Class 12 Chemistry Study Guide: CBSE Class 12 Chemistry Amines Case Study Questions Set 01
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Case Study 1
RCONH2 is converted into RNH2 by means of Hoffmann bromamide degradation. During the reaction amide is treated with Br2 and alkali to get amine. This reaction is used to descend the series in which carbon atom is removed as carbonate ion (CO32−). Hoffmann bromide degradation reaction can be written as:
(i) p-Cl–C6H4–CONH2 OH−
⟶
Br2 (ii) p-Cl–C6H4–CONH–Br
(ii) OH−
⟶ (iii) p-Cl–C6H4–CO–N−–Br (anion)
(iii) → (iv) p-Cl–C6H4–N=C=O
(iv) H2O
⟶ (v) p-Cl–C6H4–NH–COOH
(v) −CO2
⟶ (vi) p-Cl–C6H4–NH2
(a) primary amines
(b) secondary amines
(c) tertiary amines
(d) secondary aromatic amines.
Show Answer & Explanation
Answer: (a) primary amines
Explanation:
1. In Hoffmann bromamide degradation the –CONH2 group of a primary amide is converted into –NH2, so only primary amines (with one carbon atom less) are obtained.
Teacher's Note:
1. The N of –CONH2 is bonded to only one carbon, so the product can only be a primary amine.
(a) Formation of (i)
(b) Formation of (ii)
(c) Formation of (iii)
(d) Formation of (iv)
Show Answer & Explanation
Answer: (d) Formation of (iv)
Explanation:
1. The rate determining step is probably loss of Br− to form isocyanate (iv) as this is the slowest step.
2. In this step the aryl group migrates from carbon to nitrogen as Br− leaves.
Teacher's Note:
1. The isocyanate step is the rearrangement step, where the aryl/alkyl group shifts from C to N.
(a) KBr
(b) KBr + CH3ONa
(c) KBr + KOH
(d) Br2 + KOH
Show Answer & Explanation
Answer: (d) Br2 + KOH
Explanation:
1. p-Cl–C6H4–CONH2 + Br2 KOH
⟶ p-Cl–C6H4–CONHBr + KBr + H2O
Teacher's Note:
1. Br2 in alkali (hypobromite) is the reagent; KBr is a by-product, not the brominating agent.
(a) m-D–C6H4–NH2 and m-D–C6H4–15NH2
(b) m-D–C6H4–NH2 and C6H5–15NH2
(c) C6H5–15NH2 and C6H5–15NH2
(d) C6H5–15NHD and C6H6
Show Answer & Explanation
Answer: (b) m-D–C6H4–NH2 and C6H5–15NH2
Explanation:
1. Since the overall reaction is intramolecular, hence there will be no effect on product formation.
2. m-D–C6H4–CONH2 + C6H5–CO15NH2 Br2/NaOH
⟶ m-D–C6H4–NH2 + C6H5–15NH2
3. The cross products C6H5–NH2 and m-D–C6H4–15NH2 are not formed.
Teacher's Note:
1. A crossover experiment that gives no mixed products proves the migrating group never leaves its own molecule.
Case Study 2
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 +I 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 substituents such as –NO2, –CN, halogens, etc. decrease the basicity of amines. The effect of these substituents is more at p than at m-positions.
(a) Methyl amine
(b) Trimethyl amine
(c) Aniline
(d) Dimethyl amine
Show Answer & Explanation
Answer: (d) Dimethyl amine
Explanation:
1. The decreasing order of basicity of the given compounds is (CH3)2NH > CH3NH2 > (CH3)3N > C6H5NH2.
2. Due to the +I effect 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.
3. 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. Electron withdrawing groups decrease electron density on nitrogen atom and thereby decreasing basicity.
Teacher's Note:
1. In water, solvation of the protonated amine matters as much as the +I effect, which is why the 2° amine wins.
(a) Aniline > m-toluidine > o-toluidine
(b) Aniline > o-toluidine > m-toluidine
(c) o-Toluidine > aniline > m-toluidine
(d) o-Toluidine < aniline < m-toluidine
Show Answer & Explanation
Answer: (d) o-Toluidine < aniline < m-toluidine
Explanation:
1. In general, electron donating (+R) group which when present on benzene ring (–NH2, –OR, –R, etc.) increases the basicity of aniline.
