CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 9 - MCQExams.com

Which of the following is correct formula for BDC?
  •   $${ C }_{ 6 }{ H }_{ 4 }{ N }_{ 2 }Cl$$
  •   $${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }$$
  •   $${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }Cl$$
  •    $${ C }_{ 6 }{ H }_{ 4 }{ N }_{ 2 }$$
In the reaction series: $${ C }_{ 6 }{ H }_{ 5 }Cl\xrightarrow [Cu _{ 2 }O,\: 200^oC ]{NH_3} X; X\xrightarrow [ { 0-5 }^{ 0 }C ]{ { HNO }_{ 2 } } Z; X+Z\rightarrow A$$, the number of $$\sigma $$ and $$\pi $$ bonds in $$A$$ are:
  • $$25$$$$\sigma $$ and $$6$$$$\pi $$
  • $$25 \sigma $$ and $$7 \pi $$
  • $$27\sigma $$ and $$7\pi $$
  • $$27\sigma $$ and $$6\pi $$
Reaction at $${ 0-5 }^{ 0 }C$$ between aniline, $${ NaNO }_{ 2 }$$ and HCl is known as____________
  • Suphonation of benzene
  • Acylation of aniline
  • Alkylation of aniline
  • Diazotisation
Among the following which is more basic :

Statement I: Aniline on reaction with $$\mathrm{N}\mathrm{a}\mathrm{N}\mathrm{O}_{2}/\mathrm{H}\mathrm{C}l$$ at $$0^{\mathrm{o}}\mathrm{C}$$ followed by coupling with $$\beta$$-naphthol gives a dark blue coloured precipitate.
Statement II: The colour of the compound formed in the reaction of aniline with $$\mathrm{N}\mathrm{a}\mathrm{N}\mathrm{O}_{2}/\mathrm{H}\mathrm{C}l$$ at $$0^{\mathrm{o}}\mathrm{C}$$ followed by coupling with $$\beta$$-naphthol.
  • Statement I and statement II are true and statement II is the correct explanation of statement I
  • Statement I and statement II are true and statement II is not the correct explanation of statement I
  • Statement I is true and the statement II is false
  • Statement I is false and the statement II is true
A compound $$A$$ has molecular formula $$C_{7} H_{7} NO$$. On treatment with $$Br_{2}$$ and $$KOH$$, $$A$$ gives an amine $$B$$ which gives carbylamine test. $$B$$ upon diazotization and coupling with phenol gives an azo dye. $$A$$ can be:
  • $$C_{6}H_{5}CONHCOCH_{3}$$
  • $$ C_{6}H_{5} CONH_{2}$$
  • $$ C_{6}H_{5} NO_{2}$$
  • $$o,\:m$$ or $$p$$-$$ C_{6}H_{4}(NH_{2})CHO$$
The starting reagents needed to make the azo compound shown below is:
26636_ef344524a146453fab9b41a5fb9f38e8.png
Which of the following orders with regards to the basic strength of substituted aniline is correct?
  • p - methylaniline > p - chloroaniline > p - aminoacetophenone
  • p - methylaniline > p - aminoacetophenone > p - chloroaniline
  • p - aminoacetophenone > p - methylaniline > p - chloroaniline
  • p - aminoacetophenone > p - chloroaniline > p - methylaniline
Which of the following forms unstable diazonium ion when treated with $$NaNO_{2}$$ in aqueous $$HCl$$?
  • p-nitrotoluene
  • ethylamine
  • N,N-dimethyl aniline
  • All of the above
The correct order of increasing basic nature for the bases  $$NH_{3} ,CH_{3} NH_{2}  and  (CH_{3})_{2}NH$$ in gas phase is :
  • $$(CH_{3})_{2}NH < NH_{3}< CH_{3}NH_{2}$$
  • $$NH_{3} < CH_{3}NH_{2} < (CH_{3})_{2}NH$$
  • $$CH_{3}NH_{2} < (CH_{3})_{2}NH_{2} < NH_{3}$$
  • $$CH_{3}NH_{2} < NH_{3} <(CH_{3})_{2}NH $$
Amongst the following the most basic compound is :
  • p - nitroaniline
  • acetanilide
  • aniline
  • benzylamine
The major product expected, when Phthalmide is treated with NaOH, is :
Which of the following order for basic strength of amines is correct?

74900_983e8f8b3bdb4cdc93af1b54acd8ae3c.png
  • $$I > III > IV > II$$
  • $$IV > I > II > III$$
  • $$I > IV > II > III$$
  • $$I > II > IV > III$$
The increasing order of basic strength for the following compounds is:
(I) $$CH_{3}-CH_{2}-NH_{2}$$
(II) $$CH_{3}-NH-CH_{3}$$
(III) $$CH_{3}-NH-Ph$$
(IV) $$CH_{3}-NH-CHO$$
  • $$IV < III < I > II$$
  • $$III < IV < I > II$$
  • $$IV < III < II > I$$
  • $$II < I < III > IV$$
The product $$U$$ is:

