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CBSE Questions for Class 12 Medical Chemistry Aldehydes, Ketones And Carboxylic Acids Quiz 2 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Aldehydes, Ketones And Carboxylic Acids
Quiz 2
.
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0%
oxidation, chlorination
0%
reduction, chlorination
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oxidation, addition
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reduction, substtution
Explanation
The first step is oxidation with suitable oxidising agent $$(CrO_2Cl_2/CS_2)$$ that will convert alcohol to aldehyde.
Second step is chlorination that will substitute the hydrogen of aldehyde in the presence of $$(Cl_2/hv)$$.
General formula of saturated monocarboxylic acid is
:
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$$C_{n}H_{2n+1} COOH$$
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$$C_{n}H_{2n}O_{2}$$
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Both A and B
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None of the above
Explanation
Monocarboxylic acids are in the form of $$RCOOH$$.
Let's take examples,
For R is $$-H$$ or $$-CH_{3}$$,
a) $$HCOOH\to C_{1}H_{2}O_{2}$$ which is following $$C_n H_{2n} O_2$$ trend.
b) $$CH_{3}COOH\to C_{2}H_{4}O_{2}$$ which is also following $$C_n H_{2n} O_2$$ trend.
When R is other than $$-H$$ or $$-CH_{3}$$ group,
a) $$C_{2}H_{5}COOH\to C_{3}H_{6}O_{2}$$ which is following $$ C_n H_{2n+1} COOH$$ trend.
So, the general formula is both A & B, i.e., $$C_n H_{2n} O_2$$ and $$ C_n H_{2n+1} COOH$$.
Hence, both options A and B are correct.
Option C is correct.
Methyl cyanide on hydrolysis gives :
Report Question
0%
acetic acid.
0%
acetaldehyde.
0%
acetone.
0%
methyl amine.
Explanation
Methyl cynide on hydrolysis gives Acetic Acid.
Hence option A is correct.
$$CH_{3}CH_{2}OH+O_{2}\xrightarrow[]{X}$$ Vinegar
In the above reaction, $$X$$ is:
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0%
$$(CH_{3}COO)_{2}Mn$$
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$$CH_{3}COOH$$
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mycoderma aceti
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$$K_{2}Cr_{2}O_{7}/H^{\bigoplus}$$
A compound of general formula $$C_{n}H_{2n}O_{2}$$ could
be
:
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0%
an acid
0%
a diketone
0%
an ether
0%
an aldehyde
Explanation
Formula i $$C_n H_{2n} O_2$$ (H is double of C)
i) $$acid: CH_3 COOH \cdot C_n H_{2n} O_2$$
diketane:
ether: $$CH_3 OCH_3\cdot single oxygen$$
aldehyde : $$ CH_3 CHO\cdot single oxygen$$
$$CH_{3}Cl \xrightarrow[]{KCN} A \xrightarrow[]{H_{3}O^+} B$$
In this reaction, the product B is
:
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0%
$$CH_{3}COOC_{2}H_{5}$$
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$$CH_{3}COOH$$
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$$HCOOH$$
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$$CH_{3}CONH_{2}$$
Explanation
The complete reaction sequence is as follows:
$$CH_{3}Cl \xrightarrow[]{KCN} CH_3CN \xrightarrow[]{H_{3}O^+} CH_3COOH$$
Nitriles are hydrolysed to amides and then to acids in the presence of $$H^+$$ or $$OH^-$$ as the catalyst. Mild reaction conditions are used to stop the reaction at the amide stage.
Hence, $$A$$ is $$CH_3CN$$ and $$B$$ is $$CH_3COOH$$.
Option B is correct.
Toluene$$\xrightarrow[]{KMnO_{4}/KOH/H_{3}O^{\bigoplus }}$$ A.
What is A?
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0%
Acetice acid
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Benzene
0%
Benzoic acid
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Benzaldehyde
Explanation
Assertion (A) : The solubility of aldehydes and ketones in water decreases with increase of size of the alkyl group
Reason (R) Alkyl groups are electron releasing groups
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Both A and R ae true and R is the correct
explanation of A
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Both A and R are true and R is not the
correct explanation of A
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A is true but R is false
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A is false but R is true.
Explanation
B
The solubility decreases due to the increase in bulkiness as due to this steric hindrance around the carbonyl group increases so the water molecules can not interact with the carbonyl group easily and less tendency of formation of hydrogen bond.
Which of the following is not a monovalent group?
