH}$ (primary)

$\text{R-MgBr} + \text{R'CHO} \xrightarrow{\text{H}_3\text{O}^+} \text{R-CH(OH)-R'}$ (secondary)

$\text{R-MgBr} + \text{R'}_2\text{CO} \xrightarrow{\text{H}_3\text{O}^+} \text{R-C(OH)(R')}_2$ (tertiary)

Physical properties

H-bonding gives high boiling points for their molecular weight. Soluble in water up to about C4, then solubility drops. Viscosity increases with more OH groups (glycerol is very viscous).

Boiling point comparison:

Ethanol (C $_2$ H $_5$ OH, MW 46) boils at 78°C. Dimethyl ether (CH $_3$ OCH $_3$ , MW 46) boils at -25°C. Same molecular weight, but ethanol can form hydrogen bonds (O-H…O) while the ether cannot (no O-H bond). This 103°C difference is entirely due to hydrogen bonding.

Solubility trend: As the hydrocarbon chain grows, the non-polar part dominates and the molecule becomes less water-soluble. Methanol, ethanol, and propanol are fully miscible with water. Butanol is partially soluble. Pentanol and higher are essentially insoluble.

Reactions of alcohols

With Na metal → alkoxide and H $_2$ gas. With HCl/HBr → haloalkane. Oxidation — primary to aldehyde then acid; secondary to ketone; tertiary resists oxidation. Dehydration to alkene with conc H $_2$ SO $_4$ at 170°C.

Reaction with sodium metal:

2\text{R-OH} + 2\text{Na} \rightarrow 2\text{R-ONa} + \text{H}_2 \uparrow

This reaction is used to test for the presence of an OH group. The vigour decreases from primary to tertiary because steric hindrance makes it harder for Na to approach the OH group.

Oxidation reactions — this is one of the most tested areas:

Primary: $\text{R-CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{R-CHO}$ (stops at aldehyde)

Primary: $\text{R-CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{K}_2\text{Cr}_2\text{O}_7} \text{R-COOH}$ (goes to acid)

Secondary: $\text{R}_2\text{CH-OH} \xrightarrow{[\text{O}]} \text{R}_2\text{C=O}$ (ketone)

Tertiary: No oxidation under normal conditions (no $\alpha$ -hydrogen on OH carbon)

PCC (pyridinium chlorochromate) is the reagent of choice when you want to stop at the aldehyde stage. Strong oxidising agents like $\text{K}_2\text{Cr}_2\text{O}_7$ /H $^+$ push primary alcohols all the way to carboxylic acids.

Dehydration follows Zaitsev’s rule — the more substituted alkene is the major product:

\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \xrightarrow{170°\text{C, H}_2\text{SO}_4} \text{CH}_3\text{CH=CHCH}_3 \text{ (major)} + \text{CH}_3\text{CH}_2\text{CH=CH}_2 \text{ (minor)}

The ease of dehydration follows: tertiary > secondary > primary. This is because more substituted carbocations are more stable.

Lucas test

Differentiates primary, secondary and tertiary alcohols. Add ZnCl $_2$ + HCl. Tertiary turns cloudy immediately; secondary in 5-10 min; primary does not react at room temperature.

The cloudiness comes from the formation of an insoluble alkyl chloride. Tertiary alcohols react fastest because they form the most stable carbocation intermediate (SN1 mechanism).

Other important tests

Victor Meyer’s test: Distinguishes primary, secondary, and tertiary alcohols by colour. Primary gives red colour, secondary gives blue, tertiary gives no colour (colourless).

Iodoform test: Only alcohols with the structure $\text{R-CH(OH)-CH}_3$ give a positive iodoform test (yellow precipitate of CHI $_3$ ). Ethanol gives a positive test. Methanol does not.

\text{CH}_3\text{CH(OH)} + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{CHI}_3 \downarrow + \text{HCOONa} + 3\text{NaI} + 3\text{H}_2\text{O}

Worked Examples

Oxidation of an alcohol requires removal of the H attached to the OH-bearing carbon. Tertiary alcohols have no such H — they are fully substituted.

Ethanol has an OH group that can H-bond with water. The small hydrocarbon tail does not interfere much. Higher alcohols have longer tails and become less soluble.

What alcohol forms when CH $_3$ MgBr reacts with acetaldehyde (CH $_3$ CHO)?

The Grignard reagent adds to the carbonyl carbon. CH $_3$ - from Grignard attaches to the C=O carbon of acetaldehyde. After hydrolysis:

\text{CH}_3\text{MgBr} + \text{CH}_3\text{CHO} \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH(OH)CH}_3

Product is propan-2-ol (a secondary alcohol). The Grignard adds one carbon to the chain.

Dehydration goes through a carbocation intermediate. 2-methylpropan-2-ol forms a tertiary carbocation (very stable, no rearrangement needed). Propan-1-ol would need to form a primary carbocation (very unstable).

Stability order: 3° > 2° > 1° carbocation. Therefore dehydration rate: 3° > 2° > 1° alcohol.

Solved Problems (Exam Style)

Problem 1 (JEE Main pattern): An alcohol A gives a positive iodoform test and on oxidation gives a ketone B. Identify A.

