{"id":10253,"date":"2019-12-06T15:02:53","date_gmt":"2019-12-06T09:32:53","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10253"},"modified":"2021-07-02T15:43:52","modified_gmt":"2021-07-02T10:13:52","slug":"2nd-puc-chemistry-question-bank-chapter-13","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-13\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 13 Amines"},"content":{"rendered":"

You can Download Chapter 13 Amines 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 13 Amines<\/h2>\n

2nd PUC Chemistry Amines NCERT Textbook Questions<\/h3>\n

Question 1.
\nWrite IUPAC names of the following compounds and classify them into primary, secondary .and tertiary amines.
\n(i) (CH3<\/sub>)2<\/sub>CHNH2<\/sub>
\n(ii) CH3<\/sub>(CH2<\/sub>)2<\/sub>NH
\n(iii) CH3<\/sub>NHCH(CH3<\/sub>)2<\/sub>
\n(iv) (CH3<\/sub>)3<\/sub>CNH2<\/sub>
\n(V) C6<\/sub>H5<\/sub>NHCH3<\/sub>
\n(vi) (CH3<\/sub>CH2<\/sub>)2<\/sub>NCH3<\/sub>
\n(vii) m-BrC6<\/sub>H4<\/sub>NH2<\/sub>
\nAnswer:<\/p>\n

\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nGive one chemical test to distinguish between the following pairs of compounds,<\/p>\n

    \n
  1. Methylamine and dimethylamine<\/li>\n
  2. Secondary and tertiary amines<\/li>\n
  3. Ethylamine and aniline<\/li>\n
  4. Aniline and benzylamine<\/li>\n
  5. Aniline and N-methyl aniline.<\/li>\n<\/ol>\n

    Answer:
    \n1. Methylamine and dimethylamine
    \n(1) CH3<\/sub>NH2<\/sub> (Methyl amine; 1\u00b0 amine) – Carbyl amine test positive
    \n(2) (CH3<\/sub>)2<\/sub> NH (Dimethyl amine; 2\u00b0 amine) – Doesn\u2019t give this test
    \nCarbylamine test:-
    \n\"2nd<\/p>\n

    2. 2\u00b0 and 3\u00b0 amines:-
    \n(1) 2\u00b0 amine – Libermann Nitrosomine test positive
    \n(2) 3\u00b0 amine – Libermann Nitrosomine test negative.
    \nLibermann Nitrosomine test :
    \n\"2nd<\/p>\n

    3. Ethylamine and aniline:
    \n(1) Aromatic amines gives brilliantly coloured orange dye on diazotisation and coupling with phenol.
    \n(2) Ethylamine do not give a dye on diazotisation and coupling with phenol.
    \nDiazotisation:
    \n\"2nd<\/p>\n

    4. Aniline and benzylamine:-
    \n(1) Aromatic (1 \u00b0) amine gives coloured dye on diazotisation and coupling with phenol.
    \n(2) Other non aromatic 1\u00b0 amines and which are not an aromatic 1\u00b0 amine, will not give a dye.
    \nDiazotisation:-
    \n(a)
    \n\"2nd<\/p>\n

    (b)
    \n\"2nd<\/p>\n

    5. Aniline and N-methylaniline:-
    \naniline (1\u00b0 amine)-Carbylamine test positive
    \nN-methylaniline (2\u00b0 amine) – Negative
    \nCarbylamine test:-
    \n\"2nd<\/p>\n

    Question 3.
    \nAccount for the following:<\/p>\n

    1. pKb<\/sub> of aniline is more than that of methylamine.
    \nAnswer:
    \n(i) Aniline is weaker base than methylamine
    \nKb<\/sub> of aniline is less than Kb<\/sub> of methylamine.
    \npKb<\/sub> of aniline is more than pKb<\/sub> of methylamine.
    \nPKb<\/sub> = -logKb<\/sub>
    \n(a) Aniline is weaker base than methylamine because due to resonance, lone pair on N2<\/sub> gets delocalised over the ring, so electron pair availability is reduced.<\/p>\n

    (b) In methyl amine due to ‘+I’effect of methyl group, availability of lone pair on N2<\/sub> atom is increased So is strong base.<\/p>\n

    2. Ethylamine is soluble in water whereas aniline is not.
    \nAnswer:
    \n(i) Lower aliphatic amines are soluble in water because of their ability to form hydrogen bonds with water.
    \n(ii) But in aniline due to larger hydrocarbon part, it hinders hydrogen bond formation with water.<\/p>\n

    3. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
    \nAnswer:
    \nMethylamine + H2<\/sub>O \u21cc Methyl ammonium hydroxide
    \n\"2nd
    \nOH- can precipitate out Fe, AI, Cr etc…
    \n3 [CH3<\/sub>NH3<\/sub>]+<\/sup> OH + FeCl3<\/sub> \u2192 Fe (OH)3<\/sub>\u2193+ 3CH3<\/sub>NH3<\/sub>Cl
    \nFe (OH)3<\/sub> is unstable, so can be hydrated ferric oxide
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    4. Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
    \nAnswer:
    \nIn aniline NH2<\/sub> increases e density at o- and p- positions due to +R effect. But when it is nitrated by nitrating mixture, substantial amount of m-nitro aniline is formed.
    \nBut in acidic medium (Nitrating mixture – HNO3<\/sub> (c) + H2<\/sub>SO4<\/sub> (c)).
    \nNH2<\/sub> + H+<\/sup> \u21cc NH3<\/sub>+<\/sup>, NH3<\/sub>+<\/sup> is e–<\/sup> withdrawing group and so is m – directing. So gives larger amount of m- nitroaniline.<\/p>\n

