MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 6 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Organic Compounds Containing Nitrogen
Quiz 6
Which is most basic in aqueous solution?
Report Question
0%
$${ CH }_{ 3 }-{ NH }_{ 2 }$$
0%
$${ (CH }_{ 3 }{ ) }_{ 2 }{ NH }$$
0%
$${ (CH }_{ 3 }{ ) }_{ 3 }N$$
0%
$$ph-{ NH }_{ 2 }$$
Explanation
$${ CH }_{ 3 }\rightarrow \begin{matrix} N \\ \uparrow \\ { CH }_{ 3 } \end{matrix}\leftarrow { CH }_{ 3 }$$
$$+I $$ effects of Three methyl group
Here, it is 3 amine and three methyl group attached with N-atom. The inductive effect of three methyl group and lypez conjugation effect of $$9-\alpha -H$$ atom, So this is more stable and basic in nature.
Which of the following is most basic ?
Report Question
0%
0%
0%
0%
Explanation
Here, lone pair of electrons are available on the $$N-$$ atom and it can easily donate electrons and $$pK_b$$ value is also low.
Which of the following diazonium salt is relatively stable at $$0-{ 5^0 }^{ }C$$?
Report Question
0%
$${ CH }_{ 3 }-N\equiv N{ \} }^{ \oplus }C{ l }^{ - }$$
0%
$${ \left( { CH }_{ 3 } \right) }_{ 2 }CH-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$$
0%
$${ C }_{ 6 }{ H }_{ 5 }-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$$
0%
$${ \left( { CH }_{ 3 } \right) }_{ 3 }C-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$$
Explanation
The $$+I$$ effect which stabilizes the $${ N }^{ + }$$ion is in the following order $${ C }_{ 6 }{ H }_{ 5 }-<{ CH }_{ 3 }-<{ \left( { CH }_{ 3 } \right) }_{ 2 }-<{ \left( { CH }_{ 3 } \right) }_{ 3 }$$
$$\therefore $$ The $${ \left\{ { \left( { CH }_{ 3 } \right) }_{ 3 }C-N\equiv N \right\} }^{ + }{ Cl }^{ - }$$ is most stable
Select the basic strength order of following molecules.
Report Question
0%
$$III > II > I$$
0%
$$II > III > I$$
0%
$$I > III > II$$
0%
$$III > I > II$$
Explanation
Since in (III) lone pair is only involved in delocalisation, in (II) nitrogen is attached to the carbonyl (withdrawing group) and one alkyne group (donating group+I effect) which have the more basic character them the (I) in which (N) is
attached to the two group i.e benzene and carbonyl.
Which of the following conditions is most suitable for the above reaction?
Report Question
0%
strongly acidic
0%
strongly alkaline
0%
Slightly alkaine
0%
neutral
Explanation
In this reaction, primary aromatic amine is treated with nitrous acid ($${ HNO }_{ 2 }$$).
$${ HNO }_{ 2 }$$ can be obtained by dissolving solid
$${ NaNO }_{ 2 }$$ in aqueous acidic solution.
Which of the following will accept $$H^{+}$$ from $$NH_{4}^{+}$$ ion?
Report Question
0%
0%
0%
0%
$$CH_{3}-CH_{2}-NH_{2}$$
Explanation
Since the aromatic amines (aniline) are less basic than aliphatic amines.
Because lone pair of nitrogen in aromatic amines are involved in resonance.
$$H-COOH > Ph-COOH > CH_{3}-COOH$$ (acidic nature)$$CF_{3}-COOH > CCI_{3}-COOH > CH_{2}-COOH$$ (Acidic nature)$$N- methylmethanamine > methanamine > N, N- dimethylmethanamine $$ (Basic nature in aqueous phase)$$N,N- diethlethanamine > N-ethylethanamine > Ethanamine $$ (Basic nature in aqueous phase)
Then the correct order is:
Report Question
0%
$$i,ii,iv$$
0%
$$ii,iii,iv$$
0%
$$i,ii,iii$$
0%
$$i,ii,iii,iv$$
Explanation
In aqueous phase, the basicity of amines depends upon:
$$(i)$$ Inductive effect: which fovours tertiary amine to be more basic due to greater $$+I$$ effect of $$3$$alkyl group directly attached to nitrogen.
$$(ii)$$ Solution Effect: Which favours primary amine to be more basic due to greater solution of its cation.
$$(iii)$$ Steric Factor: Which cause tertiary amine to be less basic due to steric hinderance .