Positive iodoform test means A has the $\text{CH}_3\text{CH(OH)-}$ group. Oxidation to a ketone means A is a secondary alcohol.

Combining both: A must be a secondary alcohol with a methyl group adjacent to OH. The simplest such alcohol is propan-2-ol (CH $_3$ CH(OH)CH $_3$ ). Oxidation gives acetone (CH $_3$ COCH $_3$ ).

Problem 2 (NEET pattern): Arrange in order of increasing acid strength: ethanol, phenol, water.

Acid strength depends on how well the conjugate base is stabilised.

Ethoxide (C $_2$ H $_5$ O $^-$ ): negative charge on O, no stabilisation. Weakest acid.
Hydroxide (OH $^-$ ): no stabilisation beyond the O atom. Middle.
Phenoxide (C $_6$ H $_5$ O $^-$ ): negative charge delocalised into the benzene ring (resonance). Strongest acid.

Order: ethanol < water < phenol

$pK_a$ values: ethanol ~16, water ~15.7, phenol ~10.

Common Mistakes

Writing that tertiary alcohols give aldehydes on oxidation. They do not react.

Confusing Lucas test results. Tertiary is fastest, primary slowest.

Saying NaBH $_4$ reduces acids. It reduces only aldehydes and ketones, not carboxylic acids (need LiAlH $_4$ ).

Forgetting Zaitsev’s rule in dehydration. The more substituted alkene is the major product. Students often draw the less substituted alkene as the answer.

Confusing the iodoform test. Only alcohols with $\text{CH}_3\text{CH(OH)-}$ structure work. Methanol does NOT give iodoform test. Ethanol DOES (it has $\text{CH}_3\text{CH(OH)H}$ , which fits the pattern).

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

JEE Main 2024 had a Grignard synthesis question. NEET 2023 tested the Lucas test distinction. CBSE 2024 boards asked a five-mark question on preparation and reactions of alcohols. This topic appears in almost every exam — it is genuinely high-weightage in organic chemistry.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Memorise the three Lucas test outcomes and the oxidation products of each alcohol class. Covers most PYQs.

Practice Questions

Q1. What happens when ethanol reacts with acetic acid in the presence of conc. H $_2$ SO $_4$ ?

An esterification reaction occurs. Ethanol + acetic acid → ethyl acetate + water. $\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$ . The fruity smell of ethyl acetate confirms the reaction.

Q2. Why is the boiling point of ethanol higher than that of ethane?

Ethanol (bp 78°C) has an OH group that forms intermolecular hydrogen bonds. Ethane (bp -89°C) has only weak London dispersion forces. The hydrogen bonding in ethanol requires significantly more energy to break, resulting in a much higher boiling point despite similar molecular weights (46 vs 30).

Q3. An alcohol does not react with Lucas reagent at room temperature. What can you conclude?

The alcohol is primary. Primary alcohols do not form a stable carbocation at room temperature, so the SN1 reaction with ZnCl $_2$ /HCl does not proceed. To confirm, heat the mixture — primary alcohols will eventually react.

Q4. Write the product of hydroboration-oxidation of propene.

Hydroboration-oxidation gives anti-Markovnikov addition. BH $_3$ adds to the less substituted carbon. After oxidation with H $_2$ O $_2$ /NaOH, the product is propan-1-ol (CH $_3$ CH $_2$ CH $_2$ OH), not propan-2-ol.

Q5. How would you convert ethanol to ethene in one step?

Heat ethanol with excess concentrated H $_2$ SO $_4$ at 170°C. This causes dehydration: $\text{C}_2\text{H}_5\text{OH} \xrightarrow{170°\text{C, H}_2\text{SO}_4} \text{CH}_2\text{=CH}_2 + \text{H}_2\text{O}$ . At 140°C, the product would be diethyl ether instead (intermolecular dehydration).

FAQs

What is the difference between ethanol and methanol? Methanol (CH $_3$ OH) is toxic — even 10 mL can cause blindness and 30 mL can kill. It is metabolised to formaldehyde and formic acid, which damage the optic nerve. Ethanol (C $_2$ H $_5$ OH) is the alcohol in beverages. It is metabolised to acetaldehyde and then acetic acid, which are less toxic. The antidote for methanol poisoning is actually ethanol — it competes for the same enzyme (alcohol dehydrogenase).

Why does the OH group make alcohols acidic? The O-H bond is polar (oxygen is more electronegative). The hydrogen can be donated as H $^+$ . However, alcohols are very weak acids ( $pK_a$ ~16) compared to carboxylic acids ( $pK_a$ ~5) because the alkoxide conjugate base has no resonance stabilisation.

What is denatured alcohol? Denatured alcohol is ethanol with additives (methanol, isopropanol, or denatonium benzoate) that make it undrinkable. This allows it to be sold without the excise duty applicable to potable alcohol. It is used as a solvent and fuel.

Can phenol be classified as an alcohol? Technically, phenol has an OH group, but it is not classified as an alcohol. In phenol, the OH is directly attached to a benzene ring, which changes its properties dramatically — phenol is much more acidic than alcohols and does not undergo typical alcohol reactions like Lucas test or esterification with carboxylic acids under normal conditions.

Alcohols are a playground of functional group chemistry. Primary, secondary and tertiary each tell a different story — that is what examiners love to test.