{"id":10479,"date":"2019-12-07T15:51:54","date_gmt":"2019-12-07T10:21:54","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10479"},"modified":"2021-07-02T15:43:22","modified_gmt":"2021-07-02T10:13:22","slug":"2nd-puc-chemistry-question-bank-chapter-11","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-11\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 11 Alcohols, Phenols and Ethers"},"content":{"rendered":"

You can Download Chapter 11 Alcohols, Phenols and Ethers Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Chemistry Question Bank Chapter 11 Alcohols, Phenols and Ethers<\/h2>\n

2nd PUC Chemistry Alcohols, Phenols and Ethers NCERT Textbook Questions<\/h3>\n

Question 1.
\nWrite IUPAC names of the following compounds:<\/p>\n

\"2nd
\n\"2nd
\n\"2nd
\nAnswer:
\n(i) 2, 2,4 – Trimethyl pentan – 3 – ol
\n(ii) 5 – Ethyl heptane – 2,4 – diol
\n(iii) Butane – 2,3 – diol
\n(iv) Propane – 1, 2, 3 – triol
\n(v) 2 – Methyl phenol
\n(vi) 4 – Methyl phenol
\n(vii) 2,5 – Dimethyl phenol
\n(viii) 2,6 – Dimethyl phenol
\n(ix) 1 – Methoxy – 2 – Methyl propane
\n(x) Ethoxy benzene
\n(xi) 1 – Phenoxyheptane
\n(xii) 2 – Ethoxy butane<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nWrite structures of the compounds whose IUPAC names are as follows:
\n(i) 2-Methyl butan-2-ol
\n(ii) phenyl propan-2-ol
\n(iii) 3,5-Dimethylhexane -1, 3, 5-trioi
\n(iv) 2,3 – Diethyl phenol
\n(v) 1 – Ethoxy propane
\n(vi) 2-Ethoxy-3-methyipentane
\n(vii) Cyclohexylmethanol
\n(viii) 3-CycIohexyipentan-3-oI
\n(ix) Cyclopent-3-en-l-ol
\n(x) 3-Chloromethylpentan-l-ol.
\nAnswer:<\/p>\n

\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\n1. Draw the structures of all isomeric alcohols of molecular formula C5<\/sub>H12<\/sub>O and give their IUPAC names.
\n2. Classify the. isomers of alcohols in question 3 (i) as primary, secondary and tertiary alcohols.
\nAnswer:
\n1. The structures of all isomeric alcohols of molecular formula, C5<\/sub>H12<\/sub>O are shown below
\n(a)<\/p>\n

\"2nd
\n\"2nd
\n\"2nd<\/p>\n

2. Primary alcohol:
\nPentan -1 -ol; 2-Methylbutan-1- ol;
\n3-Methylbutan- 1-ol; 2,2\u2014Dimethylproparv4-ol;<\/p>\n

Secondary alcohol:
\nPentan -2 -ol; 3 -Methylbutan- 2 – ol; Pentan-3-ol<\/p>\n

Tertiary alcohol:
\n2 – Methylbutan -2 -ol<\/p>\n

Question 4.
\nExplain why propanol has higher boiling point than that of the hydrocarbon, butane?
\nAnswer:
\nPropand undergoes inter molecular H – bonding because of the presence of – OH group. On the other hand butane does not.
\n\"2nd
\nTherefore, extra energy is required to break hydrogen bonds. For this reason, propand has a higher boiling point than hydrocarbon butane.<\/p>\n

Question 5.
\nAlcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
\nAnswer:
\nAlcohols form H-bonds with water due to the presence of -OH group. However, hydrocarbons cannot form H-bonds with water.
\n\"2nd
\nAs a result, alcohols are comparatively more soluble in water than hydrocarbons, of comparable molecular masses.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhat is meant by hydroboration- oxidation reaction? Illustrate it with an example.
\nAnswer:
\nThe addition of borane followed by oxidation is known as the hydroboration – oxidation reaction. For example, propan- l-ol is produced by the hydroboration – oxidation reaction of propene. In this reaction, propene reacts with deborane (BH3<\/sub>)2<\/sub> to form trialkyl borane as an addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.<\/p>\n

\"2nd<\/p>\n

Question 7.
\nGive the structures and IUPAC names of monohydric phenols of molecular formula, C7<\/sub>H8<\/sub>O.
\nAnswer:<\/p>\n

