MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 10 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Haloalkanes And Haloarenes
Quiz 10
The product obtained on reaction of $$C_2 H_5 Cl$$ with hydrogen over palladium carbon is:
Report Question
0%
$$C_3H_8 $$
0%
$$ C_4H_{10} $$
0%
$$ C_2H_6 $$
0%
$$ C_2H_4 $$
Chloroform on treatment with phenol in presence of caustic alkali forms salicylaldehyde. This reaction is known as:
Report Question
0%
Carbylamine reaction
0%
Cannizzaro's reaction
0%
Wurtz-Fittig reaction
0%
Reimer-Tiemann reaction
Explanation
D. Reimer-Tiemann Reaction
The lowest molecular weight alkanes, which are optically active; are
Report Question
0%
3-methylhexane
0%
2,3-dimethylpentane
0%
2, 3, 3-trimethylbutane
0%
2-methylhexane
Explanation
lowest weight is 3-methylhexane and 2,3-dimethylpentane
Which of the following molecules will not show optical activity?
Report Question
0%
0%
0%
0%
0%
Explanation
A)
$$\text{has two chiral carbon but optically inactive due to presence of plane}$$
$$\text{ of symmetry.}$$
$$;$$
$$;$$;
$$;$$
.
all are
optically
active due to the presence of chiral centre and have no $$COS\ or POS$$
thus, A is the correct answer.
The IUPAC name of Westron is:
Report Question
0%
1,1,2,2-tetrachloroethane
0%
1,1,2,2-tetrachloroethene
0%
1,2-dichloroethyne
0%
1,3,3,3-tetrachloroprop-l-yne
Explanation
The IUPAC name of Westron is 1,1,2,2- Tetrachloroethane.
Since, option (A)
is
correct
.
The given reaction is:
Report Question
0%
$$ S_{N^1} $$
0%
$$ S_{N^2} $$
0%
$$ E_1 $$
0%
$$ E_2 $$
Explanation
The given reaction is $$S_N1$$ reaction.
The $$S_N1$$
reaction
is a substitution
reaction
in organic chemistry. $$'S_N'$$
stands for 'Nucleophilic substitution', and the $$'1'$$ says that the rate-determining step is unimolecular. Thus, the rate equation is often shown as having first-order dependence on electrophile and zero-order dependence on nucleophile. Hence correct answer is option A.
2-Chloro-2methylpropane on reaction with alc. KOH gives X as the product. X is.
Report Question
0%
but-2-ene
0%
2-methylbut-1-ene
0%
2-methylprop-1-ene
0%
2-methylbutane-2-ol
Explanation
Reaction of
2-Chloro-2methylpropane with alc. KOH:
Hence, Option "C" is the correct answer.
The correct order of reactivity of following alkyl halides for $$ S_{N^1} $$ reaction is:
Report Question
0%
$$ R-I> R-Br>R-Cl> R-F $$
0%
$$ R-F>R-Cl>R-Br>R-I$$
0%
$$ R-F>R-Br>R-Cl >R-I$$
0%
$$R-Cl>R-I>R-Br>R-F$$
Explanation
Reactivity of alkyl halides towards $$SN_1$$ nucleophilic substitution reaction is $$3^o > 2^o > 1^o$$ because, in $$SN_1$$ nucleophilic reaction, the first and the slow step is the formation of a carbocation. A tertiary carbocation is more stable than a secondary carbocation which is more stable than a primary carbocation.
The reactivity of the alkyl halide is decided by the ease with which the halide leaves the substrate. As per the leaving ability, the order is $$I>Br>Cl>F$$. So, the order of the reaction is $$R-I>R-Br>R-Cl>R-F$$.
$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (A)}$$
Which of the following reactions will yield $$2, 2-dibromo-propane$$?
Report Question
0%
$$H_{2}C = CHBr + HBr$$
0%
$$CH_{3}C \equiv CCH_{3} + 2HBr$$
0%
$$CH_{3}C \equiv CH + 2HBr$$
0%
$$CH_{3}CH = CHBr + HBr$$
Allyl chloride is more reactive than vinyl chloride.
