CBSE Class 12 Chemistry Amines MCQs Set E

Question. A secondary amine is
(a) a compound with two carbon atoms and an –NH₂ group.
(b) a compound containing two –NH₂ groups.
(c) a compound in which hydrogens of NH₃ have been replaced by two alkyl groups.
(d) a compound with an –NH₂ group on carbon atom in number two position.
Answer: c

Question. High basicity of Me₂NH relative to Me₃N is attributed to:
(a) effect of solvent
(b) inductive effect of Me
(c) shape of Me₂NH
(d) shape of Me₃N
Answer: a

 Question. The conjugate base of (CH₃)₂NH+2 is
(a) (CH₃)₂NH
(b) (CH₃)₂N⁺
(c) (CH₃)₃N⁺
(d) (CH₃)₂N^–
Answer: a

Question. Amines have
(a) Garlic odour
(b) Fishy odour
(c) Jasmine odour
(d) Bitter almonds odour
Answer: c

Question. Which of the following statements about primary amines is ‘False’ ?
(a) Alkyl amines are stronger bases than aryl amines
(b) Alkyl amines react with nitrous acid to produce alcohols
(c) Aryl amines react with nitrous acid to produce phenols
(d) Alkyl amines are stronger bases than ammonia
Answer: c

Question. Intermediates formed during reaction of RCONH₂ with Br₂ and KOH are
(a) RCONHBr and RNCO
(b) RNHCOBr and RNCO
(c) RNHBr and RCONHBr
(d) RCONBr₂
Answer: a

Question. Amongst the given set of reactants, the most appropriate for preparing 2° amine is ___________.
(a) 2°R–Br + NH₃
(b) 2°R–Br + NaCN followed by H₂/Pt
(c) 1°R–NH₂ + RCHO followed by H₂/Pt
(d) 1°R–Br (2 mol) + Potassium phthalimide followed by H₃O+/heat
Answer: c

Question. For alkylation of ammonia which of the following is not used?
(a) CH₃–X
(b) CH₃–CH₂–X
(c) (CH₃)₂CH–X
(d) (CH₃)₃C–X
Answer: d

 Question. The correct IUPAC name for CH₂ = CHCH₂NHCH₃ is
(a) Allylmethylamine
(b) 2-amino-4-pentene
(c) 4-aminopent-1-ene
(d) N-methylprop-2-en-1-amine
Answer: d

Question. Carbylamine reaction is used for the detection of
(a) aliphatic 2° amines
(b) aliphatic 1° amines
(c) aromatic 1° amines
(d) Both (b) and (c)
Answer: d

Question. Treatment of ammonia with excess of ethyl iodide will yield
(a) diethylamine
(b) ethylamine
(c) triethylamine
(d) tetraethylammonium iodide
Answer: d

Question. The total number of electrons around the nitrogen atom in amines are
(a) 8
(b) 7
(c) 4
(d) 3
Answer: c

Question. An alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes
(a) electrophilic substitution reaction
(b) nucleophilic substitution reaction.
(c) free radical mechanism.
(d) nucleophilic addition reaction.
Answer: b

Question. The reduction of nitro compounds is most preferred in the presence of
(a) Pd/H₂ in ethanol
(b) Sn + HCl
(c) finely divided Ni
(d) iron scrap and HCl.
Answer: d

Question. Secondary amines could be prepared by
(a) reduction of nitriles
(b) Hofmann bromamide reaction
(c) reduction of amides
(d) reduction of isonitriles
Answer: d

Question. The number of primary amines of formula C₄H₁₁N is :
(a) 1
(b) 3
(c) 4
(d) 2
Answer: c

 Question. The compound obtained by heating a mixture of a primary amine and chloroform with ethanolic potassium hydroxide (KOH) is
(a) an alkyl cyanide
(b) a nitro compound
(c) an alkyl isocyanide
(d) an amide
Answer: c