    5. Aniline does not undergo Friedel-Crafts reaction.
    \nAnswer:
    \nFriedel-Craft reaction:-
    \n\"2nd
    \nBut aniline undergoes salt formation with A1C13 (Lewis acid used as catalyst). So N acquires positive charge and hence acts as strong deactivating group for further reaction.<\/p>\n

    6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
    \nAnswer:
    \nIn aromatic diazonium salts, due to resonance their is dispersal of positive charge on benzene ring.
    \nBut in aliphatic diazonium salts, resonance is not possible, so aliphatic diazonium salts are less stable than aromatic diazonium salts.<\/p>\n

    7. Gabriel phthalimide synthesis is preferred for synthesising primary amines.
    \nAnswer:
    \nGabriel phthalimide synthesis
    \n\"2nd
    \nReaction between alkyl halide (R – X) and ammonia gives mixture, of 1\u00b0, 2\u00b0 and 3\u00b0 amines with tetraalkyl ammonium halide. Mixture is not separable.
    \nTo get pure 1\u00b0 amine, Gabriel phthalimide synthesis is preferred.<\/p>\n

    Question 4.
    \nArrange the following:
    \n1. In decreasing order of the pKb values:
    \nC2<\/sub>H5<\/sub>NH2<\/sub>, C6<\/sub>H5<\/sub>NHCH3<\/sub>, (C2<\/sub>H5<\/sub>)2<\/sub>NH and CH5<\/sub>NH2<\/sub>
    \nAnswer:
    \nC6<\/sub>H5<\/sub>NH2<\/sub> > C6<\/sub>H5<\/sub>NHCH3<\/sub> > C2<\/sub>H5<\/sub>NH2<\/sub> > (C2<\/sub>H5<\/sub>)2<\/sub> NH
    \n\"2nd<\/p>\n

    2. In increasing order of basic strength:
    \nC6<\/sub>H5<\/sub>NH2<\/sub>, C6<\/sub>H5<\/sub>N(CH3<\/sub>)2<\/sub>, (C2<\/sub>H5<\/sub>)2<\/sub>NH and CH3<\/sub>NH2<\/sub>
    \nAnswer:
    \nC6<\/sub>H5<\/sub>NH2<\/sub> < C6<\/sub>H5<\/sub>N(CH3<\/sub>)2<\/sub> < CH3<\/sub>NH2<\/sub> < (C2<\/sub>H5<\/sub>)2<\/sub> NH<\/p>\n

    3. In increasing order of basic strength:
    \n(a) Aniline, p-nitroaniline and p-toluidine
    \nAnswer:
    \n\"2nd<\/p>\n

    (b) C6<\/sub>H5<\/sub>NH2<\/sub>, C6<\/sub>H5<\/sub>NHCH3<\/sub>, C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub>.
    \nAnswer:
    \nC6<\/sub>H5<\/sub>NH2<\/sub> < C6<\/sub>H5<\/sub>NHCH3<\/sub> < C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub><\/p>\n

    4. In decreasing order of basic strength in gas phase:
    \nC6<\/sub>H5<\/sub>NH2<\/sub>, (C2<\/sub>H5<\/sub>)2<\/sub>NH, (C2<\/sub>H5<\/sub>)3<\/sub>N and NH3<\/sub>,
    \nAnswer:
    \n(C2<\/sub>H5<\/sub>)3<\/sub>N > (C2<\/sub>H5<\/sub>)2<\/sub>NH > C2<\/sub>H5<\/sub>NH2<\/sub> > NH3<\/sub>
    \nNote: In GAS PHASE, 3\u00b0 > 2\u00b0 > 1\u00b0 Ammonia [No Hydrogen bending and \u2018+I’ effect].<\/p>\n

    5. In increasing order of boiling point:
    \nC2<\/sub>H5<\/sub>OH, (CH3<\/sub>)2<\/sub>NH, C2<\/sub>H5<\/sub>NH2<\/sub>
    \nAnswer:
    \n(CH3<\/sub>)2<\/sub> NH < C2<\/sub>H5<\/sub>NH2<\/sub> < C2<\/sub>H5<\/sub>OH
    \nNote: Stronger H-bonding, more b.p more solubility
    \nAlcohol > Amine In amines on H attached to N takes part in H-bond.<\/p>\n

    6. In increasing order of solubility in water:
    \nC6<\/sub>H5<\/sub>NH2<\/sub>, (C2<\/sub>H5<\/sub>)2<\/sub>NH, C2<\/sub>H5<\/sub>NH2<\/sub>.
    \nAnswer:
    \nC6<\/sub>H5<\/sub>NH2<\/sub> < (C2<\/sub>H5<\/sub>)2<\/sub>NH < C2<\/sub>H5<\/sub>NH2<\/sub>
    \nmore H-bonding, more solubility<\/p>\n

