Column-A Column-B Column-C (1) Alcohol (2) Aldehyde (3) Ketone (4) Carboxylic acid (a) $$-CO-$$ (b) $$-COOH$$ (c) $$-CHO$$ (d) $$-OH$$ (i) Propanal (ii) Propanone (iii) Propanol (iv) Propanoic acid Direction : Match the functional group with their formulae and examples. Column-A Column-B Column-C (1) _______ _______ (2) _______ _______ (3) _______ _______ (4) _______ _______

(1) - (d) (iii) ; (2) - (c) (i) ; (3) (a) (ii) ; (4) - (b) (iv)

(1) - (d) (iii) ; (2) - (c) (i) ; (3) (b) (ii) ; (4) - (a) (iv)

(1) - (d) (iii) ; (2) - (b) (ii) ; (3) (c) (i) ; (4) - (a) (iv)

(1) - (d) (iii) ; (2) - (a) (i) ; (3) (a) (iv) ; (4) - (c) (ii)

$$1$$-phenyl-$$1$$-acetyloxyethane

MCQ Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 4

The functional group in methanol and methanal respectively are :

0%
-OH, -CHO
0%
-CHO, -OH
0%
-OH, -COOH
0%
-CHO, -COOH

How would you name the following compounds?

(i) ${ CH }_{ 3 }-{ CH }_{ 2 }-{ Br }$

(ii) $H-CHO$

$(iii)CH_3-(CH_2)_3-CH=CH_2$

0%
(i) Bromomethane (ii) Methanol (iii) Hexene
0%
(i) Bromoethane (ii) ethanol (iii) Hexane
0%
(i) Bromopropane (ii) Methanol (iii) Hexene
0%
(i) Bromoethane (ii) Methanal (iii) Hexene

Explanation

${ CH }_{ 3 }-{ CH }_{ 2 }-{ Br }$ is bromoethane or ethyl bromide.

$H-CHO$ is methanal or formaldehyde.

$CH_3-(CH_2)_3-CH=CH_2$ is

$1-$ hexene.

Note: Methanol is

$CH_3-OH$ . 1-bromopropane is

${ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ Br }$ . Hexane is

$CH_3-(CH_2)_3-CH_2-CH_3$

The name of the compound CH

$_3$ -CH

$_2$ -CHO is :

0%
Propanal
0%
Propanone
0%
Ethanol
0%
Ethanal

Explanation

Propanal is

$CH_3 CH_2 CHO$ .

In Liebig's method for the estimation of C and H, if the compound also contains N, which of the following is kept near the exist of the combustion tube?

0%
Silver wire
0%
$PbCr_4$
0%
Both (a) and (b)
0%
CuO gauze

Explanation

Compounds containing nitrogen: During combustion, organic nitrogen is oxidized to its oxides (

$NO, NO_2$ ) which are also absorbed in KOH solution along with

$CO_2$ .

Alternatively, oxides of nitrogen are reduced to nitrogen by passing over preheated

$CuO$ gauze placed near the exit of combustion tube.

Option D is correct.

The number of carbon atoms in the principal chain of the given compound are :

0%
$7$
0%
$5$
0%
$4$
0%
$6$

Explanation

The number of carbon atoms in the principal chain of the given compound is five because the aldehyde group (here act as a substituent) and double bond has to be given the lowest number. Therefore, the numbering will start from the COOH group and ends in

$CH_3$ which will include a

$CHO$ group also.

What is the correct chemical formula for

$1, 2$ -dicholoro tetrafluroethane ?

0%
$Cl-\underset{Cl}{\underset{|} {\overset{F}{\overset {|}C}}}-\underset{Cl}{\underset{|} {\overset{F}{\overset {|}C}}}-H$
0%
$H-\underset{Cl}{\underset{|} {\overset{F}{\overset {|}C}}}-\underset{Cl}{\underset{|} {\overset{F}{\overset {|}C}}}-F$
0%
$F-\underset{F}{\underset{|} {\overset{Cl}{\overset {|}C}}}-\underset{F}{\underset{|} {\overset{F}{\overset {|}C}}}-Cl$
0%
$F-\underset { Cl }{ \underset { | }{ \overset { Cl }{ \overset { | }{ C } } } } -\underset { H }{ \underset { | }{ \overset { Cl }{ \overset { | }{ C } } } } -\underset { F }{ \underset { | }{ \overset { F }{ \overset { | }{ C } } } } -F$

Explanation

$F-\underset{F}{\underset{|} {\overset{\,\,Cl}{\overset {\,\,|1}C}}}-\underset{F}{\underset{|} {\overset{F}{\overset {|2}C}}}-Cl$

Ethane has 2

$C$ atoms. Each

$C$ atom has one chlorine, and two fluorine is attached.

