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Alcohols, Phenols and Ethers Complete NCERT Resource Guide
Mastering Alcohols, Phenols and Ethers is critical to scoring high marks in your CBSE Class 12 Chemistry exam. This high-weightage chapter forms the backbone of Organic Chemistry, bridging foundational reaction mechanisms with real-world applications. Whether you are prepping for your Board Exam Questions 2026 or targeting competitive exams like NEET and JEE, this complete, student-friendly guide delivers fully solved NCERT Solutions and conceptual breakdowns to boost your confidence.
Chapter NameAlcohols, Phenols and Ethers
SubjectChemistry
ClassClass 12
BoardCBSE & State Boards
Important TopicsHydroboration-Oxidation, Kolbe's Reaction, Reimer-Tiemann Reaction, Williamson Synthesis, Acidic Nature of Phenol, Lucas Test
Difficulty LevelModerate (Requires understanding of mechanisms)
Exam WeightageApproximately 6-8 Marks
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Learning Objectives
After completing this chapter, students will be able to:
- Classify alcohols, phenols, and ethers into mono-, di-, trihydric, or allylic/benzylic categories.
- Apply IUPAC nomenclature rules accurately to name complex functional structures.
- Understand crucial preparation mechanisms such as acid-catalyzed hydration and hydroboration-oxidation.
- Explain physical properties like solubility trends and anomalous boiling points due to intermolecular hydrogen bonding.
- Differentiate functional compounds using chemical diagnostics like the Lucas test and Ferric Chloride test.
- Execute key named reactions smoothly in conversions and synthetic pathways.
Key Concepts, Definitions, and Rules
- Hydroboration-Oxidation: An anti-Markovnikov addition of water to alkenes using diborane ($\text{B}_2\text{H}_6$) followed by alkaline hydrogen peroxide, yielding excellent quantities of primary alcohols.
- Lucas Test: Used to distinguish primary, secondary, and tertiary alcohols. It relies on treating the sample with Lucas reagent ($\text{conc. HCl} + \text{anhyd. ZnCl}_2$). Tertiary alcohols produce immediate turbidity, secondary alcohols produce it within 5 minutes, and primary alcohols show no turbidity at room temperature.
- Williamson Synthesis: An important laboratory method for preparing symmetrical and asymmetrical ethers by reacting an alkyl halide with a sodium alkoxide via an $\text{S}_\text{N}2$ mechanism.
- Reimer-Tiemann Reaction: Treatment of phenol with chloroform ($\text{CHCl}_3$) in the presence of sodium hydroxide ($\text{NaOH}$) introduces a Formyl group ($-\text{CHO}$) at the ortho position, producing Salicylaldehyde.
- Kolbe's Reaction: Phenoxide ion is treated with carbon dioxide ($\text{CO}_2$) under pressure followed by acidification, leading to the formation of Ortho-hydroxybenzoic acid (Salicylic acid).
- Resonance Stabilization of Phenoxide Ion: Phenols are significantly more acidic than alcohols. When phenol loses a proton, it forms a phenoxide ion that is highly stabilized by the delocalization of the negative charge across the aromatic ring.
Full NCERT Solutions (Step-by-Step)
Here are the step-by-step, board-exam-styled solutions for Alcohols, Phenols and Ethers Class 12.
Question 11.1: Write IUPAC names of the following compounds:
1. $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{OH}$
2. $\text{CH}_3-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_2-\text{CH}_3$
3. $\text{HO}-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_2-\text{OH}$
Step 1: Compound 1 Evaluation. The longest continuous carbon chain has 4 carbon atoms with the $-\text{OH}$ group at position 1. A methyl substituent sits on carbon 3. IUPAC Name: 3-Methylbutan-1-ol.
Step 2: Compound 2 Evaluation. The principal carbon chain contains 6 carbons, and numbering starts from the left side to give the hydroxyl functional group the lowest position locant (3). IUPAC Name: Hexan-3-ol.
