NCERT Exemplar Class 12 Chemistry<\/a>. Here we have given NCERT Exemplar Class 12 Chemistry Chapter 13 Amines. https:\/\/www.cbselabs.com\/ncert-exemplar-problems-class-12-chemistry-amines\/<\/p>\nNCERT Exemplar Class 12 Chemistry Chapter 13 Amines<\/h2>\n
Multiple Choice Questions<\/strong><\/p>\nSingle Correct Answer Type<\/strong><\/p>\nQuestion 1. Which of the following is a 3\u00b0 amine?<\/strong>
\n (a) 1 -Methylcyclohexylamine (c) tert-Butylamine<\/strong>
\n Solution:<\/strong> (b)
\n<\/p>\nQuestion 2. The correct IUPAC name for CH2<\/sub> = CHCH2<\/sub>NHCH3<\/sub> is<\/strong>
\n (a) allylmethylamine (b) 2-amino-4-pentene<\/strong>
\n (c) 4-aminopent-l-ene . (d) N-methylprop-2-en-l-amine.<\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 3. Amongst the following, the strongest base in aqueous medium is .<\/strong>
\n (a) CH3<\/sub>NH2<\/sub> (b) NCCH2<\/sub>NH2<\/sub> (C) (CH3<\/sub>)2<\/sub>NH (d) C6<\/sub>H5<\/sub>NHCH3<\/sub><\/strong>
\n Solution:<\/strong> (c) 2\u00b0 amine is more basic than 1\u00b0 amine, i.e., (CH3<\/sub>)2<\/sub>NH is more basic than CH3<\/sub>NH2<\/sub>. Due to -I effect of CN group, NC-CH2<\/sub>NH2<\/sub> is less basic than CH3<\/sub>NH2<\/sub>. Further C6<\/sub>H5<\/sub>NHCH3<\/sub> is less basic than both CH3<\/sub>NH2<\/sub> and (CH3<\/sub>)2<\/sub>NH due to delocalization of lone pair of electrons present on the nitrogen atom into benzene ring. Hence, the decreasing order of amines is:
\n(CH3<\/sub>)2<\/sub>NH > CH3<\/sub>NH2<\/sub> > C6<\/sub>H5<\/sub>NHCH3<\/sub> > NC – CH2<\/sub>NH2<\/sub><\/p>\nQuestion 4. Which of the following is the weakest Bronsted base?<\/strong>
\n
\n Solution:<\/strong> (a) Due to delocalization of lone pair of electrons on the N-atom into the benzene ring, C6H5NH2 is the weakest base.
\nResonating Structure of Aniline
\n<\/p>\nQuestion 5. Benzylamine may be alkylated as shown in the following equation: C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub> + R – X > C6<\/sub>H5<\/sub>CH2<\/sub>NHR<\/strong>
\n Which of the following alkyl halides is best suited for this reaction through SN1 mechanism? .<\/strong>
\n (a) CH3<\/sub>Br (b) C6<\/sub>H5<\/sub>Br (c) C6<\/sub>H5<\/sub>CH2<\/sub>Br (d) C2<\/sub>H5<\/sub>Br<\/strong>
\n Solution:<\/strong> (c) SN1 reaction occurs in two steps. In first step R – X bond is broken to produce a carbocation which is attacked by nucleophile. The greater the stability of carbocation, the greater will be the rate of reaction. Benzylic halides show high reactivity towards SN1 reaction.
\n<\/p>\nQuestion 6. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?<\/strong>
\n (a) H2<\/sub> (excess)\/Pt \u00a0 \u00a0(b) LiAlH4<\/sub> in ether<\/strong>
\n (c) Fe and HCl \u00a0 \u00a0 (d) Sn and HCl<\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 7. In order to prepare a 10 amine\u2018from an alkyl halide with simultaneous addition of one CH2<\/sub> group in the carbon chain, the reagent used as source of nitrogen is<\/strong>
\n (a) sodium amide, NaNH2<\/sub><\/strong>
\n (b) sodium azide, NaN2<\/sub><\/strong>
\n (c) potassium cyanide, KCN<\/strong>
\n (d) potassium phthalimide, C6<\/sub>H4<\/sub>(CO2<\/sub>)N K<\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 8. The source of nitrogen in Gabriel synthesis of amines is .<\/strong>
\n (a) sodium azide, NaN3<\/sub><\/strong>
\n (b) sodium nitrite, NaNO2<\/sub><\/strong>
\n (c) potassium cyanide, KCN<\/strong>
\n (d) potassium phthalimide, C6<\/sub>H4<\/sub>(CO2<\/sub>)N’KT<\/strong>
\n Solution:<\/strong> (d) Potassium phthalimide is the source of nitrogen in Gabriel\u2019s synthesis.