The correct chemical formula for 1,2-dicholoro tetrafluroethane is represented by option

$C$

Name of the compound is :

0%
$2$ -methyl- $3$ -ethyl- $1$ -pentene
0%
$3$ -ethyl- $4$ -methyl- $4$ -pentene
0%
$3$ -ethyl- $2$ -methyl- $1$ -pentene
0%
$3$ -methyl- $2$ -ethyl- $1$ -pentene

Explanation

$5\quad \quad 4\quad \quad 3\quad 2\quad \quad 1\\ { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ C }{ =CH }_{ 2 }\quad \quad \quad \quad 3-ethyl-2-methypent-1-ene\\ \quad \quad \quad \quad |\quad \quad |\\ \quad \quad \quad { C }_{ 2 }{ H }_{ 5 }\quad { CH }_{ 3 }$

The absorption of hydrogen by platinum is known as:

0%
Hydrogenation
0%
Reduction
0%
Dehydrogenation
0%
Occlusion

$CH_3-O-\underset{\displaystyle O}{\underset{\displaystyle ||}{C}}-\underset{\displaystyle OH}{\underset{\displaystyle |}{C}}H-CH=CH-\overset{\displaystyle O}{\overset{\displaystyle ||}{C}}-OH$

How many functional groups are present in given compound?

0%
$2$
0%
$3$
0%
$4$
0%
None of the above

Column-A	Column-B	Column-C
(1) Alcohol (2) Aldehyde (3) Ketone (4) Carboxylic acid	(a) $-CO-$ (b) $-COOH$ (c) $-CHO$ (d) $-OH$	(i) Propanal (ii) Propanone (iii) Propanol (iv) Propanoic acid

Direction : Match the functional group with their formulae and examples.

Column-A Column-B Column-C

(1) _______ _______

(2) _______ _______

(3) _______ _______

(4) _______ _______

0%
(1) - (d) (iii) ; (2) - (c) (i) ; (3) (b) (ii) ; (4) - (a) (iv)
0%
(1) - (d) (iii) ; (2) - (c) (i) ; (3) (a) (ii) ; (4) - (b) (iv)
0%
(1) - (d) (iii) ; (2) - (b) (ii) ; (3) (c) (i) ; (4) - (a) (iv)
0%
(1) - (d) (iii) ; (2) - (a) (i) ; (3) (a) (iv) ; (4) - (c) (ii)

Explanation

1) Alcohol-

$-OH$ Propanol
2) Aldehyde

$-CHO$ Propanal
3) Ketone

$-CO-$ Propanone
4) Carboxylic acid

$-COOH$ Propanoic acid

In which of the following compounds, nitrogen cannotbe tested by Lassaigne's test?

0%
$CH_3CONH_2$
0%
$NH_2.NH_2.H_2O$
0%
$C_6H_5NH_2$
0%
$C_6H_5NO_2$

Explanation

Lassaigne's test:
Compound containing diazo compounds can not be tested by this so

$NH_2.NH_2.H_2O$ can not be tested by this since it does not contain C. So

$CN^{\circleddash}$ ion is not obtained which is an essential condition for the test for N.

The number of

$\pi$ and

$\sigma$ bond in the following structure respectively are:

0%
$8, 5$
0%
$8, 6$
0%
$3, 8$
0%
$5, 13$

Explanation

Clearly it has 5

$\pi$ and 13

$\sigma$ .

The catalyst used in Kjeldahl's method for the estimationof nitrogen is :

0%
Copper
0%
Magnesium
0%
Mercury
0%
Sodium

Explanation

Kjeldahl's method is a faster method than Dumas' method. However, this method is used only for those organic compounds that are converted quantitatively to ammonium sulphate on heating strongly with concentrated sulphuric acid.