    \"KSEEB<\/p>\n

    Question 5.
    \nHow will you convert:
    \n1. Ethanoic acid into methanamine
    \nAnswer:
    \n\"2nd<\/p>\n

    2. Hexanenitrile into 1-aminopentane
    \nAnswer:
    \n\"2nd<\/p>\n

    3. Methonal to ethanoic acid
    \nAnswer:
    \n\"2nd<\/p>\n

    4. Ethanamine into methanamine
    \nAnswer:
    \n\"2nd<\/p>\n

    5. Ethanoic acid into propanoic acid
    \nAnswer:
    \n\"2nd<\/p>\n

    6. Methanamine into ethanamine
    \nAnswer:
    \n\"2nd<\/p>\n

    7. Nitromethane into dimethylamine
    \nAnswer:
    \n\"2nd<\/p>\n

    8. Propanoic acid into ethanoic acid
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 6.
    \nDescribe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
    \nAnswer:
    \nHinsberg test:- Benzenesulphonyl chloride (C6<\/sub>H5<\/sub>SO2<\/sub>Cl), which is also known as Hinsberg\u2019s reagent, reacts with primary and secondary amines to form sulphonamides.
    \n(a) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide.
    \n\"2nd
    \nThe hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.<\/p>\n

    (b) In the reaction with secondary amine, N,N-diethylbenzenesulphonamide is formed.
    \n\"2nd
    \nSince N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.<\/p>\n

    (c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride, in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines. However, these days benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride.
    \n1\u00b0 amine – clear solution which on acidification gives insoluble product.
    \n2\u00b0 amine – insoluble product in solution, unaffected by acid
    \n3\u00b0 amine – No reaction, dissolves an acidification.<\/p>\n

    \"KSEEB<\/strong><\/p>\n

    Question 7.
    \nWrite short notes on the following:<\/p>\n

      \n
    1. \u00a0Carbylamine reaction<\/li>\n
    2. Diazotisation<\/li>\n
    3. Hofmann\u2019s bromamide reaction<\/li>\n
    4. Coupling reaction<\/li>\n
    5. Ammonolysis<\/li>\n
    6. Acetylation<\/li>\n
    7. Gabriel phthalimide synthesis.<\/li>\n<\/ol>\n

      Answer:
      \n1. Carbylamine reaction
      \n1\u00b0 amines (aromatic\/aliphatic) On heating with chloroform and alcoholic solution of KOH gives isocyanides\/carbylamines producing unpleasant smell.
      \n\"2nd
      \n2\u00b0 and 3\u00b0 (aromatic\/aliphatic) do not give this reaction.<\/p>\n

      2. Diazotisation .
      \nAromatic 1\u00b0 amines into diazonium salts by action of HNO2<\/sub> and dil HCl at ice cold temperature as HNO2<\/sub> and product are unstable at higher temperature.
      \n\"2nd
      \n3. Hofmann\u2019s bromamide reaction
      \nInvolves heating of an amide with bromine and caustic alkali to yield an amine with one carbon less than original amide.
      \n\"2nd
      \n4. Coupling reaction
      \nDiazonium salt acts as electrophile and brings about substitution in electron rich aromatic ring such as phenol and amines. The result is formation of dye.
      \n(i) with phenol – In basic medium (9 – 10 pH) at low temperature (273 – 278 K)
      \n\"2nd<\/p>\n

      (ii) with amines – In acidic medium (ph = 4 – 5) at low temperature (273 – 278 K)
      \n\"2nd<\/p>\n

      5. Ammonolysis – Alkyl halide (1\u00b0) react with alcoholic ammonia to form 1\u00b0, 2\u00b0, 3\u00b0 amines along with 4\u00b0ammonium salts. It is carried out by heating alkyl halide with ale solution of NH3 in sealed tube at 370 K.
      \nIf ammonia is used in excess, 1\u00b0 amine is major product
      \n\"2nd
      \nIf alkyl halide is excess, it forms all 1\u00b0, 2\u00b0, 3\u00b0, 4\u00b0 ammonium salts
      \n\"2nd<\/p>\n

      6. Acetylation
      \nAliphatic amines:
      \nIntroduction of acetyl group (CH3<\/sub>CO-) into any molecule.
      \n1\u00b0 and 2\u00b0 amines react with acetyl chloride to form amides
      \nBut 3\u00b0 amines has NO H on N, so do not react.
      \n\"2nd
      \n7. Gabriel phthalimide synthesis
      \n\"2nd
      \nReaction between alkyl halide (R – X) and ammonia gives mixture, of 1\u00b0, 2\u00b0 and 3\u00b0 amines with tetraalkyl ammonium halide. Mixture is not separable.
      \nTo get pure 1\u00b0 amine, Gabriel phthalimide synthesis is preferred.<\/p>\n

