3<\/sub><\/p>\n<\/p>\n
Question 16.
\nHow do you carryout the following conversions ?
\ni) N – Ethylamine to N, N – Diethyl propanamine
\nii) Aniline to p – Aminobenzene sulphonamide
\nAnswer:
\ni) N – ethylamine is first reacted with propylchloride to convert it to N – ethyl propanamine. It is then reacted with ethyl chloride to convert it to N, N – Diethyl propanamine.
\n
\nii) Aniline is converted to acetanilide by reaction with acetylchloride. Acetanilide is treated with chlorosulphonic acid and the product on reaction with ammonia followed by hydrolysis gives sulphanilamide.
\n<\/p>\n
Question 17.
\nExplain with a suitable example how benzene sulphonylchloride can distinguish primary, secondary and tertiary amines.
\nAnswer:
\nBenzene sulphonylchloride (Hinsberg reagent) reacts with a primary amine, ethylamine, to give N – ethylbenzene sulphonyl amide.
\n
\nThis compound contains hydrogen attached to nitrogen. It is acidic in nature and hence it is soluble in sodium hydroxide solution.<\/p>\n
When benzene sulphonyl chloride reacts with a secondary amine, for example diethyl amine, to give N, N – diethyl benzene sulphonamide. It does not contain hydrogen attached to nitrogen. It is not acidic and hence it is insoluble in sodium hydroxide solution.
\n
\nTertiary amines do not react with Hinsberg reagent.<\/p>\n
<\/p>\n
Question 18.
\nWrite the reactions of
\ni) aromatic and
\nii) aliphatic primary amines with nitrous acid.
\nAnswer:
\nAromatic primary amines react with nitrous acid at low temperatures (0 – 5\u00b0C) to form diazonium salts.
\n
\nAliphatic primary amines react with nitrous acid to form alcohol and nitrogen gas.
\n<\/p>\n
Question 19.
\nExplain why amines are less acidic than alcohols of comparable molecular masses.
\nAnswer:
\nThe acidic character of alcohols is due to the polar nature of O – H bond. The polarity of the N – H bond in amines is less than that of the O – H bond in alcohols of comparable molecular mass. Hence amines are less acidic than alcohols of comparable molecular mass.<\/p>\n
Question 20.
\nHow do you prepare Ethyl cyanide and Ethyl isocyanide from a common alkyl halide. [IPE 14]
\nAnswer:
\nEthyl chloride reacts with ethanolic potassium cyanide to form ethyl cyanide as the major product. However, when ethyl chloride reacts with ethanolic silver cyanide ethyl isocyanide will be the major product.
\n<\/p>\n
Long Answer Questions (8 Marks)<\/span><\/p>\nQuestion 21.
\nAn aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2<\/sub> and KOH forms compound ‘C’ of molecular formula C6<\/sub>H7<\/sub>N. Write the structures and IUPAC names of compounds A, B and C.
\nAnswer:
\nThe final product ‘C’ with molecular formula C6<\/sub>H7<\/sub>N is Aniline. The sequence of reactions can be explained as follows.
\n
\nCompound A is Benzoic acid. On treatment with aqueous ammonia gives ammonium benzoate which on heating gives Benzamide (B). Benzamide on heating with bromine and potassium hydroxide (Hofmann hypobromite reaction) gives Aniline (C) with molecular formula C6<\/sub>H7<\/sub>N.<\/p>\n<\/p>\n
Question 22.
\nComplete the following conversations.
\ni) CH3<\/sub>NC + HgO \u2192 ?
\nii) ? + 2H2<\/sub>O \u2192 CH3<\/sub>NH2<\/sub> + HCOOH
\niii) CH3<\/sub>CN + C2<\/sub>H5<\/sub> Mg Br \u2192 ?
\niv) CH3<\/sub> CH2<\/sub> NH2<\/sub> + CHCl3<\/sub> + KOH ?
\nv)
\nAnswer:
\ni) HgO is a mild oxidising agent. It converts isocyanides to isocyanates. Thus, methyl isocyanide is converted to methyl isocyanate.
\n<\/p>\nii) Isocyanides on hydrolysis give amines and formic acid.
\n<\/p>\n
iii) Ethyl magnesium bromide adds on to methylcyanide or acetonitrile to give an addition product which on hydrolysis forms acetone.
\n<\/p>\n
iv) Primary amine on heating with chloroform and alcoholic KOH gives foul smelling isocyanide or carbylamine.
