Nitrogen containing organic compounds - Karkea.kar.nic.in/cet2014/vikasana/chemistry/day_15.pdf4 2)...
Transcript of Nitrogen containing organic compounds - Karkea.kar.nic.in/cet2014/vikasana/chemistry/day_15.pdf4 2)...
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2) An organic compound (x) containing
nitrogen on reduction with Fe / HCl
forms an amine which reacts with
NaNO2/H+ at 0°C to liberate nitrogen gas.
The compound (x) is _____
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N2 liberation is possible when 1°amine is
not aryl (eg.: aniline). So the compound
(x) on reduction that gives a 1° amine that
is not aryl is (b)
Ans: (b)
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3)
a) Amide and 1°amine
b)1°amine and nitrile
c) Amide and isonitrile
d) nitrite and isonitrile
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2
1.Li AlHCHCl +AlcKOHBr /NaOH o
2.H OP Q R 2 amine
Pand R are:
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If P is an amide we have
Hence P is an amide and R is isonitrile
Ans: (C) Amide and isonitrile
32 CHClBr /NaOH Re duction
(P) (Q) AlcKOHamide 1 a min e isonitrile 2 a min e
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4) The variation in base strength of 1°, 2°,
3° methylamines in aqueous medium is
not due to ______
a) inductive effect
b) solvation effect
c) steric hindrance
d) resonance effect
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In methylamines there is no resonance
effect to influence the base strength
Ans: (d) Resonance effect
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5) Conjugate acid of
NH2 – CH2 - CH2 - COOH is ______
a) NH2 - CH2 - CH2 - COO-
b) +NH3- CH2 – CH2 - COOH
c) –NH - CH2 - CH2 - COOH
d) +NH3 - CH2 - CH2 - COO-
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Conjugate acid can be obtained when a
basic group accepts H+.
In H2N-CH2-CH2-COOH , H2N accepts
H+ to form its conjugate acid
Ans: b)
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6) Increase in base strength of
i) benzylamine ii) cyclohexylamine
iii) o-methoxyaniline is_______
a) iii < i < ii b) iii > ii > i
c) i > ii > iii d) ii < iii > i
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7) Schotten Baumann reaction helps to
convert
a) 1° amine to 2°amine
b) 1°amine into N-substituted amide
c) amide into 1°amine
d) 1° amine into a quaternary salt
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Benzoylation in presence of alkali is
Schotten Baumann reaction 1°amine react
to form N-substituted amide
Eg: C6H5COCl + C6H5NH2
C6H5CONHC6H5
Ans: b) 1° amine into N-substituted amide
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9) Ethanamine and benzenamine can be
best distinguished using______
a) CHCl3 + Alc. KOH
b) CH3I
c) dil. HCl
d) Bromine water
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C2H5NH2 and
both react with the reagents in a, b & c
but only benzenamine readily
decolourises orange colour of bromine
water.
Ans: d) Bromine water
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10) Gabriel phthalimide synthesis is
used to prepare
a) amide
b) 1° aliphatic amine
c) amine
d) 1° aromatic amine
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12)Alkyl halide
A and B are:
a) position isomers
b) functional isomers
c) metamers
d) chain isomers
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X is R-CN R-CH2NH2 (A)
Y is R-NC R-NH-CH3 (B)
A and B are functional isomers
Ans: (b) functional isomers
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13) N-methylbenzenamine reacts with
ethyliodide to form a salt that______
a) is achiral
b) is chiral
c) exhibits metamerism
d) exhibits geometric isomerism
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14) p-nitroaniline
gives
a) p-diiodobenzene
b) p-iodobenzenamine
c) 2, 4-dinitrobenzene
d) p-iodonitrobenzene
21.NaNO /HCl 0 C
2.KI
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15) 0.3g of organic compound containing
C,O,H,N, ammonia
absorbed in 100ml of 0.1 M H2SO4.
