NCERT Solution Chemistry Class 12 Chapter 11 Alcohols, Phenols and Ethers
NCERT Solution Chemistry Class 12 Chapter 11 Alcohols, Phenols and Ethers all questions and answers. Chemistry Class 12 11th Chapter Alcohols, Phenols and Ethers exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Chemistry Class 12 Chapter 11. This solution is designed to help students who are looking to brush up on their Chemistry concepts on Chapter 11 Alcohols, Phenols and Ethers.
Questions And Answers
(1) Write IUPAC names of the following compounds:
Ans-
(i) 2, 2, 4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2, 4-diol
(iii) Butane-2, 3-diol
(iv) Propane-1, 2, 3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2, 5-Dimethylphenol
(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) 1-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane –1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-1-ol
(x) 4-Chloro-3-ethylbutan-1-ol.
Ans-The following structure for given Compounds.
(i) 2-Methylbutan-2-ol
(ii) 1-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane –1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(viii) 3-Cyclohexylpentan-3-ol
(ix) Cyclopent-3-en-1-ol
(x) 4-Chloro-3-ethylbutan-1-ol
(3) (i) Draw the structures of all isomeric alcohols of molecular formula C5H12Oand give their IUPAC names.
(ii) Classify the isomers of alcohols in question(i) as primary, secondaryand tertiary alcohols.
Ans- structure of possible isomers of given compound are follows-
Classification –
Primary | Secondary |
Tertiary |
Pentane-1-ol | pentane-2-ol
|
2-methylbutan-2-ol
|
3-methyl-butan-1-ol
|
3-methyl-butan-2-ol
|
|
2-methyl-butan-1-ol
|
pentane-3-ol
|
|
2,2-dimethylpropan-1-ol
|
(4) Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Ans-Temperature at which liquids starts to boil is termed as boiling point of compound. Intermolecular force of attraction in compound is one the factors which affect on boiling point of compound. boiling point is directly proportional to the intermolecular forces in molecule. In propanol molecules are held by hydrogen bonding and Van der Waal’s forces. While in butane only weak van der Waal’s force of attraction is present in molecules. hence, propanol has highest boiling point (392 K) as compared to that of butane (308 K).
(5) Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Ans-Hydrogen bonding is partially electrostatic force which hydrogen carries partially positive charge and another electronegative donor atom carries partiality negative charge. Solubility of water depends upon formation of hydrogen bonding. In alcohols, −OH group is present which forms hydrogen bonds with water. Therefore alcohols are more soluble in water. on the other hand , there is absence of -OH group in hydrocarbons hence hydrocarbons can not form hydrogen bonds. therefore, hydrocarbon are comparatively less soluble in water than alcohol of comparable molecular masses.
(6) What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Ans-Hydroboration-oxidation reaction is the chemical reaction in which alkenes are converted into alcohols by addition of borane and further oxidation.
E g. 2-methyl-prop-1-ene on hydroboration-oxidation gives 2-methylpropylborane.
Oxidation by hydrogen peroxide in presence of alkali medium gives 2-methylpropan-1-ol.
(7) Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Ans-o-cresol , m-cresol and p-cresol are three possible structure of compound having formula C7H8O. The structures of three monohydric phenols are as follows-
(8) While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Ans-Atoms ino-nitrophenol is held by intramolecular hydrogen bonding. Therefore it has lower melting point and can easily convert into vapour during separation of mixture of ortho and para nitrophenols by steam distillation. On the other hand, intermolecular hydrogen bonding is present in p-nitrophenol, due to which p-nitrophenol has higher melting point. Therefore , p-nitrophenol is not steam volatile.
(9) Give the equations of reactions for the preparation of phenol from cumene.
Ans- When Cumene undergoes Oxidation in the presence of air gives cumene hydro-peroxide. again cumene hydroxide treated with dilute acid it gives phenol and acetone.
(10) Write chemical reaction for the preparation of phenol from chlorobenzene.