\"2nd<\/p>\n

Question 8.
\nWhile separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
\nAnswer:
\nIntramolecular H-bonding is present in o-hitro phenol and p-nitrophenol. In p- nitrophenol, the molecules are strongly associated due to the presence of intramolecular bonding. Hence, o-nitrophenol is steam volatile<\/p>\n

\"2nd<\/p>\n

Question 9.
\nGive the equations of reactions for the preparation of phenol from cumene.
\nAnswer:
\nTo prepare phenol, cumene is first oxidized in the presence of air of cumene hydroperoxide.<\/p>\n

\"2nd
\nThen cumene hydroperoxide is treated with dilute acid to prepare phenol and acetone as by-products.<\/p>\n

\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nWrite chemical reaction for the preparation of phenol from chlorobenzene.
\nAnswer:<\/p>\n

\"2nd
\nChlorobenzene is fused with NaOH to produce sodium phenoxide, which gives phenol on acidification.<\/p>\n

Question 11.
\nWrite the mechanism of hydration of ethene to yield ethanol.
\nAnswer:
\nThe mechanism of hydration of ethane to form ethanol involves three steps.
\nStep 1:
\nProtonation of ethane to form carbocation by electrophilic attack of H3<\/sub>O+<\/sup>
\nH2<\/sub>O + H+<\/sup> \u2192 H3<\/sub>O+<\/sup>
\n\"2nd<\/p>\n

Step 2:
\nNucleophilic attack of water on carbocation:<\/p>\n

\"2nd<\/p>\n

Step 3:
\nDeprotonation to form ethanol<\/p>\n

\"2nd<\/p>\n

Question 12.
\nYou are given benzene, cone. H2<\/sub>SO4<\/sub> and NaOH. Write the equations for the preparation of phenol using these reagents.
\nAnswer:<\/p>\n

\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nShow how will you synthesise:
\n(i) 1-phenylethanol from a suitable alkene.
\n(ii) cyclohexylmethanol using an alkyl halide by an SN2 reaction.
\n(iii) pentan-l-ol using a suitable alkyl halide?
\nAnswer:<\/p>\n

\"2nd<\/p>\n

Question 14.
\nGive two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
\nAnswer:
\nThe acidic nature of phenol can be represented by the following two reactions
\n1. Phenol reacts with sodium to give sodium phenoxide, liberating H2<\/sub><\/p>\n

\"2nd<\/p>\n

2. Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by products<\/p>\n

\"2nd
\nThe acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas ethoxide ion does not.<\/p>\n

\"2nd<\/p>\n

Question 15.
\nExplain why is ortho nitrophenol more acidic than ortho methoxyphenol?
\nAnswer:<\/p>\n

\"2nd
\nThe nitro-group is art electron withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond. Also, the O-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, O-nitrophenol is a stronger acid.<\/p>\n

On the other hand, methoxy group is an electron releasing group. Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily. For th is reason, O-nitrophenol is more acidic than O-methoxyphenol.<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nExplain how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
\nAnswer:
\nThe – OH group is an electron donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.<\/p>\n

\"2nd
\nAs a result the benzene ring is activated towards electrophilic substitution<\/p>\n

Question 17.
\nGive equations of the following reactions:
\n(i) Oxidation of propan-l-ol with alkaline KMnO4<\/sub> solution.
\n(ii) Bromine in CS2<\/sub> with phenol.
\n(iii) Dilute HNO3<\/sub> with phenol,
\n(iv) Treating phenol with chloroform in presence’ of aqueous NaOH.
\nAnswer:<\/p>\n

\"2nd
\n\"2nd<\/p>\n

Question 18.
\nExplain the following with an example.
\n1. Kolbe\u2019s reaction.
\n2. Reimer-Tiemann reaction
\n3. Williamson ether synthesis.
\n4. unsymmetrical ether.
\nAnswer:
\n1. Kolbe\u2019s reaction:
\nWhen phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon-di-oxide, followed by acidification, undergoes electrophilic substitution to give o-hydroxipenzoic acid as the main product. This reaction is known as Kolbe\u2019s reaction.
\n\"2nd<\/p>\n

2. Reimer-Tiemann reaction: When phenol is treated with chloroform (CHCl3<\/sub>) in the presence of sodium hydroxide, a – CHo group is introduced at the ortho position of the benzene ring.<\/p>\n

\"2nd
\nThis reaction is known as the Reimer – Timann reaction.
\nThe intermediate is hydrolyzed in the presence of alkalis to produce salicyclal dehyde.<\/p>\n