Question. Which of the following is not characteristic of amines?
(a) They smell like ammonia
(b) They are inflammable in air
(c) They show the property of hydrogen bonding
(d) They are amphoteric in nature
Answer: d

Question. Which of the following methods of preparation of amines will give same number of carbon atoms in the chain of amines as in the reactant?
(a) Reaction of nitrite with LiAlH₄.
(b) Reaction of amide with LiAlH₄ followed by treatment with water.
(c) Heating alkylhalide with potassium salt of phthalimide followed by hydrolysis.
(d) Treatment of amide with bromine in aquesous solution of sodium hydroxide.
Answer: c

Question. The basic character of amines is due to
(a) presence of nitrogen atom
(b) lone pair of electrons on nitrogen atom
(c) tetrahedral structure
(d) high electronegativity of nitrogen
Answer: b

Question. The general formula of quaternary ammonium compound is
(a) R–NH₂
(b) R₃N
(c) R₄N⁺ X^–
(d) NH₄X
Answer: c

Question. The correct order of basicity in amines
(i) C₂H₅NH₂
(ii) CH₃NH₂
(iii) (CH₃)₂NH
(iv) (CH₃)₃N
(a) (i) < (iv) < (ii) < (iii)
(b) (iv) < (ii) < (iii) < (i)
(c) (i) < (ii) < (iii) < (iv)
(d) (ii) < (iii) < (iv) < (i)
Answer: b

Question. Which of the following will give primary amine only ?
(i) ammonia + propylchloride
(ii) potassium pthalimide + ethylchloride
(iii) potassium pthalimide + chlorobenzene
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (i), (ii) and (iii)
Answer: a

ASSERTION-REASON TYPE QUESTIONS

Directions : Each of these questions contain two statements,Assertion and Reason. Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

Question. Assertion : Acetanilide is less basic than aniline.
Reason : Acetylation of aniline results in decrease of electron density on nitrogen.
Answer: d

Question. Assertion : Nitrating mixture used for carrying out nitration of benzene consists of conc. HNO₃ + conc. H₂SO₄.
Reason : In presence of H₂SO₄, HNO₃ acts as a base and produces NO₂⁺ ions.
Answer: a

Question. Assertoin : Aniline does not undergo Friedel-Crafts reaction.
Reason : –NH₂ group of aniline reacts with AlCl₃ (Lewis acid) to give acid-base reaction.
Answer: a

Question. Assertion : Amines are basic in nature.
Reason : Amines have lone pair of electrons on nitrogen atom.
Answer: a

Question. Assertion : Aromatic 1°amines can be prepared by Gabriel phthalimide synthesis.
Reason : Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.
Answer: a

Question. Assertion : Nitration of aniline can be conveniently done by protecting the amino group by acetylation.
Reason : Acetylation increases the electron-density in the benzene ring.
Answer: c

Question. Assertion : Only a small amount of HCl is required in the reduction of nitro compounds with iron scrap and HCl in the presence of steam.
Reason : FeCl₂ formed gets hydrolysed to release HCl during the reaction.
Answer: d

Question. Assertion : Acylation of amines gives a monosubstituted product whereas alkylation of amines gives polysubstituted product.
Reason : Acyl group sterically hinders the approach of further acyl groups
Answer: c

Very Short Answer questions: 

Question. What is Hinsberg’s reagent?
Answer. Benzene sulphonyl chloride

Question. Why is aniline acylated before its nitration?
Answer. To prevent it from oxidation

Question. Ethylamine is soluble in water but aniline is not, why?
Answer. Ethylamine forms intermolecular H‐bond with water, but aniline does not form H‐bond to a very large extent due to the presence of large hydrophobic –C₆H₅ group.

Short Answer questions:

Question. Which one is more basic CH₃NH₂ or (CH₃)3N in gaseous phase and why ?
Answer. (CH₃)₃N is more basic because greater number of alkyl groups increase the magnitude of +I effect so increase the basicity .

