{"id":37141,"date":"2022-12-01T16:22:20","date_gmt":"2022-12-01T10:52:20","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=37141"},"modified":"2022-12-01T16:22:58","modified_gmt":"2022-12-01T10:52:58","slug":"ts-inter-2nd-year-chemistry-study-material-chapter-12a","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-chemistry-study-material-chapter-12a\/","title":{"rendered":"TS Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material<\/a> Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers.<\/p>\n

TS Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers<\/h2>\n

Very Short Answer Questions (2 Marks)<\/span><\/p>\n

Question 1.
\nExplain why propanol has higher boiling point than that of the hydrocarbon butane.
\nAnswer:
\nPropanol and butane are of comparable molecular mass. However, the boiling point of propanol is higher than that of butane, due to the presence of intermolecular hydrogen bonding.<\/p>\n

Question 2.
\nAlcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
\nAnswer:
\nAlcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses, due to their ability to form hydrogen bonds with water molecules.<\/p>\n

\"TS<\/p>\n

Question 3.
\nGive the structures and 1UPAC names of monohydric phenols of molecular formula, C7<\/sub>H8<\/sub>O.
\nAnswer:
\n\"TS
\nCommon Name:
\no – Cresol
\nm – Cresol
\np – Creol<\/p>\n

IUPAC Name:
\n2 – Methylphenol
\n3 – Methylphenol
\n4 – Methylphenol<\/p>\n

Question 4.
\nGive the reagents used for the preparation of phenol from chlorobenzene.
\nAnswer:<\/p>\n

    \n
  1. NaOH<\/li>\n
  2. HCl.<\/li>\n<\/ol>\n

    Question 5.
    \nPreparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
    \nAnswer:
    \nThis is because elimination competes over substitution and as a consequence, alkenes are easily formed.<\/p>\n

    \"TS<\/p>\n

    Question 6.
    \nWrite the mechanism of the reaction of HI with methoxymethane.
    \nAnswer:
    \n\"TS<\/p>\n

    Question 7.
    \nName the reagents used in the following reactions.<\/p>\n

      \n
    1. Oxidation of primary alcohol to carboxylic acid.<\/li>\n
    2. Oxidation of primary alcohol to aldehyde.<\/li>\n<\/ol>\n

      Answer:<\/p>\n

        \n
      1. Acidified potassium permanganate.<\/li>\n
      2. Pyridinium chlorochromate (PCC) or CrO3<\/sub> in anhydrous medium.<\/li>\n<\/ol>\n

        Question 8.
        \nWrite the equations for the following reactions.
        \ni) Bromination of phenol to 2, 4, 6-tribromophenoI.
        \nii) Benzyl alcohol to benzoic acid.
        \nAnswer:
        \ni) A white precipitate of 2,4,6 – tribromophenol is formed when phenol is treated with bromine water.
        \n\"TS
        \nii) Benzyl alcohol is oxidised to benzoic acid by acidified KMn04 or acidic solution of sodium dichromate.
        \n\"TS<\/p>\n

        Question 9.
        \nIdentify the reactant needed to form t-butyl alcohol from acetone.
        \nAnswer:
        \nMethyl magnesium halide (Grignard reagent) is needed to form t-butyl alcohol from acetone.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 10.
        \nWrite the structures for the following compounds.
        \ni) Ethoxyethane
        \nii) Ethoxybutane
        \niii) Phenoxyethane
        \nAnswer:
        \ni) C2<\/sub>H5<\/sub>OC2<\/sub>H5<\/sub> – Ethoxyethane
        \nii) CH3<\/sub>CH2<\/sub>CH2<\/sub>CH2<\/sub>OCH2<\/sub>CH3<\/sub> – Ethoxybutane
        \niii) C2<\/sub>H5<\/sub>OC6<\/sub>H5<\/sub> – Phenoxyethane<\/p>\n

        Short Answer Questions (4 Marks)<\/span><\/p>\n

        Question 11.
        \nDraw the structures of all isomeric alcohols of molecular formula C5<\/sub>H12<\/sub>O and give their IUPAC names and classify them as primary, secondary and tertiary alcohols.
        \nAnswer:
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 12.
        \nWhile separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
        \nAnswer:
        \nO-nitrophenol will be steam volatile due to intramolecular hydrogen bonding while p-nitro- phenol will be less volatile due to intermolecular hydrogen bonding which causes association of molecules.
        \n\"TS<\/p>\n

