Tiêu đề DPP - 9 Solutions.pdf ✅

Class : XIIth Subject : CHEMISTRY Date : DPP No. : 9 1 (b) There are four β- hydrogens, in this quaternary ammonium salt. On heating quaternary ammonium salt gives Hofmann elimination (abstraction of most acidic hydrogen which is β 1 ). Hence, major product is CH2 = CH2. (Least substituted alkene). 2 (b) CH3NH2 + CHCI3 + 3KOH→CH3NC + 3KCI + 3H2O CH3NC or CH3 ― N + ≡ C ―methyl isocyanide or methyl carbylamine. This reaction is an example of carbylamine reaction and it is used for the distinction of p- amines from s- and t-amines or identification of p-amino group. 5 (c) Roulle first isolated urea (in 1773) from urine and named it as urea. 6 (d) Reduction of NO2 group to NH2group is taking place by Fe/HCI. OH H N H H 4 Me CH2 CH3 1 CH2 CH2 CH2 CH3 2 3 NO NH2 2 Fe/HCl NO2 NO2 Topic :- Amines Solutions
8 (d) Sulphanilic acid exists as a dipolar ion which has acidic and basic groups in the same molecule. Such ions are called Zwitter ions or inner salts 10 (a) For detection of secondary amines Liebermann’s nitroso test is used. 11 (c) C2H5NH2 + Na⟶C2H5NHNa + 1 2 H2 12 (b) Only 1° aromatic amine (primary aromatic amine) from diazonium salts at low temperature (0° ― 5°C). A reaction in which –NH2 group is converted into diazo group ( ― N + ≡ N) is called diazotization. Diazotized salts are stable in cold aqueous solution. Amines, so undergo diazotization but C6H5CH2NH2 (aliphatic amine) will not undergo diazotisation. NH2 H2SO4 NH2 .H2SO4 180 oC -H2O NHSO3H Heat NH2 SO3H sulphanilic acid NH3+ SO3 - C6H5NH2+HCl 0 o -5 oC C6H5NH2Cl - + NaNO2 +HCl 0 o -5 oC HNO2 +NaCl C6H5NCl - +HNO2 0 o -5 oC C6H5N2Cl - + 2H2O + + C6H5NH2, H2N H3C C6H4 ,H2N O2N C6H4
13 (b) Aniline is prepared by the reduction of nitrobenzene in acidic medium. 14 (b) Amines possess fishy smell. 16 (a) Electrons donors are bases. Greater the stabilisation of cation formed by loss of electron more will be basicity of amine. 2° amine is more basic than 3° amine because 2° amine is stabilized by hydrogen bonding with solvent molecule. 17 (c) RCONH2 + NaOBr→RNH2 + NaBr + Na2CO3 + 2H2O ′X′ 18 (d) Benzyl amine (C6H5CH2NH2 ) is more basic than aniline (C6H5NH2 ) because N-atom of aniline is delocalized over the benzene ring. However in benzyl amine the lone pair of electrons on the N-atom is not conjugated with the benzene ring and therefore it is not delocalized. Hence, the lone pair of electrons on the N-atom in benzyl amine is more readily available for protonation than that on the N-atom of aniline. Thus, the benzyl amine is a stronger base than aniline. 19 (d) Tertiary amines react as, (CH3 )3N + HNO2⟶(CH3 )3NHNO2 NO2 + 6H Fe/HCl NH2 +2H2O aniline nitrobenzene
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. B B D A C D B D D A Q. 11 12 13 14 15 16 17 18 19 20 A. C B B B C A C D D B