CBSE 12th Standard Chemistry Subject Amines Chapter Case Study Questions With Solution 2021
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CBSE 12th Standard Chemistry Subject Amines Case Study Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Chemistry
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Read the passage given below and answer the following questions:
When the mixture contains the three amine salts (1°, 2° and 3°) along with quaternary salt, it is distilled with KOH solution. The three amines distill, leaving the quaternary salt unchanged in the solution. Then the mixture of amines is separated by fractional distillation, Hinsbergs method and Hoffmann's method.
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Hinsberg reagent is(a) aliphatic sulphonyl chloride (b) phthalamide (c) aromatic sulphonyl chloride (d) anhydrous ZnCl2 + conc. HCl. (ii) Primary amine with Hinsberg's reagent forms
(a) N-alkyl benzene sulphonamide soluble in KOH solution (b) N-alkyl benzene sulphonamide insoluble in KOH solution (c) N, N-dialkyl benzene sulphonamide soluble in KOH solution (d) N, N-dialkyl benzene sulphonamide insoluble in KOH solution. (iii) To separate amines in a mixture Hoffmann's method is used. The Hoffmann's reagent is
(a) benzenesulphonyl chloride (b) diethyloxalate (c) benzeneisocyanide (d) p-toulenesulphonic acid. (iv) 3o amines with Hinsberg's reagent give
(a) no reaction (b) product which is same as that of 10 amine (c) product which is same as that of 2° amine (d) products which is a quaternary salt. -
Read the passage given below and answer the following questions:
A mixture of two aromatic compounds (A) and (B) was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. The organic layer containing compound (A), when heated with alcoholic solution of KOH produce C7H5N (C) associated with unpleasant odour.
The following questions are multiple choice questions. Choose the most appropriate answer:
The reaction of (A) with alcoholic solution of KOH to produce (C) of unpleasant odour is called(a) Sandmeyer reaction (b) Carbylamine reaction (c) Ullmann reaction (d) Reimer-Tiemann reaction (ii) The alkaline aqueous layer (B) when heated with chloroform and then acidified give a mixture of isomeric compounds of molecular formula C7H6O2. (B) is
(a) C6H5CHO (b) C6H5COOH (c) C6H5CH3 (d) C6H5OH (iii) In the chemical reaction, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{CHCl}_{3}+3 \mathrm{KOH} \longrightarrow(A)+(B)+3 \mathrm{H}_{2} \mathrm{O},\) the compounds (A) and (B) are respectively
(a) C2H5NC and KCI (b) C2H5CN and KCI (c) CH3CH2CONH2 and KCI (d) C2H5NC and K2CO3 (iv) Direct nitration of an aromatic compound (A) is not feasible because
(a) the reaction cannot be stopped at the mononitration stage (b) a mixture of o, m and p-nitroaniline is always obtained (c) nitric acid oxidises most of the aromatic compound to give oxidation products along with only a small amount of nitrated products (d) all of the above
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Read the passage given below and answer the following questions:
The amines are basic in nature due to the presence of a lone pair of electron on N-atom of the -NH2 group, which it can donate to electron deficient compounds. Aliphatic amines are stronger bases than NH3 because of the +1 effect of the alkyl groups. Greater the number of alkyl groups attached to N-atom, higher is the electron density on it and more will be the basicity. Thus, the order of basic nature of amines is expected to be 3° > 2° > 1°, however the observed order is 2° > 1° > 3°. This is explained on the basis of crowding on N-atom of the amine by alkyl groups which hinders the approach and bonding by a proton, consequently, the electron pair which is present on N is unavailable for donation and hence 3° amines are the weakest bases. Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron -donating groups such as -CH3 , -OCH3 , etc. increase the basicity while electron-withdrawing substitutes such as -NO2 , -CN, halogens, etc. decrease the basicity of amines. The effect of these substituents is more at p than at m-positions.
The following questions are multiple choice questions. Choose the most appropriate answer :
(i) Which one of the following is the strongest base in aqueous solution?(a) Methyl amine (b) Trimethyl amine (c) Aniline (d) Dimethyl amine (ii) Which order of basicity is correct?
(a) Aniline> m-toluidine > o-toluidine (b) Aniline> o-toluidine > m-toluidine (c) o-toluidine> aniline> m-toluidine (d) o-toluidine < aniline < m-toluidine (iii) What ts the decreasing order of basicity of primary, secondary and tertiary ethylamines and NH3?
