The chapter on Alcohols, Phenols, and Ethers in the 12th standard NCERT Chemistry textbook explores the structure, nomenclature, preparation, properties, and reactions of these important classes of organic compounds. Here’s a summary:
Alcohols:
- Structure: Contain a hydroxyl (-OH) group attached to a saturated carbon atom. Classified as primary, secondary, or tertiary depending on the number of alkyl/aryl groups attached to the carbon bearing the -OH group.
- Nomenclature: IUPAC names involve identifying the longest carbon chain containing the -OH group and using “-anol” as the suffix. Common names are often used, especially for simpler alcohols.
- Preparation: Several methods are used, including:
- Acid-catalyzed hydration of alkenes.
- Hydroboration-oxidation of alkenes.
- Reduction of aldehydes, ketones, carboxylic acids, and esters.
- Grignard reaction of aldehydes and ketones with alkyl halides.
- Physical Properties: Lower alcohols are liquids at room temperature, with increasing boiling points due to hydrogen bonding.
- Chemical Properties:
- Reactions involving the cleavage of the O-H bond (acidity, esterification).
- Reactions involving the cleavage of the C-O bond (reaction with hydrogen halides, dehydration).
- Oxidation (different products depending on the type of alcohol and oxidizing agent).
Phenols:
- Structure: Contain a hydroxyl (-OH) group directly attached to an aromatic ring.
- Nomenclature: Common names are widely used.
- Preparation:
- From chlorobenzene (Dow process).
- From cumene.
- Physical Properties: Phenols are crystalline solids, slightly soluble in water.
- Chemical Properties:
- Electrophilic aromatic substitution reactions (due to the activating effect of the -OH group).
- Reactions with zinc dust (reduction to benzene).
- Kolbe’s reaction and Reimer-Tiemann reaction (introduction of -COOH and -CHO groups, respectively, ortho and para to the -OH group).
Ethers:
- Structure: Contain an oxygen atom bonded to two alkyl/aryl groups (R-O-R).
- Nomenclature: Common names are used, naming the two alkyl/aryl groups alphabetically followed by “ether.”
- Preparation:
- Williamson ether synthesis (reaction of an alkyl halide with an alkoxide).
- Acid-catalyzed dehydration of alcohols.
- Physical Properties: Ethers are generally liquids at room temperature, with low boiling points due to the absence of hydrogen bonding.
- Chemical Properties: Ethers are relatively unreactive. They undergo cleavage with strong acids (like HI) to form alkyl halides and alcohols.
Exercise
1. Write IUPAC names of the following compounds:
Ans :
(i) 2,2,4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-dioI
(iii) Butane-2,3-diol
(iv) Propane-1,2,3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2,5-DimethylphenoI
(viii) 2,6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane
2. Write structures of the compounds whose IUPAC names are as follows:
(i)2-Methylbutan-2-ol
(ii)l-Phcnylpropan-2-ol
(iii)3,5-DimethyIhexane-l,3,5-triol
(iv)2,3-Dicthylphenol
(v)1-Ethoxypropane
(vi)2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpcntan-3-ol
(ix)Cyclopcnt-3-en-l-ol
(x)4-ChIoro-3-ethylbutan-l-ol
Ans :
3. (a) Draw the structural formulas and write IUPAC names of all the isomeric alkanols with the molecular formula CsH12O
(b) Classify the isomers of alcohols given in part (a) as primary, secondary and tertiary alcohols.
Ans :
Primary Alcohols: The carbon atom bearing the -OH group is bonded to only one other carbon atom.
- 1-Pentanol
- 2-Methyl-1-butanol
- 3-Methyl-1-butanol
- 2,2-Dimethyl-1-propanol
Secondary Alcohols: The carbon atom bearing the -OH group is bonded to two other carbon atoms.
- 2-Pentanol
- 3-Methyl-2-butanol
Tertiary Alcohols: The carbon atom bearing the -OH group is bonded to three other carbon atoms.
- 2-Methyl-2-butanol
- 2,2-Dimethyl-2-propanol
4. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Ans : Propanol (CH₃CH₂CH₂OH) has a significantly higher boiling point than butane (CH₃CH₂CH₂CH₃) due to the presence of hydrogen bonding in propanol, which is absent in butane.
