CBSE Class 12 Chemistry Organic Chemistry Amines Worksheet

Question. What is the correct IUPAC name of H₂N-(CH₂)₅-NH₂?
a. Pentan-1,5-diamine
b. 1,5-Diaminopentane
c. Pentamethylenediamine
d. Pentane-1,5-diamine
Answer. D

Question. Which of the following does not react with Hinsberg reagent?
a. C₂H₅NH₂
b. (CH₃)₂NH
c. (CH₃)₃N
d. CH₃ CH(NH₂)CH₃
Answer. C

Question. Which of the following amines are insoluble in water?
a. Methanamine
b. Ethanamine
c. Propanamine
d. Benzenamine
Answer. D

Question. In this reaction acetamide is converted to methanamine
a. Gabriel phthalimide synthesis
b. Carbylamine reaction
c. Stephen’s reaction
d. Hoffmann bromamide reaction
Answer. A

Question. Which of the following is not a final product of the reaction between propylamine and nitrous acid?
a. CH₃CH₂CH₂N₂Cl
b. CH₃CH₂CH₂OH
c. N₂ gas
d. HCI
Answer. A

Question. Hinsberg’s reagent is
a. Benzenesulphonic acid
b. Benzenesulphonyl chloride
c. p-toluenesulphonyl chloride
d. Chlorosulphuric acid
Answer. B

Question. Starting from propanoic acid, the following reactions were carried out, what is the compound Z?
a. CH₃-CH₂−Br
b. CH₃-CH₂−NH₂
c. CH₃-CH₂-COBr
d. CH₃−CH₂−CH₂−NH₂
Answer. A

Question. Aniline in a set of reactions yielded a product D. The structure of D would be
a. C₆H₅CH₂OH
b. C₆H₅CH₂NH₂
c. C₆H₅NHOH
d. C₆H₅NHCH₂CH₃.
Answer. A

Question. The hybridisation state of N of R2NH
a. sp³
b. sp²
c. sp
d. dsp²
Answer. A

ASSERTION -REASON TYPE QUESTIONS

Choose the correct answer from the following choices
a Both assertion and reason are correct statements and reason is correct explanation of assertion
b Both assertion and reason are correct statements but reason is not correct explanation of assertion
c Assertion is correct statement but reason is wrong statement
d Assertion is wrong statement but reason is correct statement

Question. Assertion: Alkylation of amines gives polysubstituted product where as acylation of amines gives a monosubstituted product
Reason: Steric hindrance of an acyl group prevents the approach of further acyl groups.
Answer. C

Question. Assertion: Anilinium chloride is more acidic than ammonium chloride
Reason: Anilinium ion is resonance stabilized.
Answer. C

Question. Assertion: Gabriel phthalimide reaction can be used to prepare aryl and alkyl amines
Reason: Aryl halides have same reactivity as alkyl halides towards nucleophilic substitution reactions.
Answer. D

Question. Assertion: Aniline does not undergo Friedel -Crafts reaction
Reason: Friedel-Crafts reaction is electrophilic substitution reaction
Answer. B

Question. Assertion: CuCl₂ gives a deep blue colored solution with ethyl amine
Reason: Ethylamine molecules coordinate with cupric ions forming a blue coloured complex.
Answer. A

Question. Assertion: The order of boiling points of isomeric amines is Primary>Secondary>Tertiary
Reason: Intermolecular association is more in primary ,then in secondary and least in tertiary amines.
Answer. A

Question. Assertion: Aliphatic amines are weaker base than ammonia
Reason:+I effect of alkyl group results in high electron density on nitrogen atom.
Answer. D

Question. Assertion: Pkb value of aniline is low,
Reason: The unshared pair on nitrogen atom to be in conjugation with the benzene ring making it less available
Answer. D

Question. Write the structure of N-methylethanamine.
Answer. CH₃CH₂NHCH₃(N-methylethanamine)

Question. Give the IUPAC name of
H₂N — CH₂ — CH₂ — CH = CH₂.
Answer. But-3-en-1-amine

Question. Give reasons for the following :
Primary amines have higher boiling point than tertiary amines. 
Answer. Primary amines (R – NH₂) have two hydrogen atoms on nitrogen which can undergo intermolecular hydrogen bonding whereas no such hydrogen bonding is present in tertiary amines (R₃N). So, primary amines boil at a higher temperature than tertiary amines.

Question. Arrange the following in the increasing order of their boiling point :
C₂H₅NH₂, C₂H₅OH, (CH₃)₃N
Answer. Increasing order of boiling points :
(CH₃)₃N < C₂H₅NH₂ < C₂H₅OH
Tertiary amine does not have hydrogen to form hydrogen bonding and hydrogen bonding in alcohol is stronger than that of amines because oxygen is more electronegative than nitrogen.