Unreacted acid reacted with 20ml of
0.5 M NaOH. The percentage of nitrogen
in the compound could be____
a) 46 b) 56
c) 36 d) 66
kjeldhal 's
method
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H2SO4 +2NaOHNa2SO4+ 2H2O
2NH3 + H2SO4 (NH4)2SO4
Initial amount of H2SO4 is 10 millimole
NH3 consumes 5millimole H2SO4
Amount NH3 =10millimole
% N =
Ans: a) 46
310 10 14 10046.6%
0.3
-
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16) Coupling reaction between
benzene diazoniumchloride and phenol
involves the electrophile_________
6 5 6 5 2
6 5 2
a)C H b)C H N
c)C H O d) N
-
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Zwitter ion is formed when there is transfer of H+ from acidic to basic group within a molecule. For this acidic or basic groups must be fairly strong. Options a, c, d, have either a strong acidic or strong basic group In option (b) NH2 is very weak, so also is COOH.
Ans: (b) HOOC NH2
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18) CCl2 is an electrophile for_______
a) Hoffmann’s bromamide reaction
b) Carbylamine reaction
c) Sandmeyer’s reaction
d) Hinsberg test
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In carbylamine reaction CHCl3 reacts
with alc. KOH to form CCl2 a neutral
electrophile
CHCl3 CCl2(dichlorocarbene)
Ans: (a) Carbylamine reaction
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In the molecule NH group is e-
releasing, ring activating, o,p directing,
but CO group is e- withdrawing, ring
deactivating, m – directing. Nitration
happens faster for the ring attached to
NH group. The NO2 group mainly goes
to para position (c).
Ans: (a) c
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20)
X & Y have functional groups as
a) amine and phenol
b) amide and phenol
c) diazonium and phenol
d) nitrile and phenol
3n FeClin water warmX Y violet colour-
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Y must be phenol to give violet colour
with n-FeCl3. Phenol is obtained from X
by hydrolysis. X must be a benzene
diazonium salt.
Ans: (c) X - diazonium group
Y - phenol
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All types of amines (1°, 2°, 3°) have a
lone pair of electrons on nitrogen.
Hence by definition they fit into a, b, c
but not d.
Ans: (d) Arrhenius base.
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22) Ethanamine is not obtained by the
reduction of
a) methylcyanide
b) nitroethane
c) acetamide
d) ethylcyanide
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Re NHCHCHduction
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Re NHCHCHduction
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Re NHCHCHduction
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Re NHCHCHCHduction
a) Methylcyanide – CH3CN
b) Nitroethane- C2H5NO2
c) Acetamide – CH3CONH2
d) Ethylcyanide–C2H5CN
Ans: (d) ethylcyanide
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23) Which one of the following is most
reactive towards electrophilic
substitution reaction?
a) aniline
b) Acetanilide
c) benzamide
d) nitrobenzene
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b, c, d have electron withdrawing groups,
which is ring deactivating. (a) has
electron releasing, ring activating group
Ans: (a) aniline
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24) A tribromo derivative is obtained
when X is treated with bromine water. X
could be_______
a) acetophenone
b) acetanilide
c) o-cresol
d) m-toluidine
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To get a tribromo derivative, the functional group must be strongly electron releasing group.(c) and (d) has it. But in (c) ortho position is blocked by CH3 group but not in (d). All o & p positions are free, hence m-toluidine on bromination forms tribromo derivative.
Ans: (d) m-toluidine
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25)
C is ---------
a) C6H5COONH4 b) C6H5CH2NH2
c) C6H5NHCH3 d) C6H5COOH
2
4
2
HNO 0 C CuCN
6 5 2HCN
1.Li AlH
2.H O
C H NH A B
C
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NH2 is electron releasing due to
resonance (+R) and electron
withdrawing due to high
electronegativity of N (-I).
Ans: c) –I, +R
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27. In detection of nitrogen in an
organic compound via sodium fusion
extract, nitrogen is converted into
a) NH3
b) NaCN
c) NaNO3
d) N2
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When nitrogen containing organic
compound is fused with sodium, the
compound formed is NaCN.