Decreasing order of basicity of the three isomers of nitro aniline is:
Report Question
0%
$$p-nitroaniline> o-nitroaniline> m-nitroaniline$$
0%
$$p-nitroaniline> m-nitroaniline> o-nitroaniline$$
0%
$$m-nitroaniline> p-nitroaniline> o-nitroaniline$$
0%
$$m-nitroaniline> o-nitroaniline> p-nitroaniline$$
Explanation
Nitro group is an electron-withdrawing group. It mainly withdraws electrons from the ortho and para positions. Therefore para isomer is less basic than meta isomer.
Ortho substituted anilines are generally weaker bases than aniline irrespective of the electron releasing or electron-withdrawing nature of the substituent.
Hence decreasing order of basicity of the isomers of nitroaniline:
$$m-nitroaniline>p-nitroaniline>o-nitroaniline$$
The order of basic strength of the given basic nitrogen atoms is :
Report Question
0%
$$III > II > I > IV$$
0%
$$III > I > II > IV$$
0%
$$I > III > II > IV$$
0%
$$II > III > I > IV$$
Explanation
Since $$(II)$$ is more basic then $$(III)$$ due $$(II)$$ nitrogen get negative charge due resonance hence it is more basic.
There for correct order is $$II$$>$$III$$>$$I$$>$$IV$$
D)
What is the major organic product of the following reaction?
$$ { CH }_{ 3 }{ CH }_{ 2 }{ NHCH }_{ 2 }{ CH }_{ 3 }\longrightarrow $$?
Report Question
0%
0%
0%
0%
0%
The correct sequence regarding base strength of aliphatic amines in aqueous solution is:
Report Question
0%
$${R}_{3}N> {R}_{2}NH> R{NH}_{2}> {NH}_{3}$$
0%
$${R}_{2}NH> R{NH}_{2}> {R}_{3}N> {NH}_{3}$$
0%
$${R}_{2}NH> {R}_{3}N> R{NH}_{2}> {NH}_{3}$$
0%
$$R{NH}_{2}> {R}_{2}NH> {R}_{3}N> {NH}_{3}$$
Explanation
Since the bascities of amine depend upon the no of alkyl group directly attached to the nitrogen directly due to $$+I$$ effect to alkyl group increased the bascities of amine.
The correct order of basic strength of the compounds?
Report Question
0%
$$I > II > III$$
0%
$$II > I > III$$
0%
$$III > II > I$$
0%
$$II > III > I$$
Explanation
the strength of basic nature of compounds depends on the ability to easily donate the lone pair of electrons.
IN $$(i)$$ lone pair is easily approachable so it has more basicity . in $$(ii)$$ the lone pair is in different plane. while in $$(iii)$$ the lone pair is in conjugation . hence order of basicity is $$I>II>III$$
The product 'Y' in the following reaction sequence is:
Report Question
0%
0%
0%
0%
Which of the following Statements is/are correct?
Report Question
0%
$$(SiH_3)_3N $$ is weaker base than $$(CH_3)_3N$$.
0%
Amongst $$Cl-Cl$$, $$N-F$$,$$C-F$$ And $$O-F$$ bond is most polar
0%
$$OF_2$$ has higher dipole moment than $$H_2O$$
0%
An alchohol is less volatile than other having same molecular formula on account of intermolecular hydrogen bonding
Explanation
$$A.$$ $${(SiH_3)}_3N$$ is a weaker base than $${(CH_3)}_3N$$.