Question. Why ethylamine is soluble in water whereas aniline is not ?
Answer. Ethylamine when added to water forms intermolecular H−bonds with water. Hence, it is soluble in water.

Question. What is the hybridisation of N in (CH₃)₃N and shape of (CH₃)₃N ?
Answer. Hybridisation of N in (CH₃)₃N is sp³ and shape of (CH₃)₃N is trigonal pyramidal.

Question. Which one is more basic CH₃NH₂ or (CH₃)3N in gaseous phase and why ?
Answer. (CH₃)₃N is more basic because greater number of alkyl groups increase the magnitude of +I effect so increase the basicity .

Question. Out of Butan-1-ol and butan-1-amine ,which will be more soluble in water and why?
Answer. Butan-1-ol are more polar than amines and forms stronger intermolecular hydrogen bonds with water molecules than amines.

Question. Which diazonium salt is insoluble in water at room temprature?
Answer. Benzene diazonium fluoro borate (C₆H₅N₂ F).

Question. Write the structure and IUPAC name of t-butylamine.
Answer. 2-Methylpropan-2-amine

Question. Which diazonium salt is stable at room temprature.
Answer. Benzene diazonium fluoro borate(C₆H₅N₂ F).

Question. Out of CH₃NH₂ and CH₃CH₂NH₂ which has higher boiling point and why ?
Answer. CH₃CH₂NH₂ because bigger the alkyl group more is the surface area higher is the magnitude of van der Waals force.

Question. Why are primary amines higher boiling than tertiary amines ?
Answer. Primary amines have two hydrogen atoms on the N atom and therefore form intermolecular hydrogen bonding. Tertiary amines do not have hydrogen atoms on the N atom and therefore,these do not form hydrogen bonds.

Question. Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(a) C₆H₅NHCH₃
Answer. (a) N‐Methyl aniline 20
(b) (CH₃CH₂)₂NCH₃
(b) N‐Ethyl‐N‐methylethanamine 30

Question. Arrange the following:
(a) In decreasing order of the pKbvalues:
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂
(b) In increasing order of basic strength: Aniline, p‐nitroaniline and p‐toluidine
Answer. (a) C₆H₅NH₂, C₆H₅NHCH₃,C₂H₅NH₂,(C₂H₅)₂NH
(b) p‐Nitroaniline , aniline, p‐toluidine

Question. How can you convert an amide into an amine having one carbon less than the starting compound?
Name the reaction.
Answer. By using Hoffmann bromamide reaction

Question. Give one chemical test to distinguish between the following pairs of compounds.
(a) Methylamine and dimethylamine
(b) Ethylamine and aniline
(c) Aniline and benzylamine
Answer. (a) Carbylamine reaction
(b) Azo dye Test
(c) Azo dye Test

 Question. Give plausible explanation for each of the following:
(a) Why do primary amines have higher boiling point than tertiary amines?
(b) Why are aliphatic amines stronger bases than aromatic amines?
Answer. (a) Due to strong intermolecular H‐bonding in primary amines.
(b) In aromatic amines lone pair is engaged with benzene in resonance.

Question. Account for the following:
(a) Although amino group is o, p− directing in aromatic electrophilic substitution reactions,aniline on nitration gives a substantial amount of m‐nitroaniline.
(b) Aniline does not undergo Friedel‐Crafts reaction.
(c) Gabriel phthalimide synthesis is preferred for synthesizing primary amines.
Answer. (a) Because nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion which is meta‐directing.
(b) Friedel‐Craft reaction is carried out in the presence of AlCl₃. But AlCl₃ acts as a Lewis acid, while aniline acts as a Lewis base. Thus, aniline reacts with AlCl₃ to form a salt.
(c) Gabriel phthalimide synthesis results in the formation of primary amines only. Secondary and tertiary amines are not formed in this synthesis. Thus, a pure primary amine can be obtained.Therefore,Gabriel phthalimide synthesis is preferred for synthesizing primary amines.