Step 3: Compound 3 Evaluation. A three-carbon chain where each carbon atom holds one hydroxyl functional group. IUPAC Name: Propane-1,2,3-triol (commonly known as glycerol).
Step 2: Compound 2 Evaluation. The principal carbon chain contains 6 carbons, and numbering starts from the left side to give the hydroxyl functional group the lowest position locant (3). IUPAC Name: Hexan-3-ol.
Step 3: Compound 3 Evaluation. A three-carbon chain where each carbon atom holds one hydroxyl functional group. IUPAC Name: Propane-1,2,3-triol (commonly known as glycerol).
Question 11.2: Explain why alcohols have higher boiling points than hydrocarbons of comparable molecular masses.
Step 1: Structural Polarity Analysis. Alcohols possess a highly polar $-\text{OH}$ group due to the large electronegativity difference between oxygen and hydrogen atoms. This structural polarity allows alcohol molecules to form extensive intermolecular hydrogen bonding with each other.
Step 2: Intermolecular Forces Comparison. Hydrocarbons, on the other hand, are non-polar or weakly polar and are held together only by weak Van der Waals dispersion forces.
Step 3: Thermal Energy Requirement. Breaking strong intermolecular hydrogen bonds requires significantly more thermal energy, resulting in much higher boiling points for alcohols compared to analogous hydrocarbons.
Step 2: Intermolecular Forces Comparison. Hydrocarbons, on the other hand, are non-polar or weakly polar and are held together only by weak Van der Waals dispersion forces.
Step 3: Thermal Energy Requirement. Breaking strong intermolecular hydrogen bonds requires significantly more thermal energy, resulting in much higher boiling points for alcohols compared to analogous hydrocarbons.
Question 11.3: Explain the fact that in aryl alkyl ethers, the alkoxy group activates the benzene ring towards electrophilic substitution.
Step 1: Resonance Mechanism. In aryl alkyl ethers (such as anisole), the lone pairs of electrons on the ether oxygen atom are in direct conjugation with the $\pi$-system of the aromatic benzene ring.
Step 2: Electron Delocalization. Through the positive resonance effect (+R effect), this lone pair is delocalized into the ring structure. As shown by the resonance states, the electron density increases significantly at the ortho and para positions.
Step 3: Electrophilic Attack. This high electron concentration activates the aromatic ring, making it highly susceptible to attacks by incoming electrophiles specifically at these ortho and para positions.
Step 2: Electron Delocalization. Through the positive resonance effect (+R effect), this lone pair is delocalized into the ring structure. As shown by the resonance states, the electron density increases significantly at the ortho and para positions.
Step 3: Electrophilic Attack. This high electron concentration activates the aromatic ring, making it highly susceptible to attacks by incoming electrophiles specifically at these ortho and para positions.
Question 11.4: Write the mechanism of acid-catalyzed hydration of ethene to yield ethanol.
The industrial synthesis of ethanol via acid-catalyzed hydration of ethene proceeds through three distinct steps:
Step 1: Protonation of alkene. Forms a carbocation intermediate via electrophilic attack of $\text{H}_3\text{O}^+$. $$\text{H}_2\text{O} + \text{H}^+ \rightleftharpoons \text{H}_3\text{O}^+$$ $$\text{CH}_2=\text{CH}_2 + \text{H}_3\text{O}^+ \rightleftharpoons ^+\text{CH}_2-\text{CH}_3 + \text{H}_2\text{O}$$ Step 2: Nucleophilic attack. Water attacks the unstable carbocation. $$^+\text{CH}_2-\text{CH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3-\text{CH}_2-\text{OH}_2^+$$ Step 3: Deprotonation. Yields the final alcohol product. $$\text{CH}_3-\text{CH}_2-\text{OH}_2^+ + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3-\text{CH}_2-\text{OH} + \text{H}_3\text{O}^+$$
Step 1: Protonation of alkene. Forms a carbocation intermediate via electrophilic attack of $\text{H}_3\text{O}^+$. $$\text{H}_2\text{O} + \text{H}^+ \rightleftharpoons \text{H}_3\text{O}^+$$ $$\text{CH}_2=\text{CH}_2 + \text{H}_3\text{O}^+ \rightleftharpoons ^+\text{CH}_2-\text{CH}_3 + \text{H}_2\text{O}$$ Step 2: Nucleophilic attack. Water attacks the unstable carbocation. $$^+\text{CH}_2-\text{CH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3-\text{CH}_2-\text{OH}_2^+$$ Step 3: Deprotonation. Yields the final alcohol product. $$\text{CH}_3-\text{CH}_2-\text{OH}_2^+ + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3-\text{CH}_2-\text{OH} + \text{H}_3\text{O}^+$$
Question 11.5: How is 1-phenylethanol prepared from a suitable Grignard reagent? Show the chemical steps.