2. Ortho substituted anilines are weaker bases than aniline due to ortho effect.
3. So, m-toluidine > aniline > o-toluidine.
Teacher's Note:
1. Almost any group at the ortho position lowers the basicity of aniline (ortho effect), even an electron-releasing one like –CH3.
(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
Show Answer & Explanation
Answer: (d) (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
Explanation:
1. In case of ethylamines, the combined effect of inductive effect, steric effect and solvation effect gives the order of basic strength as
2. (C2H5)2NH (2°) > (C2H5)3N (3°) > C2H5NH2 (1°) > NH3
Teacher's Note:
1. Do not copy the methylamine order for ethylamines: with ethyl groups the 3° amine is stronger than the 1° amine.
(a) CH3NH2 > (CH3)3N > (CH3)2NH
(b) (CH3)3N > (CH3)2NH > CH3NH2
(c) CH3NH2 > (CH3)2NH > (CH3)3N
(d) (CH3)3N > CH3NH2 > (CH3)2NH
Show Answer & Explanation
Answer: (b) (CH3)3N > (CH3)2NH > CH3NH2
Explanation:
1. In non-aqueous solvents the basic strength of alkyl amines follows the order:
2. tertiary amines > secondary amines > primary amines.
Teacher's Note:
1. Benzene cannot solvate the ammonium ions, so only the +I effect counts and more alkyl groups mean a stronger base.
Case Study 3
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, Hinsberg’s method and Hoffmann’s method.
(1°, 2° and 3° amines in mixture) (i) C6H5SO2Cl, (ii) KOH, (iii) Distillation
⟶ Distillate (Amine A) + Mother liquor (Amine B and amine C)
Mother liquor Filtered
⟶ Filtrate 70% H2SO4
⟶ (Amine B); Residue 70% H2SO4
⟶ (Amine C)
(a) aliphatic sulphonyl chloride
(b) phthalamide
(c) aromatic sulphonyl chloride
(d) anhydrous ZnCl2 + conc. HCl.
Show Answer & Explanation
Answer: (c) aromatic sulphonyl chloride
Explanation:
1. Hinsberg reagent is benzenesulphonyl chloride, C6H5SO2Cl, which is an aromatic sulphonyl chloride.
Teacher's Note:
1. p-Toluenesulphonyl chloride is sometimes used in place of benzenesulphonyl chloride and works the same way.
(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.
Show Answer & Explanation
Answer: (a) N-alkyl benzene sulphonamide soluble in KOH solution
Explanation:
1. A primary amine forms N-alkylbenzene sulphonamide with Hinsberg’s reagent because of the presence of an acidic hydrogen on the N-atom, dissolves in aqueous KOH.
2. C6H5SO2Cl + RNH2 → C6H5SO2NHR + HCl
Teacher's Note:
1. In the scheme above, amine B (in the filtrate as the soluble potassium salt) is therefore the 1° amine.
(a) benzenesulphonyl chloride
(b) diethyl oxalate
(c) benzeneisocyanide
(d) p-toluenesulphonic acid.
Show Answer & Explanation
Answer: (b) diethyl oxalate
Explanation:
1. In Hoffmann’s method the mixture of amines is treated with diethyl oxalate, (COOC2H5)2. The 1° amine gives a solid dialkyl oxamide, the 2° amine gives a liquid dialkyl oxamic ester and the 3° amine does not react.
Teacher's Note:
1. Do not confuse the two methods: Hinsberg uses C6H5SO2Cl, Hoffmann’s separation method uses diethyl oxalate.
(a) no reaction
(b) product which is same as that of 1° amine
(c) product which is same as that of 2° amine
(d) products which is a quaternary salt.
