Download 13. Amines. - Mathematic.in

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1. Classify the following amines as primary, secondary or tertiary:
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(i)
(ii)
(iii)
(iv)
.
Solution:
(i) Primary
(ii) Tertiary
(iii) Primary
(iv) Secondary.
2. (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N.
(ii) Write IUPAC names of all the isomers.
Solution:
(i)
in
(ii)
(iii)
.
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3. How will you convert
(i) Benzene into aniline
(ii) Benzene into N, N-dimethylaniline.
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Solution:
(i) Benzene into aniline
(ii) Benzene into N,N-dimethylaniline
4. Arrange the following in increasing order of their basic strength:
(i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2
(iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2.
Solution:
(i) C6H5NH2 < NH3 < C6H5CH2NH2 < C2H5NH2 < (C2H5)2NH
(ii) C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH
(iii) C6H5NH2 < C6H5CH2NH2 < (CH3)3N < CH3NH2 < (CH3)2NH
in
5. Complete the following acid-base reactions and name the products:
(i) CH3CH2CH2NH2 + HCl
(ii) (C2H5)3N + HCl.
Solution:
(i) CH3CH2CH2NH2 + HCl → CH3CH2CH2NH3+Cl- ( Propanamine hydrochloride)
(ii) (C2H5)3N + HCl→ [(C2H5)3+ NH]Cl-.(N,N-diethyl ethanamine hydrochloride).
2
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6. Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence
Solution:
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7. Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Solution:
8. Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC nam
nitrogen gas on treatment with nitrous acid.
Solution:
.
9. Convert
(i) 3-Methylaniline into 3-nitrotoluene.
(ii) Aniline into 1,3,5 - tribromobenzene.
Solution:
in
3
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(i)
(ii)
.
10. Write IUPAC names of the following compounds and classify them into primary, secondary and tertia
(i) (CH3)2CHNH2
(ii) CH3(CH2)2NH2
(iii) CH3NHCH(CH3)2
(iv) (CH3)3CNH2.
i) 2-amino propane or propan-2-amine (primary)
in
Solution:
(ii) 1-amino propane (primary)
(iii) N-Methyl-propan-2-amine (secondary)
(iv) 2-methyl-propan-2-amine (primary).
11. Write IUPAC names of the following compounds and classify them into primary,
secondary and tertiary amines.
4
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(i) C6H5NHCH3
(ii) (CH3CH2)2NCH3
(iii) m–BrC6H4NH2.
•
Solution:
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(i) N-Methyl benzenamine (secondary)
(ii) N-Ethyl-N-methyl-ethanamine(tertiary)
(iii) 3-bromo aniline (primary).
12. Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Aniline and benzylamine
(iii) Aniline and N-methylaniline.
•
Solution:
(i) Heat both these amines with chloroform and alchohol KOH solution. The compound which gives
unpleasant smell is methylamine while the compound which does not give any smell is diethylamine.
(ii) Add NaNO2 and HCl to each separately. Cool it to 0-5C. Then add alkaline solution of phenol.
Orange azo dye is formed in aniline. Benzylamine (C6H5CH2NH2) does not form azo dye.
in
(iii) Heat both the compounds separately with chloroform and alcoholic KOH. The compound which
gives an unpleasant or offensive smell in aniline while the compound which does not give any smell
is N-Methylaniline
13.
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Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iii) Aniline does not undergo Friedel-Crafts reaction.
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(iv) Gabriel phthalimide synthesis is preferred for synthesising primary amines
(v) Ethylamine is soluble in water whereas aniline is not.
(vi) Although amino group is o,p directing in aromatic substitution reactions,aniline on
nitration gives a substantial amount of meta nitro aniline.
(vii) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
•
Solution:
(i) It is because in aniline, the NH2 group is attached directly to the benzene ring. It results in the
unshared electron pair of nitrogen atom to be in conjugation with the benzene ring and thus making
it less available for protonation.
(ii) Methyl amine is a base and dissolves in water to produce hydroxide ions.
