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CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 2 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Haloalkanes And Haloarenes
Quiz 2
Ethyl iodide and n-propyl iodide are allowed to undergo Wurtz reaction. The alkane which will not be obtained in this reaction is:
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butane
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propane
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pentane
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hexane
Explanation
$$CH_3CH_2I+CH_3CH_2CH_2I$$
(Wurtz reaction) $$\downarrow Na$$
$$\underset {butane}{CH_3CH_2CH_2CH_3}+\underset {hexane}{CH_3CH_2CH_2CH_2CH_2CH_3}+\underset {pentane}{CH_3CH_2CH_2CH_2CH_3}$$
Ethyl iodide and n-propyl iodide are allowed to undergo Wurtz reaction and they form butane and hexane as self-addition products and pentane as cross addition product.
$$\therefore$$ So, propane will not be formed.
The Major products obtained when this substance is subjected to $${ E }_{ 2 }$$ reaction will be:
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both $$1$$ and $$3$$
Explanation
For $$E_{2}$$ elimination, the hydrogen atom adjacent to the $$Cl$$ atom needs to be in the opposite side of the ring.
Given reaction is an example of Nucleophilic aromatic substitution. Which of the following halide $$(-X)$$ is most readily replaced.
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$$-F$$
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$$-Cl$$
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$$-Br$$
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$$-I$$
Explanation
Flourine is readily replaced in nucleophilic aromatic substitution reaction as it has higher $$-I$$ effect compared to other halogens.
$$CH_3 - \underset{CH_3\!\!\!\!\!}{\underset{|}{\overset{CH_3\!\!\!\!\!}{\overset{|}{C}}}} - Br \xrightarrow[\Delta]{KOH (alcoholic)} (A) \xrightarrow{HBr/R_2O_2} \underset{(Major)}{(B)}$$ Product $$B$$ is:
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$$CH_3 - \underset{H}{\underset{|}{\overset{\,\,\,\,\,CH_3}{\overset{|}{C}}}} - CH_2 - Br$$
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$$CH_3 - \underset{\,\,\,Br}{\underset{|}{\overset{\,\,\,\,\,CH_3}{\overset{|}{C}}}} - CH_3$$
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$$CH_3 - \underset{\,\,\,\,\,\,\,\,\,\,\,CH_2 - Br}{\underset{|\,}{\overset{\,\,\,\,\,\,CH_3}{\overset{|\,}{C}}}} - OH$$
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$$CH_3 - \underset{\,\,\,\,\,\,CH_3}{\underset{|}{\overset{\,\,\,\,\,\,\,\,\,\,\,\,CH_2 - H}{\overset{|}{C}}}}-Br$$
Explanation
Find out the major product.
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Explanation
Correct answer- B
The reaction of toluene with chlorine in the presence of light gives:
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Explanation
Reaction of methylbenzene i.e. toluene with chlorine in the presence of sunlight is a free radical substitution .
in this reaction hydrogen atoms from side chain get replaced get replaced by chlorine .on monochlorination benzyl chloride is formed on dichlorination benzal chloride and on trichlorination benzochloride is formed.
Which of the following alkyl halides will undergo $$S_N1$$ reaction most readily?
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$$(CH_3)_3C-F$$
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$$(CH_3)_3C-Cl$$
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$$(CH_3)_3C-Br$$
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$$(CH_3)_3C-I$$
Explanation
All options given are alkyl halide (tertiary) so all will form tertiary carbocation.
But order of leaving group ability Is $$I^-$$ > $$Br^-$$ > $$Cl^-$$ > $$F^-$$
$$S_{N}{1}$$ mechanism goes through formation of carbocation.
More easily leaving group leaves more easily the carbocation will form.
Option D is correct.
Product D in the reaction is:
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optically active cyanide
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optically inactive cyanide
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optically active acid
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optically active aldehyde
In which of the following reactions, it proceeds via free radical mechanism?
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Halogenation of benzene
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Nitration of benzene
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Sulphonation of benzene
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Chlorination of toluene at high temperature
But-2-yne on chlorination gives:
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1-chlorobutabe
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1,2-dichlorobutabe
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1,1,2,2-tetrachloro butane
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2,2,3,3-tetrachloro butane
The product of reaction between chloroform and ethylamine in presence of alcoholic $$KOH$$ is allowed to react with $$Cl_2$$.
The final product is:
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ethyl chloride
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ethylene dichloride
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ethyl iminocarbonyl chloride
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acetaldehyde
Which one of the following compound will be most reactive for $$S_N1$$ reaction ?
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Which of the compound gives a mixture of the substitution products on reaction with $$NaNH_{2}$$ in liquid $$NH_{3}$$?
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Correct order of $$S_N1$$ rate 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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$$I > IV > III > II$$
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$$III > IV > I > II$$
Which of the following is most likely to show optical isomerism?
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n-Propyl chloride and isopropyl chlordie mixture is subjected to the Wurtz reaction, which one of the following compounds is not formed.