(a) NH3 > C2H5NH2 > (C2H5)2NH > (C2H5)3N (b) (C2H5)3N> (C2H5)2NH > C2H5NH2 > NH3 (c) (C2H5)2NH >C2H5NH2 > (C2H5)3N > NH3 (d) (C2H5)2NH> (C2H5)3N > C2H5NH2 > NH3 (iv) Choose the correct statement.
(a) Methylamine is slightly acidic. (b) Methylamine is less basic than ammonia. (c) Methylamine is a stronger base than ammonia. (d) Methylamine forms salts with alkalie -
Read the passage given below and answer the following questions:
Amines are alkyl or aryl derivatives of ammonia formed by replacement of one or more hydrogen atoms. Alkyl derivatives are called aliphatic amines and aryl derivatives are known as aromatic amines. The presence of aromatic amines can be identified by performing dye test. Aniline is the simplest example of aromatic amine. It undergoes electrophilic substitution reactions in which - NH2 group strongly activates the aromatic ring through delocalisation oflone pair of electrons of N-atom. Aniline undergoes electrophilic substitution reactions. Ortho and para positions to the -NH2 group become centres of high electrons density. Thus, -NH2 group is ortho and para-directing and powerful activating group. The following questions are multiple choice questions.
Choose the most appropriate answer:
(i) Cyclohexylamine and aniline can be distinguished by(a) Hinsberg test (b) carbylamine test (c) Lassaigne test (d) azo dye test (ii) Which of the following compounds gives-dye test?
(a) Aniline (b) Methyl amine (c) Diphenyl amine (d) Ethyl amine (iii) Oxidation of aniline with manganese dioxide and sulphuric acid produces
(a) phenylhydroxylamine (b) nitrobenzene (c) p-benzoquinone (d) phenol. (iv) Aniline when treated with conc, HNO3 and H2SO4 gives
(a) phenylhydroxylamine (b) m-nitroaniline (c) p-benzoquinone (d) nitrobenzene.
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CBSE 12th Standard Chemistry Subject Amines Case Study Questions With Solution 2021 Answer Keys
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(i) (c)
(ii) (a): A primary amine forms N-alkylbenzene sulphonamidewhich because ofthe presence of an acidic hydrogen on the N-atom dissolves in aqueous KOH.
(iii) (b)
(iv) (a): Tertiary amine does not contain a replaceable hydrogen on the nitrogen atom. So, 3o amine does not react with Hinsberg's reagent. -
(i) (b) : Carbylamine reaction
C6H5NH2 + CHCl3 + 3KOH (alc.) ➝ C6H52NC + 3KCI + 3H2O
Aniline Phenyl isocyanide (C)
(A)
(ii) (d): Alkaline layer on treating with CHCl3 followed by acidification gives two is?mers having formula (C7H6O2). This is Reimer-Tiemann reaction and thus (B) is C6H5OH.
\( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CHCl}_{3}+\mathrm{KOH} \stackrel{\mathrm{H}^{+}}{\longrightarrow}\\ \text { Phenol }(B) \)
(iii) (a): CH3CH2NH2 + CHCl3 + 3KOH ➝ C2H5NC + 3KCl + 3H2O
This is called carbylamine reaction.
(iv) (c): Direct nitration of aniline is not a feasible process because nitric acid oxidises most of aniline to give oxidation products along with only a small amount of nitrated products. -
(i) (d): The increasing order of basicity of the given compounds is (CH3)2NH > CH3NH2 > (CH3)3N > C6H5NH2 .Due to the +1effect of alkyl groups, the electron density on nitrogen increases and thus, the availability of the lone pair of electrons to proton increases and hence, the basicity of amines also increases. So, aliphatic amines are more basic than aniline. In case of tertiary amine (CH3)3N, the covering of alkyl groups over nitrogen atom from all sides makes the approach and bonding by a proton relatively difficult, hence the basicity decreases. Electrop withdrawing groups decrea e electron density on nitrogen atom and thereby decreasing basicity.
(ii) (d): In general, electron donating (+ R) group which when present on benzene ring (-NH2 , -OR, -R, etc.) at the para position increases the basicity of aniline.
Ortho substituted anilines are weaker bases than aniline due to ortho effect.
(iii) (d): In case of ethylamines, the combined effect of inductive effect, steric effect'and solvation effect gives the order of basic strength as
(C2H5)3N> (C2H5)2NH > C2H5NH2 > NH3
(2°) (3°) (1°)
(iv) (c) : Methyl amine is stronger base than ammonia due to electron releasing inductive effect of methyl group. -
(i) (d)
(ii) (a): Aromatic primary amines give dye test.
(iv) (b): In acidic medium aniline gets protonated to anilinium ion which is meta-directing.