The higher boiling point of propanol compared to butane is a direct result of the strong hydrogen bonding present in propanol, which is absent in nonpolar butane, where only weak London dispersion forces exist.
5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Ans : Alcohols are comparatively more soluble in water than corresponding hydrocarbons because the polar -OH group in alcohols can form hydrogen bonds with water molecules, facilitating dissolution.
Sources and related content
6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Ans : The hydroboration-oxidation reaction is a two-step chemical process that converts alkenes into alcohols. It’s a valuable tool in organic chemistry because it adds water across the double bond of an alkene in an anti-Markovnikov fashion, meaning the hydroxyl (-OH) group attaches to the less substituted carbon atom.
7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Ans :
8. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Ans : The ortho isomer’s ability to form intramolecular hydrogen bonds makes it less polar and more volatile, allowing it to be separated from the para isomer via steam distillation.
9. Give the equations of the reaction for the preparation of phenol from cumene.
Ans :
Step 1: Oxidation of Cumene to Cumene Hydroperoxide
Cumene (isopropylbenzene) is reacted with oxygen (from air) in the presence of a catalyst (usually a metal catalyst or an initiator) to form cumene hydroperoxide.
C₆H₅CH(CH₃)₂ + O₂ → C₆H₅C(CH₃)₂OOH
(Cumene) (Cumene Hydroperoxide)
Step 2: Acid-Catalyzed Cleavage of Cumene Hydroperoxide
Cumene hydroperoxide is then treated with dilute sulfuric acid (H₂SO₄) as a catalyst. This results in the cleavage of the hydroperoxide, producing phenol and acetone as the major products.
C₆H₅C(CH₃)₂OOH + H₂SO₄ → C₆H₅OH + (CH₃)₂CO
(Cumene Hydroperoxide) (Phenol) (Acetone)
Overall Reaction:
The overall reaction can be summarized as:
C₆H₅CH(CH₃)₂ + O₂ + H₂SO₄ → C₆H₅OH + (CH₃)₂CO + H₂O
(Cumene) (Phenol) (Acetone)
This cumene process is an important industrial method for the production of phenol, as it also yields acetone as a valuable byproduct.
10. Write chemical reaction for the preparation of phenol from chlorobenzene.
Ans :
11. Write the mechanism of hydration of ethene to yield ethanol.
Ans :
12. You are given benzene, cone. H2S04and NaOH. Write the equations for the preparation of phenol using these reagents.
Ans :
13. Show how will you synthesise
(i) 1 -phenylethanol from a suitable alkene.
(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) Pentan-l-ol using a suitable alkyl halide?
Ans :
14. Give two reactions that show the acidic nature of phenol. Compare its acidity with that of ethanol.
Ans :
(a) Reaction with sodium: Phenol reacts with active metals like sodium to liberate H, gas.
(b) Reaction with NaOH: Phenol dissolves in NaOH to form sodium phenoxide and water.
15. Explain why is orthonitrophenol more acidic than orthomethoxyphenol?
Ans :
1. Electron-Withdrawing Effect of the Nitro Group (-NO₂):
- The nitro group (-NO₂) is a strong electron-withdrawing group. It pulls electron density away from the benzene ring through both inductive (-I effect) and resonance (+M effect) effects.
- When ortho-nitrophenol loses a proton (H⁺) to form the phenoxide ion, the negative charge on the oxygen atom is delocalized and stabilized by the nitro group. This stabilization of the conjugate base makes it easier for the phenol to lose a proton, thus increasing its acidity.
2. Intramolecular Hydrogen Bonding in ortho-Nitrophenol:
- While it might seem counterintuitive, the nitro group in the ortho position can also participate in intramolecular hydrogen bonding with the -OH group.
- This hydrogen bond, though it exists, doesn’t significantly hinder the release of the proton as an H+ ion.
- More importantly, the intramolecular hydrogen bonding does not hinder the stabilization of the negative charge on the phenoxide ion formed after the loss of the proton. The hydrogen bond is retained even in the phenoxide ion, so the charge can still be delocalized into the nitro group.