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Hexane
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2,5-Dimethyl hexane
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2,3-Dimethyl butane
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2-Methyl pentane
Which is most reactive for $$SN^{1}$$ reaction:-
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The reaction of tertiary butyl bromide with sodium methoxide gives
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Isobutane
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Tertiary butyl methyl ether
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Isobutylene
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Butene
Identify the compound that give product on reaction with Zn dust and heat.
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Which of the following compound give solvolysis reaction with slowest rate?
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$$P-\overset { \oplus }{ N } { R }_{ 3 }-{ C }_{ 6 }{ H }_{ 4 }-C{ H }_{ 2 }-Br$$
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$$P-CCl_{3}-C_{6}H_{4}-CH_{2}-Br$$
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$$p-N$$$$\equiv C- $$$$C_{6}H_{4}-CH_{2}-Br$$
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$$C_{6}H_{5}-CH_{2}-Br$$
Which of the following compounds are optically active ?
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$$CH_3-CHOH-CH_2-CH_3$$
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$$H_2C = CH-CH_2-CH=CH_2$$
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Arrange the following in order of decreasing rate in an $$E_2$$ reaction
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$$1 > 2 > 3$$
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$$3 > 1 > 2$$
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$$3 > 2 > 1$$
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$$2 < 3 < 1$$
The major product of the following reaction is:
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Which of the these compounds represents the major monobromination isomer formed in the following reaction?
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Product (D) in the reaction is:
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optically active cyanide
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optically inactive acid
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optically active acid
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optically active aldehyde
Explanation
In the following reaction first step is Wurtz reaction and then Bromine addition which further react with $$KCN$$ and after hydrolysis it gives the optically inactive carboxylic acid.
Option B is correct.
The reagent used in dehydrohalogenation process is
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alcoholic KOH
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$$NaNH_2$$
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$$C_2H_5ONa$$
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All of these
Which one of the following compound is most reactive towards $$SN^1$$ reaction
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Explanation
The compound which forms most stable carbocation is most reactive towards $$SN^1$$ reaction.
Here, is most reactive towards $$SN^1$$ reaction.
($$\because Br$$ is better leaving group compared to Cl)
The IUPAC name of this compound is-
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2- Flouro-4-chloro-2,4-diethyl-1-pentane
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3-Flouro-5-chloro-3-methyl-5-ethyl-1-hexane
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3-Chloro-5-fluoro-3,5-dimethyl-1-heptane
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3,5-Dimethyl-5-fluoro-3-chloro-heptane
2-methylbut -3-en-2-ol reacts with $$HBr$$ and
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Produces $$1^\circ$$ - Bromide as major product through $$S_N 1$$ mechanism
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Produces $$3^\circ$$ - Bromide as major product through $$S_N 1$$ mechanism
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Produces $$1^\circ$$ - Bromide as major product through $$S_N 2$$ mechanism
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Produces $$3^\circ$$ - Bromide as major product through $$S_N 2$$ mechanism
Mark the true statement concerning mesomeric effect?
a) It occurs in system having conjugate double bonds in compounds
b) It involves electrons of $$\pi $$ bonds
c) The electron pair is transferred completely
d) It involeves lone pair of elctrons
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True
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False
Explanation
All the statement concerning mesomeric effect are true, other name is Resonance effect. It is of two types $$ + M$$ or $$ + R$$, $$ - M$$ or $$ - R$$
Identify 'Z' in the following reaction series , $$CH_{3}CH_{2}CH_{2}Br \xrightarrow {NaOH} (X) \xrightarrow [ \triangle] {Al_{2}O_{3}} (Y) \xrightarrow {HOCl} (Z) $$
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Explanation
Option B is correct.
Select the incorrect statement (s)
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$${{\text{S}}_{\text{N}}}{\text{1}}$$ reactions also take place with some elimination.
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$${{\text{S}}_{\text{N}}}{\text{2}}$$ reactions with chiral compounds gives racemic mixture.
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$${{\text{S}}_{\text{N}}}{\text{1}}$$ reactions is faster in more polar aprotic solvents.
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All of the above are incorrect statements.
The herbicide 2,4-dichlorophenoxyacetic acid, generally called 2,4-D, is synthesized from 2,4-dichlorophenol and chloroacetic acid. The synthesis involves which of these reactions:
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$$\rm{S_N1}$$
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$$\rm{ArS_N}$$
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$$\rm{S_N2}$$
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$$\rm{S_Ni}$$
Which alcohol will give only $$E_{1}$$ reaction?
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Identify the following species in which elimination of $$HBr$$ of two $$Br$$ atoms is stereoselective.
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$$CH_{3}CH_{2}CHBr_{2}$$
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$$CH_{3}CH_{2}CH_{2}Br$$
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$$CH_{3}CH_{2}\underset{Br}{\underset{|}{CH}}CH_{2}$$
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$$CH_{3}\underset{Br}{\underset{|}{CH}}\underset{Br}{\underset{|}{CH}}CH_{3}$$
Given the correct order of initial $$T$$ or $$F$$ for the following statements. Use $$T$$ if the statement is true and use $$F$$ if the statement is false.