In $${(CH_3)}_3N$$ the lone pair is concentrated on the $$N-atom$$. But in case of Trisillyl amine, the lone pair on nitrogen gets delocalised on the $$3$$ silicon atoms bonded to it.
Since, base strength is a measure of tendency to donate lone pair, trimethylamine is stranger base.
$$D.$$ In ethers there is no intermolecular hydrogen bond, where as in alcohol, intermolecular hydrogen bond is present which makes them less volatile and have high boiling point.
This reaction is example of:
Report Question
0%
Intermolecular $$C-N$$ coupling
0%
Intramolecular $$C-N$$ coupling
0%
Intermolecular $$N-N$$ coupling
0%
Intramolecular $$N-N$$ coupling
Explanation
So this is example of $$N-N$$ intra molecular coupling.
$$sp^3-$$hybridised nitrogen is present in:
Report Question
0%
0%
0%
$$CH_2 = CH - \overset{\oplus}{N}H_3$$
0%
Explanation
A) $$Sp^{2}$$ hybridized Nitrogen
B) $$S p^{2}$$ hybridized Nitrogen due to participation
of lone pair into ring.
c) $$C H_{2}=C H-NH_3+ \quad\left(S p^{3}\right.$$ hybridised nitrogen $$)$$
D) (sp $$^{2}$$ hybridised) as nitrogen is in conjugation with double bond.
So, option C) is correct.
The increasing order of basicity of the given compounds is:
Report Question
0%
(b)< (a)< (c)< (d)
0%
(b)< (a)< (d)< (c)
0%
(d)< (b)< (a)< (c)
0%
(a)< (b)< (c)< (d)
Explanation
Order of base nature depends on electron donation tendency.
In compound (b) nitrogen is $$sp^{ 2 }$$ hybridized so least basic among all given compound
Compound (C) is a very strong nitrogeneous organic base as a lone pair of one nitrogen delocalize in resonance and make another nitrogen negatively charged and conjugate acid have two equivalent resonating structure.
Thus it is most basic in given compound.
Compound (d) is secondary amine more basic than compound (a) which is a primary amine.
Therefore, increasing order of basicity is
(b) < (a) < (d) < (c)
The correct order of increasing basic nature for the bases $$NH_3,CH_3NH_2$$ and $$(CH_3)_2NH $$ is?
Report Question
0%
$$(CH_3)_2NH < NH_3 < CH_3NH_2$$
0%
$$NH_3 < CH_3 NH_2 <(CH3)_2NH $$
0%
$$CH_3NH_2 <(CH_3)_2NH < NH_3$$
0%
$$CH_3NH_2 < NH_3 < (CH_3)_2NH$$
Explanation
$$(H_3C)_2NH$$ , $$CH_3NH_2$$ , $$NH_3$$
Among the above molecules $$(CH_3)_2 NH$$ is most basic as two electron donating group are attached to it which increase its basicity is
$$NH_3 < CH_3NH_2 < (CH_3)_2NH$$
The end product (Z) of the following reaction is?
Report Question
0%
A cyanide
0%
A carboxylic acis
0%
An amine
0%
An arene
By heating ammonium chloride with two equivalents of formaldehyde it forms:
Report Question
0%
dimethylamine
0%
ethylamine
0%
methylamine
0%
ammonium formate
Explanation
$$6HCHO + 4N{H}_{4}Cl \rightarrow ({C{H}_{2}}_{6}){N}_{4} + 4HCl + 6{H}_{2}O$$
So Methylamine is formed
Find product in the following reaction is Y. What is Y?