Ans- chlorobenzene treated with NaOH at 623 K temperature and 320 atm pressure,it gives sodium phenoxide. Again sodium phenoxide treated with hydrochloric acid, phenol is produced.
(11) Write the mechanism of hydration of ethene to yield ethanol.
Ans- Mechanism for the given chemical reaction is as as follows –
(1) The protonation of ethane is first step of the reaction mechanism. there is formation of carbocation by the electro phillic attack of hydronium ion.
(2) In second step, water acts as nucleophile and attacks on carbocation.
(12) You are given benzene, conc. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Ans- For the preparation of phenol , benzene is heated with conc. sulphuric acid , it gives benzene sulphonic acid. Again benzene sulphonic acid is heated with sodium hydroxide, it gives sodium phenoxide. sodium phenoxide undergoes hydrolysis with sulphuric acid to give phenol.
(13) Show how will you synthesise:
(i) 1-phenylethanol from a suitable alkene.
(ii) cyclohexyl methanol using an alkyl halide by an SN2 reaction.
(iii) pentan-1-ol using a suitable alkyl halide?
Ans- (i) When ethylbenzene undergoes acid catalyzed hydration, it forms1-phenylethanol.Water is added to break pi bond in ethylbenzene in presence of acidic medium.
(ii) SN2 reaction undergoes for the preparation of cyclohexyl methanol. chloro methyl cyclohexane is treated with NaOH to give cyclohexyl methanol. OH- ion acts as nucleophile.
iii) SN2 reaction undergoes for the preparation of Pentan-1-ol . 1-chloropentane treated with NaOH to givePentan-1-ol and sodium chloride as byproduct. OH- ion acts as nucleophile.
CH3CH2CH2CH2CH2Cl + NaOH→┴CH3CH2CH2CH2CH2OH + NaCl
(14) Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Ans-Two reactions that show the acidic nature of phenol are as follows:
(i) When phenol reacts with sodium itforms sodium phenoxide. hydrogen gas is liberted during formation of sodium phenoxide. When acid reacts with metal hydrogen gas is released. Therefore phenol is acidic in nature.
C6H5OH+Na→C6H5−ONa +H2↑
(ii) When acid and base reacts with eath other it forms salt and water. this reaction is also known as neutralization reaction. When phenol reacts with sodium hydroxide itgives sodium phenoxide and water. Therefore it can conclude that phenol is acidic in nature.
C6H5−OH+NaOH→C6H5−ONa + H2O
While comparing phenol with ethanol in terms of acidity, ethwnol is less acidic than phenol .because after deprotonation of phenol, it gives phenoxide ion. Phenoxide ion is resonance stabilized. On the other hand, after losing proton ,ethanol forms ethoxide ion which is not resonance stabilized.
(15) Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?
Ans-Methoxy group is electron releasing group present in ortho position of o-methoxyphenol. Due to this the electron density of O−H bond is increases. hence the process of removal of proton becomes difficult. Hence, ortho methoxyphenol is less acidic. Nitro group is electron withdrawing group which is present in ortho position in o-nitrophenol. Due to this the electron density of O−H bond is decreases. Hence process of removal of proton becomes easy. After removal of proton o-nitrophenoxide ion becomes resonance stabilized. therefore, ortho nitrophenol is a stronger acid.
(16) Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Ans- The hydroxy group (−OH) is an electron donating group.OH- group introduce resonance effecton benzene ring in presence of attacking electrophiles. therefore the electron density in the benzene ring increases. therefore, hydroxyl group activates the benzene ring towards electrophilic substitution.
(17) Give equations of the following reactions:
(i) Oxidation of propan-1-ol with alkaline KMnO4 solution.
(ii) Bromine in CS2 with phenol.
(iii) Dilute HNO3 with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Ans-
(i) When propan-1-ol treated with alk.KMnO4 solution gives, propanoic acid.
(ii) When Phenol is treated with bromine in CS2 , it gives ortho bromophenol and p-bromophenol .