Report Question
0%
True
0%
False
In the following reaction, the major product $$'X'$$ is
Report Question
0%
0%
0%
0%
Identify $$'B'$$ in the following reaction
Report Question
0%
0%
0%
0%
None of these
Explanation
Answer is C, as :
$$CH_3-C_6H_5-Cl \xrightarrow{KMnO_4} COOH-C_6H_5-Cl \xrightarrow{C_2H_5OH} COOC_2H_5-C_6H_5-Cl$$
In the halogenation of aromatic nucleus, the halogen carrier,
is used to generate the species.
Report Question
0%
$$Cl\bullet$$
0%
$$Cl^{+}$$
0%
$$Cl^{-}$$
0%
$$Cl$$
Explanation
It generates $$Cl^+$$
The product of the following reaction $$C_{6}H_{6} + Cl_{2} \xrightarrow {Sunlight} ?$$; is
Report Question
0%
$$C_{6}H_{5}Cl$$
0%
Ortho $$C_{6}H_{4}Cl_{2}$$
0%
$$C_{6}H_{6}Cl_{6}$$
0%
Para $$C_{6}H_{4}Cl_{2}$$
In Raschig's process for the preparation of chlorobenzene,
the reactants are
Report Question
0%
$$C_{6}H_{6}$$ and $$Cl_{2}$$
0%
$$C_{6}H_{5}OH$$ and $$PCl_{5}$$
0%
$$C_{6}H_{6} + HCl + O_{2}$$
0%
$$C_{6}H_{5}OH$$ and $$HCl$$
In the presence of iron catalyst, benzene reacts with chlorine
to form
Report Question
0%
Chlorobenzene
0%
Benzene hexachloride
0%
Hexachlorobenzene
0%
None of these
Halogenation of alkanes is
Report Question
0%
A reductive process
0%
An oxidative process
0%
An isothermal process
0%
An indothermal process
Explanation
The reaction of a halogen with an alkane in the presence of ultraviolet (UV) light or heat leads to the formation of a haloalkane (alkyl halide).
This is a oxidative process
Which of the following is the example of $$SN^2$$ reaction
Report Question
0%
$$CH_3 Br + OH^- - CH_3OH + Br^-$$
0%
$$CH_3 CH \underset{Br}{\underset{|}{CH_3}} + OH^- \rightarrow CH_3 \underset{OH}{\underset{\!\!\!\!\!|}{CH}}CH_3 + Br^-$$
0%
$$CH_3 CH_2 OH \xrightarrow{-H_2O} CH_2 = CH_2$$
0%
$$CH_3 - \underset{Br}{\underset{|}{\overset{CH_3\!\!\!\!\!}{\overset{|}{C}}}} - CH_3 + OH^- \rightarrow CH_3 - \underset{H}{\underset{|}{\overset{CH_3\!\!\!\!\!}{\overset{|}{C}}}} - O - CH_3 + Br^-$$
Explanation
Only $$1^\circ$$ alkyl halides, i.e. $$CH_3Br$$ undergoes $$S_{N}2$$ reaction. Lesser the bulkier group attached to the attacking carbon higher is its reactivity towards $$S_N2$$ reaction.
Option A is correct.
Stilbene $$(PhCH=CHPh)$$ can exist in two diastereomeric forms $$(X)$$ and $$(Y)$$, and $$(X)$$ is found to be more soluble in water than $$(Y)$$. Predict which of the following statement is correct.
Report Question
0%
$$X$$ is trans isomer
0%
Stability of $$X >$$ stability of $$Y$$
0%
Melting point of $$X <$$ melting point of $$Y$$
0%
Boiling point of $$X <$$ boiling point of $$Y$$
The reaction
$$CH_3CH=CH_3\xrightarrow[H^{+}]{CO+H_{2}O}CH_{3}-CH-CH_{3}$$
$$COOH$$
is known as
Report Question
0%
Wurtz reactions
0%
Koch reaction
0%
Clemensons reduction
0%
Kolbes reaction
When ethyl chloride and alcoholic $$KOH$$ are heated , the compound obtained is
Report Question
0%
$$C_2H_4$$
0%
$$C_2H_2$$
0%
$$C_6H_6$$
0%
$$C_2H_6$$
Explanation
We know that
$$CH_3CH_2Cl + KOH \rightarrow CH_2 = CH_2 + KCl + H_2O$$
Thus in reaction ethene $$(C_2H_4)$$ is produced.