It involves $CuSO_4$ or Hg as catalyst.

Kjeldahl's method cannot be used for the organic compounds,

Containing nitrogen in the ring, e.g., pyridine, quinoline etc.

Containing nitro ( $-NO_2$ ) and diazo (-N = N-) groups.

A compound which does not give a positive test inLassaigne's testfor nitrogen is :

0%
Glycine
0%
Hydrazine
0%
Urea
0%
Phenyl hydrazine

Explanation

Positive Lassaigne's test is given by containing both C and N
a.

$NH_2CONH_2$ (Contains both C and N)
b.

$NH_2-NH_2$ (Does not contain C). So, the test is not given by (b).
c. Glycine

$(NH_2-CH_2-COOH)$ . (Contains both C and N)
d.

$Ph-NH-NH_2$ (Contains both C and N)

State True or False.

$\pi$ bond is formed by the colateral overlapping of atomic orbitals.

0%
True
0%
False

Explanation

Sideways overlapping or co lateral overlapping of the two atomic orbitals take place is known as lateral overlap of their 'p' orbital, and the bond formed is known as covalent pi $(\pi)$ bond. The electron density is concentrated in the region perpendicular to the bond axis in a pi bond.

So answer is A.

In Dumas method for the estimation of nitrogen in anorganic compound, nitrogen is determined in the form of :

0%
Gaseous nitrogen
0%
Sodium cyanide
0%
Ammonium sulphate
0%
Gaseous ammonia

Explanation

Dumas' method
A known mass of the organic compound is heated with cupric oxide in an atmosphere of carbon dioxide. The carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively, while nitrogen is set free. Any oxide of nitrogen produced during this process, is reduced back to free nitrogen by heated copper gauze. The gaseous mixture consisting of

$CO_2, H_2O$ and

$N_2$ is collected over an aqueous solution of potassium hydroxide. All the gases except nitrogen are absorbed by the solution. The volume of gas (nitrogen) collected is measured. From the volume of nitrogen obtained the percentage of nitrogen in the compound is calculated.

lUPAC name of

$(CH_3)_2CHCH(CH_3)_2$ is:

0%
$2.2$ -dimethylbutane
0%
$2.3$ -dimethylbutane
0%
$2.4$ -dimethylbutane
0%
$1$ -methylpentime

Explanation

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ { CH }_{ 3 }-CH-HC-{ CH }_{ 3 }\quad\\ \quad \quad \quad |\quad \quad \quad \ \ |\\ { \quad \quad \quad CH }_{ 3 }\quad \ { CH }_{ 3 }$

Methyl substituent is present at 2nd and 3rd carbon in parent chain so prefix would be

$2,3$ -dimethyl and the parent chain contains

$4$ carbon therefore suffix would be butane.

Purity of organic liquid can be checked by its characteristic :

0%
boiling point
0%
volume
0%
solubility in water
0%
solubility in alcohol

Westrosol is also known as :

0%
ethylene dichloride
0%
acetylene tetrachloride
0%
acetylene tricholoride
0%
none of the above

Explanation

Westrosol is trichloroethylene. It is also known as acetylene trichloride.

$\underset {Westron}{CHCl_2CHCl_2} \xrightarrow {Ca(OH)_2} \underset {Westrosol}{CHCl=CCl_2}$

The common and IUPAC names of the group of

$(CH_3)_2CHCH_2-$ respectively are :

0%
Isobutyl, 2 - Methylpropyl
0%
Isobutyl, 1 - Methylpropyl
0%
Tert-butyl, 1, 1 - Dimethylethyl
0%
Sec-butyl, 2 - Methylpropyl

The IUPAC name of:

$HC \equiv C - CH_2 - \overset{\quad \,\,\,\,CH_2=CH}{CH_2 - \overset{|}{C}} = CH - \overset{H}{\overset{|}{C}} = O$

0%
3-(1-butynyl)-2,4-pentadienal
0%
5-Ethynyl-3-vinyl-2-pentenal
0%
3-ethenyl-2-hepten-6-ynal
0%
5-Acetyl-3-ethenyl-2-pentenal

Explanation

$HC \equiv C - CH_2 - \overset{\quad \,\,\,\,CH_2=CH}{CH_2 - \overset{|}{C}} = CH - \overset{H}{\overset{|}{C}} = O$

7 6 5 4 3 2 1

IUPAC name for the above compound is 3-ethenyl-2-hept-2-en-6-ynal.

