Class 11 Chemistry – Unit 18: Alcohols (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Unit 18 – Alcohols strictly according to the Federal Board (FBISE) syllabus. The unit focuses on structure, classification, properties, and reactions of alcohols.
Major topics include classification of alcohols, preparation methods, physical and chemical properties, hydrogen bonding, oxidation and dehydration reactions, and important uses of alcohols. Reaction mechanisms and exam-oriented comparisons are included for effective learning.
For strong conceptual clarity and exam preparation, students can access video lectures, MCQs, reaction-based questions, numericals, test series, and live revision classes on our YouTube channel and stay connected through our WhatsApp channel.
18.1 Classification of Monohydric Alcohols
Monohydric alcohols are classified into three families based on the number of alkyl groups attached to the carbon atom bearing the hydroxyl ($$-OH$$) group.
- Primary alcohols: The carbon atom of the hydroxyl group is attached to only one alkyl group (or is methanol).Example: Ethyl alcohol ($$CH_3-CH_2-OH$$)
- Secondary alcohols: The carbon atom of the hydroxyl group is attached to two alkyl groups.Example: Isopropyl alcohol ($$CH_3-CH(OH)-CH_3$$)
- Tertiary alcohols: The carbon atom of the hydroxyl group is attached to three alkyl groups.Example: Ter-butyl alcohol ($$(CH_3)_3C-OH$$)
18.2 Preparations of Alcohols
(a) By electrophilic addition of steam to alkenes
Alkenes react with steam in the presence of phosphoric acid ($$H_3PO_4$$) as a catalyst at $$300^\circ C$$and$$60-70 \text{ atm}$$ pressure.
$$CH_2=CH_2 + H_2O \longrightarrow CH_3-CH_2-OH$$
(b) By reaction of alkenes with potassium manganate(VII)
Cold dilute acidified $$KMnO_4$$ reacts with alkenes to produce diols (1,2-ethandiol).
$$5CH_2=CH_2 + 2H_2O + 2MnO_4^{1-} + 6H^+ \longrightarrow 5CH_2(OH)-CH_2(OH) + 2Mn^{2+}$$
(c) By substitution of haloalkanes
Heating a haloalkane with an aqueous solution of $$NaOH$$ produces an alcohol.
$$R-Br + OH^- \longrightarrow R-OH + Br^-$$
(d) By reduction of carbonyl compounds
Reducing agents like $$NaBH_4$$or$$LiAlH_4$$ reduce aldehydes to primary alcohols and ketones to secondary alcohols.
- Methanal + $$2[H]$$$$\xrightarrow{LiAlH_4}$$ Methanol
- Ketone + $$2[H]$$$$\xrightarrow{LiAlH_4}$$ Secondary Alcohol
(e) By reduction of carboxylic acids
Carboxylic acids are reduced to primary alcohols using $$LiAlH_4$$.
$$R-COOH + 4[H] \longrightarrow R-CH_2-OH + H_2O$$
(f) By hydrolysis of Esters
- Acid Hydrolysis: Yields a carboxylic acid and an alcohol.$$CH_3COOC_2H_5 \xrightarrow{H_3O^+} CH_3COOH + C_2H_5OH$$
- Basic Hydrolysis: Yields a carboxylate salt and an alcohol.$$CH_3COOC_2H_5 \xrightarrow{NaOH} CH_3COONa + C_2H_5OH$$
Review Questions & Answers
Q1: Distinguish between a secondary and tertiary alcohol based on structure.
A: In a secondary alcohol, the carbon bonded to the $$-OH$$ group is attached to two other alkyl groups (e.g., 2-propanol). In a tertiary alcohol, that carbon is attached to three alkyl groups (e.g., 2-methyl-2-propanol).
Q2: What are the specific conditions required for the industrial preparation of ethanol from ethene and steam?
A: The reaction requires a phosphoric acid ($$H_3PO_4$$) catalyst, a temperature of $$300^\circ C$$, and a pressure of $$60-70 \text{ atm}$$.
Q3: What is the difference in products when an ester undergoes acid hydrolysis versus basic hydrolysis?
A: Acid hydrolysis produces a free carboxylic acid ($$R-COOH$$) and an alcohol, whereas basic hydrolysis (saponification) produces a carboxylate salt ($$R-COO^-Na^+$$) and an alcohol.
Q4: Which reagent is typically used to reduce a ketone to a secondary alcohol?
A: Reducing agents such as $$NaBH_4$$(Sodium borohydride) or$$LiAlH_4$$ (Lithium aluminum hydride) are used.
1. Acidity of Alcohols
Alcohols and water both contain the $-OH$ group. However, alcohols are weaker acids than water. This is because the alkyl group ($R$) has an electron-releasing nature, which decreases the polarity of the $O-H$ bond.
- Reaction with Metals: Alcohols react with active metals like Sodium ($Na$) or Potassium ($K$) to release hydrogen gas and form alkoxides.
$2R-OH + 2Na \rightarrow 2R-ONa + H_2$
2. Reactions of Alcohols
A. Combustion
Alcohols are highly flammable and burn in air to produce $CO_2$, $H_2O$, and heat. They are often used as high-octane fuels.
Example (Ethanol): $CH_3-CH_2-OH + 3O_2 \rightarrow 2CO_2 + 3H_2O + \text{heat}$
B. Formation of Alkyl Halides
Alcohols react with various reagents to replace the $-OH$ group with a halogen ($X$):
- Thionyl Chloride ($SOCl_2$): Reacts in the presence of pyridine to give alkyl chlorides, $SO_2$, and $HCl$.