Step 1: Reactant Identification. 1-Phenylethanol is a secondary alcohol containing a phenyl group and a methyl group on the carbon bearing the $-\text{OH}$ group. It can be synthesized by reacting Ethanal (Acetaldehyde) with Phenylmagnesium bromide (Grignard Reagent).
Step 2: Nucleophilic Addition. The Grignard reagent attacks the carbonyl carbon to form an adduct. $$\text{C}_6\text{H}_5\text{MgBr} + \text{CH}_3\text{CHO} \rightarrow \text{CH}_3-\text{CH}(\text{OMgBr})-\text{C}_6\text{H}_5$$ Step 3: Acid Hydrolysis. The adduct is hydrolyzed to yield the final alcohol. $$\text{CH}_3-\text{CH}(\text{OMgBr})-\text{C}_6\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3-\text{CH}(\text{OH})-\text{C}_6\text{H}_5 + \text{Mg}(\text{OH})\text{Br}$$
Step 2: Nucleophilic Addition. The Grignard reagent attacks the carbonyl carbon to form an adduct. $$\text{C}_6\text{H}_5\text{MgBr} + \text{CH}_3\text{CHO} \rightarrow \text{CH}_3-\text{CH}(\text{OMgBr})-\text{C}_6\text{H}_5$$ Step 3: Acid Hydrolysis. The adduct is hydrolyzed to yield the final alcohol. $$\text{CH}_3-\text{CH}(\text{OMgBr})-\text{C}_6\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3-\text{CH}(\text{OH})-\text{C}_6\text{H}_5 + \text{Mg}(\text{OH})\text{Br}$$
Question 11.6: Give structures of the products expected from the following reactions:
1. Catalytic reduction of Butanal.
2. Hydration of Propene in the presence of dilute sulfuric acid.
Step 1: Catalytic Reduction of Butanal. Reduction of an aldehyde yields a primary alcohol. The product is Butan-1-ol.
$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} + \text{H}_2 \xrightarrow{\text{Pd/Ni}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$$
Step 2: Hydration of Propene. The addition of water in the presence of dilute acid follows Markovnikov's rule, where the hydrogen atom adds to the carbon with more hydrogens. The product is Propan-2-ol.
$$\text{CH}_3-\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3$$
Question 11.7: Why are phenols more acidic than alcohols?
Phenols are stronger acids than alcohols due to two main factors:
Step 1: sp² Hybridization of Carbon. In phenol, the hydroxyl $-\text{OH}$ group is attached to an $sp^2$ hybridized carbon atom of the benzene ring, which is more electronegative than the $sp^3$ hybridized carbon in alcohols. This polarizes the $\text{O}-\text{H}$ bond further, facilitating easier proton release.
Step 2: Resonance Stability of Phenoxide Ion. Deprotonation of phenol produces a phenoxide ion. The negative charge on the oxygen atom is delocalized over the ortho and para positions of the aromatic ring through resonance, stabilizing the ion. In contrast, deprotonation of an alcohol forms an alkoxide ion, where the negative charge remains localized on the oxygen atom and is further destabilized by the electron-donating inductive effect (+I effect) of the alkyl group.