      \"KSEEB<\/p>\n

      Question 8.
      \nAccomplish the following conversions:<\/p>\n

        \n
      1. Nitrobenzene to benzoic acid<\/li>\n
      2. Benzene to m-bromophenol<\/li>\n
      3. Benzoic acid to aniline<\/li>\n
      4. Aniline to 2,4,6-tribromofluorobenzene<\/li>\n
      5. Benzyl chloride to 2-phenylethanamine<\/li>\n
      6. Chlorobenzene to p-chloroaniline<\/li>\n
      7. Aniline to p-bromoaniline<\/li>\n
      8. Benzamide to toluene<\/li>\n
      9. Aniline to benzyl alcohol<\/li>\n<\/ol>\n

        Answer:
        \n1. Nitrobenzene to benzoic acid
        \n\"2nd<\/p>\n

        2. Benezene to m-bromophenol
        \n\"2nd<\/p>\n

        3. Benzoic acid to aniline
        \n\"2nd<\/p>\n

        4. Aniline to 2,4,6-tribromofluorobenzene
        \n\"2nd<\/p>\n

        5. Benzyl chloride to 2-phenylethanamine
        \n\"2nd<\/p>\n

        6.\u00a0Chlorobenzene to p-choloroaniline<\/p>\n

        \"2nd<\/p>\n

        7. Aniline to p-bromoaniline
        \n\"2nd<\/p>\n

        8. Benzeamide to toluene<\/p>\n

        \"2nd<\/p>\n

        9. Aniline to benzeyl alcohal
        \n\"2nd<\/p>\n

        \"KSEEB<\/p>\n

        Question 9.
        \nGive the structures of A,B and C in the following reactions:<\/p>\n

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

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

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

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

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

        \"2nd
        \nAnswer:
        \n\"2nd
        \nQuestion 10.
        \nAn aromatic compound \u2018A\u2019 on treatment with aqueous ammonia and heating forms compound \u2018B\u2019 which on heating with Br2 and KOH forms a compound \u2018C\u2019 of molecular formula C6<\/sub>H7<\/sub>N. Write the structures and IUPAC names of compounds A, B and C.
        \nAnswer:
        \n\"2nd
        \nAs \u2018C\u2019 is expected as a 1\u00b0 amine obtained by Hoffmann bromamide degradation of \u2018B\u2019 an amide obtained from \u2018A\u2019 a carboxylic acid on treatment with NH3<\/sub> and heat.
        \nSo
        \n\"2nd<\/p>\n

        Question 11.
        \nComplete the following reactions:
        \n1. C6<\/sub>H5<\/sub>NH2<\/sub> + CHCI3<\/sub> + alc.KOH \u2192
        \nAnswer:
        \nC6<\/sub>H5<\/sub>NH2<\/sub> + CHCl3<\/sub> + alc.KOH \u2192 C6<\/sub>H5<\/sub>NC + 3 KCl + 3H2<\/sub>O
        \n\"2nd<\/p>\n

        2. C6<\/sub>H5<\/sub>N2<\/sub>Cl + H3<\/sub>PO2<\/sub> + H2<\/sub>O \u2192
        \nAnswer:
        \n\"2nd<\/p>\n

        3. C6<\/sub>H5<\/sub>NH2<\/sub> + H2<\/sub>SO4<\/sub> (conc.) \u2192
        \nAnswer:
        \n\"2nd<\/p>\n

        4. C6<\/sub>H5<\/sub>N2<\/sub>Cl + C2<\/sub>H5<\/sub>OH \u2192
        \nAnswer:
        \n\"2nd<\/p>\n

        \"KSEEB<\/p>\n

        5. C6<\/sub>H5<\/sub>NH2<\/sub> + Br2<\/sub>(aq) \u2192
        \nAnswer:
        \n\"2nd<\/p>\n

        6. C6<\/sub>H5<\/sub>N2<\/sub> + (CH3<\/sub>CO)2<\/sub> \u2192
        \nAnswer:
        \n\"2nd<\/p>\n

        \"2nd<\/p>\n

        Question 12.
        \nWhy cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
        \nAnswer:
        \nIn this reaction, potassium phthalimide gives phthalimide anion which acts as nucleophile and attacks electrophilic carbon of alkyl halide.<\/p>\n

        But due to much less reactivity of aryl halide (nucleophilic substitution reaction), 1\u00b0 aromatic amines can\u2019t be prepared.<\/p>\n

        Question 13.
        \nWrite the reactions of
        \n1. aromatic and
        \n2. aliphatic primary amines with nitrous acid.
        \nAnswer:
        \n1. Aromatic amines with nitrous acid
        \nAromatic 1\u00b0 amines react with HNO2<\/sub> at 273 – 278 K to form diazonium compound.
        \n\"2nd<\/p>\n

        2. Aliphatic amines with HNO2<\/sub> at 273 – 278 K to give corresponding alcohol, N2<\/sub>, along with other products.
        \n\"2nd<\/p>\n

