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NCERT Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ethers
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Alcohols, Phenols and Ethers
Chapter: 11
| Part – II |
1. Write IUPAC names of the following compounds:
(i)
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(ii)
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(iii)
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(iv) HOCH2-CHOH-CH2OH
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(v)
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(vi)
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(vii)
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(viii)
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(ix)
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(x) C6H5OC2H5
Ans: C6H5-O-C7H15
Ethoxy benzene
(xi) C6H5-O-C7H15(n-)
Ans: C6H5-O-C7H15
1-phenoxy heptane
(xii)
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2. Write structures of the compounds whose IUPAC names are as follows:
(i) 2-Methylbutan-2-ol.
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(ii) 1-Phenylpropan-2-ol.
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(iii) 3,5-Dimethylhexane –1, 3, 5-triol.
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(iv) 2,3 – Dimethylphenol.
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(v) 1 – Ethoxypropane.
Ans: HOCH2-CHOH-CH2OH.
(vi) 2-Ethoxy-3-methylpentane.
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(vii) Cyclohexylmethanol.
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(viii) 3-Cyclohexyl Pentan-3-ol.
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(ix) Cyclopent-3-en-1-ol.
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(x) 4-Chloro-3-methylbutan-1-ol.
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3. (i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.
Ans: The structures of all isomeric alcohols of molecular formula C5H12O alongwith their IUPAC names are as shown.
(i) CH3CH2CH2CH2CH2OH, Pentan-1-0I
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ii) Primary alcohols – (i),(ii),(iii),(iv)
Secondary alcohols – (v),(vi),(vii).
Tertiary alcohols – (viii).
4. Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Ans: Due to presence of – OH group, propanol undergoes intermolecular hydrogen bonding. However, butane cannot form such bonds since it has no −OH bonds.Breaking hydrogen bonds requires additional energy.
Therefore, propanol has a higher boiling point(391k) as compared to that of butane (309k).
5. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Ans: Water and alcohols both are polar in nature. When alcohols dissolve in water, they break the hydrogen bonds between water molecules to form new hydrogen bonds with alcohol.In contrast, non-polar hydrocarbons do not interact with water or form hydrogen bonds, making them insoluble in water.
Hence, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
6. What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Ans: The Hydroboration Oxidation reaction is an organic chemical reaction which is employed for the conversion of alkenes into alcohols that are neutral. Hydroboration-oxidation is a two-step reaction that converts alkenes into alcohols. In the first step, borane (BH₃) adds to the alkene, forming an organoborane. In the second step, oxidation with hydrogen peroxide (H₂O₂) and sodium hydroxide (NaOH) converts the organoborane to an alcohol.
Example:
7. Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Ans: There are only three possible monohydric phenols with the formula C7H8O. Each of these phenols is methyl substituted phenol.
8. While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give a reason.
Ans: Ortho-nitrophenol is steam volatile. The reason is chelation due to intramolecular H-bonding. Therefore, it can be separated from p-nitrophenol using steam distillation. Para-nitrophenol is not steam volatile because of the intermolecular hydrogen bonding present.
9. Give the equations of reactions for the preparation of phenol from cumene.
Ans: Oxidation of Cumene in the presence of air gives cumene hydro-peroxide. The reaction of cumene hydroperoxide with a dilute acid yields phenol and acetone (as a byproduct).
The cumene required in the above reaction is prepared from benzene and propene by Friedel- crafts reaction.
10. Write chemical reaction for the preparation of phenol from chlorobenzene.
Ans: When chlorobenzene is fused with sodium hydroxide (NaOH) at 623 K and 320 atm pressure, it forms sodium phenoxide. Upon acidification, sodium phenoxide is converted into phenol.
11. Write the mechanism of hydration of ethene to yield ethanol.
Ans: Three steps are involved in the mechanism of hydration of ethene to yield ethanol:
(i) Hydronium ion attacks ethene electrophilic ally, resulting in the protonation of ethene and the formation of a carbocation.
(ii) Water (as a nucleophile) attacks carbocation.