Show Answer & Explanation
Answer: (a) no reaction
Explanation:
1. Tertiary amine does not contain a replaceable hydrogen on the nitrogen atom. So, 3° amine does not react with Hinsberg’s reagent.
Teacher's Note:
1. That is why amine A, which distils off unchanged in the scheme, is the 3° amine.
Case Study 4
Amines are basic in nature. The basic strength of amines can be expressed by their dissociation constant, Kb or pKb.
RNH2 + H2O ⇌ RNH3+ + OH−
\( K_b = \frac{[RNH_3^+][OH^-]}{[RNH_2]} \) and pKb = −log Kb
Greater the Kb value or smaller the pKb value, more is the basic strength of amine. Aryl amines such as aniline are less basic than aliphatic amines due to the involvement of lone pair of electrons on N-atom with the resonance in benzene. In derivatives of aniline, the electron releasing groups increase the basic strength while electron withdrawing groups decrease the basic strength. The base weakening effect of electron withdrawing group and base strengthening effect of electron releasing group is more marked at p-position than at m-position. o-Substituted aniline is less basic than aniline due to ortho effect and is probable due to combination of electronic and steric effect.
(a) C6H5NH2
(b) m-O2N–C6H4–NH2
(c) C6H5N(CH3)2
(d) C6H5NHCH3
Show Answer & Explanation
Answer: (c) C6H5N(CH3)2
Explanation:
1. Lowest pKb means the strongest base. Each methyl group on nitrogen increases the electron density on N by +I effect, so C6H5N(CH3)2 is the most basic and has the lowest pKb; the –NO2 group lowers the basicity.
Teacher's Note:
1. Approximate pKb values: C6H5N(CH3)2 8.92, C6H5NHCH3 9.30, C6H5NH2 9.38 – small differences, so reason from effects rather than memory.
(a) C6H5NH2
(b) p-NO2–C6H4NH2
(c) m-NO2–C6H4NH2
(d) C6H5CH2NH2
Show Answer & Explanation
Answer: (d) C6H5CH2NH2
Explanation:
1. In benzylamine, C6H5CH2NH2, the –NH2 group is attached to an sp3 carbon, so the lone pair on nitrogen is not in resonance with the ring. It behaves like an aliphatic amine and is the strongest base here.
Teacher's Note:
1. Benzylamine is not an aryl amine – the CH2 group separates N from the ring, so it is far more basic than aniline.
(a) C6H5–NH–C6H5
(b) C6H5NHCH3
(c) (CH3CH2)2NH
(d) (CH3)2NH
Show Answer & Explanation
Answer: (a) C6H5–NH–C6H5
Explanation:
1. Maximum pKb means the weakest base. In diphenylamine the lone pair on nitrogen is delocalised over two benzene rings, so it is the weakest base and has the highest pKb.
Teacher's Note:
1. Each aryl group on nitrogen pulls the lone pair into resonance, so basicity falls sharply from aniline to diphenylamine.
(a) Methylamine is more basic than NH3.
(b) Primary amines form hydrogen bonds.
(c) Ethylamine has higher boiling point than propane.
(d) Dimethylamine is less basic than methylamine.
Show Answer & Explanation
Answer: (d) Dimethylamine is less basic than methylamine.
Explanation:
1. Dimethylamine is more basic than methylamine. Hence statement (d) is not correct.
Teacher's Note:
1. In water the basicity order of methylamines is (CH3)2NH > CH3NH2 > (CH3)3N > NH3.
Case Study 5
Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of basicity of aliphatic amines should be: primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like –CH3 group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than CH3 group, there will be steric hinderance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from –CH3 to –C2H5 results in change of the order of basic strength.
Assertion (A): (CH3)2NH > CH3NH2 > (CH3)3N > NH3 is the order of basic strength in case of methyl substituted amines.
Reason (R): The inductive effect, solvation effect and steric hindrance of the alkyl group decide the basic strength of alkyl amines in the aqueous state.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. (CH3)2NH > CH3NH2 > (CH3)3N > NH3 is the order of basic strength in case of methyl substituted amines as the inductive effect is suppressed due to solvation effect and steric hinderance of the alkyl group. Hence H-bonding dominates over +I effect.