Which of the following is liquid at room temperature
Report Question
0%
$$CH_3I$$
0%
$$CH_3Br$$
0%
$$C_2H_5Cl$$
0%
$$CH_3F$$
Explanation
$$CH_3F , CH_3Cl , CH_3Br$$ and $$C_2H_5Cl$$ are gases at room temperature. $$CH_3I$$ is a liquid at room temperature and solidifies at $$-66.5^\circ C.$$
Option A is correct.
When benzene is heated with chlorine in the presence of sunlight , it forms
Report Question
0%
$$B.H.C.$$
0%
Cyclopropane
0%
p - dichlorobenzene
0%
None of these
Explanation
When benzene is heated with chlorine in the presence of sunlight , it form benzene hexachloride.
Which metal is used in Wurtz synthesis
Report Question
0%
$$Ba$$
0%
$$Al$$
0%
$$Na$$
0%
$$Fe$$
Explanation
$$CH_3Br + 2Na + Br - CH_3 \xrightarrow[Ether]{Dry} CH_3CH_3 + 2NaBr$$
Sodium metal is used.
If the light waves pass through a nicol prism then all the oscillations occur only in one plane such beam of light is called as
Report Question
0%
Non polarised light
0%
Plane polarised light
0%
polarised light
0%
Optical light
Explanation
Correct option is $$B$$.
What is the major product of the following reaction $$CH_{3}C\equiv C-CH_{2}-CH_{3}\xrightarrow[]{1 \,mole \,of \,Cl_{2}}$$
Report Question
0%
0%
$$CH_{3}-CH_{2}-\overset{Cl}{\overset{|}{\underset{Cl}{\underset{|}{C}}}}-CH_{2}CH_{3}$$
0%
0%
$$CH_{3}-\overset{Cl}{\overset{|}{\underset{Cl}{\underset{|}{C}}}}-\overset{Cl}{\overset{|}{\underset{Cl}{\underset{|}{C}}}}-CH_{2}CH_{3}$$
Explanation
$$CH_{3}C\equiv C-CH_{2}-CH_{3}\xrightarrow[]{1 \,mole \,of \,Cl_{2}}$$
$$CH_{3}-\overset{Cl}{\overset{|}{\underset{Cl}{\underset{|}{C}}}}-\overset{Cl}{\overset{|}{\underset{Cl}{\underset{|}{C}}}}-CH_{2}CH_{3}$$
Option D is a correct answer.
The reaction of benzene with chlorine in the presence of iron gives
Report Question
0%
Benzene hexachloride
0%
Chlorobenzene
0%
Benzyl chloride
0%
Benzoyl chloride
Explanation
The reaction between benzene and chlorine in the presence of either aluminium chloride or iron gives chlorobenzene.
(i) Chlorobenzene and (ii) benzene hexachloride are obtained from benzene by the reaction of chlorine, in the presence of
Report Question
0%
(i) Direct sunlight and (ii) anhydrous $$AlCl_{3}$$
0%
(i) Sodium hydroxide and (ii) sulphuric acid
0%
(i) Ultraviolet light and (ii) anhydrous $$FeCl_{3}$$
0%
(i) Anhydrous $$AlCl_{3}$$ and (ii) direct sunlight
The major amount of $$(C)$$ is:
Report Question
0%
0%
0%
0%
What would be the product formed when $$1$$-bromo-$$3$$-chloro cyclobutane reacts with two equivalents of metallic sodium in ether ?
Report Question
0%
0%
0%
0%
Explanation
$$ (C - Br) $$ bond is weaker than $$ (C - Cl) $$ bond ; Hence, Wurtz reaction will take place with $$ (C - Br) $$ bond.