(iii) When phenol undergoes nitration in presence of dil.HNO3 it gives o-nitrophenol and p-nitrophenol.
(iv) When phenol treated with chloroform in presence of sodium hydroxide it gives sodium-2-formylphenolate which further treated with water in acidic medium to give 2-hydroxybenzaldehyde.
(18) Explain the following with an example.
(i) Kolbe’s reaction.
(ii) Reimer-Tiemann reaction.
(iii) Williamson ether synthesis.
(iv) Unsymmetrical ether.
Ans-
(i) In Kolbe’s reaction,sodium phenoxide is produced when phenol reacts with sodium hydroxide. This sodium phenoxide again treated with carbon dioxide, followed acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid (salicylic acid). Phenoxide ion is more reactive towards electrophilic aromatic substitution reaction.
(ii) Reimer-Tiemann reaction.
In reimer-tiemannre action involves conversion of phenols into to salicylaldehyde.When phenol treated with chloroform in presence of sodium hydroxide it gives sodium-2-formylphenolate which further treated with water in acidic medium to give 2-hydroxybenzaldehyde(salicylaldehyde
(iii) Williamson ether synthesis.
The williamson ether sythesis is the chemical reaction in which Sodium alkoxide reacts with alkyl halide to form an ether.
R-ONa+ R`-Cl→R−O−R`+NaCl
e.g when sodium ethoxide reacts with ethyl chloride gives diethyl ether with sodium chloride as byproduct .
C2H5ONa+C2H5Cl→C2H5−O−C2H5+NaCl
(iv) Unsymmetrical ether.
Unsymmetrical ether in which two different alkyl groups are attached to O atom. The general formula of unsymmetrical ether is R−O−R′.
e.g. ethyl propyl ether – CH3−CH2−O−CH3−CH2−CH3
(19) Write the mechanism of acid dehydration of ethanol to yield ethene.
Ans-The reaction of dehydration of ethanol is as follows –
The steps of mechanism of given reaction is as follows –
(i) First step involves addition of H+ion i.e. the protonation of oxygen atom inhydroxyl group.
(ii) In Second step ,there is formation of carbonium ion by removal of water molecule.
(iii) In third step, H+ ion gets removed i.e. deprotonation to which there is formation of carbon carbon double bond. Hence Ethene is produced by dehydration of ethanol.
(20) How are the following conversions carried out?
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol
Ans-
(i) Propene → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium chloride → Propan-1-ol
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
(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) Bromination of phenol to 2,4,6-tribromophenol.
(iv) Benzyl alcohol to benzoic acid.
(v) Dehydration of propan-2-ol to propene.
(vi) Butan-2-one to butan-2-ol.
Ans-
Sr. No. |
Reaction |
Reagents |
1 | Oxidation of a primary alcohol to carboxylic acid | Acidified K2Cr2O7 / acidified KMnO4 |
2 | Oxidation of a primary alcohol to aldehyde | Pyridinium chlorochromate (PCC) in CH2 Cl2 or Pyridinimium dichromate (PDC) in CH2Cl2 |
3 | Bromination of phenol to 2,4,6-tribromophenol | Aq. Br2 (bromine water) |
4 | Benzyl alcohol to benzoic acid | Acidified / alkaline KMnO4 followed by hydrolysis with dil sulphuric acid |
5 | Dehydration of propan-2-ol to propene | 85% phosphoric acid |
6 | Butan-2-one to butan-2-ol | NaBH4 or Ni/H2
|
(22) Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Ans- Methoxymethane and ethanol having same molecular formula C2H6O. In ethanol hydrogen atom is directly attached to electronegative oxygen atom. Therefore intermolecular Hydrogen bonding is occurs between ethanol molecules.
Therefore more heat energy required to separate molecules and therefore ethanols possesses higher boiling point. On the other hand methoxymethane which does not OH- group therefore it cannot possesses hydrogen bonding.
(23) Give IUPAC names of the following ethers:
Ans-The IUPAC names of the given ethers are as given.