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0%
Aldehydic
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Ketonic
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Carboxylic
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Hydroxy
Explanation
Valency is the number of chemical bonds formed by the functional group. Aldehyde, carboxylic acid and hydroxy groups form one chemical bond, whereas ketone can form two chemical bonds. Therefore, ketonic group is not monovalent.
$$R-CO-R$$ is ketone functional group, which is bivalent.
Oxidation product of X with molecular formula $$C_{2}H_{4}O$$ is Y with molecular formula $$C_{2}H_{4}O_{2}$$.
The compound Y is :
Report Question
0%
acetic acid
0%
formic acid
0%
propionic acid
0%
buteric acid
Explanation
A
X:- Ethanal
Y:- Ethanoic acid
In the manufacture of acetic acid from aerial oxidation of acetaldehyde, the catalyst used is :
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acidified $$K_{2}Cr_{2}O_{7}$$
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$$(CH_{3}COO)_{2}Mn$$
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$$HgSO_{4}$$
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$$Ni$$
Explanation
B
Preparation of carboxylic acid from acetaldehyde
$$CH_3CHO + \dfrac{1}{2} O_2 \xrightarrow {Mn^{2+}}CH_3COOH$$
Acetaldehyde Acetic Acid
Isobutyraldehyde on oxidation gives:
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2-Methylpropanoic acid
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2-Methylpropanone
0%
Propanoic acid
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2-Methyl butanoic acid
Explanation
The aldehydes gets oxidised to carboxylic acid.
Which of the following compounds does not have a carboxyl group?
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Methanoic acid
0%
Ethanoic acid
0%
Picric acid
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Benzoic acid
Assertion (A) : $$CH_{3}CN$$ on hydrolysis gives Acetic Acid
Reason (R) : Cyanides on hydrolysis liberates $$NH_{3}$$ gas
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0%
Both A and R ae true and R is the correct explanation of A
0%
Both A and R are true and R is not
the correct explanation of A
0%
A is true but R is false
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A is false but R is true.
Explanation
A
R is the correct explanation of A.
When acetonitrile is hydrolysed the product formed is ________.
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$$CH_{3}COOH$$
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$$C_{2}H_{5}COOH$$
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$$CH_{3}CONH_{2}$$
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$$C_{2}H_{2}NH_{2}$$
Explanation
Acetonitrile is $$CH_{3}CN$$.
An aldehyde on oxidation gives
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a carboxylic acid
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an alcohol
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an ether
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a ketone
Explanation
Alcohols on oxidation give aldehydes while aldehydes on further oxidation give carboxylic acids.
In aldehydes and ketones, carbonyl carbon shows _______ hybridisation.
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$$sp$$
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$${ sp }^{ 3 }$$
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$${ sp }^{ 2 }$$
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$${ sp }^{ 3 }d$$
Explanation
The carbonyl carbon in aldehydes and ketones forms $$sp^2$$ hybridization, as the Carbon and the oxygen atoms form a sigma bond in the molecule by overlapping the $$sp^2$$ hybrid orbitals. The pi bond between the carbon and oxygen is the 2p-2p overlap.
$$X$$ is:
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0%
0%
0%
0%
Explanation
It is an example of oxidative cleavage. The $$C=C$$ double bond breaks and C atoms attached to double bond are oxidised to carboxylic groups.
Consider the reaction:
$$RCHO + NH_{2}NH_{2} \rightarrow RCH = N NH_{2}$$
What sort of reaction is it?
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0%
Free radical addition elimination reaction
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Electrophilic substitution-elimination reaction
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Nucleophilic addition elimination reaction
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Electrophilic addition elimination reaction
Explanation
$$R-CHO+N{ H }_{ 2 }-N{ H }_{ 2 }\longrightarrow RC{ H }=NN{ H }_{ 2 }$$
The above reaction is Wolf Kirchner reduction.
Here $$N{ H }_{ 2 }-{ NH }^{ \ominus }$$ is nucleophile added to aldehyde and water molecule is eliminated from the compound hence it is nucleophilic addition elimination reaction.
The correct structural formula of butanoic acid is
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$$H-\overset{H}{\overset{|}{\underset{H}{\underset{|}{C}}}}-\overset{H}{\overset{|}{C}}=\overset{H}{\overset{|}{C}}-\overset{O}{\overset{||}{C}}-OH$$
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$$H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{O}{\overset{||}{C}}}}-OH$$
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$$H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-OH$$
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$$H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\overset{O}{\overset{||}{C}}-OH$$
Explanation
Butanoic acid is a carboxylic
acid
with the
structural
formula CH
3
CH
2
CH
2
-COOH.
where $$-COOH$$ is the functional group.
$$H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\overset{O}{\overset{||}{C}}-OH$$
So, the correct representation of butanoic acid is $$D$$
$$CH_{3}CH_{2}CH_{2}CHO\overset{x}{\rightarrow}CH_{3}CH_{2}CH_{2}COOH$$
In the above reaction X is an oxidising agent and X is__________.