FeCl3 combines with OH- ions to give reddish brown precipitate of Fe(OH)3
(iii) Aniline does not undergo Friedel Craft reaction due to the salt formation with aluminium chloride,
the Lewis acid which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge
and hence acts as a strong deactivating group for further reaction.
(iv) Using excess of ammonia in ammonolysis of alkyl halide is not a very practicable method for the
synthesis of primary amines. Alternately, phthalimide is alkylated with alkyl or benzyl halide and
then hydrolysed to get pure primary amine. In this method phthalic acid is produced which can be
again converted into phthalimide and used over and over again.
(v)Ethyl amine can form hydrogen bonds with water and so it dissolves. In aniline the large benzene
ring decreases the extent of hydrogen bonding.
in
(vi) In the presence of acids used for nitration , aniline gets protonated to anilinium ion.It is meta
directing and so a considerable amount of meta nitroaniline is also got.
(vii)In aromatic diazonium salts, the positive charge on the nitrogen is dispersed over the benzene
ring due to resonance.Thus it is stabilized.
14. Arrange the following:
(i) In increasing order of boiling point:
C2H5OH, (CH3)NH, C2H5NH2
(ii) In increasing order of solubility in water:
C6H5NH2, (C2H5)2NH, C2H5NH2.
Edumass
Edumass
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(iii) In the decreasing order of pKb values:C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2
(iv) In the increasing order of basic strength: C6H5NH2, C6H5N(CH3)2 , ( C2H5)2NH and
CH3NH2
(v) In increasing order of basic strength: a. aniline,para nitraniline ,and para toludine
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b.C6H5NH2, C6H5NHCH3, C6H5CH2NH2
(vi) In increasing order of basic strength in gas phase.:C2H5NH2 (C2H5)2NH, (C2H5)3N and
NH3.
•
Solution:
(i) (CH3)NH < C2H5NH2 < C2H5OH
(ii) C6H5NH2 < (C2H5)2NH < C2H5NH2.
(iii) C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH
(iv) C6H5NH2 < C6H5 N(CH3)2, < CH3NH2 < (C2H5)2NH
(v) (a) p-nitroaniline < aniline < p-toludine
(b) C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2
(vi)NH3 < (C2H5)NH2 < (C2H5)2NH < (C2H5)3N
15.
How will you convert:
(i) Ethanoic acid into methanamine
(ii) Methanol to ethanoic acid
(iii) hexanenitrile to 1-aminopentane
(iv) Ethanamine the methanamine
(vi) Methanamine to ethanamine
in
(v) Ethanoic acid to propanoic acid
(vii) nitromethane to dimethylamine
(viii) propanoic acid to ethanoic acid
•
Solution:
(i) Ethanoic acid into methanamine
7
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(ii) Methanol to ethanoic acid
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.
(iii)
(iv)
(v)
(vi)
in
(vii)
(viii)
8
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16.
Write short notes on the following:
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(i) Diazotisation
(ii) Hoffmannbromamide reaction
(iii) Coupling reaction.
(iv)Carbylamine reaction
(v)Ammonolysis
(vi)Gabriel’s pthalimide reaction
(vii)Acetylation
•
Solution:
(i) Diazotization
It is a process of treating primary aromatic amine with nitrous acid at 273-278 K to get diazonium
salts which are very useful compounds.
They are used to prepare benzene, phenol, haloarenes, cyanobenzene, dyes, etc.,
(ii) Hoffmann Bromamide Degradation Reaction
Reaction of amine with solution of bromine and alkali results in the formation of primary amine with
one less carbon.
in
(iii) Coupling Reaction
It is a reaction of diazoniuim salts with aromatic amines, phenols, etc. to get azo dyes.
9
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(iv) It is a reaction of primary amines with chloroform and alcoholic KOH to give carbylamine or
isocyanide .
C2H5NH2 + CHCl3 +3 KOH → C2H5 NC + 3 KCl +3 H2O
(v) It is the reaction of alkyl halides with ammonia to give primary, secondary and tertiary amines
and quarternary ammonium salt.