Question. Account for the following :
Ethylamine is soluble in water whereas aniline is not.
Answer. Ethylamine is soluble in water due to formation of intermolecular hydrogen bonds with water molecules. However, in aniline due to large hydrophobic aryl group the extent of hydrogen bonding decreases considerably and hence aniline is insoluble in water.

Question. Account for the following :
Nitro compounds have higher boiling points than the hydrocarbons having almost the same molecular mass.
Answer. The nitro compounds are highly polar molecules. Due to this polarity they have strong intermolecular dipole – dipole interactions which causes them to have higher boiling points in comparison to the hydrocarbons having almost same molecular mass.

Question. Give a simple chemical test to distinguish between the following pair of compounds :
(CH₃)₂NH and (CH₃)₃N
Answer. When treated with benzenesulphonyl chloride (Hinsberg’s reagent), (CH₃)₂NH forms insoluble N, N-dialkylbenzene sulphonamide which is insoluble in KOH whereas tertiary amine does not react at all.

Question. Arrange the following compounds in increasing order of solubility in water :
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂
Answer. C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂ 1° amines are more soluble in water than 2° amines.
Aniline due to large hydrophobic benzene ring is least soluble.

Question. Arrange the following in the decreasing order of their basic strength in aqueous solutions :
CH₃NH₂, (CH₃)₂NH, (CH₃)₃N and NH₃
Answer. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

Question. Arrange the following in increasing order of basic strength :
C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
Answer. C₆H₅NH₂ < C₆H₅NHCH₃ < C6H₅CH₂NH₂ C₆H₅NH₂ and C₆H₅NHCH₃ are less basic than aliphatic amine C₆H₅CH₂NH₂ due to lone pair of nitrogen is in conjugation with benzene ring. But due to +I effect of —CH₃ group in C_6H5NHCH₃, it is more basic than C₆H₅NH₂.

Question. Why cannot primary aromatic amines be prepared by Gabriel phthalimide synthesis?
Answer. Aromatic amines cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Question. Write the chemical equation involved in the following reaction :
Hofmann bromamide degradation reaction
Answer. R — CONH₂ + Br₂ + 4NaOH →
Acid amide
R — NH₂ + Na₂CO₃ + 2 NaBr + 2H₂O
1° amine

Question. Give the chemical tests to distinguish between the following pairs of compounds :
(i) Methylamine and dimethylamine
(ii) Aniline and N-methylaniline
Answer. (i) Methyl amine gives carbylamine test, i.e., on treatment with alc. KOH and chloroform, followed by heating it gives offensive odour of methyl isocyanide. Dimethyl amine does not give this test.
(ii) Aniline gives carbylamine test, i.e., on treatment with alc. KOH and chloroform followed by heating it gives offensive odour of phenylisocyanide but N-methylaniline being secondary amine, does not show this test.

CASE BASED QUESTIONS

1 The basicity of amines of different classes do not follow a simple pattern because the number of groups bonded to nitrogen affects the electron density at the nitrogen atom. And, the stability of the conjugate acid in the solvent has a major effect on basicity. Thus, the basicity of amines can be explained only for amines with similar structures at the nitrogen atoms.
The basicity of an amine is increased by electron-donating groups and decreased by electron- withdrawing groups. Aryl amines are less basic than alkyl-substituted amines because some electron density provided by the nitrogen atom is distributed throughout the aromatic ring. Basicity is expressed using Kb values measured from the reaction of the amine with water. An alternate indicator of basicity is pKb, which is −log Kb. A strong base has a large Kb and a small pKb. The basicity of amines is also expressed by the acidity of their conjugate acids. A strong base has a weak conjugate acid, as given by a small value of Ka and a large pKa. 

Question. pKb values for NH₃ , CH₃NH₂,(CH₃)₂NH and (CH₃)3N has 4.75,3.38,3.27 and 4.22 respectively.
Write them in the decreasing order of basic strength. Usually the order of basicity of amines will be different from the expected order.
Answer. (CH₃)₂NH> CH₃NH₂>(CH₃)₃N>NH₃

Question. Which are the three main factors affecting basicity of amines?
Answer. +I effect ,extent of hydrogen bonding with water molecules and steric effects of the alkyl group

Question.Write the decreasing order of basicity for CH₃CH₂NH₂,(CH₃CH₂)₂NH and (CH₃CH₂)₃N
Answer. (CH₃CH₂)₂NH>(CH₃CH₂)₃N> CH₃CH₂NH₂>NH₃

Question. Compare the basicity of m-toluidine and Aniline.
Answer. m-Toluidine is more basic than aniline due to +I effect from meta position

SHORT ANSWER TYPE QUESTIONS:

