MCQ Questions for Class 12 Chemistry Chapter 13 Amines with Answers
MCQ Questions for Class 12 Chemistry with Answers
Que 1. Out of the following compounds, which is the most basic in aqueous solution?
(a) CH3NH2
(b) (CH3)2NH
(c) (CH3)3N
(d) C6H5NH2
Ans 1. (b)
Explanation: Basicity is the ability of the atom to donate its lone pair of electrons. In aniline nitrogen is directly attached with the benzene ring which leads to the delocalization of lone pair of electrons of nitrogen atom in the benzene ring. This leads to the less availability of the electron for donation.
In case of (CH₃)₃N, lone pair of electrons did not take place in resonance. But while taking steric hindrance and inductive effect into considerations (CH₃)₂NH is more basic than (CH₃)₃N.
Que 2. Aniline undergoes condensation to form Schiff base on reacting with:
(a) Acetyl chloride
(b) Ammonia
(c) Acetone
(d) Benzaldehyde
Ans 2. (d)
Explanation: Imines are also called as Schiff base. A primary amine reacts with aldehydes to form Schiff bases.
Reaction for the question is given below
RNH2 + C6H5CHO → RN = CHC6H5
Que 3. Hinsberg’s reagent is:
(a) Benzene sulphonyl chloride
(b) Benzene sulphonic acid
(c) Phenyl isocyanide
(d) Benzene sulphonamide
Ans 3. (a)
Explanation: Hinsberg’s reagent is known by the other name benzene sulphonyl chloride. It is categorized in the organosulfur compounds. It is soluble in organic solvents but insoluble in aprotic solvents. It is used in Hinsberg’s test that is to distinguish between primary, secondary, and tertiary amine. Primary amine, such as methylamine (a primary amine) will dissolve in an alkali solution after it has undergone a reaction with Hinsberg’s reagent.
Secondary amine forms alkyl sulphonamide which does not form a salt with alkali and hence remains insoluble (solid) in alkali solution.
Tertiary amine does not react with Hinsberg’s reagent.
Que 4. Silver chloride is soluble in methylamine due to the formation of:
(a) Ag(CH3NH2)CI
(b) Ag + CH3CI + NH4CI
(c) [Ag(CH3NH2)2CI]
(d) AgOH
Ans 4. (c)
Explanation: Silver chloride dissolves in methylamine to form a complex.
AgCl + 2CH₃NH₂ → [Ag(NH₂CH₃)₂]Cl (soluble complex)
Que 5. Gabriel phthalimide reaction is used for the preparation of:
(a) Primary aromatic amine
(b) Secondary amine
(c) Primary aliphatic amine
(d) Tertiary amine
Ans 5. (c)
Explanation: Gabriel phthalimide reaction is used for the preparation of primary amines.
MCQ Questions for Class 12 Chemistry with Answers
Que 6. The correct statement regarding the basicity of arylamines is:
(a) Aryl amines are generally more basic than alkyl amines because of the aryl group.
(b) Aryl amines are generally more basic than alkyl amines because the nitrogen atom in arylamines is sp-hybridized.
(c) Aryl amines are generally less basic than alkyl amines because the nitrogen lone-pair electron is delocalized by interaction with the aromatic ring π- electron system.
(d) Aryl amines are generally more basic than alkyl amines because the nitrogen lone-pair electron is not delocalized by interaction with the aromatic ring π- electron system.
Ans 6. (c)
Explanation: Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring π electron system which makes electrons less available. While in alkyl amines, the +I electron releasing effect of the alkyl group makes it strongly basic.
Que 7. Which of the following compound is least basic?
(a) Aniline
(b) p-Methoxyaniline
(c) o-Nitroaniline
(d) o-toluidine
Ans 7. (c)
Explanation: Because of the electron-withdrawing NO2 group the lone pair electrons of Nitrogen in -NH2 is less available for donation.
Que 8. The basicity of aniline is weaker in comparison to that of methylamine due to:
(a) Hyper conjugative effect of Me group in MeNH2.
(b) Resonance effect of the phenyl group in aniline
(c) The lower molecular weight of methylamine as compared to that of amine.
(d) Resonance effect of –NH2 group in MeNH2.
Ans 8. (b)
Explanation: The lone pair of electrons, present in N-atom in aniline involve in-ring resonance. So it is less basic than methylamine.
Que 9. Benzylamine is a stronger base than aniline because:
(a) The lone pair of electrons on the nitrogen atom in benzylamine is delocalized
(b) The lone pair of electrons on the nitrogen atom in aniline is delocalized
(c) The lone pair of electrons on the nitrogen atom in aniline is not involved in resonance.
(d) Benzylamine has a higher molecular mass than aniline.
Ans 9. (b)
As we know a base is the one that has the availability of electrons to donate to an acid. Now, here we need to find the stronger base among benzylamine and aniline. A stronger base will be the one whose electrons are easily available for donation. The chemical formula of aniline is C6H5NH2 and its structure is as given below:
In this molecule, a lone pair of electrons is present on the nitrogen atom (N) and this nitrogen atom is attached to an sp2 hybridized carbon of the benzene ring. The lone pair of electrons present on nitrogen will involve in resonance with the benzene ring as:
Thus, it is quite clear that due to resonance present in aniline, the lone pair of electrons available are delocalized and hence will be less available for donation. This makes aniline a weaker base.
Now, the chemical formula of benzylamine is C6H5CH2NH2 and its structure is as given below:
In this molecule, a lone pair of electrons is again present on nitrogen (N). But here, nitrogen is attached to an sp3 hybridized carbon and not to the benzene ring. Hence, lone pairs of electrons on nitrogen are not involved in any resonance with the benzene ring. Therefore, lone pairs of electrons are localized here on nitrogen atoms and will be readily available for donation. Thus, benzylamine is a stronger base. Hence, benzylamine is a stronger base than aniline because the lone pair of electrons on the nitrogen atom in aniline is delocalized.
Que 10. Which of the following is the most stable diazonium salt?
(a) C6H5CH2N2+X–
(b) CH3N2+X –
(c) CH3CH2N2+X–
(d) C6H5N2+X–
Ans 10. (d)
Explanation: Diazonium salt-containing aryl group directly linked to the nitrogen atom is most stable due to resonance stabilization between the benzene nucleus and N-atom. Diazonium ion act as a electrophile.
MCQ Questions for Class 12 Chemistry with Answers
Assertion Reason and Matchup Questions of Amines