\n<\/p>\n
v) Aniline on treatment with bromine water gives a white precipitate of 2,4,6 – tribromoaniline.
\n<\/p>\n
Question 23.
\ni) Write the structures of different isomeric amines corresponding to the molecular formula C9<\/sub>H13<\/sub>N.
\nii) What reducing agents can bring about reduction of nitrobenzene ?
\niii) Write the product formed when benzyl chloride is reacted with ammonia followed by treatment with methyl and ethyl chlorides.
\nAnswer:
\ni)
\n
\nii) The following reagents can bring about reduction of nitrobenzene.
\na) Sn \/ HCl (or)Fe\/HCl
\nb)Zn\/NaOH
\nc) Zn \/ NH4<\/sub>Cl\/ H2<\/sub>O
\nd)H2<\/sub>\/Pd<\/p>\niii)
\n<\/p>\n
Question 24.
\ni) Identify the amide and cyanide which on reduction with appropriate reducing agent give n – butylamine.
\nii) Write the mechanism of Hoffmann bromamide reaction.
\nAnswer:
\ni) Butanamide, CH3<\/sub>CH2<\/sub>CH2<\/sub>CONH2<\/sub> on reduction with lithium aluminium hydride yields n – butylamine
\n
\nn – propylcyanide on reduction LiAlH4<\/sub> or Na(Hg)\/C2<\/sub>H5<\/sub>OH gives n – Butylamine.
\n<\/p>\n <\/p>\n
ii) Hoffmann bromamide reaction mechanism :
\n<\/p>\n
<\/p>\n
Question 25.
\nHow do you make the following conversions ?
\ni) Chlorophenylmethane to phenylacetic acid
\nii) Chlorophenylmethane to 2 – phenylethanamine
\nAnswer:
\ni)
\n
\nChlorophenyl methane is reacted with potassium cyanide and converted to benzylcyanide which on hydrolysis gives phenylacetic acid.<\/p>\n
ii)
\n
\nChlorophenylmethane is reacted with KCN and converted to benzylcyanide which on reduction gives 2 – phenylethanamine.<\/p>\n
Question 26.
\nIdentify the starting amide which gives p – methyl aniline on reaction with bromine and sodium hydroxide and write all the steps involved in the reaction.
\nAnswer:
\np – methylbenzamide on reaction with bromine and sodium hydroxide (Hoffmann bromamide degradation reaction) gives p – methylaniline.
\n
\nThe following steps are involved in the reaction.
\n2NaOH + Br2<\/sub> \u2192 NaBr + NaOBr + H2<\/sub>O
\n<\/p>\n<\/p>\n
Question 27.
\nExplain wiy the order of basicity methylamine, N, N – dimethylamine and N, N, N – trimethylamine changes in gaseous and aqueous medium.
\nAnswer:
\nIn the gaseous state the basicity of the methyl substituted amines follows the order.
\nN, N, N – trimethylamine > N, N – dimethylamine > methylamine<\/p>\n
It can be explained as follows. The methyl group has electron releasing nature. It pushes electrons towards nitrogen and thus makes the unshared electron pair on nitrogen more available for sharing with the proton of the acid.<\/p>\n
Thus the methyl substituted ammonium gets stabilised due to to the dispersal of the positive charge by the +1 effect of thecilkyl group. Thus the basic nature of the methyl substituted amines increases with the increase in the number of methyl groups. This trend is followed in the gaseous phase.<\/p>\n
In the aqueous phase, the substituted ammonium cations get stabilised not only by electron releasing effect of the methyl group but also by solvation with water molecules. Another factor that decides the basic strength of the alkylamines in aqueous state is steric hindrance of the alkyl groups:<\/p>\n
Hence, due to the presence of two electron releasing methyl groups attached to the nitrogen atom, dimethylamine is a stronger base than methylamine. If so, trimethyl amine having three methyl groups attached to nitrogen should be expected to be more basic them dimethyl amine. But actually trimethylamine is considerably less basic than dimethyl amine.<\/p>\n
Why so ? In methyl amine and dimethylamine the ‘electronic effect’ increases the basic strength of the amine. However, in trimethylamine the over crowding of the three methyl groups attached to nitrogen causes the ‘steric effect’ – to dominate over the ‘electronic effect’. This steric effect retards the protonation of nitrogen which results in an appreciably lower basic strength of trimethylamine. Hence the basic strength of the amines in the aqueous phase follows the order :
\n(CH3<\/sub>)2<\/sub> NH > CH3<\/sub>NH2<\/sub> > (CH3<\/sub>)3<\/sub>N > NH3<\/sub>.<\/p>\nQuestion 28.