Report Question
0%
benzamide
0%
benzophenone
0%
benzoic acid
0%
benzaldehyde
Explanation
In the given reaction, reduction of the cyanide group takes place followed by steam distillation to form benzaldehyde as a product.
$$CH_3Cl \xrightarrow{AgCN} A$$ (major) $$\xrightarrow{H-2O^+} C + D$$ here '$$C$$' is formic acid '$$D$$ is:'
Report Question
0%
$$2^o$$ amine
0%
Carboxylic acid
0%
Alkyl Cyanide
0%
$$1^o$$ amine
Explanation
$$\text { Hence correct Answer - D }$$
Reductive amination of $$A$$ forms:
Report Question
0%
0%
0%
0%
The reactivities of these ions in azo-coupling reactions will be in the order:
Report Question
0%
$$3<1<4<2$$
0%
$$1<4<2<3$$
0%
$$1<2<3<4$$
0%
$$3<1<2<4$$
Explanation
In azo-coupling, the diazonium ion should be more electrophilic for more reactivity. Presence of electron-withdrawing groups increase the reactivity.
Considering these factors, the order of reactivity should be:
$$1<2<3<4$$
In the above reaction sequence (D) is?
Report Question
0%
0%
0%
0%
Suitable explanation for the order of basic character, $$(CH_{3})_{3}N < (CH_{3})_{2}NH$$, is:
Report Question
0%
steric hindrance by bulky methyl group
0%
higher volatility of $$3^{\circ}$$ amine
0%
stability of conjugate acid due to solvation
0%
all of these
Explanation
$${ ({ CH }_{ 3 }) }_{ 3 }$$ N is less basic due to steric hindrance by bulky methyl group prevent any acid to take a lone pair of nitrogen.
With which one of the following substrate do diazonium salt reacts to form a colourful compound?
Report Question
0%
0%
0%
0%
Explanation
The answer is refer to image.
In the reaction
$$Me_{3}N \xrightarrow{BrCN} A \xrightarrow[ii)\Delta]{i)H_{2}O} B$$.
B is
Report Question
0%
$$MeNH_{2}$$
0%
$$Me_{3}NO$$
0%
$$Me_{2}NH$$
0%
$$Me_{2}NCOOH$$
Explanation
$$Me_3N\xrightarrow {BrCN}{}\underset{A}{Me_2N}-CN$$
$$Me_2N-Cn\xrightarrow[\triangle ]{H_2O}Me_2NCOOH$$
The reagents/conditions K, L and M are respectively:
Report Question
0%
$$Ba(OH)_{2}$$, $$KCN$$ and $$CHCl_{3} / NaOH$$ heat, $$H_{3}O^{+}$$
0%
$$H_{2}O$$ (boil), $$CO_{2} / KOH$$ and $$Ac_{2}O - AcONa$$, heat
0%
Steam ; $$CHCl_{3} / NaOH$$, heat, $$H_{3}O^{+}$$ and $$AC_{2}O - AcONa$$, heat
0%
$$Cu_{2}Cl_{2} / HCl$$, $$NaOH$$ / high P and $$CO_{2}|NaOH$$, heat
Explanation
$$ \text { Honre option } A \text { is correct } $$
The correct order of boiling points of the following amines $${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 },\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH,{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$ is :
Report Question
0%
$${ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$
0%
$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$
0%
$${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2} >\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$
0%
$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$
Explanation
Because $${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$ is primary amine and can form hydrogen bonding more than secondary amine $$({ { C }_{ 2 }{ H }_{ 5 } })_{ 2 }NH$$ and tertiary amine $${ C }_{ 2 }{ H }_{ 5 }N{ ({ CH }_{ 3 }) }_{ 2 }$$ .More hydrogen bonding leads to strong bonding between the molecules, hence increases the boiling point.
Which of the following compounds is an imine?
Report Question
0%
0%
0%
0%
Explanation
The structure of imine is the given figure.
The nitrogen atom can be attached to hydrogen or an organic group.