Refer to the lesson for equations.
(vi) Refer to the lesson
(vii) Acetylation is the reaction of introducing CH3CO - group into a molecule.
The reagent for this is acid chloride or acid anhydride.
C6H5 NH2 + CH3COCl → C6H5 NH CO CH3 + HCl.
Aniline
acetanilide
17. Accomplish the following conversions:
(i) Aniline to 2,4,6 –tribromo aniline
(ii) Benzene to m-bromophenol.
(iii) nitrobenzene to benzoic acid
(iv) benzoic acid to aniline
(v) benzyl chloride to 2-phenylethanamine
in
(vi) chlorobenzene to p-chloroaniline
(vii) aniline to p-bromoaniline
(viii)Aniline to benzyl alcohol
(ix)benzamide to toluene
• Solution:
(i)
Edumass
Edumass
10
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(ii)
(iii)
(iv)
(v)
(vi)
in
(vii)
11
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(viii)
(ix)
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18. Give the structures of A, B and C in the following reactions:
(i)
(ii)
(ii)
(iii)
(iv)
(v)
(vi)
Solution:
(i)
in
•
12
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(ii)
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•
(iii)
•
(iv)
•
•
(v)
(vi)
•
•
19. An aromatic compound A on treatment with aqueous ammonia and heating forms
compound B which on heating with Br2 and KOH forms a compound C of molecular formula
C6H7N. Write the structures and IUPAC names of compounds A, B and C.
•
Solution:
in
20. Complete the following reactions:
(i)
(ii)
(iii)C6H5NH2 +H2SO4 (conc)
1
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(iv)C6H5N2Cl +C2H5OH
(v)C6H5NH2 +Br2
(vi) C6H5NH2+(CH3CO)2
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(vii)C6H5N2Cl
•
Solution:
(i)
(ii)
(iii)
(iv)
(v)
in
(vi)
(vii)
21. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
•
Solution:
14
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Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because due to low
reactivity of aryl halide. As C-X bond in aryl halide is very strong , n-aryl phthalimide is not formed.
22. Write the reaction of aliphatic and aromatic primary amines with nitrous acid.
•
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Solution:
Primary aliphatic amines react with nitrous acid at low temperature to form primary alchohol and
nitrogen gas accompanied by risk effervescence.
For example,
Primary aromatic amines react with nitrous acid to form benzene diazonium salt which is unstable.
C6H5NH2 +HNO2(NaNO2+HCl)
C6H5 N2Cl + H2O (not balanced)
23. Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why are primary amines higher boiling than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
•
Solution:
(i) Oxygen in alcohol is more electronegative than nitrogen in amines. RO-H bond in alcohol is more
polar δ+ charge on H atom i.e. ROδ- Hδ+ as compared to RN-H bond in RNH2. Alcohols can loose
proton to some extent but amine are proton acceptors.
Unstable
in
(ii) Amines are polar compounds and cal form H-bonds as proton acceptors. Only primary and
secondary amines can form intermolecular H-bonding due to presence H atom at N atom. Tertiary
amines cannot form intermolecular H-bonding. Primary amines are higher boiling due to the presence
of stronger intermolecular force of attraction due to H-bonding.
15
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(iii) In aliphatic amines , the alkyl groups are electron releasing and make the N atom better electron
donors. But in aromatic amines, the lone pair of electrons on the N atom is involved in resonance
with the benzene ring and so is less available.
24. Describe the method of identification of primary secondary and tertiary amines. Also
write the chemical equations of the reactions involved
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•
Solution:
Hinsberg test can be used to identity primary, secondary and tertiary amines. The amine is shaken
with benzene sulphonyl chloride in the presence of excess of KOH solution.
Primary amine gives clear solution which on acidification gives an insoluble compound that separates
out.
A secondary amine forms N, N - diethyl benzene sulphonamide which does not dissolve in KOH
does not dissolve (no reaction)
A tertiary amine +
no reaction
in
16
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