        Question 14.
        \nGive plausible explanation for each of the following:
        \n1. Why are amines less acidic than alcohols of comparable molecular masses?
        \nAnswer:
        \n\"2nd
        \nBut as \u2018O\u2019 is more electronegative than \u2018N\u2019. So \u2018O\u2019 can accommodate more positive charge and is more stable than \u2018N\u2019. So alcohol liberates H+ easily than amines.
        \nSo alcohols are more acidic than amines.<\/p>\n

        2. Why do primary amines have higher boiling point than tertiary amines?
        \nAnswer: 1\u00b0 amines (NH2) has stronger and extensive H – bonding.
        \n3\u00b0 amines (> N-) has no H, so no H – bonding.
        \nSo 1\u00b0 amines have higher b.p than 3\u00b0 amines.<\/p>\n

        3. Why are aliphatic amines stronger bases than aromatic amines?
        \nAnswer:
        \n\"2nd
        \nAlkyl amines has positive I effect\/electron donating effect, so high electron density on N and stable alkyl ammonium ion formation.
        \nAromatic amines due to resonance effect and lower stability of aryl ammonium ion than aryl amine, e- density on N is less.
        \nSo Alkyl amine is stronger base than aromatic amines.<\/p>\n

        \"KSEEB<\/p>\n

        2nd PUC Amines Intext Questions<\/h3>\n

        Question 15.
        \nClassify as primary, secondary or tertiary amines.
        \n1.
        \n\"2nd
        \n2.
        \n\"2nd<\/p>\n

        3. (C2<\/sub>H5<\/sub>)2<\/sub> CHNH2<\/sub>
        \n4. (C2<\/sub>H5<\/sub>)2<\/sub> NH
        \nAnswer:
        \n1.
        \n\"2nd
        \n\u2192 Primary(As only one of H is replaced by other group, contains – NH2<\/sub> group)<\/p>\n

        2.
        \n\"2nd
        \n\u2192 No H attached to N, so teritary ( As, > N group is teritary)<\/p>\n

        3. (C2<\/sub>H5<\/sub>)2<\/sub> CHNH2<\/sub> \u2014> Primary (contains – NH2<\/sub> group)
        \n4. (C2<\/sub>H5<\/sub>)2<\/sub>NH \u2014> Secondary (contains -NH group)<\/p>\n

        Question 16.
        \n(i) Write a structure of different isomeric amines corresponding to C4<\/sub>H11<\/sub>N
        \n(ii) Write IUPAC names of all isomers
        \n(iii) What type of isomerism is exhibited by different pair of amines
        \nAnswer:
        \nC5<\/sub>H11<\/sub>N
        \nPrimary amines (-NH2<\/sub>)
        \n\"2nd
        \n\"2nd<\/p>\n

        Question 17.
        \nConvert:
        \n(a) Benzene into aniline
        \nAnswer:
        \n\"2nd<\/p>\n

        (b) Benzene into N,N – dimethyl aniline
        \nAnswer:
        \n\"2nd<\/p>\n

        (c) Cl – [CH2<\/sub>]4<\/sub> – Cl into hexane-1,6 – diamine
        \n\"2nd<\/p>\n

        \"KSEEB<\/p>\n

        Question 18.
        \nArrange following in increasing order of their Basic character:
        \n1. C2<\/sub>H5<\/sub>NH2<\/sub>, C6<\/sub>H5<\/sub>NH2<\/sub>, NH3<\/sub>, C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub>
        \nAnswer:
        \nC6<\/sub>H5<\/sub>NH2<\/sub> < NH3<\/sub> < C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub> < C2<\/sub>H5<\/sub>NH2<\/sub> < (C2<\/sub>H5<\/sub>)2<\/sub>NH<\/p>\n

        2. C2<\/sub>H5<\/sub>NH2<\/sub>, (C2<\/sub>H5<\/sub>)NH, (C2<\/sub>H5<\/sub>)3<\/sub>N, C6<\/sub>H5<\/sub>NH2<\/sub>
        \nAnswer:
        \nC6<\/sub>H5<\/sub>NH2<\/sub> < C2<\/sub>H5<\/sub>NH2<\/sub> < C2<\/sub>H5<\/sub>)3<\/sub>N<\/sub> < (C2<\/sub>H5<\/sub>)2<\/sub>NH<\/p>\n

        3. CH3<\/sub>NH2<\/sub>, (CH3<\/sub>)2<\/sub>NH, (CH3<\/sub>)3<\/sub>N, C6<\/sub>H5<\/sub>NH2<\/sub>, C6<\/sub>H5<\/sub>CH2<\/sub>NH2
        \n<\/sub>
        \nAnswer:
        \nC6<\/sub>H5<\/sub>NH2<\/sub> < C2<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub> < (CH3<\/sub>)3<\/sub>N < CH3<\/sub>NH2<\/sub> < (CH3<\/sub>)2<\/sub>NH
        \n\"2nd<\/p>\n

        Question 19.
        \nCompliete following acid-base reaction and name the products:
        \nAnswer:
        \n\"2nd<\/p>\n

        Question 20.
        \nWrite reactions of final alkylation product of aniline with excess of methyl iodine is presence of solution
        \nAnswer:
        \n\"2nd<\/p>\n

        Question 21.
        \nWrite chemical reaction of aniline with benzoyl chloride and write product name.
        \nAnswer:
        \n\"2nd<\/p>\n

