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CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 1 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Organic Compounds Containing Nitrogen
Quiz 1
Which of the following is the most stable diazonium salt?
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$$C_6H_5CH_2N_2^+X^-$$
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$$CH_3N_2^+X^-$$
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$$CH_3CH_2N_2^+X^-$$
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$$C_6H_5N_2^+X^-$$
Explanation
Aliphatic amines do not give diazo compounds. The most stable diazonium salt is $$C_6H_5 N_2^+ X^-$$. Diazonium salt containing aryl group directly linked to the nitrogen atom is most stable due to resonance stabilization between the benzene nucleus and N-atom.
Which of the following amines yields foul smelling product with haloform and alcoholic $$KOH$$?
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Ethylamine
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Diethyl amine
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Triethylamine
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Ethyl methylamine
Explanation
Primary amine react with haloform and $$alc.KOH$$ and forms carbylamine which gives foul smell.
Which of the following is a secondary amine ?
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Aniline
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Diphenyl amine
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Sec. butyl amine
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Tert. butyl amine
Explanation
Diphenyl amine is aromatic secondary amine. N atom is attached to two aryl groups and one H atom. Aniline is aromatic primary amine.
Sec. butyl amine and tert. butyl amine are aliphatic primary amines.
Which of the following will not undergo diazotisation ?
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m-toluidine
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Aniline
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p-aminophenol
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Benzyl amine
Explanation
Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.
Which of the N atoms is least basic in the given compound ?
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a
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b
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c
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d
Explanation
Electron pair of $$NH_{2}$$ attached to benzene undergoes resonance i.e lone pair on Nitrogen is delocalised and hence less basic than other amine groups.
Which of the following is the least basic?
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0%
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All are equally basic
Explanation
Since $$NO_{2}$$ is strong withdrawing group than $$OCH_{3}$$ and $$C_{6}H_{5}$$ , hence
withdraw more electron density from benzene and leads to lone pair of $$NH_{2}$$ to be dicolocalised at faster rate.
Which is most basic?
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Which of the following compound is expected to be most basic
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Aniline
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Methylamine
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Hydroxylamine
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Ethylamine
Explanation
Due to +ve I.E. of alkyl group, N-atom of amines acquires patrial -ve charge and thus electron pair is easily donated. Hence most basic compound is Ethylamine ($$C_2H_5NH_2$$).
Option D is correct.
Which of the following is obtained by reducing methyl cyanide with $$Na+ C_2H_5OH$$?
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Methyl alcohol
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Acetic acid
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Ethyl amine
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Methane
Explanation
Sodium in alcohol is a strong reducing agent and reduces nitriles to amines.
$$CH_3-CN+Na+C_2H_5OH\rightarrow C_2H_5-NH_2$$
This reaction is also known as Mendius Reduction.
What is the_decreasing order of basicity of pri., sec., tert., ethyl amines and $$NH_3?$$
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$$NH_3>C_2H_5NH_2>(C_2H_5)_2NH>(C_2H_5)_3N$$
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$$(C_2H_5)_3N>(C_2H_5)_2NH>C_2H_5NH_2>NH_3$$
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$$(C_2H_5)_2NH>C_2H_5NH_2>NH_3>(C_2H_5)_3N>NH_3$$
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$$(C_2H_5)_2NH>C_2H_5NH_2>NH_3>(C_2H_5)_3N$$
Explanation
$$(C_2H_5)_2NH>C_2H_5NH_2>NH_3>(C_2H_5)_3N$$
this option is correct answer
The correct increasing order of basic strengths in,
$$CH_3CH_2CN$$,
$$CH_3CH_2NH_2$$,
$4CH_3N=CHCH_3$$ is:
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$$CH_3N=CHCH_3,CH_3CH_2NH_2,CH_3CH_2CN$$
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$$CH_3CH_2NH_2,CH_3N=CHCH_3,CH_3CH_2CN$$
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$$CH_3CH_2CN,CH_3N=CHCH_3,CH_3CH_2NH_2$$