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0%
Tollens reagent
0%
Fehlings reagent
0%
$$H NO_{3}$$
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All the above
Explanation
D
The functional group present in carboxylic acids is ?
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-COOH
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-CO-
0%
-OH
0%
-CHO
Explanation
Carboxylic acids are represented by -COOH- functional group. It ionizes to give carboxylate ions and hydrogen ions. This group present in a compound makes it an acid and if present with other groups in the compound it is always present at position no.1 and carbon is included in the carbon chain atoms.
The acid in which $$-COOH$$ group is (are) present ?
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ethanoic acid
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picnic acid
0%
lactic acid
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palmitic acid
Explanation
Formula of Ethanoic acid is: $$CH_{3}COOH$$
Formula of Lactic acid is: $$C_{2}H_{4}OHCOOH$$
Formula of Palmitic acid is: $$CH_{3}(CH_{2})_{14}COOH$$
All have -COOH as the functional group and are thus acids.
On vigorous oxidation by permanganate solution $$(CH_{3})_{2}C=CH-CH_{2}CHO$$ gives :
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$$(CH_{3})_{2}CO$$ and $$OHC-CH_{2}-CHO$$
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$$(CH_3)_3COH$$ and $$HCHO$$
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$$(CH_{3})_{2}CO$$ and $$OHC-CH_{2}-COOH$$
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$$(CH_{3})_{2}CO$$ and $$CH_{2}(COOH)_{2}$$
Explanation
On vigorous oxidation by permanganate solution $$(CH_3)_2C=CH-CH_2CHO$$ gives $$(CH_3)_2CO$$ and $$CH_2(COOH)_2$$.
The $$C=C$$ bond is cleared and oxidised to $$-COOH, -CHO$$ group is also oxidised to $$-COOH$$.
$$(CH_3)_2C=CH-CH_2CHO\xrightarrow[KMnO_4]{[O]} (CH_3)_2CO+COOH-CH_2-COOH$$
Due to vigorous oxidation the $$\pi$$ bond breaks and oxidation takes place forming ketone and acid.
The IUPAC name for the compound $$CH_3-C(CH_3)=CH-COOH$$ is:
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2-Methyl-2-butenoic acid
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3-Methyl-3-butenoic acid
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3-Methyl-2-butenoic acid
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2-Methyl-3-butenoic acid
Explanation
The numbering is done from right-hand side as carboxylic acid is present in right-hand side. There is double bond present at the C-2 position and methyl group at C-3 position.
Thus, the IUPAC name is: 3
-methyl-2-butenoic acid
What is the common name of propanone?
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0%
Oxobutane
0%
Acetone
0%
Ethyl acetone
0%
Ethyl methyl ketone
Explanation
The IUPAC name of $$CH_{3}COCH_{3}$$ is propanone and the common name is acetone.
The general formula of carboxylic acids is:
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$$C_{n}H_{n}COOH$$
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$$C_{n}H_{2n}COOH$$
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$$C_{n}H_{2n-2}COOH$$
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$$C_{n}H_{2n+1}COOH$$
Explanation
General formula of acids is: $$C_{n}H_{2n+1}COOH$$.
For example: for n=1 we get $$CH_3-COOH$$ which is acetic acid.
IUPAC name of the compound $$CH_3-CH_2-\overset{\overset{O}{||}}C-CH_2-\overset{\overset{O}{||}}C-COOH$$ is:
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0%
hexanedione - 2, 4-ic-1-oic acid
0%
2, 4- dioxohexanoic acid
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propionyl-dioxo-oic acid
0%
none of these
Explanation
For naming the given compound ; $$H_{3}C-CH_{2}-CO-CH_{2}-CO-COOH$$, the following steps are followed:
i) The parent chain consists of six carbon atoms and the primary functional group is -COOH, thus the suffix -oic acid.
ii) The two secondary functional groups are -CO, thus the prefix -oxo is used for them.
iii) Numbering the compound is done from right to left, as lowest number is given to the -COOH group and also in accordance with the least sum rule for secondary functional groups.
Thus the name of the compound is
2,4-dioxohexanoic acid.
The secondary suffix, '-one' indicates the following functional group in the compound.