In this compound, CHO is a functional group.

So answer is C.

The IUPAC name of the compound is:

0%
$1.1$ -dimethylbutane- $1,3$ -diol
0%
$1.3.3$ -trimethylpropane- $1.3$ -diol
0%
$2$ -methylpentane- $2.4$ -diol
0%
$1.3,3$ -trimethyl- $1,3$ -propanediol

Explanation

$2$ -Methylpentane

$-2,4-$ diol

IUPAC name of the following compound is:

0%
N, N-dimethyl cyclopropane carboxamide
0%
N-methyl cyclopropanamide
0%
cyclopropionamide
0%
none of the above

Explanation

IUPAC name of the following compound is

$N, N-$ Dimethyl cyclopropane carboxamide.
It is an amide which is named as carboxamide.

$N$ atom has two methyl substituents which are named as dimethyl. The substituent on amide carbonyl is cyclopropane.

The IUPAC name is:

0%
$1$ -phenyl- $1$ -acetyloxyethane
0%
$1$ -acetoxyeihylbenzene
0%
$1$ -phenylethyl ethanoate
0%
$1$ -methylbenzyl acetate

The IUPAC name of (image) :

0%
N- Cyclohexylbenzamide.
0%
N-Phenyl-N-cyclohexylmethanamide
0%
N-Phenylcyclohexanecarboxamide
0%
N-Cyclohexenyl-N-phenylmethanamid

IUPAC name of the compound is:

$CH_3-\underset{Cl}{\underset{|}{C}}=\underset{CH_3}{\underset{|}{C}}-\underset{C_2H_5}{\underset{|}{C}H}-CH_2-C\equiv CH$

0%
$6$ -Chloro- $4$ -ethyl- $5$ -methyl hept- $5$ -en- $1$ -yne
0%
$6$ -Chloro- $4$ -ethyl- $5$ -methyl hept- $1$ -yn-Hne
0%
$2$ -Chloro- $4$ -ethyl- $3$ -methyl hept- $2$ -en- $6$ -yne
0%
$2$ -Chloro- $4$ -ethyl- $3$ -methyl hept- $4,3$ -yn- $2$ -ene

Explanation

C. 2-Chloro-4-ethyl-3-methylhept-2-en-6-yne
By the functional group priority alkene and chloro group gets priority and by lowest sum rule positions are given to

$7$ carbon atoms. The naming starts from alphabetical order of chloro group then ethyl group,substituent and then positions of double and triple bond.

Amongest the following, the acceptable name as per IUPAC nomenclature is :

0%
pent-4-en-1-yne
0%
pent-1-en-4-yne
0%
2, 2-dimethyl-4-ethylpentane
0%
3-hydroxybutan-1-ol

$CH_3\,-\,CH\,=\,CH\, -\,C\, \equiv \,CH$ has lUPAC name :

0%
pent - 2 - en- 4 - yne
0%
pent - 4 - yn - 2 - ene
0%
pent - 1 - yn - 3 - ene
0%
pent - 3 - en - 1 - yne

What is IUPAC name of

$CH_3-C(CH_3)(OH)-C(O)-CH_3$ ?

0%
Heptane-1,3,5-tricarboxylic acid
0%
Pentane-1,3,5-tricarboxylic acid
0%
3-Hydroxy-3-methylbutane-2-one
0%
3-butyl-1 ,4-hexadien-6-al

The IUPAC name of the compound

$\begin{matrix} CH_ 2&-&CH&-&CH_2\\|& &|&&| \\CN& &CN& &CN \end{matrix}$ is:

0%
$1,2,3$ -Tricyanopropane
0%
$3$ -Cyanopentane- $1 ,5$ -dinitrile
0%
propane- $1,2,3$ -tricarbonitrile
0%
propan- $1,2,3$ -tricarbylamine.

Explanation

The IUPAC name of the compound

$\displaystyle \begin{matrix} CH_ 2&-&CH&-&CH_2\\|& &|&&| \\CN& &CN& &CN \end{matrix}$ is 1,2,3-Tricyanopropane.

