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CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 8 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Organic Compounds Containing Nitrogen
Quiz 8
$$ {C}_{2} {H}_{5} {NH}_{2} $$ is neutral to litmus.
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True
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False
Explanation
Ethylamine is a primary amine with availability of one lone pair on nitrogen atom, hence it is basic to litmus.
Amines are Lewis acids.
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True
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False
Explanation
Lewis base is an electron-pair donor and Lewis acid is an electron pair acceptor.
Hence, amines are Lewis bases.
Amides are the oxidation products of nitriles (cyanides).
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True
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False
Explanation
Nitriles can be converted to amides. This reaction can be acid or base catalyzed. In the case of acid catalysis the nitrile becomes protonated. Protonation increases the electrophilicity of the nitrile so that it will accept water, a poor nucleophile.
Hence, amides are the oxidation product of nitriles.
Mark the correct statement:
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methylamine is slightly acidic
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methylamine is less basic than ammonia
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methylamine is more basic than ammonia
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methylamine forms salts with alkalies
Explanation
methylamine is more basic than ammonia because methyl group are present inductive effect
Primary amines are less soluble than tertiary amines.
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True
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False
Explanation
Primary amines are more soluble than tertiary amines.
Which of the following is the least basic amine?
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Methylamine
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Ethylamine
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Diethylamine
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Aniline
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Benzylamine
Explanation
following compound aniline is least basic amine
$$CH_3CH_2NH_2$$ contains a basic $$NH_2$$ group, but $$CH_3CONH_2$$ does not because:
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in $$CH_3CONH_2$$, the lone pair of electron on N-atom is delocalised due to resonance
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$$CH_3CONH_2$$ is amphoteric in nature
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in $$CH_3CH_2NH_2$$, the lone pair of electrons on N-atom is delocalised due to resonance
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$$CH_3CONH_2$$ is an acidic derivative
Explanation
in $$CH_3CONH_2$$, the lone pair of electron on N-atom is delocalised due to resonance
Comparing basic strength of $$NH_3$$, $$CH_3NH_2$$ and $$C_6H_5NH_2$$, it may be concluded that:
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basic strength remains unaffected
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basic strength of $$NH_3$$ is highest
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basic strength of alkylamine is lowest
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basic strength of arylamine is lowest
Explanation
basic strength of arylamine is lowest
because of present in amine bond
Which one of the following on reduction with $${ LiAlH }_{ 4 }$$ yields a secondary amine?
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Methyl cyanide
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Nitroethane
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Methyl isocyanide
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Acetamide
Explanation
On catalytic reduction or with lithium aluminium hydride (LiAlH4) or with nascent hydrogen, alkyl isocyanide yield 2∘ amine whereas cyanide gives 1∘ amine on reduction.
hence c is correct answer
Reduction of alkyl nitrile produces:
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secondary amine
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primary amine
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tertiary amine
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amide
Explanation
The nitrile reacts with the lithium tetrahydridoaluminate in solution in ethoxyethane (diethyl ether, or just "ether") followed by treatment of the product of that reaction with dilute acid. Overall, the carbon-nitrogen triple bond is reduced to give a primary amine. Primary amines contain the -NH2 group.
hence B is correct answer
An isonitrile on reduction gives:
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$$3^oamine$$
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$$2^oamine$$
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$$1^oamine$$
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quaternary ammonium salts
Explanation
Reduction of isocyanide by LiAlH4 gives secondary amines, in which nitrogen is directly attached to one hydrogen and two alkyl groups.
Potassium phthalimide reacts with 'A' which on hydrolysis gives isopentylamine, whiat is 'A'?
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0%
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0%
Explanation
hence A is correct answer
Which on reduction does not give primary amine?
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$${ CH }_{ 3 }CN$$
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$${ C }_{ 2 }{ H }_{ 5 }NC$$
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$${ CH }_{ 3 }{ CONH }_{ 2 }$$
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All of these
Explanation
$$C_2H_5NC$$ is after reaction not give primary amine
The basicity of compounds I, II, III and IV varies in the order:
$$\underset { I }{ { CH }_{ 3 }{ NH }_{ 2 } } $$, $$\underset { II }{ { (CH }_{ 3 }{ )_{ 2 }NH } } $$, $$\underset { III }{ { (CH }_{ 3 }{ )_{ 3 }N } } $$, $$\underset { IV }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 } } $$
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I>II>III>IV
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II>I>III>IV
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III>I>II>IV
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IV>I>II>III
Explanation
$$(i)$$ The order of basicity of amines in aqueous solution is :
secondary $$>$$ primary $$>$$ tertiary
$$(ii)$$ In case of benzylamine, the phenyl ring $$(C_6H_5)$$ is slightly electron withdrawing. So, the electron density on the nitrogen atom is less in benzylamine in comparison to that of primary amine.