(iii) The deprotonation lastly generates the ethanol.
The mechanism is illustrated above.
12. You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Ans: Benzene is heated with conc. sulphuric acid to form benzene sulphonic acid. It is then heated with sodium hydroxide to give sodium phenoxide. Hydrolysis using sulfuric acid produces phenol. The reaction sequence is described above.
13. Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
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(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
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(iii) pentan-1-ol using a suitable alkyl halide?
Ans: CH3(CH2)4 + NaOH → CH3(CH2)4OH + NaCI
Pentan-1-ol.
14. Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Ans: Following reactions show the acidic nature of phenol.
(i) The reaction of phenol with sodium to form sodium phenoxide with liberation of hydrogen gas.
(ii) The reaction of phenol with sodium hydroxide to give sodium phenoxide and water.
Phenol is more acidic than ethanol because after losing a proton (H+) phenol forms phenoxide ion which is stabilised by resonance whereas ethoxide ion is not.
15. Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
Ans:
In ortho nitro phenol, an electron withdrawing nitro group is present in the ortho position, the electron density of the O−H bond is reduced. This facilitates the removal of a proton. After deprotonation, the ortho-nitrophenoxide ion is resonance stabilised. As a result, ortho-nitrophenol is a stronger acid.In ortho methoxyphenol, electron releasing methoxy group is present in ortho position, the electron density of O−H bond is increased. This makes removal of protons difficult. Hence, ortho methoxyphenol is less acidic.
16. Explain how the –OH group attached to a carbon of benzene ring activates it towards electrophilic substitution?
Ans: Since there is -ve charge at o and p-position.”Hence, the -OH group enhances the reactivity of the benzene ring towards electrophilic substitution reactions.”
17. Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.
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(ii) Bromine in CS2 with phenol.
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(iii) Dilute HNO3 with phenol.
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(iv) Treating phenol with chloroform in presence of aqueous NaOH.
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18. Explain the following with an example.
(i) Kolbe’s reaction.
Ans: Kolbe Reaction: When sodium phenoxide is treated with carbon dioxide under pressure (4 to 7 atoms) and at 400 K, sodium salicylate is formed. The Kolbe reaction is a radical reaction that is essentially a decarboxylative dimerisation. An aqueous solution of sodium or potassium salt of carboxylic acid is electrolyzed in this reaction, resulting in the dissociation of the salt into carboxylate ion and sodium or potassium ions.
(ii) Reimer-Tiemann-Reaktion.
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(iii) Williamson ether synthesis.
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(iv) Unsymmetrical ether.
Ans: Unsymmetrical ether: Ethers in which the groups R and R’ are different are called unsymmetrical ethers.
Examples: Methyl ethyl ether (C₂H₅-O-CH₃) consists of a methyl group and an ethyl group.
19. Write the mechanism of acid dehydration of ethanol to yield ethene.
Ans: The mechanism of acid dehydration of ethanol to yield ethene is shown above.
(i) Due to two lone pairs on oxygen, alcohols act as weak bases and react with strong acids to form oxonium salts.
(ii) The presence of a positive charge on the highly electronegative oxygen atom weakens the C-O bond. Thus, the protonated ethanol readily eliminates a molecule of H₂O to form ethyl carbocation.
(iii) The ethyl carbocation produced in Step 2 is highly reactive and quickly loses a proton to form an ethene molecule.
20. How are the following conversions carried out?
(i) Propene → Propan-2-ol.
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(ii) Benzyl chloride → Benzyl alcohol.
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(iii) Ethyl magnesium chloride → Propan-1-ol.
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(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
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21. Name the reagents used in the following reactions:
(i) Oxidation of a primary alcohol to carboxylic acid.
Ans: Acidified K2Cr2O7 or acidified KMnO4
(ii) Oxidation of a primary alcohol to aldehyde.
Ans: Acidified potassium permanganate.
(iii) Bromination of phenol to 2,4,6-tribromophenol.
Ans: Aqueous bromine (or bromine water).
(iv) Benzyl alcohol to benzoic acid.