Ans: b) NaCN
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28. Which of the following statements
is correct?
a) Amides are more basic than aliphatic amines
b) Electron withdrawing group increases the base strength of aniline
c) Diphenyl amine is less basic than aniline
d) Amines are stronger bases than ammonia
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29. The name of
a) ethylmethyl propylamine
b) sec-butyl methylamine
c) ethylmethyl isopropylamine
d) ethyl propylamine
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It is a tertiary amine with alkyl groups
methyl, ethyl and isopropyl.
Ans: c) ethylmethyl isopropylamine
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30. Benzene is converted into
The best sequence of reactions to do
this is
a) Methylation, nitration, reduction
b) Nitration, reduction, methylation
c) Methylation, reduction, nitration
d) Reduction, nitration, methylation
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31. Which one of these is neither a test
nor a method to prepare amines?
a) Hoffmann reaction
b) Carbylamine
c) Methylation
d) Sulphonation
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34. For maximum activity of sulpha
drugs, the minimum standard feature of
the drug should be
a)
b)
c)
d)
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35. An aromatic compound with the
formula C7H9N cannot be
a) 1° amine
b) 2° amine
c) toluidine
d) 3° amine
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36. Aniline can be obtained by the
reduction of _____
a) benzonitrile b) benzamide
c) anilide d) nitrobenzene
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C6H5CN C6H5CH2NH2
(benzonitrile)
C6H5CONH2 C6H5CH2NH2
(benzamide)
R-CONH-Ar RCH2NHAr
(anilide)
C6H5NO2 C6H5NH2
(nitrobenzene)
Ans: d) Nitrobenzene
reduction
reduction
reduction
reduction
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37. p-hydroxy azobenzene is obtained
when
a) Phenol is coupled with azobenzene
b)p-hydroxyaniline is coupled with benzene
c)Phenol is coupled with bezenediazonium ion
d) Aniline is coupled with phenol
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is obtained when phenol couples with
a benzene diazonium salt or ion
Ans: c) Phenol is coupled with
bezenediazonium ion
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38. 1,3,5-tribromobenzene can be
obtained from aniline by a series of
reactions in the order
a) bromination, reduction, diazotisation b) bromination, diazotisation, boil with
ethanol c) carbylamine reaction, diazotisation,
boil with Br2 water d) bromination, heat with Zn dust, boil
with ethanol
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39. Enantiomers are possible for
a) ethylmethyamine
b) secondary butylamine
c) N, N-dimethyl aniline
d) trimethylammonium ion
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Enantiomers are possible if the given
compound is chiral or has a chiral
carbon.
has a chiral carbon and hence is chiral.
Ans: b) sec-butylamine
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40. Acetylation of aniline, makes it a) less reactive towards ESR & more
reactive towards oxidation
b) more reactive towards both ESR and oxidation
c) less reactive towards both ESR and oxidation
d) more reactive towards ESR but less reaction towards oxidation
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Acetylation of aniline gives
which has electron withdrawing
group in the side chain
Ans:
c) less reactive towards both ESR and oxidation
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41. Among i) aniline ii) benzaldehyde
iii) acetanilide iv) toluene,
the least and the most reactive towards
Friedel Craft’s alkylation is
a) i and iv b) iv and ii
c) iii and ii d) ii and i
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42. Which one of the following gives a
product with Hinsberg reagent which is
insoluble in an alkali?