        Question 22.
        \nWrite structure of isomers of C3<\/sub>H9<\/sub>N. Write IUPAC names of isomers which will liberate N2<\/sub> gas on treatment with nitrous acid. .
        \nAnswer:
        \n\"2nd<\/p>\n

        Question 23.
        \nConvert
        \n1. 3-methylaniline into 3-nitrotolune
        \nAnswer:
        \n\"2nd<\/p>\n

        2. Aniline into 1,3,5 – tribromobenzene
        \nAnswer:
        \n\"2nd<\/p>\n

        \"KSEEB<\/p>\n

        2nd PUC Amines Additional Questions<\/h3>\n

        Question 1.
        \nComplete the reaction.
        \nC6<\/sub>H5<\/sub>N2<\/sub>Cl + CH3<\/sub>CH2<\/sub>OH \u2192 (Benzene)
        \nAnswer:
        \nC6<\/sub>H5<\/sub>N2<\/sub>Cl + CH3<\/sub>CH2<\/sub>OH \u2192 C6<\/sub>H6<\/sub> +CH3<\/sub>CHO + N2<\/sub> + HCl<\/p>\n

        Question 2.
        \nWrite the IUPAC name of C6<\/sub>H6<\/sub>NHCH3<\/sub> .
        \nAnswer:
        \nN – Methyl aniline.<\/p>\n

        Question 3.
        \nWhich is more soluble in water, NH3<\/sub> or methylamine ?
        \nAnswer:
        \nNH3<\/sub> is more soluble in water due to more H-bonding.<\/p>\n

        Question 4.
        \nWhich of the following amines has largest value of pKb<\/sub>?
        \np-Toluidine, Aniline, p- Nitroaniline.
        \nAnswer:
        \nHigher the pKb<\/sub> value, weaker is the base
        \n\"2nd
        \nPresence of -CH3<\/sub> group in the ring will increase the basic character as it is an electron releasing group. Presence of -NO2<\/sub> group in the ring will decrease the basic character as it an electron withdrawing group. Thus p-nitro aniline will have largest value of pKb<\/sub>.<\/p>\n

        Question 5.
        \nState two important uses of aniline. (DSB 2000)
        \nAnswer:<\/p>\n

          \n
        1. In the manufacture of dyes and drugs.<\/li>\n
        2. In the preparation of phenyl isocyanide required for the manufacture of polyurethane plastic.<\/li>\n<\/ol>\n

          Question 6.
          \nWhy do amines behave as nucleophiles ?
          \nAnswer:
          \nAmines behave as nucleophiles. For eg: amines react with alkyl halides, acid chlorides etc.<\/p>\n

          Question 7.
          \nWhy are aqueous solution of amines basic in nature ? (DSB 2006)
          \nAnswer:
          \nAqueous solution of amines which are freely soluble in water (like CH3<\/sub>NH2<\/sub>, C2<\/sub>H5<\/sub>NH2<\/sub> etc) are basic in nature. It can be explained as follows,
          \nFor eg:-
          \n\"2nd
          \nThe aqueous solution of methyl amine will be basic in nature, due to the presence of \"2nd\u00a0ions.<\/p>\n

          Question 8.
          \nExplain why silver chloride is soluble in aqueous solution of methyl amine. (CBSE 1997, 1998,2002,2004)
          \nAnswer:
          \nDue to the formation of water soluble complex
          \nAgCl + 2CH3<\/sub>NH2 <\/sub> \u2192 [Ag(CH3<\/sub>NH2<\/sub>)2<\/sub>]+Cl<\/p>\n

          Question 9.
          \nWrite IUPAC name of. (Delhi CBSE 1999)
          \n\"2nd
          \nAnswer:
          \np-nitro-N, N dimethyl aniline.<\/p>\n

          Question 10.
          \nFor an amine RNH2<\/sub>, write the expression for Kb<\/sub> to indicate its base strength. (DBS 2000)
          \nAnswer:
          \n\"2nd<\/p>\n

          \"KSEEB<\/p>\n

          Question 11.
          \nWhat does Kb<\/sub> value for an amine stand for? (AISB 2000)
          \nAnswer:
          \nKb<\/sub> value for an amine stands for its basic strength. More the Kb<\/sub> value of an amines more is its basic strength.<\/p>\n

          Question 12.
          \nGive one example of anbident nucleophile. (CBSE 2001)
          \nAnswer:
          \n\"2nd
          \nacts as ambident nucleophiles.<\/p>\n

          Question 13.
          \nWhy do amines acts as nucleophiles ? (AISB 2007)
          \nAnswer:
          \nAliphatic amines act as nucleophiles. This is due to the availability of lone pair of electrons on nitrogen
          \n\"2nd
          \nAromatic amines do not behave as nucleophiles. This is because the lone pair of electrons on nitrogen in aromatic amines is involved in resonance with benzene ring, and as such not available for nucleophije attack.<\/p>\n

          Question 14.
          \nWrite IUPAC name of. (AISB 2004)
          \n\"2nd
          \nAnswer:
          \n4 – Methoxy benzene amine.<\/p>\n