Hence, from the above two points, the correct order of basicity of amines is : $$(CH_{3})_{2}NH>CH_{3}NH_{2}>(CH_{3})_{3}N>C_6H_5CH_{2}NH_{2}$$
$$II$$ $$I$$ $$III$$ $$IV$$
So, option (B) is correct.
Ethyl isocyanide on reduction with sodium and alcohol gives:
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ethylamine
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propylamine
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dimethylamine
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ethyl methylamine
Explanation
Ethylamine possesses two hydrogen atoms that can form hydrogen bonds wheras diethylamine having only one hydrogen atom is capable of forming only one hydrogen bond.
State, whether the following statements are True or False:
Aniline forms phenol when treated with nitrous acid.
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True
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False
Explanation
Aniline reacts with nitrous acid from $$NaNO_2/ HCl$$
at 0-5$$^oC$$ to form a diazoniumsalt.
$$C_6H_5NH_2 \xrightarrow {NaNO_2/HCl\\\ {0-5 }^oC} C_6H_5N_2^+Cl^-$$
The given statement is false.
Ethyl nitrite on reduction with $$Sn/HCI$$ gives:
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$$C_2H_5NH_2+HNO_2$$
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$$C_2H_5NH_2+H_2O$$
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$$C_2H_5OH+NH_4OH$$
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$$C_2H_5OH+NaNO_2$$
Among the following, the strongest base is
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$$C_{6}H_{5}NH_{2}$$
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$$p-NO_{2}-C_{6}H_{4}NH_{2}$$
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$$p-CH_{2} - C_{6}H_{4}NH_{2}$$
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$$C_{6}H_{5}CH_{2}NH_{2}$$
Explanation
d>a>b>c , $$EWG (-NO_2)$$ decreases basicity.
Which of the following is least basic?
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Aniline
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$$p-Methylaniline$$
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Diphenylamine
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Triphenylamine
Explanation
Triphenyl amine is least basic as it has 3 phenyl groups thus giving more chances of electron conjugation, and -M effect.
The correct order of basicities of the following compounds is
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$$2> 1> 3> 4$$
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$$1> 3> 2> 4$$
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$$3> 1> 2> 4$$
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$$1> 2> 3> 4$$
Explanation
The correct order is 2>1>3>4
Because oh the ethyl amine is most basic than amide and 2 degree amine
Which of the following compound is the strongest base?
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Ammonia
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Aniline
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Methylamine
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N-methyl aniline
Explanation
Methyl amine is the strongest base.
The order of basic strength among the following amines in benzene solution is
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$$CH_{3}NH_{2}> (CH_{3})_{3} N> (CH_{3})_{2}NH$$
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$$(CH_{3})_{2}NH> CH_{3}NH_{2}> (CH_{3})_{3}N$$
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$$CH_{3}NH_{2}> (CH_{3})_{2}NH> (CH_{3})_{3}N$$
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$$(CH_{3})_{3}N> CH_{3}NH_{2}> (CH_{3})_{2}NH$$
Explanation
$$(CH_{3})_{2} NH \quad> CH_{3}NH_{2} \quad> (CH_{3})_{3}N$$
$$K_{b} = 5.4\times 10^{-4} \ \ 4.5\times 10^{-4} \quad 0.6\times 10^{-4}$$
Order of basicity of ethyl amines in aqueous medium is
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$$Secondry> primary> Tertiary$$
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$$primary>Secondry>Tertiary$$
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$$Secondry>Tertiary> primary$$
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$$Tertiary> primary> Secondry$$
Explanation
order of basicity in aqueous medium in methylamines is secondary>primary>tertiary. But in the case of ethylamines it is secondary>tertiary>primary.