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-COOH
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0%
0%
- OH
Explanation
Suffix" -one " is used for the ketone functional group. Therefore RCOR is the correct answer.
$$O$$
$$||$$
$$X\xrightarrow[(ii)\,H_3O^{\oplus}]{(i)\, KMnO_4/H_2O/\Delta}HOOC - C-CH_2-CH_2-CH_2-CH_2-COOH$$.
$$X$$ is :
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0%
0%
0%
0%
Explanation
The acidified $$KMnO_4$$ solution oxidises the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds. Under mild conditions,
$$KMnO_4$$
can affect conversion of alkenes to glycols. That further oxidizes the glycol with cleavage of the carbon-carbon bond and forms acyclic dicarboxylic acid with substituted keto group.
Hence, X is represented by option A (Benzoic acid)
IUPAC name of $$CH_3COOH$$ is:
Report Question
0%
acetic acid
0%
formic acid
0%
methanoic acid
0%
ethanoic acid
Explanation
It has 2 carbon atoms, and functional group is carboxylic group,suffix is "oic acid"
thus IUPAC name is Ethanoic acid
The IUPAC name of carboxylic acid with three carbon atoms is:
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0%
ethanoic acid
0%
acetic acid
0%
propanoic acid
0%
propionic acid
Explanation
According to $$IUPAC$$ convention, acid group is given highest priority than saturated carbons. So the name would be propanoic acid since the number of carbons atom is three, a prefix is $$(propan-)$$
What is the IUPAC name for the following compound?
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0%
3-methylpentanoic acid
0%
isohexanoic acid
0%
$$\beta$$-methylvaleric acid
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2-methylpentanoic acid
Explanation
The IUPAC name of the given compound is 3-methylpentanoic acid.
The parent compound is a carboxylic acid with five carbon atoms.
It is known as pentanoic acid.
A methyl group is present as a substituent at a third carbon atom.
The IUPAC name of above structure is :
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0%
phenyl ethanone
0%
methyl phenyl ketone
0%
acetophenone
0%
phenyl methyl ketone
Explanation
The IUPAC name of given compound is phenyl ethanone. Its common name is acetophenone.
Alkyl aryl ketones are named as aryl alkanone.
Here, alkyl group is methyl group and aryl group is phenyl group.
Therefore, the option is A.
In addition reactions of aldehydes, the carbonyl carbon atom changes from:
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$$sp^{3}\ to\ sp^{2}$$
0%
$$sp^{2}\ to\ sp^{3}$$
0%
$$sp\ to\ sp^{3}$$
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$$sp^{3}\ to\ sp$$
Explanation
the carbonyl carbon atom has one $$\Pi$$ bond, thus one empty p-orbital, therefore $$sp^2$$ hybridisation.
on addition reaction $$\Pi$$ bond is lost and empty p-orbital is used up. Thus hybridisation is $$sp^3$$.
change is from $$sp^2$$ to $$sp^3$$.
The name of the compound is cyclohexylidene methanone.
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0%
True
0%
False
Explanation
(i)Main chain is that which contain functional group.
(ii)If side chain attached to carbon of main chain by double bond,then the name of chain end with suffix.
(iii)Functional group is tetone means suffix is one.
$$\displaystyle CH_{3}CHO \xrightarrow[( ii )HCl\Delta ]{( i )Zn / Hg}$$
Write the product :
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0%
Methane
0%
Ethane
0%
Propane
0%
Butane
Explanation
Refer to image
Heating of carboxyl compound with finely divide, amalgamated zinc in a hydroxylic solvent containing a mineral acid such as $$HCl$$. The mercury alloyed with the zinc does not participate in the reaction it reverse only to provide a clean active metal surface.
The name of the compound is Cyclopropanecarboxylic acid.
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0%
True
0%
False
Explanation
(i)First count number of carbon in ring and simply acid cyclo as a prefix.
(ii)If there is functional group then functional group has lower number >name of compound>cyclopropanecarboxylic acid.
Cyanohydrin of which compound on hydrolysis will give lactic acid?
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0%
$$\displaystyle C_{6}H_{5}CHO$$
0%
$$HCHO$$
0%
$$\displaystyle CH_{3}CHO$$
0%
$$\displaystyle CH_{3}CH_{2}CHO$$
Explanation
The hydrolysis of acetaldehyde cyanohydrin gives lactic acid.
Which of the following names are incorrect?