Step 1: sp² Hybridization of Carbon. In phenol, the hydroxyl $-\text{OH}$ group is attached to an $sp^2$ hybridized carbon atom of the benzene ring, which is more electronegative than the $sp^3$ hybridized carbon in alcohols. This polarizes the $\text{O}-\text{H}$ bond further, facilitating easier proton release.
Step 2: Resonance Stability of Phenoxide Ion. Deprotonation of phenol produces a phenoxide ion. The negative charge on the oxygen atom is delocalized over the ortho and para positions of the aromatic ring through resonance, stabilizing the ion. In contrast, deprotonation of an alcohol forms an alkoxide ion, where the negative charge remains localized on the oxygen atom and is further destabilized by the electron-donating inductive effect (+I effect) of the alkyl group.
Question 11.8: Write the chemical equations for Kolbe's reaction and Reimer-Tiemann reaction.
Step 1: Kolbe's Reaction. Sodium phenoxide is treated with $\text{CO}_2$ under pressure, followed by acidification to produce Salicylic acid.
$$\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} \xrightarrow[\text{(2) H}^+]{\text{(1) CO}_2} \text{C}_6\text{H}_4(\text{OH})(\text{COOH}) \quad \text{(Salicylic acid)}$$
Step 2: Reimer-Tiemann Reaction. Phenol is treated with chloroform and aqueous $\text{NaOH}$ to yield Salicylaldehyde.
$$\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + \text{NaOH} \rightarrow \text{C}_6\text{H}_4(\text{ONa})(\text{CHCl}_2) \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_4(\text{ONa})(\text{CHO}) \xrightarrow{\text{H}^+} \text{C}_6\text{H}_4(\text{OH})(\text{CHO}) \quad \text{(Salicylaldehyde)}$$
Question 11.9: Explain the mechanism of Williamson synthesis for the preparation of diethyl ether.
Step 1: Reaction Classification. Williamson synthesis is a nucleophilic substitution ($\text{S}_\text{N}2$) reaction where an alkoxide ion acts as a strong nucleophile to displace a halide ion from a primary alkyl halide.
Step 2: Reaction Mechanism. $$\text{CH}_3\text{CH}_2-\text{O}^- \text{Na}^+ + \text{CH}_3\text{CH}_2-\text{Cl} \rightarrow \left[\text{CH}_3\text{CH}_2-\text{O}\cdots\text{CH}_2\cdots\text{Cl} \right]^{\ddagger} \rightarrow \text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3 + \text{NaCl}$$ Step 3: Crucial Constraint. For optimal yields, the alkyl halide must be primary. If a secondary or tertiary alkyl halide is used instead, the strongly basic alkoxide ion induces elimination over substitution, yielding an alkene as the major product.
Step 2: Reaction Mechanism. $$\text{CH}_3\text{CH}_2-\text{O}^- \text{Na}^+ + \text{CH}_3\text{CH}_2-\text{Cl} \rightarrow \left[\text{CH}_3\text{CH}_2-\text{O}\cdots\text{CH}_2\cdots\text{Cl} \right]^{\ddagger} \rightarrow \text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3 + \text{NaCl}$$ Step 3: Crucial Constraint. For optimal yields, the alkyl halide must be primary. If a secondary or tertiary alkyl halide is used instead, the strongly basic alkoxide ion induces elimination over substitution, yielding an alkene as the major product.
Question 11.10: Illustrate the reaction of HI with methoxybenzene (anisole). What are the products and why?