        Question 13.
        \nGive the equations for the preparation of phenol from Cumene.
        \nAnswer:
        \n\"TS<\/p>\n

        Question 14.
        \nWrite the mechanism of hydration of ethene to yield ethanol.
        \nAnswer:
        \n\"TS
        \nMechanism : The mechanism of the reaction involves three steps.
        \nStep 1: Protonation of the alkene (ethene) to form carbonation by electrophilic attack of H3<\/sub>O+<\/sup>.
        \n\"TS
        \nStep 2 : Nucleophilic attack of water on carbocation.
        \n\"TS
        \nStep 3 : Deprotonation to form an alcohol.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 15.
        \nExplain the acidic nature of phenols and compare with that of alcohols.
        \nAnswer:
        \nThe reactions of phenol with metals (e.g. : Na, Al) and NaOH indicate its acidic nature. The hydroxyl group, in phenol is directly attached to the sp2<\/sup> hybridised carbon of benzene ring which acts as an electron withdrawing group. The charge distribution in phenol molecule as depicted in its resonance structures, causes the oxygen of – OH group to be positive.
        \n\"TS
        \nThe reaction of phenol with NaOH solution indicates that phenols are stronger acids than alcohols and water. Let us compare the acidic nature of phenol with that of alcohol.
        \nThe ionisation of an alcohol and a phenol takes place as shown below.
        \n\"TS
        \nOwing to the higher electronegativity of sp2<\/sup> hybridised carbon of phenol to which – OH is attached, electron density decreases on oxygen. This increases the polarity of – OH bond and results in an increase in ionisation of phenols than that of alcohols.<\/p>\n

        The relative stabilities of alkoxide and phenoxide ions should be considered. In alkoxide ion, the negative charge is localised on oxygen whereas in phenoxide ion, the charge is delocalised. The delocalisation of negative charge makes phenoxide ion more stable and favours the ionisation of phenol. There is delocalisation in unionised phenol also but the resonance structures have charge separation. Hence, phenol molecule is less stable them phenoxide ion.
        \n\"TS<\/p>\n

        Question 16.
        \nWrite the products formed by the reduction and oxidation of phenol. [IPE 14]
        \nAnswer:
        \nReduction of phenol: Phenol is converted to benzene when heated with zinc dust.
        \n\"TS
        \nOxidation : Phenol gives benzoquinone on oxidation with chromic acid.
        \n\"TS
        \nIn the presence of air, phenols are slowly oxidised to dark coloured mixtures containing quinones.<\/p>\n

        \"TS<\/p>\n

        Question 17.
        \nEthanol with H2<\/sub>SO4<\/sub>, at 443K forms ethane while at 413 K it forms ethoxy ethane. Explain the mechanism.
        \nAnswer:
        \n\"TS
        \nStep 2 : Formation of carbocation.
        \n\"TS
        \nMechanism : The formation of ether is a nucleophilic bimolecular reaction (SN<\/sub>2) involving the attack of alcohol molecule on a protonated alcohol.
        \n\"TS<\/p>\n

        Question 18.
        \nAccount for the statement: Alcohols boil at highest temperature than hydrocarbons and ethers of comparable molecular masses.
        \nAnswer:
        \nThe higher boiling points of alcohols are mainly due to the presence of inter molecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 19.
        \nExplain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
        \nAnswer:
        \nIn benzene derivatives the electrophile is most likely to react at the position of highest electron density. The methoxy group ih anisole is an electron releasing group. It increases the relative electron density at ortho and para positions and hence electrophilic substitution takes place at these positions and not at the meta position.
        \n\"TS<\/p>\n

        Question 20.
        \nWrite the products of the following reactions:
        \n\"TS
        \nAnswer:
        \n\"TS<\/p>\n

        Long Answer Questions (8 Marks)<\/span><\/p>\n

        Question 21.
        \nWrite the IUPAC names of the following compounds:
        \n\"TS
        \nAnswer:
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 22.
        \nWrite structures of the compounds whose IUPAC names are as follows: [IPE 14]
        \ni) 2- Methyl butan-1- ol
        \nii) 1-Phenylpropan-2-ol
        \niii) 3, 5-Dimethythexane- 1, 3, 5- triol
        \niv) 2, 3- Dlethylphenol
        \nv) 1 – Ethoxypropane
        \nvi) 2- Ethoxy -3- methylpentane
        \nvii) Cyclohexylmethanol
        \nviii) 3- Chloromethylpentan – 1 – ol
        \nAnswer:
        \n\"TS
        \n\"TS<\/p>\n