a) (CH3)3N
b) CH3CH2NH2
c) C6H5NHCH3
d) (CH3)2NC6H5
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43. 2° amine reacts with NaNO2 and HCl
at 0° C to
a) form a 2° alcohol
b) liberate nitrogen gas
c) form a diazonium salt
d) form N-nitrosoamine
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44. Aniline reacts with concentrated
H2SO4 at 450 K to give
a) phenol
b) sulphanilic acid
c) aniline hydrogensulphate
d) benzene sulphonic acid
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45. The base strengths of the following
compounds decreases as
a) o-toluidine > aniline > o-nitroaniline
b) o-toluidine > o-nitroaniline > aniline
c) Aniline > o-nitroaniline > o-toluidine
d) o-nitroaniline<aniline<o-toluidine
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46. In the gaseous state the base strengths
of the following amine increases as:
a) CH3NH2<(CH3)2NH<(CH3)3N
b) (CH3)3N<CH3NH2<(CH3)2NH
c) (CH3)3N<(CH3)2NH<CH3NH2
d) CH3NH2<(CH3)3N<(CH3)2NH
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47. Gabriel phthalimide synthesis does not
involve
a) Neutralisation
b) Nucleophilic substitution
c) Hydrolysis
d) Hydration
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48. A positive result for carbylamine test
will be obtained when the amine is
a) Aliphatic 1° amine only
b) Aromatic 1° amine only
c) 1° amine
d) 2° aliphatic amine
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49. Aniline and benzylamine can be best
distinguished using
a) NaNO2/ HCl at 0°C + b-naphthol in
NaOH
b) Hinsberg reagent
c) CHCl3 + alc.KOH
d) CH3COCl
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50. Amide can be converted into 1° amine
by _____ reaction
a) Sandmeyer
b) Carbylamine
c) Hoffman’s bromamide
d) Rosenmund’s
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51. Which of the following statements is
correct?
i) Nitration of aniline gives meta isomer.
ii)Acetylation of aniline decreases the tendency of aniline to get oxidized.
iii) anilinium ion is a strong conjugate base.
iv) Aniline is an antioxidant.
a) i, ii, iv b) ii, iii c) ii, iii, iv d) iii, iv
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54. Compounds P and Q that yield an
isocyanide when treated with alc. AgCN
and CHCl3/ alkali respectively are:
a) chlorobenzene and acetone
b) ethyl bromide and acetamide
c) methyl iodide and ethyl amine
d) ethyl alcohol and ethyl bromide
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55. An optically active compound A
(C8H11N) dissolves in HCl and answers
carbylamine test. The compound could
be
a) C6H5CH2CH2NH2
b) CH3NHCH2CH2C6H5
c) C6H5CH(NH2)CH3
d) C6H4(NH2) (C2H5)
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Lower the pKb value, stronger is the base.
Among the options dimethyl amine in (c)
is the strongest base.
Ans: c) (CH3)2NH
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59.
X, Y, Z are
a) CH3OH, CH3NH2, CH3OH
b) CH3Cl, CH3NH2, CH3NO2
c) Cl-CH2-COOH, NH2-CH2-COOH,
HO-CH2-COOH
d) CH3COCl, CH3CONH2, CH3OH
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60. A mixture of aniline and nitrobenzene
can be separated using
a) NaOH solution
b) HCl acid
c) alcohol
d) ether
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Aniline is basic hence dissolves in HCl
acid. Nitrobenzene remains undissolved
forms a separate layer, hence can be
separated.
Ans: b) HCl acid
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61.
X should be
a) o-nitrobenzenamine
b) o-nitrobenzoic acid
c) m-nitrobenzoic acid
d) p-nitrobenzoic acid
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62. The reagent that will not help to
distinguish benzylamine from N-
methylbenzenamine is
a) CHCl3/ alc.KOH
b) C6H5SO2Cl
c) HNO2
d) HCl acid
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Both of these react differently with
reagents in a, b and c but both dissolve in
HCl acid to form aqueous solution.
Ans: d) HCl acid
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63. 1.78 g of an amine reacts with nitrous
acid at 0°C to liberate 4.48 dm3 of N2 at
STP. The molar mass of the amine is
_____ g mol-1.
a) 78
b) 89
c) 68
d) 48
127
= 0.02 mol nitrogen at STP 4.22
48.4
Mass of 0.02 mole of amine = 1.78g
Molar mass of the amine =
= 89 gmol-1
Ans: b) 89
1.78
0.02
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64. Aniline can be purified by steam
distillation. Efficiency of steam
distillation can be increased by the
addition of
a) alcohol
b) HNO3
c) NaCl
d) benzene
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When NaCl is dissolved, vapour pressure
of salt solution (water + salt) is lowered,
boiling point of water increases. Aniline
does not dissolve NaCl and hence gets
distilled more efficiently.
Ans: c) NaCl