Condition for maximum yeild of $$C_2H_5Cl$$ is
Report Question
0%
$$C_2H_6 (\text{excess}) + Cl_2\xrightarrow[]{\text{UV Light}}$$
0%
$$C_2H_6 + Cl_2 \xrightarrow[\text{Room temp.}]{\text{Dark}}$$
0%
$$C_2H_6 + Cl_2$$ (excess) $$ \xrightarrow[]{\text{UV Light}}$$
0%
$$C_2H_6 + Cl_2 \xrightarrow[]{\text{UV Light}}$$
Explanation
$$C_2H_6 (\text{excess}) + Cl_2 \xrightarrow[]{\text{UV Light}} \underset{\text{(Major product)}}{\underset{\text{Ethyl chloride}}{C_2H_5Cl}} + HCl$$
The compounds are optically inactive because
Report Question
0%
Both compounds have the plane of symmetry
0%
Both compounds have the centre of symmetry
0%
Compound I has the plane of symmetry, while compound II has the centre of symmmtry.
0%
Compound I has the centre of symmetry, while compound II has the plan of symmmtry.
Explanation
The statement is self-explanatory.
Arrange the following in the decreasing order of nucleophilic acyl substitution reaction.
Report Question
0%
(I)>(II)>(III)>(IV)
0%
(IV)>(III)>(II)>(I)
0%
(I)>(III)>(II)>(IV)
0%
(IV)>(II)>(III)>(I)
Explanation
(I)>(II)>(III)>(IV)
Acyl chloride > Anhydride > Ester > Amide.
Statement 1: $$S_{N}1$$ reaction is carried out in the presence of polar protic solvents.
Statement 2: Polar protic solvent increases the stability of carbocation due to the solvation.
Read the above statements and choose the correct option regarding it.
Report Question
0%
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation for Statement 1.
0%
Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation for Statement 1.
0%
Statement 1 is correct and Statement 2 is wrong.
0%
Statement 1 is wrong and Statement 2 is correct.
Explanation
$$SN^1$$ mechanism proceeds through carbocation formation. As the stability of carbocation increases, reaction favours $$SN^1$$ mechanism. Polar protic solvent stabilises carbocation so they favour reaction. In this reaction products formed are racemic mixture.
Option A is correct.
$$CH_3-CH_3\xrightarrow[]{Br_2/hv}A\xrightarrow[]{LiAlH_4}B\xrightarrow[]{Br_2/hv}C\xrightarrow[]{Na/ether}D$$
The compound $$D$$ is:
Report Question
0%
$$CH_{3}-CH_{2}-CH_{2}-CH_{3}$$
0%
$$CH_{3}-CH_{2}-O-CH_{2}-CH_{3}$$
0%
$$CH_{3}-CH(CH_{3})-CH(CH_{3})-CH_{3}$$
0%
$$CH_{3}-CH_{2}-CH_{3}$$
Explanation
By this process, first alkyl halide forms which reduced by $$LiAlH_4$$ to form alkane and again bromination takes place to give alkyl halide which in presence of $$Na/ dry \:ether$$ forms alkane as product (Wurtz reaction).
$$CH_3CH_3\xrightarrow[]{Br_2/hv}CH_3CH_2Br\xrightarrow[]{LiAlH_4}CH_3CH_3\xrightarrow []{Br_2/hv}CH_3-CH_2-Br\xrightarrow[]{Na/dry\,ether}CH_3-CH_2-CH_2-CH_3$$
Option A is correct.
$$CH_3-CH(Br)-CH_3\xrightarrow[]{alc.\,KOH} A\xrightarrow[peroxide]{HBr}B\xrightarrow[acetone]{NaI}C$$
The compound $$C$$ is:
Report Question
0%
$$CH_{3}-CH_{2}-CH_{2}-I$$
0%
$$CH_{3}-CH(I)-CH_{3}$$
0%
$$CH_{3}-CHI-CH_{2}-I$$
0%
$$CH_{3}-CH=CH-I$$
Explanation
This process is used to form alkyl iodide as a product.
Alcohol $$KOH$$ is dehydrohalogenating agent and peroxide is used to form anti-markovnikov product.
$$CH_3-CHBr-CH_3\xrightarrow[]{alc.KOH}CH_3-CH=CH_2\xrightarrow[peroxide]{HBr} CH_3CH_2CH_2Br\xrightarrow[acetone]{Nal}CH_3CH_2CH_2I$$
Option A is correct.