Step 1: Product Identification. When methoxybenzene reacts with concentrated Hydroiodic acid ($\text{HI}$), the products are Phenol and Methyl iodide.
$$\text{C}_6\text{H}_5-\text{O}-\text{CH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5-\text{OH} + \text{CH}_3\text{I}$$
Step 2: Mechanism and Reasoning. The reaction begins with the protonation of the ether oxygen to form an oxonium ion. The cleavage of the $\text{O}-\text{C}_6\text{H}_5$ bond has a high activation energy because the bond has partial double-bond character due to resonance. Consequently, the nucleophile $\text{I}^-$ attacks the less sterically hindered methyl carbon via an $\text{S}_\text{N}2$ mechanism, cleaving the lower-energy $\text{O}-\text{CH}_3$ bond to yield phenol and methyl iodide.
Extra Important Questions (Board Style)
Boost your preparations with these highly anticipated Board Exam Questions 2026 curated by subject experts.
1. Multiple Choice Questions (MCQs)
Q1. Which of the following alcohols reacts fastest with Lucas Reagent at room temperature?
A) Butan-1-ol
B) Butan-2-ol
C) 2-Methylpropan-2-ol
D) Propan-1-ol
Correct Answer Choice: C
Step 1: Mechanism Verification. Lucas reagent reacts via a carbocation intermediate mechanism. Tertiary alcohols form highly stable tertiary carbocations and react almost instantaneously, producing immediate turbidity at room temperature. 2-Methylpropan-2-ol is a tertiary alcohol.
Step 1: Mechanism Verification. Lucas reagent reacts via a carbocation intermediate mechanism. Tertiary alcohols form highly stable tertiary carbocations and react almost instantaneously, producing immediate turbidity at room temperature. 2-Methylpropan-2-ol is a tertiary alcohol.
Q2. What is the major product obtained when t-butyl chloride is treated with sodium ethoxide?
A) Ethyl t-butyl ether
B) Isobutylene
C) But-1-ene
D) Diethyl ether
Correct Answer Choice: B
Step 1: Reaction Constraint Analysis. Sodium ethoxide is a strong nucleophile and a powerful base. When reacted with a tertiary alkyl halide like t-butyl chloride, elimination dominates over substitution, forming Isobutylene (2-methylpropene) via an E2 mechanism.
Step 1: Reaction Constraint Analysis. Sodium ethoxide is a strong nucleophile and a powerful base. When reacted with a tertiary alkyl halide like t-butyl chloride, elimination dominates over substitution, forming Isobutylene (2-methylpropene) via an E2 mechanism.
2. Assertion-Reason Questions
Directions: Choose (A) if both A and R are true and R is the correct explanation. Choose (B) if both A and R are true but R is not the correct explanation. Choose (C) if A is true but R is false. Choose (D) if A is false but R is true.
Q3. Assertion (A): The boiling point of ethanol is much higher than that of methoxymethane despite having identical molecular formulas.
Reason (R): Ethanol forms intermolecular hydrogen bonds, whereas methoxymethane cannot.
Correct Answer Choice: A
Step 1: Property Validation. Ethanol contains a hydrogen atom bonded to a highly electronegative oxygen atom, enabling strong intermolecular hydrogen bonding. Methoxymethane does not have hydrogen directly bonded to oxygen, so its molecules are held together only by weaker dipole-dipole interactions.
Step 1: Property Validation. Ethanol contains a hydrogen atom bonded to a highly electronegative oxygen atom, enabling strong intermolecular hydrogen bonding. Methoxymethane does not have hydrogen directly bonded to oxygen, so its molecules are held together only by weaker dipole-dipole interactions.
Q4. Assertion (A): Nitration of phenol with concentrated $\text{HNO}_3$ gives a poor yield of picric acid due to ring oxidation.
Reason (R): Nitric acid is a powerful oxidizing agent that can degrade the electron-rich benzene ring of phenol.
Correct Answer Choice: A
Step 1: Reagent Analysis. Because the $-\text{OH}$ group highly activates the aromatic ring, treating phenol with concentrated $\text{HNO}_3$ oxidizes much of the starting material into complex tarry mixtures, reducing the yield of Picric acid (2,4,6-trinitrophenol).