Teacher's Note:
1. For methyl amines in water, the tertiary amine drops below the primary one; learn this order separately from the ethyl series.
Assertion (A): (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 is the order of basic strength in case of ethyl substituted amines.
Reason (R): The change of nature of the alkyl group, does not result in change of the order of basic strength.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 is the order of basic strength in case of ethyl substituted amines. As alkyl group is greater than –CH3 group, then steric hindrance is there due to hydrogen bonding. Hence +I effects will dominate over H-bonding.
2. So changing the alkyl group from –CH3 to –C2H5 does change the order; R is false.
Teacher's Note:
1. Compare the two series: tertiary is above primary for ethyl amines but below primary for methyl amines.
Assertion (A): Greater is the stability of the substituted ammonium cation, stronger is the corresponding amine as a base.
Reason (R): The order of basicity of aliphatic amines is: primary > secondary > tertiary.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Greater the stability of the substituted ammonium cation, stronger is the corresponding amine as a base but the inductive effect, solvation effect and steric hinderance of the alkyl group decide the basic strength of alkyl amines in the aqueous state.
2. So the order primary > secondary > tertiary is not the actual order; R is false.
Teacher's Note:
1. In water, secondary amines are the strongest bases in both the methyl and the ethyl series.
Assertion (A): Amines behave as a Lewis base.
Reason (R): Amines have an unshared pair of electrons on nitrogen atom.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Amines behave as a Lewis base as they have an unshared pair of electrons on nitrogen atom.
Teacher's Note:
1. A Lewis base is an electron-pair donor; the lone pair on N is also why amines are nucleophiles.
Case Study 6
Benzene ring in aniline is highly activated. This is due to the sharing of lone pair of nitrogen with the ring which results in increase in the electron density on the ring and hence facilitates the electrophilic attack. The substitution mainly takes place at ortho and para positions because electron density is more at ortho and para positions. On reaction with aqueous bromine all the ortho and para positions get substituted resulting in the formation of 2,4,6-tribromoaniline. To get a monobromo compound, the amino group is acetylated before bromination. After bromination, the bromoacetanilide is acid hydrolysed to give the desired halogenated amine.
Assertion (A): Benzene ring of aniline is highly deactivated.
Reason (R): In aniline, the sharing of lone pair of nitrogen with the ring increases the electron density on the ring.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. Benzene ring in aniline is highly activated to donate electrons, as lone pair of (–NH2) nitrogen are delocalized to increase electron density in the ring.
Teacher's Note:
1. –NH2 is one of the strongest activating groups; it is –NH3+ (in acid) that deactivates the ring.
Assertion (A): In aniline –NH2 group facilitates the electrophilic attack.
Reason (R): It is due to decrease in electron density on the ring.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. In aniline, –NH2 group facilitates the electrophilic attack because the sharing of lone pair of nitrogen with the ring increases the electron density on the ring.
Teacher's Note:
1. Electrophiles are attracted to electron-rich rings, so a group that helps electrophilic attack must increase, not decrease, ring electron density.
Assertion (A): In aniline, the substitution mainly takes place at ortho and para positions.
Reason (R): The electron density is more at ortho and para positions.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. In aniline, the electron density is more at ortho and para positions than meta position, so, the substitution mainly takes place at ortho and para positions.
2. The resonating structures of aniline place a negative charge on the ortho and para carbons (with N+), showing more electron density at these positions.
Teacher's Note:
1. Draw the resonance structures of aniline: the negative charge appears only at the two ortho carbons and the para carbon, never at meta.
Assertion (A): The amino group of aniline is acetylated before bromination.
Reason (R): It is due to the strong deactivating effect of –NH2 group.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. –NH2 group of aniline is acetylated before bromination due to the strong activating effect of –NH2 group.
Teacher's Note:
1. Acetylation is a protecting step: it tones down the strong activation so that only one bromine enters, and it is removed afterwards by hydrolysis.
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