Hence compound $$B$$ is an imine.
$${ C }_{ 3 }{ H }_{ 9 }N$$ represents
Report Question
0%
primary amine
0%
secondary amine
0%
tertiary amine
0%
all of these
Explanation
this compound have all 3 type of amine
The correct order of basicity of the above compounds is:
Report Question
0%
II > I > IIl > IV
0%
I > III > II > IV
0%
III > I > II > IV
0%
I > II > III > IV
Explanation
(Refer to Image)
The order of basicity is follows
$$I >III >II >IV$$
$$\rightarrow $$ in $$IV$$ electron withdrawing is attached to amine. So, it is least basic.
Basic nature order of the given compounds is
(a) $$CH_3 - CH = \ddot{N}H $$
(b) $$CH_2= CH - CH = CH -NH_2$$
(c) $$CH_3 -NH - CH_3$$
Report Question
0%
c>a>b
0%
a>b>c
0%
a>c>b
0%
c=a=b
Explanation
$$C H_{2}=C H-C H=C H-N H_{2}(b)$$
In this compound lone pair on $$-N H_{2}$$ group
is involued in resonance. so givening lone
Pair is difficult. So this compound
has less basicity. than other two.
$$\mathrm{CH}_{3}\mathrm{CH}=\ddot{\mathrm{N}}\mathrm{H}(\mathrm{O})$$
$$ \begin{array}{l} \text { Nitrogen in }^{\prime}=\ddot{N} H^{\prime} \text { group is more electro. } \\ \text { negitiven than }-N H-\text { is } C H_{3}-N H-C H_{3}(C) \\ \text { due to double bond. } \\ \text { More electronegitive Nitrogen in ' }=\text { NH 'grap } \\ \text { cannot tolarate positive charge of it (when } \\ \text { lone pair is given it get positive charge) } \end{array} $$
$$ \begin{array}{l} \text { Compared to Nitrogen in }-\mathrm{NH} \text { - group } \\ \text { So } \mathrm{CH}_{3}-\mathrm{NH}-\mathrm{Ct}_{3} \text { is more basic than } \\ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{NH} \text { (a) } \\ \text { Hence Basisity order is } c>a>b \\ \text { so option } A \text { is correct. } \end{array} $$
Arrange these compounds in decreasing order of basic strength?
Report Question
0%
i > ii > iii > iv
0%
ii > iii > i > iv
0%
iv > i > iii > ii
0%
iv > i > ii > iii
Explanation
$$iv>i>iii>ii$$
Electron - donating substituents increase the electron density of N, thereby increasing the Lewis basicity. Amongst $$NO_2$$ and $$CN,NO_2$$ is slightly resonance stabilization.
Hence, the answer is $$iv>i>iii>ii.$$
Which of the following is the strongest base?
Report Question
0%
0%
0%
0%
Explanation
Aniline is the strongest base here. This is because, $$NH_2$$ is stabilized by conjugation and resonance stabilised . More stability implies stronger the base.
The decreasing order of the basic character at $$a, b, c, d$$ in the compound:
Report Question
0%
$$a > c > b > d$$
0%
$$d > c > b > a$$
0%
$$c > b > d > a$$
0%
$$b > c > d > a$$
Explanation
$$d>c>b>a.$$
Due to ease of delocalisation of electron pair is least in $$(d)$$ Nitrogen. For $$(a), (b), (c)$$ Nitrogen atoms , basic character order is governed by electronic effects of rings and $$-CH_3$$ group
Arrange the following in order of decreasing basic strength.
Report Question
0%
IV > I > II > III
0%
IV > I > III > II
0%
IV > II > I > III
0%
I > II > III > IV
Explanation
$$IV>I>II>III$$
Which of the following is weak base than aniline
?
Report Question
0%
0%
0%
0%
Explanation
An electron withdrawing groups decrease the availability of a lone pair of electrons on $$N-$$atom of aniline thereby decreasing its basicity.
Hence it is less basic than aniline.
What is strong base?