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$$CH_3CH_2CN,CH_3CH_2NH_2,CH_3N=CHCH_3$$
Explanation
$$CH_3CH_2CN,CH_3N=CHCH_3,CH_3CH_2NH_2$$
the correct order is
The boiling points of amines and their corresponding alcohols and acids vary in the order:
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$$RCH_2NH_2>RCOOH>RCH_2OH$$
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$$RCH_2NH_2>RCH_2OH>RCOOH$$
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$$RCH_2NH_2$$ $$<$$ $$RCOOH$$ $$<$$ $$RCH_2OH$$
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$$RCH_2NH_2$$ $$<$$ $$RCH_2OH$$ $$<$$ $$RCOOH$$
Explanation
$$RCH_2NH_2$$ $$<$$ $$RCH_2OH$$ $$<$$ $$RCOOH$$
Hence D is correct
The correct order of basic nature of $$CH_3NH_2,(CH_3)_2NH,(CH_3)_3N$$ and $$NH_3$$ is:
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$$CH_3NH_2>(CH_3)_2NH>(CH_3)_3N>NH_3$$
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$$(CH_3)_3N>(CH_3)_2NH>CH_3NH_2>NH_3$$
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$$(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3$$
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$$NH_3>(CH_3)_3N>CH_3NH_2>(CH_3)_2NH$$
Explanation
$$(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3$$
correct answer
Choose the correct order for the boiling points of amines:
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$${ CH }_{ 3 }\ddot { N } { H }_{ 2 }>({ CH }_{ 3 })_{ 2 }\ddot { N } H>(CH_{ 3 })_{ 3 }\ddot { N } $$
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$$({ CH }_{ 3 })_{ 2 }\ddot { N } H>(CH_{ 3 })_{ 3 }\ddot { N } >{ CH }_{ 3 }\ddot { N } { H }_{ 2 }$$
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$$(CH_{ 3 })_{ 3 }\ddot { N } <{ CH }_{ 3 }\ddot { N } { H }_{ 2 }<({ CH }_{ 3 })_{ 2 }\ddot { N } H$$
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$$C{ H }_{ 3 }\ddot { N } { H }_{ 2 }<({ CH }_{ 3 })_{ 2 }\ddot { N } H<(CH_{ 3 })_{ 3 }\ddot { N } $$
Explanation
$${ CH }_{ 3 }\ddot { N } { H }_{ 2 }>({ CH }_{ 3 })_{ 2 }\ddot { N } H>(CH_{ 3 })_{ 3 }\ddot { N } $$
Among following the weakest base is:
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$$C_6H_5CH_2NH_2$$
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$$C_6H_5CH_2NHCH_3$$
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$$O_2N-CH_2NH_2$$
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$$CH_3NHCHO$$
Explanation
$$O_2N-CH_2NH_2$$ No2 is electron withdrowing group so this is week base
Which has the highest $$pK_b$$ value?
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$$R_3C-NH_2$$
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$$R_2NH$$
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$$RNH_2$$
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$$NH_3$$
Explanation
when a compound attech with alkyl compound so the pKb value is less so amine NH3 have highest value of Pkb
The correct order of basic nature in aqueous solution is:
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$$C_6H_5NH_2>NH_3>CH_3NH_2>(CH_3)_2NH$$
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$$NH_3>C_6H_5NH_2>CH_3NH_2>(CH_3)_2NH$$
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$$(CH_3)_2NH>CH_3NH_2>NH_3>C_6H_5NH_2$$
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$$CH_3NH_2>(CH_3)_2NH>NH_3>C_6H_5NH_2$$
Explanation
$$(CH_3)_2NH>CH_3NH_2>NH_3>C_6H_5NH_2$$
the secondry amine is most basic than primary then amine other molecule
Which one of the following is most basic?
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$$FCH_2NH_2$$
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$$FCH_2CH_2NH_2$$
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$$C_6H_5NH_2$$
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$$C_6H_5CH_2NH_2$$
Explanation
benzene structure is more resonance and increase the basicity so ateech benzene with CH3Nh2 is more basic
hence D is correct answer
Which of the following amines would undergoes diazotisation
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Primary aliphatic amines
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Primary aromatics amines
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Both (a) and (b)
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None of these
Explanation
Only primary aromatic amines can undergo diazotization.
The correct increasing order of basic strength of the labeled N is?
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1 < 2 < 3 <4
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4 < 3 < 2 < 1
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1 < 3 < 4 < 2
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1 < 2 < 4 < 3
Amines are basic in nature because
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They produce $$OH^-$$ ions when treated with $$H_2O$$
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They have polar H atom attached to nitrogen atom
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They have lone pair of electrons on nitrogen atom
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They have alkyl group attached with nitrogen atom
The most stable diazonium salt is ........