\nWrite the equations involved in the reaction of Nitrous acid with ethylamine and aniline.
\nAnswerw:
\nEthyl amine reacts with nitrous acid to form ethyl diazonium salt which being unstable liberates nitrogen gas and forms ethyl alcohol.
\n
\nAniline reacts with nitrous acid at low temperatures (0 – 5\u00b0C) to form diazonium salt.
\n<\/p>\n
<\/p>\n
Question 29.
\nExplain with equations how methylamine, N, N – dimethylamine and N, N, N-trimethylamine react with benzene sulphonyl chloride and how this reaction is useful to separate these amines.
\nAnswer:
\nMethylamine reacts with benzene suphonyl chloride to give N – methyl benzene sulphonamide.
\n
\nThe hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali, say NaOH solution. N, N – dimethylamine reacts with benzene sulphonyl chloride to give N, N – dimethylbenzene sulphonair; le. Since this compound does not contain any hydrogen atom attached to nitrogen atom it is not soluble in alkali.
\n
\nN, N, N – trimethylamine does not react with benzene sulphonyl chloride. This property of these three methylamines reacting with benzene sulphonylchloride in a different manner is used for their separation from a mixture.<\/p>\n
Question 30.
\nExplain why aniline in strong acidic medium gives a mixture of Nitro anilines and what steps need to be take to prepare selectively p-nitroaniline.
\nAnswer:
\nIn strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.
\n
\nHowever, by protecting the -NH2<\/sub> group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product.
\n<\/p>\n<\/p>\n
Question 31.
\ni) Account for the stability of aromatic diazonium ions when compared to aliphatic diazonium ions.
\nii) Write the equations showing the conversion of aniline diazonium chloride to
\na) Chlorobenzene, b) Iodobenzene and c) Bromobenzene.
\nAnswer:
\ni) The relative stability of aromatic diazonium ions can be ascribed to the fact that its structure is a resonance hybrid of the canonical forms involving the participation of the benzene ring.
\n
\nThe hybrid structure shows that: a) benzene ring is deactivated to attack of electrophiles, (b) the C -N bond acquires some double bond character and becomes stronger. Alkyl diazonium ions cannot exhibit such resonance and hence C – N bond in them is weak. That is why they are unstable relative to their aromatic counterparts.<\/p>\n
ii) a) Aqueous solution of benzene diazonium chloride when heated with cuprous chloride gives chlorobenzene.
\n
\nb) Iodobenzene is formed when benzene diazonium chloride solution is treated with potassium iodide.
\n
\nc) Bromobenzene is formed when benzenediazonium chloride solution is treated with hydrobromic acid in the presence of copper power.
\n<\/p>\n
Question 32.
\nComplete the following conversions:
\nAniline to i) Fluorobenzene iQ Cyanobenzene iif) Benzene and iv) Phenol
\nAnswer:
\n
\n<\/p>\n
<\/p>\n
Question 33.
\nExplain the following name reactions : [AP 16, 15]
\ni) Sandmeyer reaction
\nii) Gatterman reaction.
\nAnswer:
\ni) Sandmeyer reaction: The diazonium group of a diazonium salt can be replaced by chlorine (-Cl) or bromine (-Br) by heating the aqueous solution of the diazonium salt with cuprous chloride or cuprous bromide. This reaction is called Sandmeyer reaction.
\n<\/p>\n
ii) Gatterman reaction : It is a modification of Sandmeyer reaction. The diazonium group is replaced by – Cl or – Br when the diazonium salt solution is treated with the corresponding halogen acid in the presence of copper powder.
\n<\/p>\n
Question 34.
\nWrite the steps involved in the coupling of Benzene diazoniumchloride with aniline and phenol.
\nAnswer:D
\niazonium salts react with aromatic amines and phenols to give azocompounds having the general formula Ar – N = N – Ar. The reaction is known as coupling reaction. The coupling of benzene diazonium chloride with phenols is carried out in mild alkaline solution and with amines in weakly acidic medium.
\nCoupling with aniline :
\n
\nCoupling with Phenol :
\n<\/p>\n
<\/p>\n
Question 35.
\nWrite the equations involved in the conversion of acetamide and propanaldehydeoxime to methyl cyanide and ethyl cyanide respectively.