3. Electron-Donating Effect of the Methoxy Group (-OCH₃):
- The methoxy group (-OCH₃) is an electron-donating group. It pushes electron density towards the benzene ring (primarily through the +M mesomeric effect).
- In ortho-methoxyphenol, the methoxy group destabilizes the phenoxide ion by increasing the electron density on the oxygen atom. This destabilization makes it harder for the phenol to lose a proton, reducing its acidity.
16. Explain how does the – OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Ans :
Electron Donation by Resonance: The -OH group is a powerful ortho, para-directing and activating group due to its ability to donate electron density to the benzene ring through resonance (the +M effect). The oxygen atom’s lone pair of electrons can delocalize into the pi system of the benzene ring. This creates several resonance structures, increasing the electron density, particularly at the ortho and para positions.
Increased Electron Density: The increased electron density makes the benzene ring more attractive to electrophiles (electron-deficient species). Electrophiles are seeking electron density to form a bond, and the -OH group makes the ring a more generous electron donor.
Stabilization of the Intermediate Carbocation: When an electrophile attacks the benzene ring, it forms a carbocation intermediate. The -OH group, through its resonance effect, helps to stabilize this carbocation by delocalizing the positive charge. This stabilization lowers the activation energy for the electrophilic substitution reaction, making it proceed more readily. The resonance structures involving the -OH group are particularly important when the electrophile attacks at the ortho or para positions, because it is at those positions that the -OH group can directly participate in delocalizing the positive charge of the carbocation.
Ortho/Para Direction: The enhanced electron density is greatest at the ortho and para positions due to the resonance effect. Also, the carbocation intermediate formed when the electrophile attacks at these positions is more stable due to the direct participation of the -OH group in resonance stabilization.
Therefore, electrophilic substitution reactions on phenol tend to occur preferentially at the ortho and para positions.
17. Give equations of the following reactions:
(i) Oxidation of propan-l-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 acid with phenol
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Ans :
18. Explain the following with an example
(i) Kolbe’s reaction
(ii) Reimer – Tiemann reaction –
(iii) Williamson ether synthesis
(iv) Unsymmetrical ether
Ans :
(i) Oxidation of propan-1-ol with alkaline KMnO₄ solution:
CH₃CH₂CH₂OH + [O] –KMnO₄/OH⁻–> CH₃CH₂COOH + H₂O
(Propan-1-ol) (Propanoic acid)
Propan-1-ol (a primary alcohol) is oxidized by alkaline potassium permanganate (KMnO₄) to propanoic acid. The “[O]” represents the oxidizing agent.
(ii) Bromine in CS₂ with phenol:
C₆H₅OH + Br₂ –CS₂/Low Temp–> o-BrC₆H₄OH + p-BrC₆H₄OH + HBr
(Phenol) (o-Bromophenol) (p-Bromophenol)
Phenol reacts with bromine in carbon disulfide (CS₂) at low temperature to give a mixture of ortho-bromophenol and para-bromophenol. The -OH group strongly activates the benzene ring and directs the bromine to the ortho and para positions. If excess bromine is used, further bromination can occur.
(iii) Dilute HNO₃ with phenol:
C₆H₅OH + HNO₃ –Dil. HNO₃/Low Temp–> o-HOC₆H₄NO₂ + p-HOC₆H₄NO₂ + H₂O
(Phenol) (o-Nitrophenol) (p-Nitrophenol)
Phenol reacts with dilute nitric acid (HNO₃) at low temperature to give a mixture of ortho-nitrophenol and para-nitrophenol. Again, the -OH group directs the incoming nitro group to the ortho and para positions.
(iv) Treating phenol with chloroform in the presence of aqueous NaOH (Reimer-Tiemann Reaction):
C₆H₅OH + CHCl₃ + 3NaOH → o-HOC₆H₄CHO + 3NaCl + 2H₂O (Major product)
(Salicylaldehyde)
+
p-HOC₆H₄CHO + 3NaCl + 2H₂O (Minor product)
(p-Hydroxybenzaldehyde)
This is the Reimer-Tiemann reaction. Phenol reacts with chloroform (CHCl₃) in the presence of aqueous NaOH to introduce an aldehyde (-CHO) group at the ortho position (major) and para position (minor) of the benzene ring. The major product formed is salicylaldehyde (2-hydroxybenzaldehyde).