Step 1: Reagent Analysis. Because the $-\text{OH}$ group highly activates the aromatic ring, treating phenol with concentrated $\text{HNO}_3$ oxidizes much of the starting material into complex tarry mixtures, reducing the yield of Picric acid (2,4,6-trinitrophenol).
3. Short Answer Questions (SAQs)
Q5. Name the reagents used in the following conversions:
1. Primary alcohol to an aldehyde.
2. Benzyl alcohol to benzoic acid.
Step 1: Conversion 1. Pyridinium Chlorochromate (PCC) or Pyridinium Dichromate (PDC) in $\text{CH}_2\text{Cl}_2$. Note: Strong oxidizers like $\text{KMnO}_4$ would over-oxidize it directly to a carboxylic acid.
Step 2: Conversion 2. Alkaline Potassium Permanganate ($\text{KMnO}_4$) followed by acidification ($\text{H}_3\text{O}^+$).
Step 2: Conversion 2. Alkaline Potassium Permanganate ($\text{KMnO}_4$) followed by acidification ($\text{H}_3\text{O}^+$).
Q6. Identify the products when Propan-2-ol is passed over heated copper catalyst at $573\text{ K}$.
Step 1: Dehydrogenation Process. Passing a secondary alcohol over a heated copper catalyst at $573\text{ K}$ causes dehydrogenation, converting it into a ketone.
$$\text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3 \xrightarrow{\text{Cu, } 573\text{ K}} \text{CH}_3-\text{CO}-\text{CH}_3 \quad \text{(Acetone)} + \text{H}_2$$
Q7. Out of o-nitrophenol and p-nitrophenol, which is more steam-volatile? Explain the reason behind this choice.
Step 1: Ortho-nitrophenol Analysis. Ortho-nitrophenol is more steam-volatile. It forms intramolecular hydrogen bonds (within the same molecule), creating a chelated structure that reduces its capacity to bond with surrounding molecules.
Step 2: Para-nitrophenol Analysis. In contrast, p-nitrophenol forms intermolecular hydrogen bonds with neighboring molecules, creating a linked network. As a result, p-nitrophenol has a lower vapor pressure and higher boiling point, making o-nitrophenol significantly more volatile in steam distillation.
Step 2: Para-nitrophenol Analysis. In contrast, p-nitrophenol forms intermolecular hydrogen bonds with neighboring molecules, creating a linked network. As a result, p-nitrophenol has a lower vapor pressure and higher boiling point, making o-nitrophenol significantly more volatile in steam distillation.
Q8. Convert Phenol into Benzoquinone. Write the structural equation.
Step 1: Oxidation Mechanism. Phenol undergoes oxidation when treated with chromic acid ($\text{Na}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4$) to yield a conjugated diketone known as para-benzoquinone.
$$\text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{Na}_2\text{Cr}_2\text{O}_7/ \text{H}_2\text{SO}_4} \text{O}=\text{C}_6\text{H}_4=\text{O}$$
Q9. Write the chemical test to distinguish between Phenol and Ethanol.
Step 1: Test Selection. The Neutral Ferric Chloride ($\text{FeCl}_3$) Test can differentiate the two compounds.
Step 2: Observations. Phenol reacts with neutral $\text{FeCl}_3$ solution to form a characteristic violet/purple coordination complex. Ethanol does not produce a color change when treated with neutral $\text{FeCl}_3$.
Step 2: Observations. Phenol reacts with neutral $\text{FeCl}_3$ solution to form a characteristic violet/purple coordination complex. Ethanol does not produce a color change when treated with neutral $\text{FeCl}_3$.
4. Long Answer Questions (LAQs)
Q10. An organic compound (A) with molecular formula $\text{C}_4\text{H}_{10}\text{O}$ does not react with sodium metal. On treatment with excess hot $\text{HI}$, it yields only one single alkyl halide (B). Identify compounds (A) and (B) and write all the chemical equations involved.