\"2nd<\/p>\n

3. Williamson ether synthesis:
\nWilliamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.
\nR-X + R- ONa \u2192 R-O-R + NaX
\nThis reaction involves SN<\/sub>2 attack of the alkoxides ion on the alkyl halide. Better results are obtained in case of primary alkyl halides.<\/p>\n

\"2nd
\nIfthealkyl halide is 2\u00b0 or 3\u00b0, then elimination competes over substitution.<\/p>\n

4. Unsymmetrical ether:
\nAn unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms). For e.g: ethyl methyl ether<\/p>\n

\"2nd<\/p>\n

Question 19.
\nWrite the mechanism of acid dehydration of ethanol to yield ethene.
\nAnswer:
\nThe mechanisms of acid dehydration of ethanol to yield ethene involves the following three steps.
\nSteps 1:
\nProtonation of ethanol to form ethyl oxonium ion
\n\"2nd<\/p>\n

Steps 2:
\nFormation of carbocation (rate determining step)
\n\"2nd<\/p>\n

Steps 3:
\nElimination of proton to form ethene
\n\"2nd
\nThe acid consumed in step 1 is released in step 3. After the formation of ethene, it is removed to shift the equilibrium in forward direction.<\/p>\n

\"KSEEB<\/p>\n

Question 20.
\nHow are the following conversions carried out?
\n1. Propene \u2192 Propan-2-ol.
\n2. Benzyl chloride \u2192 Benzyl alcohol.
\n3. Ethyl magnesium chloride \u2192 Propan-l-ol.
\n4. Methyl magnesium bromide \u2192 2-Methylpropan-2-ol.
\nAnswer:
\n1. If propene is allowed to react with water in the presence of an acid as a catalyst, then propan – 2-ol is obtained.<\/p>\n

\"2nd<\/p>\n

2. If benzyl chloride is treated with NaoH (followed by acidification) then benzyl alcohol is produced.<\/p>\n

\"2nd<\/p>\n

3. When ethyl magnesium chloride is treated with methanal, an adduct is produced which gives propan-l-ol on hydrolysis.<\/p>\n

\"2nd<\/p>\n

4. When methyl magnesium bromide is treated with propane, an adduct is the product which give 2- methyl propane – 2-ol hydrolysis.<\/p>\n

\"2nd<\/p>\n

Question 21.
\nName the reagents used in the following reactions:
\n1. Oxidation of a primary alcohol to carboxylic acid.
\n2. Oxidation of a primary alcohol to aldehyde.
\n3. Brumination of phenol to 2,4,6-tribromophenol.
\n4. Benzyl alcohol to benzoic acid.
\n5. Dehydration of propan-2-ol to propene.
\n6. Butan-2-one to butan-2-ol.
\nAnswer:<\/p>\n

    \n
  1. Acidified KMnO4<\/sub><\/li>\n
  2. Pyridinium chlorochromate (PCC)<\/li>\n
  3. Bromine water<\/li>\n
  4. Acidified KMnO4<\/sub><\/li>\n
  5. 85% phosphoric acid<\/li>\n
  6. NaBH4<\/sub> or Li AiH4<\/sub><\/li>\n<\/ol>\n

    Question 22.
    \nGive reason for the higher boiling point of ethanol in comparison to methoxy -methane.
    \nAnswer:
    \nEthanol undergoes intermolecular H-boding due to the presence of -OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the bp of ethanol is higher than that of methoxymethane.<\/p>\n

    \"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 23.
    \nGive IUPAC names of the following ethers:
    \n1.
    \n\"2nd
    \n2. CH3<\/sub>OCH2<\/sub>CH2<\/sub>CL
    \n3. O2<\/sub>N-C6<\/sub>H4<\/sub>-OCH3<\/sub>(p)
    \n4. CH3<\/sub>CH2<\/sub>CH2<\/sub>OCH3<\/sub>
    \n5.
    \n\"2nd
    \n6.
    \n\"2nd
    \nAnswer:<\/p>\n

      \n
    1. 1- Ethoxy – 2 – methyl propane<\/li>\n
    2. 2 – chloro -1 – methoxyethane<\/li>\n
    3. 4 – Nitroanisole<\/li>\n
    4. 1 – Methoxypropane<\/li>\n
    5. 1 – Ethoxy – 4,4 – dimethyl cyclohexane<\/li>\n
    6. Ethoxy benzene<\/li>\n<\/ol>\n