The parent hydrocarbon contains 3 C atoms (excluding C atoms of cyano groups) and is derivative of propane. Three cyano groups are present as substituents .

The functional group, present in

$CH_3COOC_2H_5$ is :

0%
ketonic
0%
aldehydic
0%
ester
0%
carboxylic

Explanation

Ester is represented by the formula

$\quad \quad O\\ \quad \quad \parallel \\ R-C-O-R'$ . It is ethyl acetate.

The IUPAC name of the compound (Refer diagram) is :

0%
$3, 3$ - dimethyl - $1$ - hydroxy cyclohexane
0%
$1, 1$ - dimethyl - $3$ - hydroxy cyclohexane
0%
$3, 3$ - dimethyl - $1$ - cyclohexanol
0%
$1, 1$ - dimethyl - $3$ - cyclohexanol

IUPAC name of the structure is :

0%
$3,3$ -di methyl pentane dioic acid
0%
$1$ -carboxy- $2,2$ -di methyl butanoic acid
0%
$1,3$ -di carbo- $2,2$ -di methyl propane
0%
$4$ -carboxy- $3,3$ -di methyl butanoic acid

Explanation

Carboxylic acids always get the priority of getting position 1 as two acid groups are present at position 1 and 5. At position no 3 two methyl groups are present on the same carbon atom. So the IUPAC name is

$3,3$ -dimethylpentanedioic acid.

Structural formula of 3-methyl-1,4-pentanediol is :

0%
0%
0%
0%
None of the above

The general formula of an ester (where R represents an alkyl group) is :

0%
R-OH
0%
R-COOH
0%
R-COOR
0%
R-H

Explanation

Ester is formed by reaction of an acid with alcohol with formula

$\quad \quad O\\ \quad \quad \parallel \\ R-C-O-R'$ . It has sweet odor giving confirmation of ester group.

The IUPAC name of the compound shown below is (refer image) :

0%
2 - Bromo - 6 - chlorocyclohex - 1 -ene
0%
6 - Bromo - 2 - chlorocyclohexene
0%
3 - Bromo - 1 - chlorocyclohex - 1 - ene
0%
1 - Bromo - 3 - chlorocyclohexene

Which of the following compounds has wrong IUPAC name?

0%
$CH_3\,-\, CH_2\,-\, CH_2\,-\, COO\,-\, CH_2CH_3\, \quad\, \rightarrow\, Ethyl butanoate$
0%
$CH_3-\underset{CH_3}{\underset{|}{C}H}-CH_2-CHO$ $\rightarrow$ 3-Methylbutanal
0%
$CH_3-\underset{OH}{\underset{|}{C}H}-\underset{CH_3}{\underset{|}{C}H}-CH_3$ $\rightarrow$ -Methyl-3-butanol
0%
$CH_3-\underset{CH_3}{\underset{|}{C}H}-\overset{O}{\overset{||}{C}}-CH_2-CH_3$ $\rightarrow$ 2-Methyl-3-pentanone

The IUPAC name of

$(CH_3)_2CHCH_3$ is:

0%
$2$ -methylpropane
0%
$2,2$ -dimethylethane
0%
trimethylmethane
0%
none of these

Explanation

The IUPAC name of

$(CH_3)_2CHCH_3$ is

$2$ -methylpropane. The parent hydrocarbon contains

$3 C$ atoms and is named as propane. A methyl substituent is present on second

$C$ atom.

Name of some compounds are given. Which one is not in IUPAC system?

The IUPAC name of the compound having the formula

$(CH_3)_3CCH=CH_2$ is :

0%
$3,3,3$ -Trimethyl $-1-$ propene
0%
$1,1,1$ -Trimethyl- $1$ -propene
0%
$3,3$ -Dimethyl- $1$ -butene
0%
$1,1$ -Dimethyl $-2$ -butene

The IUPAC name of

$CH_3-C(CH_3)(OH)CH_2-CH(CH_3)CH_3$ is :

0%
$2,4$ -Dimethylpentan $-2-$ ol
0%
$2,4$ -Dimethylpentan $-4-$ ol
0%
$2,2$ -Dimethylbutane
0%
Butanol $-2-$ one

The IUPAC name of the following compound is

$CH_2\, =\, CH\, -\, CH(CH_3)_2$ :

0%
1, 1 - Dimethyl - 2 - propene
0%
3 - Methyl - 1 - butene
0%
2 - Vinylpropane
0%
1 - lsopropylethylene

Explanation

The IUPAC name of

$CH_{2}= CH-CH(CH_{3})_{2}$ is 3-methyl- 1-butene as the parent chain has four carbon atoms with

$C_{1}= C_{2}$ (lowest numbering to the double bond) and one methyl substituent at

$C_{3}$ .