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$${ CH }_{ 3 }N_{ 2 }X$$
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$${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }X$$
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$${ C }_{ 2 }{ H }_{ 5 }{ N }_{ 2 }X$$
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$${ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }N_2X$$
Which of the following order for basic strength of amines is correct?
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$$I > III > IV > II$$
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$$IV > I > II > III$$
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$$I > IV > II > III$$
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$$I > II > IV > III$$
In the following compound the order of basicity is ?
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$$IV>I>III>II$$
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$$III>I>IV>II$$
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$$II>I>III>IV$$
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$$I>III>II>IV$$
Among the following amines, the strongest Bronsted base is
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0%
0%
0%
Which of the following order of basic character are correct ?
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$$ p-CH_3O.C_6H_4NH_2 > o-CH_3O.C_6H_4NH_2 > m-CH_3O.C_6H_4NH_2 $$
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$$ p-CH_3O.C_6H_4NH_2 > m-CH_3O.C_6H_4NH_2 > o-CH_3O.C_6H_4NH_2 $$
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$$ o-CH_3O.C_6H_4NH_2 > p-CH_3O.C_6H_4NH_2 > m-CH_3O.C_6H_4NH_2 $$
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$$ o-CH_3O.C_6H_4NH_2 > m-CH_3O.C_6H_4NH_2 > p-CH_3O.C_6H_4NH_2 $$
The correct increasing order of basic strength of the labelled N ?
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II
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IV
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I
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III
$$HCONR\xrightarrow[pyridine ]{POCl_3}(a)+H_2O$$; (a) in the above reaction is:
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$$RCH=NOH$$
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$$R-N=C=O$$
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$$R-C=N$$
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$$R-N=C$$
The correct order of decreasing basicity of the given compounds is :
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I $$>$$ II $$>$$ III $$>$$ IV
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II $$>$$ I $$>$$ IV $$>$$ III
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III $$>$$ IV $$>$$ II $$>$$ I
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II $$>$$ I $$>$$ III $$>$$ IV
Amoung the following compounds, the increasing order of their basic strength is :
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$$(I) > (II) < (IV) < (III)$$
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$$(I) > (II) < (III) < (IV)$$
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$$(II) > (I) < (IV) < (III)$$
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$$(II) > (I) < (III) < (IV)$$
What is the decreasing order of basicity ?
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$${ NH }_{ 3 }>{ C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }>\left( { C }_{ 2 }H_{ 5 } \right) _{ 2 }NH>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 3 }N$$
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$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{3 }N>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }>{ NH }_{ 3 }$$
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$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 3 }N>{ NH }_{ 3 }$$
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$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }>{ NH }_{ 3 }>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 3 }N$$
The relationship between the values of osmotic pressure of 0.1 M solutions of $${ KNO }_{ 3 }\left( { P }_{ 1 } \right) { CH }_{ 3 }COOH\left( { P }_{ 2 } \right) $$ is
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$${ p }_{ 1 }>{ p }_{ 2 }$$
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$${ p }_{ 2 }>{ p }_{ 1 }$$
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$${ p }_{ 1 }={ p }_{ 2 }$$
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$$\dfrac { { p }_{ 1 } }{ { p }_{ 1 }+{ p }_{ 2 } } =\dfrac { { p }_{ 2 } }{ { p }_{ 1 }+{ 2p }_{ 2 } } $$
$$RX$$ can be converted into $$R{CH}_{2}{NH}_{2}$$ by treatment with $$KCN$$ and
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reduction with reducing agent as $$Na$$ in alcohol
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hydrolysis with dil. $$HCl$$
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hydrolysis with alkaline $${H}_{2}{O}_{2}$$
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none of these
Which one of the following is the strongest base in aqueous solution?
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Dimethylamine
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Methylamine
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Trimethylamine
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None of these
Diazonium salts are used
:
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after storing for a few days after preparation
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immediately after its preparation
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after storing for $$2-3$$ hours
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difficult to prepare commercially
Explanation
Diazonium salts are light sensitive and break down under near UV
or violet light.This is why it is used immedietly after being prepared so that it does not loses its properties.
A primary amine {B), on treatment with a solution of $$AuCl_{3}$$ in conc. $$HCl$$ yields a compound whose formula would be?