The compound $$Y$$ is:
Report Question
0%
1, 3 - dichloro benzene
0%
benzyl chloride
0%
1, 3, 5 - trichloro benzene
0%
mixture of 1, 2 and 1, 4 dichlorobenzene
Explanation
$$C_6H_6 \xrightarrow {Cl_2,Fe} C_6H_5Cl \xrightarrow {Cl_2}$$ $$o-$$ and $$p-$$ dichlorobenzene
In first reaction chlorobenzene forms as a product and as $$Cl-$$ group is ortho-, para- directing so ortho, para dichloro benzene forms as product.
Second step is chlorination of chlorobenzene.
Option D is correct.
Which one of the following compounds is most reactive for aromatic electrophilic substitution reaction?
Report Question
0%
$$C_{6}H_{5}-F$$
0%
$$C_{6}H_{5}-Cl$$
0%
$$C_{6}H_{5}-I$$
0%
$$C_{6}H_{5}-NO_{2}$$
Explanation
Due to $$- I$$ effect of $$X-$$ group, it reduces electron density of benzene, so halobenzene is less reactive than benzene for electrophilic substitution reaction.
As electro-negativity of $$F$$ is highest among halogens, so its $$-I$$ effect is highest.
So, reactivity of halobenzene compounds towards electrophilic substitution is $$ Ph-I > Ph-Cl >Ph-F$$.
Due to $$-M$$ effect of nitrate group, it is less reactive than other compounds.
Option C is correct.
Assertion: Aryl halides undergo electrophilic substitution more readily than benzene.
Reason: Aryl halide gives a mixture of $$o -, p -$$ products when they undergo electrophilic substitution.
Report Question
0%
Assertion and Reason are correct and Reason is the correct explanation of Assertion
0%
Assertion and Reason are correct but Reason is not the correct explanation of Assertion
0%
Assertion is true and Reason is false
0%
Assertion is false and Reason is true
Explanation
$$X$$- (halogen group) has two effects$$ -I$$ & $$+M$$.
Due to $$- I$$ effect of $$X$$- group, it reduces electron density of benzene so halobenzene is less reactive than benzene for electrophilic substitution reaction.
But due to $$+M$$ effect, it is ortho, para directing and in halobenzene further substitution takes place at ortho and para positions.
The final product C, obtained in this reaction,
would be :
Report Question
0%
0%
0%
0%
Explanation
To get a monobromo compound, the amino group is acetylated before bromination. After bromination, bromoacetanilide is acid hydrolysed to yield the desired halogenated amine.
In comparison to $$C-Cl$$ bond in ethyl chloride, the $$C-Cl$$ bond in vinyl chloride is:
Report Question
0%
longer and weaker
0%
shorter and weaker
0%
longer and stronger
0%
shorter and stronger
Explanation
$$C-Cl$$ bond in vinyl chloride has partial double bond character.
Therefore, the $$C-Cl$$ bond in vinyl chloride is shorter and stronger as compound to $$C-Cl$$ bond in ethyl chloride.
Option D is correct.
For the compounds $$CH_3Cl, CH_3Br, CH_3I$$ and $$CH_3F$$, the correct order of increasing C-halogen bond length is:
Report Question
0%
$$CH_3F < CH_3Cl < CH_3Br < CH_3I$$
0%
$$CH_3F < CH_3Br < CH_3Cl < CH_3I$$
0%
$$CH_3F < CH_3I < CH_3Br < CH_3Cl$$
0%
$$CH_3Cl < CH_3Br < CH_3F < CH_3I$$
Explanation
$$C$$-halogen bond length totally depends on the electronegativity difference between Carbon and halogen. The more the difference is, the less is the bond length. The electronegativity order of halogens is $$I < Br < Cl < F$$. Hence, the correct order of increasing $$C$$-halogen bond length is $$CH_3F<CH_3Cl<CH_3Br<CH_3I$$.