- Phosphorus Halides ($PCl_3, PCl_5$): Converts alcohols to alkyl chlorides.
- Hydrogen Halides ($HX$): A substitution reaction producing alkyl halides and water ($R-OH + HX \rightarrow R-X + H_2O$).
C. The Lucas Test
Used to distinguish between primary, secondary, and tertiary alcohols using Lucas Reagent (Anhydrous $ZnCl_2$ + conc. $HCl$).
| Alcohol Type | Observation |
|---|---|
| Tertiary ($3^\circ$) | Forms an oily layer immediately. |
| Secondary ($2^\circ$) | Forms an oily layer within 5 to 10 minutes. |
| Primary ($1^\circ$) | Forms an oily layer only on heating. |
D. Dehydration of Alcohols
Alcohols undergo dehydration (removal of water) using conc. $H_2SO_4$ or $Al_2O_3$. The product depends on the temperature:
- At $180^\circ C$: Produces an alkene (Ethene). $CH_3CH_2OH \xrightarrow{180^\circ C} CH_2=CH_2 + H_2O$
- At $140^\circ C$: Produces an ether (Diethyl ether). $2CH_3CH_2OH \xrightarrow{140^\circ C} CH_3CH_2OCH_2CH_3 + H_2O$
3. Concept Assessment Solutions
Exercise 18.1: Complete the Reactions
- $CH_3-CH=CH_2 + H_2O \rightarrow CH_3-CH(OH)-CH_3$ (Isopropanol)
- $CH_3-CH_2-Br + NaOH_{(aq)} \rightarrow CH_3-CH_2-OH + NaBr$
- $2CH_3-CH_2-OH + 2Na \rightarrow 2CH_3-CH_2-ONa + H_2$
- $CH_3-COOH + 4[H] \rightarrow CH_3-CH_2-OH + H_2O$ (Reduction)
Exercise 18.2: Conversions
(a) Methanol into chloromethane:
$CH_3OH + PCl_5 \rightarrow CH_3Cl + POCl_3 + HCl$ (or use $SOCl_2$)
(b) Ethanol into bromoethane:
$CH_3CH_2OH + KBr + H_2SO_4 \rightarrow CH_3CH_2Br + KHSO_4 + H_2O$
1. Esterification (Formation of Esters)
- Reaction: Alcohols react with carboxylic acids to form esters and water.
- Conditions: Requires a catalyst such as concentrated sulfuric acid ($H_2SO_4$) or aluminum oxide ($Al_2O_3$).
- General Equation: $R-OH + R'-COOH \xrightarrow{H_2SO_4} R'-COOR + H_2O$
- Example: Acetic acid reacts with ethanol to produce ethyl acetate and water. $CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O$
2. Oxidation of Alcohols
Oxidation is typically carried out using acidified potassium dichromate ($K_2Cr_2O_7 + H_2SO_4$).
- Color Change: The dichromate ion ($Cr_2O_7^{2-}$) is orange. Upon reducing to chromium ions ($Cr^{3+}$), the solution turns green.
- Primary Alcohols: Oxidized first to an aldehyde, then further to a carboxylic acid. $CH_3CH_2OH + [O] \xrightarrow{50^\circ C} CH_3CHO + H_2O$ (Acetaldehyde) $CH_3CHO + [O] \xrightarrow{50^\circ C} CH_3COOH$ (Acetic acid)
- Secondary Alcohols: Oxidized to a ketone. They do not oxidize further easily. $CH_3-CHOH-CH_3 + [O] \to CH_3-C(=O)-CH_3 + H_2O$ (2-Propanol to Acetone)
- Tertiary Alcohols: Resistant to oxidation. Under strong conditions (heating with $H_2SO_4$), they undergo dehydration to form an alkene.
3. Iodoform Reaction
- Requirement: Characteristic of ethanol and secondary alcohols containing the $CH_3-CH(OH)-$ group.
- Reagents: Iodine ($I_2$) and Sodium Hydroxide ($NaOH$).
- Observation: Formation of a pale yellow precipitate of tri-iodomethane ($CHI_3$).
- Primary Alcohols: Only ethanol gives a positive test. $CH_3CH_2OH + 4I_2 + 6NaOH \xrightarrow{\Delta} CHI_3 + HCOONa + 5NaI + 5H_2O$
- Tertiary Alcohols: Do not give this test.
Questions & Answers
Q1: What is the visual indicator that oxidation of an alcohol using potassium dichromate has occurred?
A: The solution changes color from orange (due to $Cr_2O_7^{2-}$ ions) to green (due to the formation of $Cr^{3+}$ ions).
Q2: Why do tertiary alcohols behave differently than primary and secondary alcohols when treated with acidified potassium dichromate?
A: Tertiary alcohols are resistant to oxidation because the carbon atom holding the $-OH$ group does not have a hydrogen atom to lose. Instead of oxidizing, they undergo dehydration to form an alkene when heated with sulfuric acid.
Q3: How can you distinguish between Ethanol and Methanol using a chemical test mentioned in the text?
A: The Iodoform test can be used. Ethanol reacts with $I_2$ and $NaOH$ to form a yellow precipitate of $CHI_3$, whereas methanol does not react because it lacks the necessary $CH_3-CH(OH)-$ structure.
Q4: Write the chemical equation for the formation of ethyl acetate.
A: $CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O$
Q5: What are the products of the oxidation of a secondary alcohol like 2-propanol?
A: The oxidation of 2-propanol yields acetone (a ketone) and water.