An orange dye p-hydroxy azobenzene can be synthesized
from benzene diazonium chloride by
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Sandmeyer reaction
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Gomberg reaction
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Coupling reaction
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Gattermann reaction
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Etard reaction
Explanation
This is known as coupling reaction
Which of the following is $$ 1^0 $$ amine
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Ethylene diamine
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Dimethyl amine
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Trimethyl amine
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N-methyl aniline
Explanation
in ethylene diamine here both amine are bond with different methyl group so this is $$1^0$$ amine
A is correct answer
The basicity order of $$I, II$$ and $$III$$ is
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$$III > I > II$$
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$$I > II > III$$
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$$III > II > I$$
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$$II > III > I$$
Explanation
The basicity order will depend on the groups attached with 'N' other than benzene ring. In (III) $$-C_2H_5$$ increases the basicity order due to its +I effect. Therefore (III) has highest basicity, (II) has lowest basicity due to -I effect of $$-COCH_3$$ group.
Hence the correct order is $$(III)>(I)>(II)$$.
Option A.
Consider the following compounds, What is the correct order of basicity of the above comp
ounds?
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$$I > II > IIl$$
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$$III > I > II$$
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$$III > II > I$$
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$$I > III > II$$
Explanation
III>II>I
Compounds containing both amino and COOH groups are known as
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Diamines
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unknown
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Amino acids
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Enzymes
Explanation
An amino acid contains both an amino group, -NH2, and a carboxylic acid group, -COOH, in the same molecule. As with all acids the carbon chain is numbered so that the carbon in the -COOH group is counted as number 1.
Correct the order of increasing basicity is
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$$NH_3 < C_6H_5NH_2 < (C_2H_5)_2NH < C_2H_5NH_2 < (C_2H_5)_3N$$
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$$C_6H_5NH_2 < NH_3 < (C_2H_5)_3N < (C_2H_5)_2NH < C_2H_5NH_2$$
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$$ C_{6}H_{5}NH_{2}< NH_{3}< C_{2}H_{5}NH_{2}< (C_{2}H_{5})_{3}N< (C_{2}H_{5})_{2}NH $$
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$$C_6H_5NH_2 < (C_2H_5)_3N < NH_3 < C_2H_5NH_2 < (C_2H_5)_2NH$$
Explanation
Electron donors are bases. Electron withdrawing groups (eg, benzyl) decrease the basicity of amines and Electron donating groups (eg, alkyl) increase the acidity of amines.
so correct order is
$$C_6H_5NH_2 < (C_2H_5)_3N < NH_3 < C_2H_5NH_2 < (C_2H_5)_2NH$$
Hence D is correct answer
By the presence of a halogen atom in the ring, basic properties of aniline is
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Increased
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Decreased
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unchanged
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Doubled
Explanation
By the presence of a halogen atom in the ring, basic properties of aniline is increased because it is more electronegative so donation of electron will be easy, so basicity increases.
hence A is correct
Gabriel's phthalimide synthesis is used for the preperation of
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primary aromatic amine
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Secondary acid
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Primary aliphatic amine
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Tertiary amine
Explanation
Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. Since ethanamine is the only primary amine among the given compounds,
Hence primary aromatic amine is correct answer.
The decreasing order of the basic character of the three amines and ammonia is
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$$ NH_{3}> CH_{3}NH_{2}> C_{2}H_{5}NH_{2}> C_{6}H_{5}NH_{2} $$
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$$ C_{2}H_{5}NH_{2}> CH_{3}NH_{2}> NH_{3}> C_{6}H_{5}NH_{2} $$
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$$ C_{6}H_{5}NH_{2}> C_{2}H_{5}NH_{2}> CH_{3}NH_{2}> NH_{3} $$
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$$ CH_{3}NH_{2}>C_{2}H_{5}NH_{2}>C_{6}H_{5}NH_{2}> NH_{3} $$
Explanation
Electron donors are bases. Electron withdrawing groups (eg, benzyl) decrease the basicity of amines and Electron donating groups (eg, alkyl) increase the acidity of amines.
so The correct order of basicity of amines in
$$ C_{2}H_{5}NH_{2}> CH_{3}NH_{2}> NH_{3}> C_{6}H_{5}NH_{2} $$
Hence B is correct answer
Among the following compounds nitrobenzene , benzene, aniline and phenol, the strongest basic behaviuor in acid medium is exhibited by :
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Phenol
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Aniline
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Nitrobenzene
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Benzene
Explanation
Because the N atom in aniline has a lone pair to donate and also due to +I effect of -$$ NH_{2} $$ group
The strongest base among the following is
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0%
0%
0%
Explanation
does not have aromaticity by which the lone pair of electron of nitrogen does not delocalised in benzene ring so it will be strong base on other hand rest 3 have aromaticity i.e., the follow the huckel rule so the electron pair of Nitrogen delocalised in ring by resonance and resulting they become less basic.