Ans: Acidified or alkaline KMnO4 followed by hydrolysis with dil H2SO4
(v) Dehydration of propan-2-ol to propene.
Ans: Ortho phosphoric acid.
(vi) Butan-2-one to butan-2-ol.
Ans: Lithium aluminium hydride.
22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Ans: Hydrogen bonding is present between the ethanol molecules which is absent in the molecules of methoxy methane. A significant amount of energy is needed to break these strong bonds in ethanol.
23. Give IUPAC names of the following ethers:
(i)
Ans: 1-Ethoxy-2-methyl propane.
(ii) CH3OCH2CH2CI.
Ans: 2-Chloro-1-methoxy ethane.
(iii) O2N-C6H4-OCH3(p)
Ans: 4-Nitroanisole.
(iv) CH3CH2CH2OCH3
Ans: 1-Methoxy propane.
(v)
Ans: 1-Ethoxy-4,4-dimethylcyclohexane.
(vi)
Ans: Ethoxy benzene.
24. Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane.
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(ii) Ethoxybenzene.
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(iii) 2-Methoxy-2-methylpropane.
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(iv) 1-Methoxyethane.
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25. Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Ans: (i) If the alkyl halide is tertiary, the major product is an alkene. In this case, the alkoxide acts as a strong base for elimination rather than as a nucleophile for substitution.
(ii) Chlorobenzene does not give Williamson’s synthesis due to partial double bond character of C-Cl bond. So, aromatic ether are prepared from sodium phenoxide.
26. How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Ans: An one of the following two methods can be employed.
(i) Williamson,synthesis
(a)
(b)
Mechanism
(ii) Dehydration of 1-propanol using concentrated sulfuric acid at 413 K yields propene and water, via an elimination reaction.
27. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give a reason.
Ans: Dehydration of alcohol to form ether is a bimolecular reaction (SN2 mechanism). This is suitable for unhindered alcohols. In secondary and tertiary alcohols, the alkyl groups introduce steric hindrance, making the nucleophilic attack more challenging. Therefore, elimination to form an alkene is preferred over substitution to produce an ether. Hence, preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method.
28. Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane.
Ans: If HI is present in a limited quantity, a mixture of 1-iodopropane and 1-propanol is produced. If HI is present in excess, then the 1-propanol formed further reacts with HI, resulting in the formation of only 1-iodopropane.
(ii) methoxybenzene. and
Ans:
-CH, bond is weaker than -OCH. Therefore, CH, bond is broken easily to form phenol and iodo Dethane and no CH,OH is formed. It is also due to steric factors during protonation at methoxy benzene.
(iii) benzyl ethyl ether.
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Smaller group forms RX preferably due to steric factors.
29. Explain the fact that in aryl alkyl ethers:
(i) the alkoxy group activates the benzene ring towards electrophilic substitution. and
(ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Ans: Alkyl aryl ethers are the resonating hybrids os the the following structure:
The alkoxy group (-OR) is ortho and para directing group and it activates the aromatic ring in ortho and para positions due to conjuga-tion with A- electrons of the benzene ring. How-ever, ethers are some what less reactive than phenol in these substitution reactions. Activa-tion of benzene ring takes place as shown above.
The following reactions show that alkoxy group directs the incoming substituents to ortho and para positions in benzene ring.
30. Write the mechanism of the reaction of HI with methoxymethane.
Ans: The first step in the reaction mechanism of hydrogen iodide (HI) with methoxymethane involves the protonation of methoxymethane. Second step is the nucleophilic attack of iodide ion to form iodomethane and methanol. If HI is present in excess and if temperature is high, methanol reacts with HI to form methyl iodide.
If HI is present in excess, CH3OH formed in step II is fu
31. Write equations of the following reactions:
(i) Friedel-Crafts reaction – alkylation of anisole.
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(ii) Nitration of anisole.
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(iii) Bromination of anisole in ethanoic acid medium.
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(iv) Friedel-Crafts acetylation of anisole.
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32. 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.
(i)
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(ii)
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(iii)
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(iv)
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33. When 3-methylbutan-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.
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