19. Write the mechanism of acid dehydration of ethanol to yield ethene.
Ans :
20. How are the following conversions carried out?
(i) Propane → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl mag. chloride → Propan-1-ol
(iv) Methyl mag. bromide → 2-Methylpropan-2-ol.
Ans :
21. Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
(ii) Oxidation of a primary alcohol to aldehyde.
(iii) Brominationofphenolto2,4,6-tribromophenol
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propan-2-oI to propene.
(vi) Butan-2-one to butan-2-oL .
Ans :
(i) Oxidation of a primary alcohol to carboxylic acid: Acidified potassium dichromate (K2Cr2O7/H2SO4) or alkaline potassium permanganate solution (KMnO4/OH-)
(ii) Oxidation of a primary alcohol to aldehyde: Pyridinium chlorochromate (PCC) or Collins reagent (CrO3.2C5H5N)
(iii) Bromination of phenol to 2,4,6-tribromophenol: Bromine water (Br2 in H2O)
(iv) Benzyl alcohol to benzoic acid: Acidified potassium permanganate solution (KMnO4/H+) or Jones reagent (CrO3/H2SO4)
(v) Dehydration of propan-2-ol to propene: Concentrated sulfuric acid (H2SO4) or phosphoric acid (H3PO4) with heat
(vi) Butan-2-one to butan-2-ol: Sodium borohydride (NaBH4) or Lithium aluminum hydride (LiAlH4)
22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Ans : Ethanol has a higher boiling point than methoxymethane because ethanol molecules can form strong hydrogen bonds, while methoxymethane molecules can only interact through weaker London dispersion forces and weak dipole-dipole interactions. Hydrogen bonds require more energy to break, leading to a higher boiling point.
23. Give IUPAC names of the following ethers.
Ans :
(i) C₂H₅OCH₂CH(CH₃)₂
- Longest Carbon Chain: The longest carbon chain is the one with the isopropyl group, making it a propane base.
- Alkoxy Group: The ethoxy group (C₂H₅O-) is attached to the propane chain.
- Numbering: Number the propane chain starting from the carbon attached to the oxygen. The ethoxy group is on carbon 1, and the methyl group is on carbon 2.
- IUPAC Name: 1-Ethoxy-2-methylpropane
(ii) CH₃-O-CH₂CH₂Cl
- Longest Carbon Chain: The longest carbon chain is the two-carbon chain with the chlorine.
- Alkoxy Group: The methoxy group (CH₃O-) is attached to this chain.
- Numbering: Number the carbon chain from the carbon with the chlorine. The chlorine is on carbon 1, and the methoxy is on carbon 1 as well.
- IUPAC Name: 1-Chloro-2-methoxyethane
(iii) O₂N-C₆H₄-OCH₃ (p)
- Base Structure: This is a substituted benzene (phenol with an ether group).
- Substituents: You have a nitro group (O₂N-) and a methoxy group (CH₃O-). The methoxy group is what makes it an ether.
- Positions: The nitro group is para (p) to the ether linkage (1-methoxy-4-nitro…).
- IUPAC Name: 4-Nitroanisole (Commonly used name is acceptable) or 1-Methoxy-4-nitrobenzene
(iv) CH₃CH₂CH₂OCH₃
- Longest Carbon Chain: The longest chain is the three-carbon chain.
- Alkoxy Group: The methoxy group (CH₃O-) is attached to this chain.
- Numbering: The methoxy group is on carbon 1.
- IUPAC Name: Methoxypropane
(v) (The cyclic ether with OC₂H₅)
- Base Structure: The base is a cyclohexane ring.
- Substituents: You have an ethyl group attached to the oxygen (ethoxy group), and two methyl groups on the ring.
- Numbering: Start numbering at the carbon attached to the ethoxy group. The methyl groups are on carbons 1 and 4 (geminal and cis).