Step 1: Identification. Since compound (A) has the molecular formula $\text{C}_4\text{H}_{10}\text{O}$ but does not react with sodium metal, it cannot be an alcohol; it must be an ether. Since it yields only one single type of alkyl halide upon cleavage with excess hot $\text{HI}$, (A) must be a symmetrical ether. Therefore, compound (A) is Ethoxyethane ($\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3$).
Step 2: Cleavage Mechanism. Treating ethoxyethane with excess hot $\text{HI}$ cleaves both ether bonds, converting it into Ethyl iodide ($\text{CH}_3\text{CH}_2\text{I}$), which is compound (B).
Step 3: Chemical Equations. $$\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{I}$$ $$\text{CH}_3\text{CH}_2\text{OH} + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{I} + \text{H}_2\text{O}$$ Net Reaction with excess HI: $$\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3 + 2\text{HI} \xrightarrow{\Delta} 2\text{CH}_3\text{CH}_2\text{I} + \text{H}_2\text{O}$$
Step 2: Cleavage Mechanism. Treating ethoxyethane with excess hot $\text{HI}$ cleaves both ether bonds, converting it into Ethyl iodide ($\text{CH}_3\text{CH}_2\text{I}$), which is compound (B).
Step 3: Chemical Equations. $$\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{I}$$ $$\text{CH}_3\text{CH}_2\text{OH} + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{I} + \text{H}_2\text{O}$$ Net Reaction with excess HI: $$\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3 + 2\text{HI} \xrightarrow{\Delta} 2\text{CH}_3\text{CH}_2\text{I} + \text{H}_2\text{O}$$
Q11. Complete the following multi-step conversions:
1. Phenol to Benzene.
2. Aniline to Phenol.
3. Propene to Propan-1-ol.
Step 1: Phenol to Benzene. Distill phenol with Zinc dust.
$$\text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO}$$
Step 2: Aniline to Phenol. Diazotize aniline using $\text{NaNO}_2 + \text{HCl}$ at $0-5^\circ\text{C}$ to form Benzene diazonium chloride, then warm the solution with water.
$$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_2\text{O}, \Delta} \text{C}_6\text{H}_5\text{OH} + \text{N}_2 + \text{HCl}$$
Step 3: Propene to Propan-1-ol. Use Hydroboration-Oxidation to achieve anti-Markovnikov hydration.
$$\text{CH}_3-\text{CH}=\text{CH}_2 \xrightarrow{\text{(1) B}_2\text{H}_6, \text{ (2) H}_2\text{O}_2/\text{OH}^-} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$$
5. Case-Based Questions
Q12. Read the passage and answer the questions below:
The dehydration of alcohols to form either alkenes or ethers depends on the reaction conditions, temperature, and the structure of the alcohol. Primary alcohols undergo dehydration to form ethers at $413\text{ K}$ in the presence of concentrated sulfuric acid via a bimolecular nucleophilic substitution ($\text{S}_\text{N}2$) pathway. However, raising the temperature to $443\text{ K}$ alters the mechanism, making elimination dominant and yielding ethene as the primary product. Secondary and tertiary alcohols readily yield alkenes under milder acidic conditions due to the relative stability of their carbocation intermediates.
1. Why does the reaction of ethanol with conc. $\text{H}_2\text{SO}_4$ at $443\text{ K}$ yield an alkene instead of an ether?
2. Arrange primary, secondary, and tertiary alcohols in increasing order of their ease of dehydration.
3. Predict the major product formed when 2-Methylpropan-2-ol is warmed with $20\%$ $\text{H}_3\text{PO}_4$.
Step 1: Answer to Part 1. At the higher temperature of $443\text{ K}$, elimination has a lower activation energy barrier than substitution. The elimination of a proton from the $\beta$-carbon becomes the faster pathway, making the formation of an alkene dominant.