Question. Write the chemical equations involved when C₂H₅NH₂ is treated with
(1) CH₃COCl/Pyridine (2) CHCl₃+KOH.
Answer. (1) C₂H₅NH₂+ CH₃COCl/Pyridine → CH₃CONHC₂H₅
(2) C₂H₅NH₂+ CHCl₃+3KOH(Alcoholic) →C₂H₅NC+3KCl+3H₂O

Question. How will you convert (1)Aniline to Bromobenzene(2)Aniline to Benzene?
NaNO₂/HCl/50C CuBr/HBr
Answer. (1) C₆H₅NH₂ → C₆H₅N2Cl → C₆H₅Br
(2) NaNO₂/HCl/50C H₃PO₂/H₂O
(1) C₆H₅NH₂ → C₆H₅N₂Cl → C₆H₆

Question. pKb of aniline is more than that of methylamine. Why?
Answer. Aniline is a weaker base than methylamine since lone pair on N is not available for donation since it is involved in conjugation with pi electrons of benzene ring.

Question. Ethylamine is soluble in water whereas aniline is not. Why?
Answer. Ethylamine can form hydrogen bonding with water while aniline can not due bulky phenyl group.

Question. Give one chemical test to distinguish between the following pairs of compounds
(1) Methylamine and Dimethylamine
(2)Aniline and benzylamine
Answer. (1) Methylamine on reaction with chloroform and alcoholic KOH gives foul smelling methyl isocynide while dimethylamine does not.
(2) Aniline on treatment with nitrous acid forms Benzene diazonium chloride which on coupling with phenol forms orange dye while benzylamine does not.

Question. Give a simple chemical test to distinguish between the following pair of compounds:
(CH₃)₂NH and (CH₃)₃N 
Answer. When treated with benzenesulphonyl chloride (Hinsberg’s reagent), (CH₃)₂NH forms insoluble N, N-dialkylbenzene sulphonamide which is insoluble in KOH whereas tertiary amine does not react at all.

Question. Why do amines act as nucleophiles?
Answer. Because the electron pair of nitrogen can coordinate with the electron deficient electrophiles

Question. Give reasons for the following:
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
Answer. (i) In Friedel - Crafts reaction, AlCl₃ is added as a catalyst which is a Lewis acid. It forms a salt with aniline due to which the nitrogen of aniline acquires positive charge. This positively charged nitrogen acts as a strong deactivating group, hence aniline does not undergo Friedel - Crafts reaction.
(ii) In aqueous solution 2° amine is more basic than 3° amine due to the combination of inductive effect, solvation effect and steric reasons.

Nomenclature of Organic Compound

Question. Write the structure of N-methylethanamine. 
Answer. Structure of N-methylethanamine : H₃C—H₂C—NH—CH₃

Question. Write the IUPAC name of the following compound:
(CH₃)₂N-CH₂CH₃
Answer. IUPAC name: N,N-Dimethylethanamine

Question. Write the IUPAC name of the following compound: 
CH₃NHCH(CH₃)₂
Answer. IUPAC name: N-Methylpropan-2-amine

Question. Give the IUPAC name of H₂N — CH₂—CH₂—CH = CH₂. 
Answer. IUPAC name : But-3-ene-1-amine

Question. Write the structure of prop-2-en-1-amine.
Answer. H₂C=CH—H₂C—NH₂

Question. Write the structure of n-methylethanamine.
Answer. Structure of n-methylethanamine :- H₃C—H2C—NH—CH₂

Question. Arrange the following in increasing order of basic strength :
C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
Answer. C₆H₅NH2 < C₆H₅NHCH₃ < C₆H₅CH₂NH₂

Question. Why is an alkylamine more basic than ammonia?
Answer. Due to electron releasing inductive effect (+I) of alkyl group, the electron density on the nitrogen atom increases and thus, it can donate the lone pair of electrons more easily than ammonia.

Question. Arrange the following compounds in increasing order of solubility in water :
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂
Answer. C₆H₅NH₂ < (C₂H₅)2NH <C₂H₅NH,

Question. Write IUPAC name of the following compound:
Answer. N-Ethyl-N-methylethanamine

Question. Give the IUPAC name of H₂N — CH₂—CH₂—CH = CH₂.
Answer. IUPAC name : But-3-ene-1-amine

Question. State reasons for the following :
(i) pK_b value for aniline is more than that for methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not soluble in water.
(iii) Primary amines have higher boiling points than tertiary amines.
Answer. (i) Higher the pKb value, lower will be the basicity therefore aniline is less basic than methylamine because the lone pair of electrons on nitrogen atom gets delocalized over the benzene ring are unavailable for protonation due to resonance in aniline which is absent in case of alkylamine.
(ii) Ethylamine is soluble in water due to its capability to form H-bonds with water while aniline is insoluble in water due to larger hydrocarbon part which tends to retard the formation of H-bonds.
(iii) Due to presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of a H-atom on the N-atom, do not undergo H- bonding. As a result, primary amines have higher boiling points than 3° amines.