\n<\/p>\nQuestion 9. Amongst the given set of reactants, the most appropriate for preparing 2\u00b0 amine is .<\/strong>
\n (a) 2\u00b0 R – Br + NH3<\/sub><\/strong>
\n (b) 2\u00b0 R – Br + NaCN followed by H2<\/sub>\/Pt<\/strong>
\n (c) 10 R – NH2<\/sub> + RCHO followed by H2<\/sub>\/Pt<\/strong>
\n (d) 1\u00b0 R – Br + (2 mol) + potassium phthalimide followed by H3<\/sub>O+\/heat<\/strong>
\n Solution:<\/strong> (c)
\n<\/p>\nQuestion 10. The best reagent for converting 2-phenylpro-panamide into-2-phenyl- propanamine is .<\/strong>
\n (a) excess H2<\/sub><\/strong>
\n (b) Br2<\/sub> in aqueous NaOH<\/strong>
\n (c) iodine in the presence of red phosphorus<\/strong>
\n (d) LiAlH4<\/sub> in ether<\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 11. The best reagent for converting-2-phenylpro-panamide into 1-phenylethana- mine is .<\/strong>
\n (a) excess H2<\/sub>\/Pt (b) NaOH \/Br2<\/sub><\/strong>
\n (c) NaBH4<\/sub>\/methanol (d) LiAlH4<\/sub>\/ether<\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 12. Hoffmann bromamide degradation reaction is shown by .<\/strong>
\n (a) ArNH2<\/sub> (b) ArCONH2<\/sub> (c) ArNO2<\/sub> (d) ArCH2<\/sub>NH2<\/sub><\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 13. The correct increasing order of basic strength for the following compounds is<\/strong>
\n
\n Solution:<\/strong>(d)
\n
\nElectron withdrawing group decreases the basic strength while electron releasing groups increases the basic strength of aniline.<\/p>\nQuestion 14. Methylamine reacts with HN03 to form .<\/strong>
\n (a) CH3-0-N = 0 (b) CH3-0-CH3<\/strong>
\n (c) CH3OH (d) CH3CHO<\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 15. The gas evolved when methylamine reacts with nitrous acid is .<\/strong>
\n (a) NH3<\/sub> (b) N2<\/sub> (C) H2<\/sub> (d) C2<\/sub>H6<\/sub><\/strong>
\n Solution:<\/strong> (b) Chemical reaction takes place during reaction of methylamine with nitrous acid is as follows
\n<\/p>\nQuestion 16. In the nitration of benzene using a mixture of cone. H2<\/sub>SO4<\/sub> and cone. HNO3<\/sub>,<\/strong>
\n the species which initiates the reaction is .<\/strong>
\n (a) NO2<\/sub> (b) NCf (c) NO2<\/sub> (d) NH2<\/sub><\/strong>
\n Solution:<\/strong> (c)
\nNO2<\/sub> (Nitronium ion) electrophile initiates the process of nitration. It is obtained as:
\n<\/p>\nQuestion 17. Reduction of aromatic nitro compounds using Fe and HCl gives .<\/strong>
\n (a) aromatic oxime (b) aromatic hydrocarbon<\/strong>
\n (c) aromatic primary amine (d) aromatic amide<\/strong>
\n Solution:<\/strong> (c)
\n<\/p>\nQuestion 18. The most reactive amine towards dilute hydrochloric acid is<\/strong>
\n
\n Solution:<\/strong> (b) The greater will be the strength of base, the greater will be its reactivity towards dilute HCl. Hence, (CH3<\/sub>)2<\/sub>NH has the highest basic strength as it has the highest reactivity.<\/p>\nQuestion 19. Acid anhydrides on reaction with primary amines give .<\/strong>
\n (a) amide (b) imide<\/strong>
\n (c) secondary amine (d) imine<\/strong>
\n Solution:<\/strong>
\n<\/p>\nQuestion 20.<\/strong>
\n
\n Solution:<\/strong>
\n<\/p>\nQuestion 21. Best method for preparing primary amines from alkyl halides without<\/strong>
\n changing the number of carbon atoms in the chain is<\/strong>
\n (a) Hoffmann bromamide reaction (b) Gabriel phthalimide reaction (c) Sandmeyer reaction (d) reaction with NH3<\/sub><\/strong>
\n Solution:<\/strong> (b) Gabriel phthalimide synthesis is used to get primary amine from alkyl halide without any change in number of carbon atoms.