- IUPAC Name: 1-Ethoxy-1,4,4-trimethylcyclohexane
(vi) (The benzene ring with OC₂H₅)
- Base Structure: The base is benzene.
- Substituent: You have an ethoxy group (OC₂H₅).
- Common Name: This is commonly called phenetole.
- IUPAC Name: Ethoxybenzene
24. Write the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis :
(i) 1-Propoxypropane
(ii) 2-Methoxy-2-methylpropane
(iii) Ethoxybenzene
(iv) Methoxyethane.
Ans :
25. Illustrate with examples the limitations of Willamson synthesis for the preparation of certain types of ethers.
Ans : Williamson ether synthesis, while a versatile method for preparing ethers, has some significant limitations, particularly concerning the types of ethers it can synthesize effectively. Here are some illustrated examples:
26. How is 1-propoxypropane synthesised from propane-1-ol? Write mechanism of the reaction.
Ans : 1-propoxypropane (also known as dipropyl ether) can be synthesized from propan-1-ol through an acid-catalyzed dehydration reaction. This is a condensation reaction where two alcohol molecules react to eliminate a water molecule and form an ether.
27. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Ans : When secondary or tertiary alcohols are subjected to acid-catalyzed dehydration conditions, the formation of a stable carbocation favors elimination reactions, leading primarily to alkenes. The competing SN1 substitution reaction to form the ether is disfavored due to steric hindrance (especially in tertiary alcohols) and the dominance of elimination. Therefore, this method is not synthetically useful for preparing ethers from secondary or tertiary alcohols. Primary alcohols, which follow an SN2 mechanism, are the only alcohols where dehydration to ethers is synthetically useful.
28. Write the equation of the reaction of hydrogen iodide with (i)1-propoxypropane (ii)methoxybenzene, and (iii)benzyl ethyl ether
Ans :
29. Explain the fact that in alkyl aryl ethers, alkoxy group :
(i) activates the benzene ring towards electrophilic substitution.
(ii) directs the incoming substituents towards ortho and para positions in the ring.
Ans :
(i) Activation of the Benzene Ring:
The alkoxy group activates the benzene ring primarily through resonance (also known as the +M or mesomeric effect). The oxygen atom in the alkoxy group has lone pairs of electrons that can delocalize into the pi system of the benzene ring, increasing the electron density and making the ring more nucleophilic. This enhanced nucleophilicity makes the benzene ring more attractive to electrophiles, which are electron-deficient species seeking electrons to form bonds. The alkoxy group effectively makes the benzene ring a better electron donor, thus activating it towards electrophilic attack. Furthermore, the alkoxy group helps stabilize the carbocation intermediate formed during the reaction via resonance, lowering the activation energy and making the reaction proceed more readily.
(ii) Ortho and Para Direction:
The alkoxy group is an ortho, para-directing group. This means that when electrophilic substitution occurs, the incoming substituent preferentially goes to the ortho or para positions relative to the alkoxy group. The resonance structures resulting from the alkoxy group’s electron donation show that the electron density is highest at the ortho and para positions. These positions are where the negative charge is concentrated in the resonance hybrid. Additionally, the carbocation intermediate formed when the electrophile attacks at the ortho or para position is more stable than if the electrophile attacked at the meta position. This is because the alkoxy group can directly participate in delocalizing the positive charge of the carbocation through resonance, creating more resonance structures. The carbocation formed from meta attack cannot be similarly stabilized by the alkoxy group.
30. Write the mechanism of the reaction of HI with methoxymethane.
Ans :
31. Write equations of the following reactions:
(i) Friedel-Crafts reaction -alkylation of anisole
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium
(iv) Friedel-Craft’s acetylation of anisole.
Ans :
32. Show how will you synthesise the following from appropriate alkenes.
Ans :
(ii)
(iii)
(iv)
33. When 3-methylbutant 2-ol is treated with HBr, the following reaction takes place:
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Ans :
The reaction of 3-methylbutan-2-ol with HBr results in the formation of 2-bromo-2-methylbutane. The key step in this reaction is the carbocation rearrangement, which leads to the more stable tertiary carbocation and hence the observed product.