          Question 15.
          \nHow is phenyl aminomethane obtained from phenyl nitrile. (AISB 2004)
          \nAnswer:
          \n\"2nd<\/p>\n

          Question 16.
          \nWhy is an alkyl amine more basic than ammonia ? (CBSE Delhi 2009)
          \nAnswer:
          \nThis is because of the presence of electron releasing alkyl group (positive I effect) which increases the electron density on nitrogen.<\/p>\n

          Question 17.
          \nGive the IUPAC name of. H2<\/sub>N \u2014 CH3<\/sub>\u2014 CH2<\/sub> \u2014 CH = CH2<\/sub> (CBSE Delhi 2010)
          \nAnswer:
          \nBut-3-en-l-amine.<\/p>\n

          2nd PUC Amines Short Answers Questions<\/h3>\n

          Question 1.
          \nHow are the following conversions accomplished
          \n(1) Aniline to chlorobenzene
          \n(2) Nitrobenzene to phenol
          \n(3) Aniline to benzoic acid
          \nAnswer:
          \n(1)
          \n\"2nd<\/p>\n

          \"KSEEB<\/p>\n

          Question 2.
          \nAccount for the following observation:
          \n1. Tertiary amine does not undergo acylation reaction
          \n2. Aniline readily reacts with bromine to give 2,4,6- tribromo aniline (CBSE 2002, 2004)
          \nAnswer:
          \n1. Tertiary amine do not undergo acylation reaction because they do not contain a H-atom on the N-atom.
          \n2. This is due to the strong activating effect of the amino group, Halogenation of amines occur very fast and halogens enter at all the three (one P-and two o-positions) positions even in the absence of the catalyst.<\/p>\n

          Question 3.
          \nGive a chemical test to distinguish between aniline and N-methylaniline. (CBSE 2001, 2006)
          \nAnswer:
          \nAniline being a 1\u00b0 amine gives carbylamine test i.e. on heating with CHCl3<\/sub>, in presence of alc. KOH, it gives offensive smell of phenyl isocyanide while N-methyl aniline being a 2\u00b0 amine does not give this test.
          \n\"2nd<\/p>\n

          Question 4.
          \nWrite one chemical reaction each to illustrate the following
          \n1. Hoffmann\u2019s bromamide reaction
          \n2. Gabriel phthalimide synthesis (CBSE 2008)
          \nAnswer:<\/p>\n

          1. Carbylamine reaction
          \n1\u00b0 amines (aromatic\/aliphatic) On heating with chloroform and alcoholic solution of KOH gives isocyanides\/carbylamines producing unpleasant smell.
          \n\"2nd
          \n2\u00b0 and 3\u00b0 (aromatic\/aliphatic) do not give this reaction.<\/p>\n

          2. Diazotisation .
          \nAromatic 1\u00b0 amines into diazonium salts by action of HNO2<\/sub> and dil HCl at ice cold temperature as HNO2<\/sub> and product are unstable at higher temperature.
          \n\"2nd
          \n3. Hofmann\u2019s bromamide reaction
          \nInvolves heating of an amide with bromine and caustic alkali to yield an amine with one carbon less than original amide.
          \n\"2nd
          \n4. Coupling reaction
          \nDiazonium salt acts as electrophile and brings about substitution in electron rich aromatic ring such as phenol and amines. The result is formation of dye.
          \n(i) with phenol – In basic medium (9 – 10 pH) at low temperature (273 – 278 K)
          \n\"2nd<\/p>\n

          (ii) with amines – In acidic medium (ph = 4 – 5) at low temperature (273 – 278 K)
          \n\"2nd<\/p>\n

          5. Ammonolysis – Alkyl halide (1\u00b0) react with alcoholic ammonia to form 1\u00b0, 2\u00b0, 3\u00b0 amines along with 4\u00b0ammonium salts. It is carried out by heating alkyl halide with ale solution of NH3 in sealed tube at 370 K.
          \nIf ammonia is used in excess, 1\u00b0 amine is major product
          \n\"2nd
          \nIf alkyl halide is excess, it forms all 1\u00b0, 2\u00b0, 3\u00b0, 4\u00b0 ammonium salts
          \n\"2nd<\/p>\n

          6. Acetylation
          \nAliphatic amines:
          \nIntroduction of acetyl group (CH3<\/sub>CO-) into any molecule.
          \n1\u00b0 and 2\u00b0 amines react with acetyl chloride to form amides
          \nBut 3\u00b0 amines has NO H on N, so do not react.
          \n\"2nd
          \n7. Gabriel phthalimide synthesis
          \n\"2nd
          \nReaction between alkyl halide (R – X) and ammonia gives mixture, of 1\u00b0, 2\u00b0 and 3\u00b0 amines with tetraalkyl ammonium halide. Mixture is not separable.
          \nTo get pure 1\u00b0 amine, Gabriel phthalimide synthesis is preferred.<\/p>\n