Among the following the weakest base is
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$$ C_{6}H_{5}CH_{2}NH_{2} $$
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$$ C_{6}H_{5}CH_{2}NHCH_{3} $$
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$$ O_{2}N CH_{2} NH_{2} $$
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$$ CH_{3} NH CHO $$
Explanation
$$CH_3CONH_2>CH_3CH_2NH_2>C_6H_5CH_2NH_2>C_6H_5CH_2NHCH_3$$
So the weekest base is B
Arrange the following in increasing order of basicity
$$ CH_{3}NH_{2},(CH_{3})_{2}NH,C_{6}H_{5}NH_{2},(CH_{3})_{3}N $$
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$$ (CH_{3})_{3}N< (CH_{3})_{2}NH< CH_{3}NH_{2}< C_{6}H_{5}NH_{2} $$
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$$ (CH_{3})_{3}N> (CH_{3})_{2}NH> CH_{3}NH_{2}> C_{6}H_{5}NH_{2} $$
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$$ C_{6}H_{5}NH_{2}< (CH_{3})_{3}N< CH_{3}NH_{2}< (CH_{3})_{2}NH $$
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$$ C_{6}H_{5}NH_{2}> (CH_{3})_{3}N> CH_{3}NH_{2}> (CH_{3})_{2}NH $$
Explanation
The correct order is
$$ (CH_{3})_{3}N> (CH_{3})_{2}NH> CH_{3}NH_{2}> C_{6}H_{5}NH_{2} $$
Hence B is correct order
Decreasing order of basicity is
(1) $$ CH_3CONH_2$$
(2)$$ CH_{3}CH_{2}NH_2 $$
(3) $$ Ph-CH_{2}CONH_{2} $$
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$$ 1> 2> 3 $$
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$$ 2> 1> 3 $$
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$$ 3> 2>1 $$
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None of these
Explanation
The more basic is 2 because of the N lon pair is not conduct in resonance so
Hence the B is correct option
Which of the following is most basic
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$$ C_{6}H_{5}NH_{2} $$
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$$ (CH_{3})_{2}NH $$
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$$ (CH_{3})_{3}N $$
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$$ NH_{3} $$
Explanation
In general, methylamine $$CH_3NH_2$$) is a stronger base than ammonia (NH3) and dimethylamine $$((CH_3)_2NH$$ is a stronger base than methylamine. However, in water, triethylamine $$((CH_3CH_2)_3N)$$ which should be even stronger based on stronger inductive effect, is a weaker base than dimethylamine or methylamine.
Hence B is correct
In the diazotisation of aniline with sodium nitrite and hdrochloric acid, an excess of hydrochloric acid is used primarily to
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Suppress the concentration of free aniline available for coupling
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Suppress hydrolysis of phenol
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Insure a stoichiometric amount of nitrous acid
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Neutralize the base liberated
Explanation
Excess of HCl is used to convert free aniline to aniline hydrochloride. Other wise free aniline would undergo coupling reaction with benzene diazonium chloride
Hence A is correct option
Which one of the following is not a base
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$$ N_{2}H_{4} $$
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$$ N_{2}OH $$
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$$ (CH_{3})_{3}N $$
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$$ NH_{3} $$
Explanation
substance that accepts and H+ from water is considered a base Both $$NH_3$$ and
H2O are amphoteric
The correct order of increasing basic nature for the bases $$ NH_{3} $$,$$ CH_{3}NH_{2} $$ and $$ (CH_{3})_{2}NH $$ is
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$$ CH_{3}NH_{2}< NH_{3}< (CH_{3})_{2}NH $$
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$$ \left ( CH_{3} \right )_{2}NH< NH_{3}< CH_{3}NH_{2} $$
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$$ NH_{3}< CH_{3}NH_{2}< (CH_{3})_{2}NH $$
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$$ CH_{3}NH_{2}< (CH_{3})_{2}NH< NH_{3} $$
Explanation
Hence C is correct
If methyl is alkyl group, then which order of basicity is correct
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$$R_{2}NH> RNH_{2}> R_{3}N> NH_{3}$$
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$$R_{2}NH> R_{3}N> RNH_{2}> NH_{3}$$
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$$RNH_{2}> NH_{3}> R_{2}NH> R_{3}N$$
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$$NH_{3}> RNH_{2}> R_{2}NH> R_{3}N$$
Explanation
Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of the basicity of aliphatic amines should be primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like $$CH_3$$ group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than methyl group, there will be steric hindrance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from -$$CH_3$$
to$$C_2H_5$$ results in change of the order of basic strength. Thus, there is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state.