        Question 23.
        \nWrite the equations for the preparation of phenol using benzene, Cone. H2<\/sub>SO4<\/sub> and NaOH.
        \nAnswer:
        \nBenzene is sulphonated with oleum and benzene sulphonic acid so formed is converted to sodium phenoxide on heating with molten sodium hydroxide. Acidification of the sodium salt gives phenol.
        \n\"TS<\/p>\n

        Question 24.
        \nIllustrate hydroboration-oxidation reaction with a suitable example.
        \nAnswer:
        \nDiborane B2<\/sub>H6<\/sub> [or (BH3<\/sub>)2<\/sub>] reacts with aikenes to give trialkyl boranes as addition product. This is oxidised to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.
        \nEx : Propene gives Propan-1-ol on hydroboration – oxidation reaction.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 25.
        \nWrite the IUPAC names of the following compounds.
        \n\"TS
        \nAnswer:
        \ni) 2-methyl phenol
        \nii) 4 – methyl phenol
        \niii) 2, 5 – dimethyl phenol
        \niv) 2, 6 – dimethyl phenol<\/p>\n

        Question 26.
        \nHow will you synthesise :
        \ni) 1 – Phenylethanol from a suitable alkene ?
        \nii) Cyclohexylmethanol using an alkyl halide by an SN<\/sub>2 reaction ?
        \niii) Pentan-1-ol using a suitable alkyl halide ?
        \nAnswer:
        \ni) Styrene on acid catalysed hydration gives 1-phenyl ethanol.
        \n\"TS
        \nii) Cyclohexyl chloromethane reacts with aqueous sodium hydroxide solution to give cyclohexyl methanol.
        \n\"TS
        \niii) Pentan-1-ol is obtained by the reaction of 1-chloropentane with aqueous NaOH solution.
        \n\"TS<\/p>\n

        Question 27.
        \nExplain why –
        \ni) Ortho nitrophenol is more acidic than ortho methoxyphenol.
        \nii) OH group attached to benzene ring activates it towards electrophilic substitution.
        \nAnswer:
        \ni) Electron withdrawing groups enhance the acidic strength of phenol. This effect is more pronounced if these groups are present in ortho and para positions. It is due to the effective delocalisation of negative charge in phenoxide ion. On the other hand electron releasing
        \n‘ groups do not favour the formation of phenoxide ion resulting in decrease in acid strength.<\/p>\n

        Nitro group is an electron withdrawing group. It increases the acidic strength of phenol. Methoxy group is an electron releasing group. It decreases the acidic strength of phenol. Hence o-nitrophenol is more acidic than orthomethoxyphenol.<\/p>\n

        ii) The – OH group attached to the benzene ring activates it towards electrophilic substitution. Further, it directs the incoming group to ortho and para positions in the ring as these positions become electron rich due to the resonance effect caused by – OH group.
        \n\"TS<\/p>\n

        Question 28.
        \nWith a suitable example write equations for the following:
        \ni) Kolbe’s reaction.
        \nii) Reimer-Tiemann reaction.
        \niii) Willamson’s ether synthesis.
        \nAnswer:
        \ni) Kolbe\u2019s synthesis: When sodium salt of phenol (sodium phenoxide) is heated with carbon dioxide under pressure, a carboxylic group is introduced in the aromatic ring preferably at ortho position with respect to phenolic group. The sodium salt of o-hydroxybenzoic acid (sodium salicylate) formed, when treated with acid yields salicylic acid.
        \n\"TS<\/p>\n

        ii) Reimer-Tiemann reaction : When phenol is heated with chloroform in the presence of sodium hydroxide at 60\u00b0C, a – CHO group is introduced at the ortho position with respect to the phenolic group. This reaction is known as Reimer-Tiemann reaction.
        \n\"TS<\/p>\n

        iii) Willamson’s ether synthesis: It is an important method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.
        \n\"TS
        \nEthers containing substituted alkyl groups (2\u00b0 or 3\u00b0) may also be prepared by this method. The reaction involves SN<\/sub>2 attack of an alkoxide ion on primary alkyl halide.
        \n\"TS
        \nIn case of secondary and tertiary alkyl halides, elimination competes over substitution. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 29.
        \nHow are the following conversions carried out ?
        \ni) Benzyl chloride to Benzyl alcohol
        \nii) Ethyl magnesium bromide to Propan-1-ol
        \niii) 2-butanone to 2-butanol
        \nAnswer:
        \ni) Benzyl chloride is converted to benzyl alcohol by hydrolysis with aqueous sodium hydroxide.
        \n\"TS<\/p>\n

        ii) Ethyl magnesium bromide reacts with formaldehyde to form an adduct. Hydrolysis of the adduct yields an propan-1-ol.
        \n\"TS<\/p>\n

        iii) 2-butanone bn reduction with sodium borohydride (NaBH4<\/sub>) gives 2-butanol.
        \n\"TS<\/p>\n