          \"KSEEB<\/p>\n

          Question 5.
          \n(a) Arrange the following in an increasing order of basic strength in water:
          \nC2<\/sub>H5<\/sub>NH2<\/sub>, (C2<\/sub>H5<\/sub>)2<\/sub>NH, (C2<\/sub>H5<\/sub>)3<\/sub>N and NH3<\/sub>
          \n(b) Arrange the following in increasing order of basic strength in the gas phase
          \nC2<\/sub>H5<\/sub>NH2<\/sub>, (C2<\/sub>H5<\/sub>)2<\/sub>NH, (C2<\/sub>H5<\/sub>)3<\/sub>N and CH3<\/sub>NH2<\/sub>. (CBSE Delhi 2008)
          \nAnswer:
          \n(a) In aqueous solution, basic strength decreases in the order. .
          \n(C2<\/sub>H5<\/sub>)NH > (C2<\/sub>H5<\/sub>)3<\/sub>N > C2<\/sub>H5<\/sub>NH2<\/sub> > NH3<\/sub>
          \n(b) In gaseous phase,(basic strength decreases in the order.
          \n(C2<\/sub>H5<\/sub>)3<\/sub>N > (C2<\/sub>H5<\/sub>)2<\/sub>NH > C2<\/sub>H5<\/sub>NH2<\/sub> > NH3<\/sub>.<\/p>\n

          Question 6.
          \nGive an example for each describe the following reactions.
          \n1. Hoffmann\u2019s bromamide reaction
          \n2. Coupling reaction
          \n3. Gattermann reaction (CBSE 2009)
          \nAnswer:
          \n1.<\/p>\n

          1. Carbylamine reaction
          \n1\u00b0 amines (aromatic\/aliphatic) On heating with chloroform and alcoholic solution of KOH gives isocyanides\/carbylamines producing unpleasant smell.
          \n\"2nd
          \n2\u00b0 and 3\u00b0 (aromatic\/aliphatic) do not give this reaction.<\/p>\n

          2. Diazotisation .
          \nAromatic 1\u00b0 amines into diazonium salts by action of HNO2<\/sub> and dil HCl at ice cold temperature as HNO2<\/sub> and product are unstable at higher temperature.
          \n\"2nd
          \n3. Hofmann\u2019s bromamide reaction
          \nInvolves heating of an amide with bromine and caustic alkali to yield an amine with one carbon less than original amide.
          \n\"2nd
          \n4. Coupling reaction
          \nDiazonium salt acts as electrophile and brings about substitution in electron rich aromatic ring such as phenol and amines. The result is formation of dye.
          \n(i) with phenol – In basic medium (9 – 10 pH) at low temperature (273 – 278 K)
          \n\"2nd<\/p>\n

          (ii) with amines – In acidic medium (ph = 4 – 5) at low temperature (273 – 278 K)
          \n\"2nd<\/p>\n

          5. Ammonolysis – Alkyl halide (1\u00b0) react with alcoholic ammonia to form 1\u00b0, 2\u00b0, 3\u00b0 amines along with 4\u00b0ammonium salts. It is carried out by heating alkyl halide with ale solution of NH3 in sealed tube at 370 K.
          \nIf ammonia is used in excess, 1\u00b0 amine is major product
          \n\"2nd
          \nIf alkyl halide is excess, it forms all 1\u00b0, 2\u00b0, 3\u00b0, 4\u00b0 ammonium salts
          \n\"2nd<\/p>\n

          6. Acetylation
          \nAliphatic amines:
          \nIntroduction of acetyl group (CH3<\/sub>CO-) into any molecule.
          \n1\u00b0 and 2\u00b0 amines react with acetyl chloride to form amides
          \nBut 3\u00b0 amines has NO H on N, so do not react.
          \n\"2nd
          \n7. Gabriel phthalimide synthesis
          \n\"2nd
          \nReaction between alkyl halide (R – X) and ammonia gives mixture, of 1\u00b0, 2\u00b0 and 3\u00b0 amines with tetraalkyl ammonium halide. Mixture is not separable.
          \nTo get pure 1\u00b0 amine, Gabriel phthalimide synthesis is preferred.<\/p>\n

          2. The reaction of diazonium salt with phenols and aromatic amines to form azo compounds of the general formula Ar – N = N – Ar is called coupling reaction. The coupling with phenol takes place in mildly alkaline medium while with amines it occurs under faintly acidic condition.
          \n\"2nd
          \n\"2nd<\/p>\n

          3. Gattermann reaction
          \nWhen benzene diazonium chloride reacts with Halogen acids in presence of Cu powder, aryl halide is formed as.
          \n\"2nd<\/p>\n

          Question 7.
          \nComplete the following reaction equation ?
          \n(1)
          \n\"2nd
          \n(2) C6HSN2CI + H3PO2 + H2O \u2192
          \n(3) C6H5NH2 + Br2 (aq) 4
          \nAnswer:
          \n(1)
          \n\"2nd
          \n(2) and (3) Same as NCRT text book Q.No. 11.<\/p>\n

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

          You can Download Chapter 13 Amines 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 13 Amines 2nd PUC Chemistry Amines NCERT Textbook Questions Question 1. Write IUPAC names …<\/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\/10253"}],"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=10253"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10253\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10253"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10253"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}