      Question 24.
      \nWrite the names of reagents and equations for the preparation of the following ethers by Williamson\u2019s synthesis:
      \n(i) 1-Propoxy propane
      \n(ii) Ethoxy benzene
      \n(iii) 2-Methoxy-2-methylpropane
      \n(iv) 1-Methoxyethane
      \nAnswer:<\/p>\n

      \"2nd<\/p>\n

      \"KSEEB<\/p>\n

      Question 25.
      \nIllustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
      \nAnswer:
      \nThe reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide<\/p>\n

      \"2nd
      \nBut if secondary or tertiary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence they react with alkyl halides, which results in an elimination reaction.<\/p>\n

      \"2nd<\/p>\n

      Question 26.
      \nHow is 1-propoxypropane synthesised from propan-l-ol? Write mechanism of this reaction.
      \nAnswer:
      \n1-Propoxy propane can be synthesised from propan-l-ol by dehydration.<\/p>\n

      \"2nd
      \nThe mechanism of this rxn involves the following three steps.
      \nSteps 1: Protonation
      \n\"2nd<\/p>\n

      Step 2: Nucleophilic attack
      \n\"2nd<\/p>\n

      Step 3: Deprotonation<\/p>\n

      \"2nd<\/p>\n

      Question 27.
      \nPreparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
      \nAnswer:
      \nThe formation of ethers by dehydration of alcohol is a bimolecular reaction (SN<\/sub>2) involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates over substitution. Hence, in place of ethers, alkenes are formed.<\/p>\n

      Question 28.
      \nWrite the equation of the reaction of hydrogen iodide with:
      \n(i) 1-propoxypropane
      \n(ii) methoxy benzene and
      \n(iii) benzyl ethyl ether.
      \nAnswer:<\/p>\n

      \"2nd<\/p>\n

      Question 29.
      \nExplain the fact that in aryl alkyl ethers
      \n1. the alkoxy group activates the benzene ring towards electrophilic substitution and
      \n2. it directs the incoming substituents to ortho and para positions in benzene ring.
      \nAnswer:
      \n1.
      \n\"2nd
      \nIn aryl alkyl ethers, due to +R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure<\/p>\n

      \"2nd
      \nThus, benzene is activated towards electrophilic substitution by the alkoxy group.<\/p>\n

      2. It can be observed from the resonance structures that the electron density increases more at the ortho and para positions that at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.<\/p>\n

      \"KSEEB<\/p>\n

      Question 30.
      \nWrite the mechanism of the reaction of HI with methoxymethane.
      \nAnswer:
      \nThe mechanism of the reaction of HI with methoxymethane involves the following steps.
      \nStep 1: Protonation of methoxymethane
      \n\"2nd<\/p>\n

      Step2: Nucleophilic attack of I
      \n\"2nd<\/p>\n

      Step 3: When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide.
      \n\"2nd<\/p>\n

      Question 31.
      \nWrite equations of the following reactions:
      \n(i) Friedel-Crafts reaction – alkylation of anisole.
      \n(ii) Nitration of anisole.
      \n(iii) Bromination of anisole in ethanoic acid medium.
      \n(iv) Friedel-Craft\u2019s acetylation of anisole.
      \nAnswer:<\/p>\n

      \"2nd
      \n\"2nd<\/p>\n

      \"KSEEB<\/p>\n

      Question 32.
      \nShow how would you synthesise the following alcohols from appropriate alkenes?<\/p>\n

      \"2nd
      \nAnswer:<\/p>\n

      \"2nd
      \nAcid hydration of Pent-2-ene produces Penton – 3-ol along with penton-2-ol. Hence first rxn using Pent-l-ene is preferred.<\/p>\n

      \"2nd<\/p>\n

      Question 33.
      \nWhen 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:<\/p>\n

      \"2nd
      \nGive a mechanism for this reaction.
      \n(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom. ‘
      \nAnswer:
      \nThe mechanism of the given reaction involves the following steps:
      \nStep 1: Protonation<\/p>\n

      \"2nd<\/p>\n

      Step 2: Formation of 2\u00b0 carbocation by elimination of water molecule\"2nd<\/p>\n

      Step 3: Rearrangement by the hydride-ion shift<\/p>\n

      \"2nd<\/p>\n

      Step 4: Nucleophilic attack<\/p>\n

      \"2nd<\/p>\n

      \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

      You can Download Chapter 11 Alcohols, Phenols and Ethers Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Chemistry Question Bank Chapter 11 Alcohols, Phenols and Ethers 2nd PUC Chemistry Alcohols, Phenols and …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10479"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10479"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10479\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10479"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10479"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10479"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}