Report Question
0%
0%
0%
0%
An organic compound (A) $$C_9H_{13}N$$ dissolves in dil. HCI and releases $$N_2$$ with $$HNO_2$$ giving an optically active alcohol. Alcohol on oxidation gives dicarboxylic acid, which on heating form anhydride. The organic compound 'A' is :-
Report Question
0%
0%
0%
0%
The relative basics strenghts of $$NH_{3},CH_{3}NH_{2}$$ and $$NF_{3}$$ are in the order:
Report Question
0%
$$CH_{3}NH_{2} > NH_{3}>NF_3$$
0%
$$NF_{3} > CH_{3}NH_{2} > NH_{3}$$
0%
$$NH_{3} > CH_{3}NH_{2} > NF_{3}$$
0%
$$NF_{3} > CH_{3}NH_{2} = NH_{3}$$
Explanation
$$CH_3-$$ group is electron donating group.
$$F$$ is electron withdrawing group.
So $$CH_3NH_2$$ is more basic. $$NH_3$$ is more basic than $$NF_3$$ because $$H$$ is not a withdrawing group. $$NF_3$$ is less basic.
$$\therefore$$ Basic nature order= $$CH_3NH_2 > NH_3 >NF_3$$
Acetic anhydride with ammonia gives the product:
Report Question
0%
$$ CH_{3}CONH_{2} $$
0%
$$ CH_{3}CONHCH_{3} $$
0%
$$ CH_{3}CN$$
0%
$$ CH_{3}COONH_{4} $$
What will be the final product of the given reaction?
Report Question
0%
0%
0%
0%
Explanation
Diazonium chloride on treating with $$HF/{ BF }_{ 3 }$$ yields benzene diazonium fluoroborate. Benzene diazonium fluoroborate when heated undergoes decomposition reaction to give fluorobenzene.
Option B is correct.
which compound is Prepare propan -I- amine from?
Report Question
0%
Butaniteuie
0%
I-Nitropropane
0%
Propanamide
0%
Butanamide
Explanation
Propan-1-amine can be prepared by propanamide by treating it with an acid anhydride and then reducing the nitrile compound formed with the help of Lithium Aluminum Anhydride.
The major product of the reaction is:
Report Question
0%
0%
0%
0%
Explanation
(Refer to Image)
In presence of $$NaNO_2$$, diazotisation reaction takes place and $$NH_2$$ group leaves as $$-N_2$$. Then methyl group shift takes place, which makes the carboction more positive than former. Then oxygen stabilise the carbocation by $$+M$$ effect and ketone formation takes place.
Adrenaline and ephedrine contain:
Report Question
0%
$${1}^{o}$$ amino group
0%
$${2}^{o}$$ amino group
0%
$${3}^{o}$$ amino group
0%
quaternary amino group
Explanation
Thus Adrenaline and ephedrine contain $$2^0$$ amino group.
Sanger's reagent is used for the identification of:
Report Question
0%
N-thermal of peptide chain
0%
C-terminal of a peptide chain
0%
side chain of amino acids
0%
number of amino acids in peptide chain
Explanation
1-Fluoro-2,4-dinitrobenzene. Undergoes nucleophilic aromatic substitution (NAS) with the N-terminal amino group of a peptide or protein, resulting in a chemical tag on the terminal amino acid residue. After hydrolysis into individual amino acids, the peptide's or protein's terminal amino acid can be identified by the specific wavelength of light absorbed (i.e., colorimetric analysis).
So option $$A$$ is correct
Among the following which is not correct?
Report Question
0%
P methyl benzyl carbonyl is more stable than ethyl benzyl carbonyl
0%
Aniline is more basic than pyridine
0%
Cyclopentadienyl anion is more stable than cyclopentadienyl Carbo cation
0%
K of Ethene 1 ol is more than that of propene 2 en 1 ol
$${C}_{6}{H}_{5}-{NO}_{2}\xrightarrow [ ]{ Fe/HCl } A\xrightarrow [ H{ NO }_{ 3 } ]{ 273K } B\xrightarrow [ ]{ { C }_{ 6 }{ H }_{ 5 }OH } C$$, $$C$$ is
Report Question
0%
0%
0%
0%
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Medical Chemistry Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