\nAnswer:
\nAcetamide is converted to methyl cyanide by heating it with benzene sulphonyl chloride in pyridine at 70\u00b0C.
\nCH3<\/sub> CO NH2<\/sub> + C6<\/sub>H5<\/sub>SO2<\/sub>Cl > CH3<\/sub>CN + C6<\/sub>H5<\/sub>SO3<\/sub>H + HCl
\nPropanaldehydeoxime is converted to ethylcyanide by dehydrating with acetic anhydride.
\nCH3<\/sub> – CH2<\/sub> – CH = NOH + (CH3<\/sub>CO)2<\/sub>O \u2192 CH3<\/sub> – CH2<\/sub> – CN + 2CH3<\/sub>COOH<\/p>\nIntext Questions – Answers<\/span><\/p>\nQuestion 1.
\nClassify the following amines as primary, secondary or tertiary.
\n
\niii) (C2<\/sub>H5<\/sub>)2<\/sub> CHNH2<\/sub>
\niv) (C2<\/sub>H5<\/sub>)2<\/sub> NH
\nAnswer:
\ni) and
\niii) are primary amines
\nii) is a tertiary amine
\niv) is a secondary amine<\/p>\nQuestion 2.
\ni) Write structures of different isomeric amines to the molecular formula, C4<\/sub>H11<\/sub>N
\nii) Write IUPAC names of all the isomers.
\niii) What types of isomerism is exhibited by different types of amines ?.
\nAnswer:
\nMolecular formula C4<\/sub>H11<\/sub>N
\n<\/p>\nii) IUPAC names ;
\na) Butan-1-amine
\nb) 2-methyl propanamine
\nc) 2-methyl-propan-2-amine
\nd) N-methyl propan-1-amine
\ne) N-ethyl ethanamine
\nf) N-methyl-1-methylethanamine
\ng) N, N-Dimethylmethanamine<\/p>\n
iii) Primary amines (a), (b) and (c) exhibit chain isomerism.
\nSecondary amines (a), (e) and (b) exhibit metamerism.<\/p>\n
<\/p>\n
Question 3.
\nHow will you convert
\ni) Benzene into aniline
\nii) Benzene into N, N – dimethylaniline
\niii) Cl – (CH2<\/sub>)4<\/sub> – Cl into hexan -1, 6 – diamine ?
\nAnswer:
\ni) Benzene is first converted into nitrobenzene by nitration. Nitrobenzene on reduction with tin and hydrohloric acid gives aniline.
\n<\/p>\nii) Benzene is converted into aniline by nitration followed by reduction. Aniline on heating with excess of methyliodide gives N, N – Dimethylaniline.
\n<\/p>\n
iii) Cl – (CH2<\/sub>)4<\/sub> – Cl is converted to NC – (CH2<\/sub>)4<\/sub> – CN by reacting with ethanolic potassium cyanide. NC – (CH2<\/sub>)4<\/sub> – CN on reduction with LiAlH4<\/sub> or sodium and alcohol gives H2<\/sub>N (CH2<\/sub>)6<\/sub> NH2<\/sub>.
\n<\/p>\nQuestion 4.
\nArrange the following in increasing order of their basic strength :
\ni) C2<\/sub>H5<\/sub>NH2<\/sub>, C6<\/sub>H5<\/sub>NH2<\/sub>, NH3<\/sub>,C6<\/sub>H5<\/sub>CH2<\/sub> NH2<\/sub> and (C2<\/sub>H5<\/sub>)2<\/sub>NH
\nii) C2<\/sub>H5<\/sub>NH2<\/sub>, (C2<\/sub>H5<\/sub>)2<\/sub> NH, (C2<\/sub>H5<\/sub>)3<\/sub>N, C6<\/sub>H5<\/sub>NH2<\/sub>
\niii) CH3<\/sub>NH2<\/sub>, (CH3<\/sub>)2<\/sub> NH, (CH3<\/sub>)3<\/sub>N, C6<\/sub>H5<\/sub>NH2<\/sub>, C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub>
\nAnswer:
\ni) C6<\/sub>H5<\/sub>NH2<\/sub> < NH3<\/sub> < C6<\/sub>H5<\/sub>CH2<\/sub>NH2<\/sub> < C2<\/sub>H5<\/sub>NH2<\/sub> < (C2<\/sub>H5<\/sub>)2<\/sub> NH
\nii) C6<\/sub>H5<\/sub>NH2<\/sub> < C2<\/sub>H5<\/sub>NH2<\/sub> < (C