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 4-Nitroanisole
(iv) 1-Methoxypropane
(v) 1-Ethoxy-4,4-dimethylcyclohexane
(vi) Ethoxybenzene
(24) Write the names of reagents and equations for the preparation of the followingethers by Williamson’s synthesis:
(i) 1-Propoxypropane (ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane
Ans- The names of reagents and equations for the preparation of the ethers by Williamson’s synthesis are as follows.
(i) CH3CH2CHONa + CH3CH2CH2Br → C2H5CH2-O-CH2C2H5 + NaBr
Sodium propoxide 1-bromopropane 1-propoxypropane
(ii)
(iii)
iv) CH3CH2-ONa + CH3-Br → CH3CH2-O-CH3 + NaBr
Sodium ethoxide bromoethane 1methoxyethane
(25) Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Ans- SN2 reaction is undergoes during Williamson synthesis for the preparation of ethers. Alkyl halide is used for this reaction is must be primary alkyl halide not be secondary or tertiary alkyl halide because there is possibility of elimination with strong bases . In this reaction, alkoxide ion acts as a nucleophile then it attacks on a primary alkyl halide.
e.g. For preparation of tertiary butyl ethyl ether i.e.2-ethoxy-2-methylpropane. the reactants used are ethyl bromide and sodium tertiary butoxide, if tertiary butyl bromide i.e.2-bromo-2-methylpropane and sodium ethoxide is used for reaction there is formation of 2-methyl-prop-1-ene without formation of ether.
(26) How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.
Ans-1-propoxypropane is prepared from propan-1-ol by the action of acids which donates acid i.e. protic acid such as sulfuric acid, hydrochloric acid, etc.
In the first step of mechanism involves addition of H+ ions in hydroxyl group. i.e. −OH group is protonated.
1-propoxypropane is prepared from propan-1-ol by the action of acids which donates acid i.e. protic acid such as sulfuric acid, hydrochloric acid, etc.
(27) Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Ans- The reaction for preparation of ether from alcohol is followed by SN2 mechanism. steric hindrance created by alkyl groups in secondary and tertiary alcohols therefore the nucleo phillic attack becomes complicated. therefore, elimination reaction undergoes rather than nucleo phillic substitution. it form alkene instead of ether. Hence, secondary or tertiary alcohols is not a suitable for preparation of ether.
(28) Write the equation of the reaction of hydrogen iodide with:
(i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether.
Ans-
(i) Reaction of hydrogen iodide with1-propoxypropane
(ii) Reaction of hydrogen iodide with methoxybenzene
(iii) Reaction of hydrogen iodide with benzyl ethyl ether
(29) Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs theincoming substituents to ortho and para positions in benzene ring.
Ans-In case of aryl alkyl ethers, the alkoxy group releases electrons to remaining part of moleculeby delocalisation of electrons hence itshows +R effect. As positive resonance effect increases, the electron density also increases in the benzene ring. Consequently benzene ring activated towards electrophilic substitution.
Electron density is shown in the following structures.
As per resonance structures given above, electron density increases at ortho and para position than the meta position. therefore, 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 mechanism of the reaction of HI with methoxymethane is shown below:
In the first step of mechanism involves addition of H+ ions in methoxymethane i.e. methoxymethane is protonated.
(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- The different reactions are shown above.
(32) Show how would you synthesise the following alcohols from appropriate alkenes?
Ans- Convertion of alkenes to alcohols can be carried out by the acid catalyzed hydration of alkenes.
A molecule of water is added across C=C double bond. Addition follows Markovnikov’s rule.
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.)
Ans- The steps of mechanism for the given reaction is as follows.
(2) In the second step, there formation of secondary carbocation by losing a water molecule.
(3) In third step of mechanism is hydrogen atom undergoes 1,2-hydride shift. the rearrangement of less stable secondary carbocation to more stable tertiary carbocation by hydrogen atom.
(4) In final step involves formation of 2-bromo-2-methylbutane by the nucelophilic attack of bromide ion on tertiary carbocation.
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