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$$(BH)^{+}.AuCl_{4}^{-}$$
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$$(BH_{2})^{+}.AuCl_{4}^{-2}$$
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$$(BH)^{+}.HAuCl_{4}^{-}$$
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$$B.AuCl_{4}$$
Explanation
$$B+AuCl_3 \xrightarrow []{HCl}(BH)^+AuCl_4^-$$ .
Which of the following is NOT a correct method of the preparation of benzylamine from cyanobenzene?
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(i) $$HCl/ H_{2}O$$ (ii) $$NaBH_{4}$$
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(i) $$LiAlH_{4}$$ (ii) $$H_{3}O^{+}$$
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(i) $$SnCl_{2} + HCl(gas)$$ (ii) $$NaBH_{4}$$
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$$H_{2}/ Ni$$
Explanation
Hence option A is correct.
Considering the basic strength of amines in aqueous solution, which one has the smallest $$ pK_{b}$$ value?
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$$ (CH_{3})_{3}N$$
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$$ C_{6}H_{5}NH_{2}$$
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$$ (CH_{3})_{2}NH$$
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$$ CH_{3}NH_{2}$$
In the following reactions, the major product W is:
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0%
0%
0%
Explanation
Th major product W is (E)-1-(phenyldiazenyl)naphthalen-2-ol. Th first step is the diazotization of benzene to form benzene diazonium chloride. The second step is azo coupling.
Consider the following four compounds I, II, III, and IV.
Choose the correct statement(s).
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The order of basicity is $$II > I > III > IV$$.
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The magnitude of $$pK_b$$ difference between $$I$$ and $$ II$$ is more than that between $$ III$$ and $$IV.$$
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Resonance effect is more in $$III$$ than in $$IV.$$
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Steric effect makes compound $$ IV$$ more basic than $$III.$$
Explanation
The correct basic strength order is
(A) $$K_0 : IV > II > I > III$$:
$$\bullet$$ IV is stronger base due to SIR effect
$$\bullet$$ III is the weakest base due to $$-M$$ group of three nitro group present at Ortho and Para positions.
$$\bullet$$ II is stronger than I since III is tertiary and I primary aromatic amine
(B) IV is found to be $$40,000$$ times more basic than III while I and II differ vary little in basic strength
(C, D) Due to SIR effect in IV both $$-NO_2$$ and $$N(CH_3)_2$$ will be out of plane, hence resonance effect is more in III than in IV
The correct statement regarding the basicity of arylamines is:
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arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring $$\pi$$ electron system.
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arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring $$\pi$$ electron system.
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arylamines are generally more basic than alkylamines because of aryl group.
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arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized.
Explanation
Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring $$\pi$$ electron system which makes electrons less available. While in alkyl amines, the $$+I$$ electron releasing effect of the alkyl group makes it strongly basic. Thus, option A is correct.
Which of the following will be most stable diazonium salt $$RN_2^+X^-$$?
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$$CH_3N_2^+X^-$$
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$$C_6H_5N_2^+X^-$$
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$$CH_3CH_2N_2^+X^-$$
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$$C_6H_5CH_2N_2^+X^-$$
Explanation
Among the given diazonium salts, benzene diazonium halide $$C_6H_5N_2^+X^-$$ is most stable due to conjugation of the $$N \equiv N$$ triple bond with benzene ring.
What is the decreasing order of strengths of the following bases?
$$\overset { \circleddash }{ O } H, N{ H }_{ 2 }^{ \circleddash }, HC\equiv { C }^{ \circleddash }$$ and $$C{ H }_{ 3 } - C{ H }_{ 2 }^{ \circleddash }$$
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$$C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash } > N{ H }_{ 2 }^{ \circleddash } > H - C\equiv { C }^{ \circleddash } > \overset { \circleddash }{ O } H$$
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$$H -C\equiv { C }^{ \circleddash } > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash } > N{ H }_{ 2 }^{ \circleddash }>\overset { \circleddash }{ O } H$$
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$$\overset { \circleddash }{ O } H > N{ H }_{ 2 }^{ \circleddash } > H - C\equiv { C }^{ \circleddash } > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash }$$
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$$N{ H }_{ 2 }^{ \circleddash } > H -C\equiv { C }^{ \circleddash } > \overset { \circleddash }{ O } H > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash }$$
Explanation
The conjugate base of strong acid is weaker and conjugate base of weak acid is stronger. So, d
ecreasing order of basic strength:
$${ CH }_{ 3 }-{ CH }_{ 2 }^{ - }$$>
$${ NH }_{ 2 }^{ - }$$>$${ CH }\equiv { C }^{ - }$$>
$${ OH }^{ - }$$
Which of the following compounds is most basic?