        Question 30.
        \nWrite the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
        \ni) 1-Propoxypropane
        \nii) Ethoxybenzene
        \niii) 2-Methoxy-2-methylpropane
        \niv) 1-Methoxyethane
        \nAnswer:
        \ni) Sodium propoxide and propyl bromide.
        \nCH3<\/sub>CH2<\/sub>CH2<\/sub>ONa + CH3<\/sub>CH2<\/sub>CH2<\/sub> – Br \u2192 CH3<\/sub>CH2<\/sub>CH2<\/sub>OCH2<\/sub>CH2<\/sub>CH3<\/sub> + NaBr<\/p>\n

        ii) Chlorobenzene and sodium ethoxide.
        \n\"TS<\/p>\n

        iii) Sodium tertiary butoxide and methylbromide.
        \n\"TS<\/p>\n

        iv) Sodium ethoxide and methyl bromide.
        \n\"TS<\/p>\n

        Question 31.
        \nHow is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.
        \nAnswer:
        \n1-rpropoxypropane is synthesised from propan-1-ol by dehydration in the presence of sulphuric acid.
        \nMechanism :
        \nThe formation of 1-propoxypropane is an SN<\/sub>2 reaction involving the attack of alcohol molecule on a protonated alcohol.
        \n\"TS
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 32.
        \nExplain the fact that in aryl alkyl ethers the alkoxy group activates the benzene ring towards electrophilic substitution.
        \nAnswer:
        \nAlkoxy group \"TS is an electron releasing group. When attached to benzene ring alkoxy group activates the ring towards electrophilic substitution. Further, it directs the incoming group to ortho and para positions in the benzene ring as these positions become electron rich due to the resonance effect caused by the – OR group.
        \n\"TS<\/p>\n

        Question 33.
        \nWrite equations of the below given reactions :
        \ni) Alkylation of anisole
        \nii) Nitration of anisole
        \niii) Friedel-Crafts acylation of anisole
        \nAnswer:
        \ni) Alkylation of anisole: Anisole undergoes Friedel-Crafts alkylation reaction with alkyl halide in the presence of anhydrous AlCl3<\/sub> as catalyst. The alkyl group is introduced in the ortho and para positions.
        \n\"TS<\/p>\n

        ii) Nitration of anisole : Anisole on nitration with a mixture of concentrated H2<\/sub>SO4<\/sub> and HNO3<\/sub> yields a mixture of ortho and para nitroanisole.
        \n\"TS<\/p>\n

        iii) Friedel-Crafts acylation of anisole : Anisole undergoes Friedel-Crafts acylation with acyl halide in the presence of anhydrous AlCl3<\/sub>.
        \n\"TS
        \nThe acyl group is introduced in the ortho and para positions.<\/p>\n

        \"TS<\/p>\n

        Question 34.
        \nShow how you would synthesize Hie following alcohols from appropriate aikenes ?
        \n\"TS
        \nAnswer:
        \nBy acid – catalysed hydration
        \n\"TS<\/p>\n

        Question 35.
        \nExplain why phenol with bromine water forms 2, 4, 6-tribromophenol while on reaction with bromine in CS2 at low temperatures forms para-bromophenol as the major product.
        \nAnswer:
        \nThe hydroxyl group (- OH) is a very powerful activating substituent, and electrophilic substitution in phenols occurs faster, and under milder conditions, than in benzene.<\/p>\n

        Bromination of phenol occurs readily even in the absence of a catalyst at low temperature; Substitution occurs primarily at the para position to the hydroxyl group. When the para position is blocked, ortho substitution is observed. The reaction is carried out in a non-polar solvent CS2<\/sub> or ClCH2<\/sub>CH2<\/sub>Cl. In polar solvents such as water it is difficult to limit the bromination of phenols to monobromination. With bromine water all three positions that are ortho or para to the hydroxyl undergo rapid substitution.
        \n\"TS<\/p>\n