\n<\/p>\nQuestion 22. Which of the following compound will not undergo azo coupling reaction with benzene diazonium chloride?<\/strong>
\n (a) Aniline (b) Phenol (c) Anisole (d) Nitrobenzene<\/strong>
\n Solution:<\/strong> (d) Diazonium cation is a weak electrophile and hence reacts with electron rich compounds containing electron donating groups such as -OH, -NH2<\/sub> and -OCH3<\/sub> groups and not with compounds containing electron withdrawing groups such as -NO2<\/sub>, etc.<\/p>\nQuestion 23. Which of the following compounds is the weakest Bronsted base?<\/strong>
\n
\n Solution:<\/strong> (c) Amines (a, b) have stronger tendency to accept a proton and hence are stronger Bronsted bases than phenol (c) and alcohol (d). Since phenol is more acidic than alcohol, therefore, phenol (c) has the least tendency to accept a proton and hence it is the weakest Bronsted base.<\/p>\nQuestion 24. Among the following amines’ the strongest Bronsted base is<\/strong>
\n
\n Solution:<\/strong> (d) Aniline is weaker base than NH3<\/sub> due to delocalization of lone pair of electrons on the N-atom into the benzene ring. Pyrrole (c) is not at all basic because the lone pair of electrons on N-atom is donated towards aromatic sextet formation. Therefore, pyrrolidine (d) has a strong tendency to accept a proton and is hence, the strongest base.<\/p>\nQuestion 25. The correct decreasing order of basic strength of the following species is<\/strong>
\n H2<\/sub>O, NH3<\/sub>, OH , NH;<\/strong>
\n (a) NH3<\/sub> > OH > NH3<\/sub> > H2<\/sub>O (b) OH > NH2<\/sub> > H2<\/sub>O > NH3<\/sub><\/strong>
\n (c) NH3<\/sub> > H2<\/sub>O > NH2<\/sub> > OH (d) H2<\/sub>O > NH3<\/sub> > OH > NH2<\/sub><\/strong>
\n Solution:<\/strong> (a) NH2<\/sub> > OH > NH3<\/sub>\u00a0> H2<\/sub>O. Due to higher electronegativity of O than N atom, O – H bond is more polar than N – H bond. Hence, O – H is more acidic in nature than N – H bond. Now, NH2<\/sub> and OH have negative charge due to which they are more basic than NH3 and H20.<\/p>\nQuestion 26. Which of the following should be most volatile?<\/strong>
\n
\n Solution:<\/strong> (b) 1\u00b0 and 2\u00b0 amines have higher boiling points due to intermolecular H-bonding (and hence less volatile than 3\u00b0 amines and hydrocarbons of comparable molecular mass).<\/p>\nQuestion 27. Which of the following methods of preparation of amines will not give same number of carbon atoms in the chain of amines as in the reactant?<\/strong>
\n (a) Reaction of nitrile with LiAlH4<\/strong>
\n (b) Reaction of amide with LiAlH4 followed by treatment with water.<\/strong>
\n (c) Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis<\/strong>
\n (d) Treatment of amide with bromine in aqueous solution of sodium hydroxide.<\/strong>
\n Solution:<\/strong> (d) Aliphatic and arylalkyl primary amines can be prepared by the reduction of the corresponding nitriles with LiAlH4.
\n
\nHeating alkyl halide with primary, secondary and tertiary amine can be prepared by reduction of LiAlH4 followed by treatment with water.
\n
\nHeating alkyl halide with potassium salt of phthalimide followed by hydrolysis produces primary amine. This process is known as Gabriel phthalimide reaction. The number of carbon atoms in the chain of amines of product is same as reactant.
\n<\/p>\nMore than One Correct Answer Type<\/strong><\/p>\nQuestion 28. Which of the following cannot be prepared by Sandmeyer\u2019s reaction?<\/strong>