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0%
0%
0%
Explanation
$$\bf{Hint:}$$ Compound in which $$N$$ lone pair is not involved in conjugation are more basic.
$$\bf{Correct \ answer:}$$ Option $$B$$
$$\bf{Explanation \ for \ correct \ option:}$$
The basic strength of an a-amino group-containing compound depends on how easily $$N$$ atom of $$-NH_{2}$$ group share its lone pair electrons.
If these lone pairs are taking part in conjugation with a ring or an electron-withdrawing group attached with the compound then it becomes least basic because the lone pair is not available for sharing to act as a base.
In compound $$B$$, the lone pair of $$N$$ atom of $$-NH_{2}$$ group not involved in resonance with ring neither an electron-withdrawing group present there, hence, it is the most basic compound among all given compounds.
$$\bf{Explanation \ for \ incorrect \ options:}$$
In compound $$A$$, the lone pair of $$N$$ atom of $$-NH_{2}$$ group involved in resonance with the ring as well as an electron-withdrawing $$-NO_{2}$$ group present there, hence, its basic strength reduced.
In compound $$C$$, the lone pair of $$N$$ atom of $$-NH_{2}$$ group involved in resonance with the ring as well as adjacent $$CO$$, hence, its basic strength is reduced. and it is the least basic of all.
In compound $$D$$, the lone pair of $$N$$ atom of $$-NH_{2}$$ group involved in resonance with the ring, hence, its basic strength is reduced.
The decreasing order of basicity is shown as below;
$$B > D > A > C $$
In the following reaction, the product (A) is:
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0%
0%
Explanation
In the given coupling reaction, compound A is yellow dye represented by the structure given in option D.
It is an electrophilic substitution reaction.
The coupling reaction of aniline takes place at the para-position to $$NH_2$$ group in the benzene nucleus gives azodye.
The major product of the following reaction is:
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0%
0%
0%
The correct order of the basic strength of methyl substitited amines in aqueous solution is :
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$$(CH_3)_2NH> CH_3NH_2> (CH_3)_3N$$
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$$(CH_3)_3N> CH_3NH_2>(CH_3)_2NH$$
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$$(CH_3)_3N> (CH_3)_2NH>CH_3NH_2$$
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$$CH_3NH_2>(CH_3)_2NH> (CH_3)_3N$$
Explanation
In aqueous solution, electron donating inductive effect, solvation effect (H-bonding) and steric hindrance all together affect basic strength of substituted amines.
Basic character:
$$\underset{2^0}{(CH_3)_2NH}> \underset{1^0}{CH_3NH_2}>\underset{3^0}{(CH_3)_3N}$$
In the given replacement reaction, the reaction will be most favourable if $$M$$ happens to be:
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$$K$$
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$$Rb$$
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$$Li$$
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$$Na$$
Explanation
Reaction is faster with $$Rb$$ because lattice of $$RbF$$ is less than $$LiF, NaF, KF$$e energy
Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?
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$${H}_{2}(excess)/Pt$$
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$$LiAl{H}_{4}$$ in ether
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$$Fe$$ and $$HCl$$
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$$Sn$$ and $$HCl$$
Explanation
$$LiAl{ H }_{ 4 }/ether$$ reduces aryl nitro compounds to azo compounds
$$2{C}_{6}{H}_{5}{NO}_{2} \xrightarrow [ ]{ LiAl{ H }_{ 4 } } {C}_{6}{H}_{5}N=N-{C}_{6}{H}_{5}$$
nitrobenzene azobenzene
The correct decreasing order of $$pK_{b}$$ is:
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$$I > II > III > IV$$
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$$III > IV > III > I$$
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$$II > III > IV > I$$
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$$IV > II > I > III$$
Explanation
Substituent with strong $$+R$$ effect, $$+I$$ effect and weaker $$-I$$ effect increases the basicity.
Hence $$pK_{b}$$ decreases.
Also, alkylamines are stronger bases than arylamines.
So, the order is option D.
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Practice Class 12 Medical Chemistry Quiz Questions and Answers
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