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Published on: 02/06/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
Arrange the following
i. In increasing order of solubility in water, C6H5 NH2, (C2H5)2NH, C2H5NH2
ii. In increasing order of basic strength
a) aniline, p- toludine and p – nitroaniline
b) C6H5 NH2, C6H5 NHCH3, C6H5NH2, p-Cl-C6- H4-NH2
iii. In decreasing order of basic strength in gas phase
(C2H5)NH2, (C2H5)NH, (C2H5)5N and NH3
iv. In increasing order of boiling point
C6H5OH, (CH3)2NH, C2H5NH2
v. In decreasing order of the pKb values
C2H5NH2, C6H5NHCH3.(C2H5)2 NH and CH3NH2
vi. Increasing order of basic strength
C2H5NH2,C6H5N(CH3)2, (C2H5)2 NH and CH3NH2
vii. In decreasing order of basic strength
2.
Account for the following
i. Aniline does not undergo Friedel – Crafts reaction
ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines
iii. pKb of aniline is more than that of methylamine
iv. Gabriel phthalimide synthesis is preferred for synthesising primary amines.
v. Ethylamine is soluble in water whereas aniline is not
vi. Amines are more basic than amides
vii.Although amino group is o – and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m – nitroaniline.
3.
How will you distinguish between primary secondary and tertiary alphatic amines.
4.
Identify A,B,and C
CH3- NO2 \(\overset { { L }_{ 1 }{AlH }_{ 4 } }{ \underset { {} }{ \longrightarrow } }\) A \(\overset { { 2CH_3 }{Ch_2Br } }{ \underset { {} }{ \longrightarrow } }\) B \(\overset { {H}_{ 2 }{SO}_{ 4 } }{ \underset { {} }{ \longrightarrow } } \) C
5.
How will you prepare propan – 1- amine from
i) butane nitrile
ii) propanamide
ii) 1- nitropropane
1.
i) In increasing order of solubility in water:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\)
ii) In increasing order of basic strength:
a. Aniline, p - toluidine and p - nitro aniline
p - toluidine > aniline >p - nitro aniline
b. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, \mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{p}-\mathrm{Cl}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NH}_{2}\\ 2^{o} \text { amine } \quad \quad \quad \quad e^{\ominus} \text { with drawing (group) }\)
(iii) In decreasing order of basic strength in gas phase:
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right) \mathrm{NH},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N} \text { and } \mathrm{NH}_{3} \)
\(\mathrm{NH}_{3}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}<\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NH}\\ \quad \quad \quad 1^{0} \text { amine } \quad 2^{0} \text { amine } \quad \quad \quad 3^{0} \text { amine }\)
(iv) In increasing order of boiling point:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \)
\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}>\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \\ 2^{0} \text { amine } \quad \quad1^{0} \text { amine }\)
Generally amines have lower boiling point than alcohol. Due to comparable molecular mass and weaker H-bonds in Amines.
(v) In decreasing order of the pKb values:
\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}<\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}<\mathrm{CH}_{3} \mathrm{NH}_{2}<\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} \)
\(\quad \quad \quad 2^{0} \text { amine } \quad 1^{0} \text { amine } 1^{0} \text { amine } \quad 2^{0} \text { amine } \)
\(\mathrm{PK}_{b}: \quad 3.00\quad < \quad 3.29 \quad < \quad 3.38 \quad<\quad 9.30\)
pKb ,Due to + 1 effect of C2H5 group. Higher the value of pKb lower is the basicity
(vi) Increasing Order of basic strength:
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2},\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \text { and } \mathrm{CH}_{3} \mathrm{NH}_{2} \)
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}\left(\mathrm{CH}_{3}\right)_{2}>\mathrm{CH}_{3} \mathrm{NH}_{2}>\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH} \\ \left(\mathrm{pK}_{6}: 9.38 \quad > \quad \quad 8.92 \quad \quad > \quad 3.38 \quad > \quad \quad 3.00\right)\)
Due to + 1 effect of C2H5 group.
In decreasing order of basic strength:
2.
Aniline does not undergo Friedel - Craft's reaction:
Aniline does not undergo Friedel - Craft's reaction (alkylation and acetylation). Aniline is basic in nature and it donates its lone pair of electrons to the lewis acid AlCl3 to form an adduct which inhibits further electrophilic substitution reaction.
Diazonium salts of aromatic amines are more stable than those of aliphatic amines:
This is due to resonance
Resonance Structure:
The stability of arene diazonium salt is due to the dispersal of the positive charge over the benzene ring.
pKb of aniline is more than that of methylamine:
pKb - methylamine -3.35
pKb - aniline -9.376
In aniline the lone pair of electrons on N - atom is delocalized over the benzene ring. So, the electron density on the N - atom decreases. In methylamine + 1 effect to CH3 group increases the electron density on the nitrogen atom Hence aniline is a weaker base than methylamine. Due to this, the pKb value for aniline is more than that of methylamine.
(iv) Gabriel phthalimide synthesis is preferred for synthesising primary amines:
In this method alkyl halides react with pottassium phthalimide to give pure primary amine by nucleophilic substitution. In contrast, Aniline (Aromatic primary amine) can not be prepared by this method because Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. Therefore, this method used for the Aliphatic primary. amines only. Aryl halides do not undergo SN2 mechanism with the ion formed by the phthalimide.
(v) Ethylamine is soluble in water whereas aniline is not:
(a) Ethylamine is soluble in water, as it can form intermolecular H - bonds with water molecules. In aqueous solution, the substituted ammonium cation get stabilized not only by electron releasing (+I) effect of the alkyl group but also by solvation with water molecules. The greater the size of the ion, the lower will be the solvation.
(b) Amiline doesn't form H - bond with water to a very large extent due to the presence of a large hydrophobic -C6H5 group.
(vi) Amines are more basic than amides:
This is because, in amides, the carbonyl group is highly electro negative It has a greater power to attract the electrons towards it. It makes the lone pair of electrons on amide nitrogen (-CONH2) less available to accept a proton.
(vii) Although amino group is o - and p - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m - nitro aniline:
In strong acid medium, aniline is protonated to form anilinium ion which is m - directing and hence m - nitro aniline is formed.
3.
| S.No | Reagents or Reaction | Primary amine RNH2 | Secondary amine R2NH | Tertiary amine R3N |
|---|---|---|---|---|
|
1. |
Carbylamine reaction or with CHCl3/KOH |
Carbylamine is formed (unpleasant smell) |
- | - |
| 2. | Mustard oil reaction or CS2/HgCl2 (Hoffmann's mustard oil test) |
Alkyl isothiocyanate is formed (Mustard oil odour) |
- | - |
| 3. | HNO2 (or) NaNO2 / HCl |
Alcohol is formed +H2 | Yellow oily nitrosoamine is formed, insoluble in water. (Liberman's Test) |
Forms nitrite in cold, soluble in water. |
| 4. | CH3COCl | N-acetyl derivative is formed | N,N- diacetyl derivative is formed |
- |
| 5. | Diethyl oxalate Hoffmann's method |
Solid oxamide is formed | Liquid oxamic ester is formed |
- |
| 6. | Benzene sulphonyl chloride in presence of excess. KOH (Hinsberg's reaction) |
N- alkyl benzene sulphonamide is formed (soluble) |
N, N - dialkyl benzene sulphonamide is formed (Insoluble). |
- |
| 7. | With RX | 1 mol → 2o amine 2 mol → 3o amine 3 mol → Quarternary salt |
1 mol → 3o amine 2 mol → Quarternary salt |
1 mol → Quarternary salt |
4.
5.
+ 2H2O
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