72533_45ed6e42e3af44fb953433b72f7f74b9.jpg
Which compound does not give positive test in Lassaigne's test for nitrogen?
  • Urea
  • Hydrazine
  • Azobenzene
  • Phenyl hydrazine
The correct order of basicity of the above compounds is given.
Identify the correct statement(s).
43841.PNG
  • In III, lone pair of electrons is involved in delocalisation but not in II.
  • In IV, lone pair of electrons is involved in delocalisation but not in III.
  • I is more basic than II because six membered ring is more stable than five membered ring.
  • In III, lone pair of electrons is present in sp orbital but in I, II and IV lone pair of electrons are present in $$sp^{2}$$ orbital.
Aniline when diazotized in cold and then treated with dimethylaniline gives a coloured product. Its structure would be:
$$3HC\equiv CH \xrightarrow {polymerisation} X \xrightarrow[conc.\ HNO_{3}]{conc.\ H_{2}SO_{4}} Y \xrightarrow {Sn/HCl}Z\xrightarrow[NaNO_{2},HCl]{(0-5^{0}C)} V\xrightarrow[Phenol]{dil\ NaOH}W$$
Product W is :
Which of the following is weakest base?
  • $$H_2N-CH_2-NO_2$$
  • $$CH_3NH-CH=O$$
Which compound is soluble in water :
  • $$\displaystyle \left [ \left ( CH_{3} \right )_{2}NH_{2}\right ]^{+}Cl^{-}$$
  • $$\displaystyle \left [ CH_{3}NH_{3} \right ]^{+}Cl^{-}$$
  • $$\displaystyle \left [ \left ( CH_{3} \right )_{3}NH\right ]^{+}Cl^{-}$$
  • $$\displaystyle PhNO_{2}$$
What is the principle product when aniline is treated with $$NaNO_2$$ and $$HCl$$ at $$0-5^{\circ}C$$, and this mixture is added to p-toludine (p-methyl aniline)?
The following product(s) will be obtained from the above reaction:
121309_d6d1d31057454058810397464eb9aaac.png
N $$\displaystyle -$$ Ethyl pthalimide on hydrolysis gives:
  • methyl alcohol
  • ethyl amine
  • dimethyl amine
  • diethyl amine
Which compound will liberate $$\displaystyle CO_{2}$$ from $$\displaystyle NaHCO_{3}$$ solution :
  • $$\displaystyle CH_{3}CO\:NH_{2}$$
  • $$\displaystyle CH_{3}NH_{2}$$
  • $$\displaystyle \left ( CH_{3} \right )_{4}N^{+}OH^{-}$$
  • $$\displaystyle CH_{3}N^{+}H_{3}Cl^{-}$$
Which of the following diazonium salt is relatively stable of $$0-5^\circ C$$- :
  • $$CH_{3}CH_2N\equiv N\left. \right \} ^\oplus Cl^{-} $$
  • $$CH_{3}-C(CH_{3})_{2}-N\equiv N\left. \right \} ^\oplus Cl^{-} $$
  • $$CH_{3}-N\equiv N\left. \right \} ^\oplus Cl^{-} $$
  • $$(CH_{3})_{3}C-N\equiv N\left. \right \} ^\oplus Cl^{-} $$
Which of the following diazonium salt is relatively stable at $$\displaystyle 0-5^{\circ}$$C?
  • $$\displaystyle CH_{3}-N\equiv N^{\oplus }Cl^{-}$$
  • $$\displaystyle CH_{3}CH\left ( CH_{3} \right )-N\equiv N ^{\oplus }Cl^{-}$$
  • $$\displaystyle C_{6}H_{5}-N\equiv N ^{\oplus }Cl^{-}$$
  • $$\displaystyle \left ( CH_{3} \right )_{3}C-N\equiv N ^{\oplus }Cl^{-}$$
Of the following statements :
$$(P)\:C_{6}H_{5}N=CH-C_{6}H_{5}$$ is a Schiff's base.
$$(Q)\:$$A dye is obtained by the reaction of aniline and $$C_{6}H_{5}N^+ \equiv NCl$$.
$$(R)\:C_{6}H_{5}CH_{2}NH_{2}$$ on treatment with $$[NaNO_{2}+HCl]$$ gives diazonium salt.
$$(S)\:p-$$Toluidine on treatment with $$[HNO_{2}+HCl]$$ gives diazonium salt.
  • only $$(P)\:and\:(Q)$$ are correct
  • only $$(P)\:and\:(R)$$ are correct
  • only $$(R)\:and\:(S)$$ are correct
  • $$(P),\:(Q)\:$$and$$\:(S)$$ are correct
Arrange the following in the increasing order of their basicities.
I.  p - Toluidine
II.  N, N - Dimethyl - p - toluidine
III.  p - Nitroaniline
IV.  Aniline
  • III < IV < I < II
  • II < IV < I < III
  • III < II < I < IV
  • I < IV < III < II
Which one(s) is/are true?
244984_4a91c8421a5843ef99c951718adf0721.png
  • (I) and (III) are modest Bronsted bases whereas (II) is not
  • In (III) $${N}^{a}$$ is more basic than $${N}^{b}$$
  • When (II) is protonated in the presence of a strong acid, protonation occurs at $$C - 2$$
  • All the nitrogen present in (I), (II), and (III) is $$s{p}^{2}$$ hybridised
Arrange the following in their decreasing order of acidity.