I. $$Me-CH=C=C=CH-Br$$ is optically active.
II. All optically active compounds are chiral.
III. All chiral pyramidal molecules are optically inactive.
IV. $$CH_{3}-CH_{2}-CH_{2}-COOH$$ and $$CH_{3}-CH_2-CH_{3}$$ are positional isomers.
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$$TTTF$$
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$$FTFT$$
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$$FTFF$$
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$$TFTT$$
Which of the following compound may be optical active ?
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$$Co{(en)_2}C{I_2}$$
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$${\left[ {Rh{\rm{ CI Br N}}{{\rm{H}}_3}{\rm{ P}}{{\rm{H}}_3}} \right]^ - }$$
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$$Cu{(en)_2}C{I_2}$$
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All of these
Explanation
Optical activity requires chirality
$${\left[ {RhCIBrN{H_3}P{H_3}} \right]^ - }$$
Which of the following compound will rotate the plane polarized light at room temperature?
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Alkyl halides can be obtained by all methods except?
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$$CH_{3}-CH_{2}-CH_{3} + Cl_2 \xrightarrow [ UV\ light ]{ } $$
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$$CH_{3}CH_{2}OH+HCl/ZnCl_{2}\rightarrow$$
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$$C_{2}H_{5}OH+NaCl\rightarrow$$
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$$CH_{3}COOAg+Br_{2}/CCl_{4}\rightarrow$$
Compound $$C_4H_8Cl_2 \,(A)$$ on hydrolysis gives a compound $$C_4H_8O(B)$$ which reacts with hydroxylamine and does not give any test with Tollens regent . What are (A) and (B)?
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$$1,1-$$ Dichlorobutane and butanal
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$$2,2-$$ Dichlorobutane and butanal
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$$1,1-$$ Dicholorobutane and butan $$-2-$$ one
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$$2,2-$$ Dichlorobutane and butan $$-2-$$ one
$$Compound (refer\ image)\overset { alc. }{ \underset { KOH }{ \longrightarrow } } (A)\overset { NBS }{ \longrightarrow } (B)\overset { alc. }{ \underset { KOH }{ \longrightarrow } } (C)\overset { NBS }{ \longrightarrow } (D)\overset { alc. }{ \underset { KOH }{ \longrightarrow } } (E)$$ The product E is
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Provide the structure of the major organic product which results in the following reaction.
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Which one is not prepared by Wurtz reaction?
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$$C_4H_8$$
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$$CH_4$$
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The major product of the following reaction is :
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Explanation
When an alkyl bromide is treated with $$\ C_2H_5ONa/ C_2H_5-OH$$, it undergoes dehydrobromintion
. A molecule of $$\displaystyle HBr$$ is eliminated and $$\displaystyle C=C$$ double bond is formed. In present example, $$ C=C$$ double bond is conjugated with aromatic ring. This leads to stability.
Choose the correct options (s) that gives (s) aromatic compound as the major product:
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Explanation
Hence option A and B are correct.
The compound $$C_{7}H_{8}$$ undergoes the following reactions:
$$C_{7}H_{8}\xrightarrow {3Cl_{2}/\triangle}A \xrightarrow {Br_{2}/ Fe} B \xrightarrow {Zn/HCl}C$$. The product $$C$$ is:
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$$3-$$bromo$$-2,4,6-$$trichlorotoluene
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$$m-$$bromotoluene
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$$p-$$bromotoluene
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$$o-$$bromotoluene
Explanation
Option B is correct
Among the following, the reaction that proceeds through an electrophilic substitution is:
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Explanation
Hence option B is correct.
In an $$S_N1$$ reaction on chiral centres, there is:
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$$100$$% retention
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$$100$$% inversion
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$$100$$% racemization
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Inversion more than retention leading to initial racemization
Explanation
In an $$S_N1$$ reaction on chiral centres, there is inversion more than retention leading to initial racemization.
Thus if we start with a pure enantiomer and carry out $$S_N1$$ substitution on chiral carbon, the product will be racemic. This is because a planar carbocation intermediate is obtained. The nucleophile can attack from either side of this intermediate.
Option D is correct.
Which of the following is the most correct electron displacement for a nucleophilic reaction to take place?
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Explanation
The most correct electron displacement for a nucleophilic reaction to take place is as given above.
Which of the following alkane cannot be made in good yield by Wurtz reaction?
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2,3-Dimethylbutane
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n-Heptane
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n-Butane
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n-Hexane
Explanation
Wurtz reaction is a reaction in which alkyl halide is converted to alkane by Na/ether. It is limited to the synthesis of symmetrical alkanes.
As we know that n-Heptane involves the odd number of carbon hence it won't be produced by this method. Rest molecules can be formed by the Wurtz reaction.
Option B is correct.
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