Step 2: Answer to Part 2. Primary < Secondary < Tertiary. (This trend follows the stability of the carbocation intermediate formed during elimination).
Step 3: Answer to Part 3. 2-Methylpropene (Isobutylene). Since it is a tertiary alcohol, it undergoes rapid dehydration even with mild, dilute acids.
Step 2: Answer to Part 2. Primary < Secondary < Tertiary. (This trend follows the stability of the carbocation intermediate formed during elimination).
Step 3: Answer to Part 3. 2-Methylpropene (Isobutylene). Since it is a tertiary alcohol, it undergoes rapid dehydration even with mild, dilute acids.
Common Mistakes Students Make
- Misapplying Williamson Synthesis Limits: Students often try to synthesize t-butyl ethyl ether using sodium t-butoxide and ethyl chloride. However, reversing the components—using sodium ethoxide and t-butyl chloride—induces elimination instead of substitution, producing an alkene rather than the target ether. Always ensure your alkyl halide is primary.
- Confusing Temperature Thresholds in Dehydration: Mixing up $413\text{ K}$ (which produces ethers) and $443\text{ K}$ (which produces alkenes) in ethanol dehydration reactions is a frequent error. Memorize these specific conditions carefully.
- Neglecting Ring Disruption by Strong Oxidizers: Writing $\text{KMnO}_4$ for oxidizing phenol to benzoquinone is incorrect. Phenol requires chromic acid ($\text{Na}_2\text{Cr}_2\text{O}_7 / \text{H}_2\text{SO}_4$) for proper conversion.
Exam Preparation Tips
- Review Key Named Mechanisms: The CBSE board regularly tests specific organic mechanisms. Dedicate time to mastering processes like Hydroboration-Oxidation, the Reimer-Tiemann reaction, and Kolbe's reaction.
- Understand Solubility Trends: Remember that phenol contains a polar $-\text{OH}$ group that forms intermolecular hydrogen bonds with water molecules. Chlorobenzene is non-polar and cannot form hydrogen bonds, making it insoluble in water.
- Clarify Reagent Uses: Continuous distillation removes the ether as it forms because its boiling point is lower than that of the source alcohol. This prevents over-reaction and shifts the equilibrium forward according to Le Chatelier's principle.
Frequently Asked Questions (FAQs)
1. Why do alcohols have higher boiling points than hydrocarbons?
Alcohols contain a polar -OH group that forms strong intermolecular hydrogen bonds between molecules. Hydrocarbons are non-polar and are held together only by weak Van der Waals forces, which require less energy to break.
2. What limits the Williamson ether synthesis?
The reaction requires a primary alkyl halide. If a secondary or tertiary alkyl halide is used with the strongly basic alkoxide ion, elimination dominates over substitution, yielding an alkene instead of an ether.
3. What is the role of continuous distillation in the synthesis of ethers from alcohols?
Continuous distillation removes the ether as it forms because its boiling point is lower than that of the source alcohol. This prevents over-reaction and shifts the equilibrium forward according to Le Chatelier's principle.
Conclusion: Mastering Alcohols, Phenols and Ethers comes down to understanding reaction mechanisms, identifying key structural trends, and practicing conversions regularly. Do not just memorize reactions—write them out, understand the core concepts, and track electron flows to ensure maximum retention before your board exams!
More Class 12 Chemistry Chapters
Ch 1: The Solid State
Ch 2: Solutions
Ch 3: Electrochemistry
Ch 4: Chemical Kinetics
Ch 5: Surface Chemistry
Ch 6: General Principles
Ch 7: The p-Block Elements
Ch 8: The d- & f-Block Elements
Ch 9: Coordination Compounds
Ch 10: Haloalkanes
Ch 11: Alcohols & Phenols
Ch 12: Aldehydes & Ketones
Ch 13: Amines
Ch 14: Biomolecules
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