Which one less of alkaline
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0%
0%
0%
All of these
Explanation
NO2 is electron withdrawing group so the stability is decreasing
Hence A is correct answer
Complete the following reaction
$$ R NH_{2}+H_{2}SO_{4}\rightarrow $$
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$$ \left [ R NH_{3} \right ]^{+}HSO^{-}_{4} $$
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$$ \left [ R NH_{3} \right ]^{+}_{2}SO^{2-}_{4} $$
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$$ R NH_{2}.H_{2}SO_{4} $$
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No reaction
Explanation
Bases react with acid to form salt.
∵
Amines are basic in nature
∵
It forms salt on reaction with $$H_2SO_4$$
Azo dye is prepared by the coupling of pehnol and :
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0%
Diazonium chloride
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o-nitro aniline
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Benzoic acid
0%
Chlorobenzene
Explanation
Azo dye is prepared by the coupling of phenol and diazonium chloride.
Hence A is correct option
Which one of the following compound is most basic
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(A)
0%
(B)
0%
(C)
0%
All are equally basic
Explanation
In B structure the attached ethyl group is less electron withdrawing so this is most basic
B is correct option
The correct order of basicity of amines in water is
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$$ (CH_{3})_{2}NH> (CH_{3})_{3}N> CH_{3}NH_{2} $$
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$$ CH_{3}NH_{2}> (CH_{3})_{2}NH> (CH_{3})_{3}N $$
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$$ (CH_{3})_{3}N> (CH_{3})_{2}NH> CH_{3}NH_{2} $$
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$$ (CH_{3})_{3}N> CH_{3}NH_{2}> (CH_{3})_{2}NH $$
Explanation
$$(CH_3)_2NH >(CH_3)_3N>CH_3NH_2$$
As basicity of amines in water is determined by the tendency of donating electrons and 3degree amines will have the highest +I effect Leading to greater electron density followed by 2degree and primary amines
Which of the following is/are correct?
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$$PhOH$$ has a higher boiling point than $$PhSH$$ (benzenthiol).
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$$p-Hydroquinone$$ has a higher melting point than catechol.
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$$o-Nitrophenol$$ has a lower boiling point than $$m-$$ and $$p-isomers$$.
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$$o-Hydroxybenzaldehyde$$ is more water soluble than $$m-$$ and $$p-isomers$$.
Explanation
$$a.$$ $$PhOH$$ has higher boiling point due to $$H-bonding$$.
$$b.$$ Molecules of $$p-benzonoquinone$$ can fit closer in solid state causing it to have higher melting point.
$$c. and \,d.$$ Intramolecular H-bonding (chelation) in the o-isomers inhibits the intermolecular attraction, lowers the boiling point, reduces H-bonding with $$H_2O$$, and decreases water solubilities. Intra-molecular chelation dots not occur in m- and $$p-isomers$$.
Hence, Options "A" ,"B" & "C" are correct answers.
Which is the weakest base among the followings?
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0%
0%
0%
Explanation
Lewis base = The entity that has the tendency to donate electrons,
Here in $$4th$$ compound, lone pairs of $$N$$ is not going in resonance but two $$Methyl(-CH_3)$$ group are sterically hindered it to donate electrons, that's why it is weakest base.
The correct most basicity in amines is
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$$ C_{4}H_{5}NH_{2} $$
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$$ CH_{3}NH_{2} $$
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$$ (CH_{3})_{2}NH $$
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$$ (CH_{3})_{3}N $$
Explanation
The observed order in the case of lower members is found to be as secondary > primary > tertiary. This anomalous behavior of tertiary amines is due to steric factors i.e. crowding of alkyl groups cover nitrogen atom from all sides and thus makes it unable for protonation. Thus the relative strength is in order CH3NH2<(CH3)2NH>(CH3)3N
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