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0%
2-chloropentanoic acid
0%
2-methylhex-3-enoic acid
0%
$$\displaystyle CH_{3}CH_{2}CH=CHCOCH_{3}$$ ; hex-3-en-2-one
0%
1-methylpentanal
Explanation
1-methylpentanal is not possible but 2
-methylpentanal is
The IUPAC name of $$CH_{3}COCH(CH_{3})_{2}$$ is:
Report Question
0%
4-methylisopropyl ketone
0%
3-Methyl-2-butanone
0%
Isopropylmethyl ketone
0%
2-methyl-3-butanone
Explanation
The IUPAC name of this structure is 3-Methyl-2-butanone.
The longest chain is of four carbon. The functional group here is keto group.
The numbering will start from left carbon as the functional group should also have the lowest numbering.
The substituent methyl group will be at third carbon.
Hence t
he IUPAC name of this structure is 3-Methyl-2-butanone.
IUPAC name of $$CH_3CH(CH_3)COOH$$ is :
Report Question
0%
butyric acid
0%
propionic acid
0%
2-methylpropanoic acid
0%
dimethylacetic acid
Explanation
As it contains 4 carbon and functional group is carboxylic acid so its name is 2-methyl propanoic acid.
$$\displaystyle PhCHO\xrightarrow{SF_{4}}B$$
This reaction is complete at $$160^o C$$
so product B will be ?
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0%
$$\displaystyle PhCHF_{3}$$
0%
$$\displaystyle PhCHF_{4}$$
0%
$$\displaystyle PhCHF_{2}$$
0%
none
Which of the following is most reactive in a hydrolysis reaction to make butanoic acid?
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0%
Butanoic anhydride
0%
Methyl butanoate
0%
N-methylbutanamide
0%
Butanenitrile
Explanation
Due to the higher level of unsaturation, n
itriles are highly reactive to hydrolysis reaction and forms acid as shown above :
Hence, the correct option is D.
The correct statement regarding hydrolysis of a nitrile to a carboxylic acid is (are)
:
$$RCN + H_2O \longrightarrow RCOOH$$
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the reaction passes through an amide intermediate
0%
both acid and base catalyzes the hydrolysis reaction.
0%
intermediate product is impractical to separate
0%
the reaction does not work for the preparation of an aromatic acid.
What is /are true regarding the following reaction?
$$CH_3CH_2CHO \overset{HCN}{\longrightarrow} \quad \xrightarrow{Dil\, H_2SO_4} Acid$$
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The product is an $$\alpha$$-hydroxy acid with one carbon greater than the starting aldehyde.
0%
The product is racemic mixture of $$\alpha$$-hydroxy acid.
0%
Adding a small amount of NaCN in the first step catalyzes the reaction.
0%
The first step can also be affected using base as catalyst followed by acidification of product
Out of the following the one that undergoes reaction with 50% NaOH solution to give corresponding alcohol and acid is:
Report Question
0%
phenol
0%
benzaldehyde
0%
butanol
0%
benzoic acid
Explanation
The reaction of benzaldehyde to form alcohol and acid can be written as:
$$C_{6}H_{5}CHO \xrightarrow[H_{3}O^{-}]{NaOH} C_{6}H_{5}COOH + C_{6}H_{5}CH_{2}OH$$
The reaction is also called as Cannizzaro's reaction. This reaction can be given by only those aldehyde which does not contain alpha carbon.
Consider the following reaction:
The m-isomer of D and E is called :
Report Question
0%
Phthalic acid
0%
Isophthalic acid
0%
Terephthalic acid
0%
None
Explanation
Alkyl group containing Benzene generally get oxidized on treatment with $$KMnO_4$$ solution.
The reaction
only works
if there is a hydrogen attached to the benzylic carbon.
The position directly adjacent to an aromatic group is called the benzylic position. If we treat m-methyl toluene and p-methyl toluene with $$KMnO_4$$ we will get dicarboxy acid. m-isomer of the two compounds will be Isophthalic acid.
An organic compound $$A(C_8H_6O_2)$$ on treatment with $$NaOH(aq)$$ followed by acidifying the products gives $$B(C_8H_8O_3)$$. B on oxidation gives benzoic acid as the important organic product. What is $$A$$?
Report Question
0%
$$C_6H_5COCHO$$
0%
$$C_6H_5CH(OH)-COOH$$
0%
0%
Identify the compound A among the following compounds.
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0%
o-terephthalic acid
0%
o-terephthalaldehyde
0%
Salicylic acid
0%
None of these
Explanation
$$Cr_2O_7^{2-} / H^+$$ is an oxidizing agent which oxidizes the compound and oxidative cleavage takes place to form o-terephthalic acid and acetic acid.
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