        Intext Questions – Answers<\/span><\/p>\n

        Question 1.
        \nClassify the following as primary, secondary and teritary alcohols :
        \n\"TS
        \nAnswer:
        \ni) Primary alcohol
        \nii) Primary alcohol
        \niii) Primary alcohol
        \niv) Secondary alcohol
        \nv) Secondary alcohol
        \nvi) Tertiary alcohol<\/p>\n

        \"TS<\/p>\n

        Question 2.
        \nIdentify allylic alcohols in the above examples.
        \nAnswer:
        \nIn the examples given above (under question 1), (ii) and (vi) are allylic alcohols.<\/p>\n

        Question 3.
        \nName the following compounds according to IUPAC system.
        \n\"TS
        \nAnswer:
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 4.
        \nShow how the following alcohols are prepared by the reaction of a suitable Grignard reagent on methanal ?
        \n\"TS
        \nAnswer:
        \n\"TS<\/p>\n

        Question 5.
        \nWrite structures of the products of the following reactions.
        \n\"TS
        \nAnswer:
        \n\"TS<\/p>\n

        Question 6.
        \nWrite the structures of the major products expected from the following reactions :
        \na) Mononitration of 3-methylphenol
        \nb) Dinitration of 3-methylphenol
        \nc) Mononitration of phenyl methanoate.
        \nAnswer:
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 7.
        \nGive structures of the products you would expect when each of the following alcohol reacts with
        \na) HCl – ZnCl2<\/sub>,
        \nb) HBr and
        \nc) SOCl2<\/sub>.
        \ni) Butan-1-ol
        \nii) 2-Methylbutan-2-ol
        \nAnswer:
        \n\"TS
        \n\"TS
        \n\"TS<\/p>\n

        Question 8.
        \nPredict the major product of acid catalysed dehydration of
        \ni) 1-methylcyclohexanol and
        \nii) butan-1-ol
        \nAnswer:
        \n\"TS
        \nA mixture of but-l-ene and but-2-ene will be obtained. The 1 \u00b0 carbocation formed as intermediate will undergo rearrangement to give a more stable 2\u00b0 carbocation. Loss of proton results in the mixture of but-1-ene and but-2-ene. However, but-l-ene will be the major product.<\/p>\n

        \"TS<\/p>\n

        Question 9.
        \nOrtho and para nitrophenols are more acidic titan phenol. Draw the resonance structures of the corresponding phenoxide ions.
        \nAnswer:
        \nElectron delocalization in o-nitrophenoxide ion.
        \n\"TS
        \nElectron delocalization in p-nitrophenoxide ion.
        \n\"TS<\/p>\n

        Question 10.
        \nWrite the equations involved in the following reactions:
        \ni) Reimer – Tiemann reaction
        \nii) Kolbe’s reaction
        \nAnswer:
        \ni) Reimer-Tiemann reaction :
        \n\"TS
        \nii) Kolbe’s reaction
        \n\"TS<\/p>\n

        Question 11.
        \nWrite the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol
        \nAnswer:
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 12.
        \nWhich of the following is an appropriate set of reactants for the preparation of 1-methoxy- 4-nitrobenzene and why ?
        \n\"TS
        \nAnswer:
        \nSet (ii) is appropriate.<\/p>\n

        Question 13.
        \nPredict the products of the following reactions:
        \ni) CH3<\/sub>-CH2<\/sub>-CH2<\/sub>-O-CHS + HBr \u2192
        \nAnswer:
        \nCH3<\/sub> – CH2<\/sub> – CH2<\/sub> – OH + CH3<\/sub>Br<\/p>\n

        ii)
        \n\"TS
        \nAnswer:
        \n\"TS<\/p>\n

        iii)
        \n\"TS
        \nAnswer:
        \n\"TS<\/p>\n

        iv)
        \n\"TS
        \nAnswer:
        \n(CH3<\/sub>)3<\/sub>C – I + C2<\/sub>H5<\/sub>OH<\/p>\n","protected":false},"excerpt":{"rendered":"

        Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers. TS Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Very Short Answer Questions (2 Marks) Question 1. Explain why propanol has higher boiling point than that of the hydrocarbon butane. Answer: Propanol and … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-chemistry-study-material-chapter-12a\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers. TS Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Very Short Answer Questions (2 Marks) Question 1. Explain why propanol has higher boiling point than that of the hydrocarbon butane. Answer: Propanol and ... 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TS Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Very Short Answer Questions (2 Marks) Question 1. Explain why propanol has higher boiling point than that of the hydrocarbon butane. Answer: Propanol and ... 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