Which of the following is correct about the number of bonds formed by atoms of different elements in acetaldehyde?

(Report your answer in the ratio of

$H:O:C$ )

0%
4:2:2
0%
4:1:2
0%
4:1:3
0%
3:3:3

Explanation

${ CH }_{ 3 }CHO-Acetaldehyde\\ No\quad of\quad atoms\quad of\quad H:O:C=4:1:2$

The IUPAC name of

$(CH_3)_2\, CHCH_2CH_2Br$ is :

0%
1 - Bromopentane
0%
2 - Methyl - 4 - bromopentane
0%
1 - Bromo - 3 - methylbutane
0%
2 - Methyl - 3 - bromopentane

Explanation

$(CH_{3})_{2}CH-CH_{2}-CH_{2}Br$ , the primary substituent is Br and thus numbering is done from right to left.Also the parent chain has four carbon atoms with a methyl substituent at

$C_{3}$ . Thus the name of the compound is; 1 - Bromo - 3 - methylbutane.

The IUPAC name of the following compound is :

$CH_2=CH-CH(CH_3)_2$

0%
$1,1$ -Dimethyl $-2-$ propene
0%
$3$ -Methyl $-1-$ butene
0%
$2$ -Vinyl propane
0%
$1$ -Isopropyl ethylene

The IUPAC name of

$(CH_3)_2CHCH_2CH_2Br$ is :

0%
$1$ -Bromopentane
0%
$2-$ Methyl $-4-$ bromo pentane
0%
$1$ -Bromo $-3-$ methyl butane
0%
$2$ -Methyl $-3-$ bromo pentane

The IUPAC name of the following compound is :

0%
3 - Ethyl - 2 - methylbut - 3 - ene
0%
2 - Ethyl - 3 - methyl but - 1 - ene
0%
2 - Methyl - 3 - ethylbut - 3 - ene
0%
3 - Methyl - 2 - ethylbut - 1 - ene

Explanation

$2-Ethyl-3-methyl but-1-ene$

Double bond gets priority than alkyl groups in the carbon chain.

The functional group in an alcohol is :

0%
$-\overset{\overset{\displaystyle O}{||}}{C}-O-$
0%
$-\overset{\overset{\displaystyle O}{||}}{C}-OH$
0%
$-OH$
0%
$-\overset{\overset{\displaystyle O}{||}}{C}-H$

The IUPAC name of the compound having the formula

$(CH_3)_3CCH\, =\, CH_2$ is :

0%
3,3,3-Trimethyl-1-propane
0%
1, 1, 1-Trimethyl-1-butene
0%
3, 3-Dimethyl-1-butene
0%
1,1- Dimethyl-3-butene

Explanation

The IUPAC name of

$(CH_{3})_{3}CCH= CH_{2}$ is 3, 3 - Dimethyl - 1 - butene as the longest parent chain has four carbon atoms with a

$C_{1}= C_{2}$ (lowest numbering to double bond) and two methyl substituents at

$C_{3}$ .

$CH_3-\underset{COOH\!\!\!\!\!}{\underset{|}{\overset{H}{\overset{|}{C}}}-}CH_2-CH_3$

The IUPAC name of compound is :

0%
Butan-3-oic acid
0%
Butan -2-oic acid
0%
3 - Methylbutanoic acid
0%
2 - Methylbutanoic acid

Explanation

$(CH_{3})(COOH)CH-CH_{2}-CH_{3}$ , the functional group is carboxylic acid;-COOH and thus lowest numbering is given to it. Thus the parent chain has four carbon atoms and a methyl substituent at

$C_{2}$ .The IUPAC name can be written as;2 - Methylbutanoic acid.

CBSE Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 4 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 4 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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