244989_b3f46e7b98d94e01bbdc8ed803a45d2c.png
  • $$III > IV > I > II$$
  • $$I > II > III > IV$$
  • $$IV > III > II > I$$
  • $$II > III > I > IV$$
Which of the following is the product of the reaction shown above?
206794.png
  • $$(CH_3)_2CHCH_2CH_2NHNH_2$$
  • $$[(CH_3)_2CHCH_2CH_2]_2NH$$
  • $$(CH_3)_2CHCH_2CH_2NH_2$$
  • $$(CH_3)_2CHCH_2CH_2CONH_2$$
The decreasing order of basic characters of the following is :
245041_86a21bdb1e854fd1a74329652e535599.png
  • III > IV > I > II
  • II > I > IV > III
  • IV > III > II > I
  • I > II > III > IV
Which of the following statement(s) is/are correct?
244985_4a1a58265d4940aab28debbcc7a7befa.png
  • (I) and (II) are aromatic and have equal basic strength
  • (I) is aromatic and (II) is anti-aromatic, but (II) is stronger base than (I)
  • The order of basicity of the above compounds is (IV) > (III) > (II) > (I)
  • The conjugate acid of (IV) is more stabilised than the conjugate acid of (II)
Identify the most basic amine among the above compounds :
246541_5a5d451b1658448ebf39783fa8bd5e68.png
  • i
  • ii
  • iii
  • iv
Alkyl isocyanides ($$R N^{\bigoplus} \equiv C^{\ominus}$$) are reduced to 2$$^{\circ}$$ amines($$R-NH-CH_3$$) with:
  • $$LAH$$
  • $$NaBH_4$$
  • $$HI + P$$
  • $$H_2 / Pt$$
Which statements are correct? 
  • Phenol and aniline give coupling reaction with diazonium salt.
  • Phenol couples with diazonium salt in mild basic conditions ($$\displaystyle pH = 8-10$$).
  • Aniline couples with diazonium salt in mild acidic condition ($$\displaystyle pH = 4-6$$).
  • Both phenol and aniline couple with diazonium salt in neutral condition ($$ pH = 7$$).
Compound $$(E)$$ is:
Compound $$(A)$$ is:
  • Ethylbenzene
  • $$p-Xylene$$
  • $$m-Xylene$$
  • $$o-Xylene$$
Compound $$(F)$$ is:
Compound $$(D)$$ is:
The following reaction is:
253878_ce022c735cb140a394a3fab0eb63a738.png
  • benzidine rearrangement
  • pinacol-pinacolone rearrangement
  • fries rearrangement
  • benzil-benzilic acid arrangement
The decreasing order of the rate of bromination of the following compounds is:
$$I. Ph\overset { \oplus  }{ N{ Me }_{ 3 } } $$  
$$II. Ph{CH}_{3}\overset { \oplus  }{ N{ Me }_{ 3 } } $$  
$$III. PhMe$$  
$$IV. PhN{Me}_{2}$$
  • $$(I)>(II)>(III)>(IV)$$
  • $$(IV)>(III)>(II)>(I)$$
  • $$(III)>(IV)>(I)>(II)$$
  • $$(III)>(IV)>(II)>(I)$$
Gabriel synthesis is used for the preparation of: 
  • $$\displaystyle { 1 }^{ o }$$ amine
  • $$\displaystyle { 2 }^{ o }$$ amine
  • $$\displaystyle { 3 }^{ o }$$ amine
  • all of the above
Which of the following orders regarding the basic strength of substituted aniline is correct?
  • p-methylaniline> p-chloroaniline > p-aminoacetophenone
  • p-methylaniline> p-aminoacetophenone> p-chloroaniline
  • p-aminoacetophenone> p-methylaniline> p-chloroaniline
  • p-aminoacetophenone> p-chloroaniline> p-methylaniline
Most basic compound in gaseous phase is :
  • $${NH}_{3}$$
  • $${CH}_{3}{CH}_{2}{NH}_{2}$$
  • $$({CH}_{3}{CH}_{2})_2NH$$
  • $${({CH}_{3}{CH}_{2})}_{3}N$$
Benzenediazonium chloride on reaction with aniline in a weakly basic medium gives :
  • diphenyl ether
  • p-aminoazobenzene
  • chlorobenzene
  • benzene
Which of the following compounds has the lowest boiling point?
  • 2-propanamine
  • ethylmethylamine
  • 1-propanamine
  • N,N-dimethylmethanamine
Among the following, the strongest base is :
  • $$C_6H_5NH_2$$
  • $$p-NO_2C_6H_4NH_2$$
  • $$m-NO_2C_6H_4NH_2$$
  • $$C_6H_5CH_2NH_2$$
Which of the following statement(s) is/are correct?
  • Primary amines show intermolecular hydrogen bonding.
  • Secondary amines intermolecular hydrogen bonding.
  • Tertiary amines show intermolecular hydrogen bonding.
  • Amines have lower boiling points as compared to those of alcohols and carboxylic acid of comparable molar masses.
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