Hence, the correct option is $$(A)$$
$$CH_3-CH_2-CH_2-O-\underset{S}{\underset{||}{C}}-S-CH_3$$ undergoes elimination on strong heating giving $$CH_3-CH=CH_2, CH_3SH$$ and $$COS$$. This reaction involves :
Report Question
0%
formation of cyclic intermediate
0%
syn elimination
0%
anti elimination
0%
first order kinetics
Explanation
Pyrolysis of xanthate esters is known as Chugaev reaction. The reaction involves cyclic intermediate. It follows syn elimination. The reaction follows first order kinetics.
Which of the following method would be better for the synthesis of 1-bromo-3-ethylbenzene?
Report Question
0%
0%
0%
0%
Explanation
Hence option C is correct.
One of the iosmers of $$C_{4}H_{9}Cl$$ by Wurtz reaction forms $$2,2,3,3$$- tetramethylbutane. The structure of that isomer of $$C_{4}H_{9}Cl$$:
Report Question
0%
$$1$$-Chloro - $$2$$- methylpropane
0%
$$2$$-Chloro - $$2$$- methylpropane
0%
$$2$$-Chloro butane
0%
$$1$$-Chlorobutane
Explanation
Wurtz reaction: In this process alkyl halide forms alkane in presence of sodium and dry ether.
Reaction:
$$CH_3C(CH_3)ClCH_3+2Na+CH_3C(CH_3)ClCH_3\rightarrow CH_3C(CH_3)_2C(CH_3)_2CH_3$$
$$2$$-Chloro - $$2$$- methylpropane is the answer.
$$1-$$bromo$$-3-$$chlorocyclobutane when treated with two equivalents of $$Na$$, in the presence of ether which of the following will be formed?
Report Question
0%
0%
0%
0%
Explanation
$$1-$$bromo$$-3-$$chlorocyclobutane when treated with two equivalents of $$Na$$, in the presence of dry ether(D.E.) will form the product represented by option D. It is an intramolecular Wurtz reaction and involves formation of $$C-C$$ single bond between two $$C$$ atoms that are attached to halogen atoms.
Select the best combination of regents which will bring the following conversion.
Report Question
0%
$$Br_2/FeBr_3; \;HNO_3/H_2SO_4;\ NaOH; \;H_2/Pt$$
0%
$$CH_3COCl/AlCl_3; \;Br_2/FeBr_3;\; NaBH_4/CH_3OH$$
0%
$$CH_3Cl/AlCl_3; \;KMnO_4; \;CH_3Cl/AlCl_3; \;NBS/hv;\;LiAiH_4/H_2O;\;NaOH$$
0%
$$CH_3Cl/AlCl_3; \;HNO_3/H_2SO_4; \;NBS/hv/NaOH$$
Explanation
Benzene is first acetylated with acetyl chloride in the presence of anhydrous aluminum chloride to obtain acetophenone (Friedel-crafts acylation reaction).
Then, it is brominated with bromine in the presence of ferric bromide. Acetyl group is a meta directing group.
Hence, m-bromo acetophenone is obtained. The carbonyl group is then reduced to alcoholic group by using sodium borohydride in methanol.
Option B is correct.
The major product of the above shown reaction is
:
Report Question
0%
0%
0%
0%
Explanation
Both bromine and methoxy group are ortho and para directing groups.
But, bromine deactivates the aromatic ring and methoxy group activates the ring.
Hence, the effect of methoxy group is more pronounced.
Therefore, the substitution occurs in the position indicated in option C.
Option C is correct.
The most stable conformation of the product of following reaction is:
Report Question
0%
0%
0%
0%
Explanation
Conformation $$D$$ is most stable. This conformer corresponds to anti-peri-planar orientation (conformation) of two largest substituents $$Br$$ in the system. Due to the steric size the antiperiplanar conformer is the most stable.
Which of the following reaction will give Iodobenzene as the major organic product?
Report Question
0%
0%
0%
0%
Explanation
Benzene undergoes electrophilic aromatic substitution with iodine in presence of anhydrous $$FeCl_3$$ catalyst to formed iodobenzene. The by product $$HI$$ can reverse the reaction. To prevent this,$$HgO$$ is used which consumes $$HI$$. When $$ICl$$ and $$AlCl_3$$ are used, iodination competes with chlorination.
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 Engineering Chemistry Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
