Match the column I with column II. Columns I Columns II A Carbon tetrachloride i Paint remover B Methylene chloride ii Refrigerators and air conditioners C DDT iii Fire extinguisher D Freons iv Non biodegradable insectiside

A-(iii), (B)-(i), (C)-(iv), (D)-(ii)

A-(ii), (B)-(iii), (C)-(i), (D)-(iv)

A-(i), (B)-(ii), (C)-(iii), (D)-(iv)

A-(iv), (B)-(iii), (C)-(ii), (D)-(i)

MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 12

Following is an example of an E2 reaction.

0%
True
0%
False

Explanation

The reaction is an acid-catalyzed enol dehydration—an elimination reaction(Involves removal of a water molecule or an alcohol molecule). In the reaction, the removal of an

$HOH$ substituent results in the formation of a double bond. Therefore, it is an

$E1$ elimination reaction.

Consider the following anions.
When attached to

$sp^{3-}$ hybridized carbon, their leaving group ability in nucleophilic substitution reactions decreases in the order.

0%
$\;I>II>III>IV$
0%
$\;I>II>IV>III$
0%
$\;IV>I>II>III$
0%
$\;IV>III>II>I$

Explanation

When attached to

$sp^3$ −hybridized carbon, their leaving group ability in nucleophilic substitution reactions decreases in the order

$\displaystyle I>II>IV>III$

$I$ is best leaving group due to electron withdrawing

$(-I)$ effect of trifluoromethyl group and delocalization of negative charge over

$3\: O$ atoms.

$III$ is worst leaving group because negative charge is delocalized over

$C$ atoms.

The stabilization effect of resonance for the anion in which negative charge is delocalized over

$O$ atoms is more pronounced over the delocalization in which the negative charge is delocalized over

$C$ atoms.

II is better leaving group than

$IV$ because in

$II$ , the extent of delocalisation of negative charge is much more than the extent of delocalization in

$IV$ .

Allyl chloride is more reactive than vinyl chloride.

0%
True
0%
False

Explanation

$CH_{2}=CH-Cl$ (Vinyl chloride) and

$CH_{2}=CH-CH_{2}-Cl$ (Allyl chloride)

In allyl chloride, chlorine is attached to

$sp^3$ carbon atom.
In vinyl chloride, chlorine is attached to

$sp^2$ carbon atom.

$sp^2$ carbon atom is more electronegative than

$sp^3$ carbon atom.

Hence, the bond pair of electrons of C-Cl bond in allyl chloride are less strongly held than that in vinyl chloride and the removal of chlorine is easier in allyl chloride than in vinyl chloride.
Hence, allyl chloride is more reactive than vinyl chloride.

When phenol is treated with excess bromine water, it gives:

0%
m-Bromophenol
0%
o- and p-Bromophenol
0%
2,4-Dibromophenol
0%
2,4,6-Tribromophenol

Explanation

Hint: Activating group detects the incoming nucleophile at ortho and para positions.

Explanation:
In phenol the hydroxyl group in ring is a activating group. It activates the ring to a great extent and bromination means the addition of bromine take place at ortho and para positions.

So, the $Br$ groups attack at $2,4,6$ positions and give $2,4,6-tribromophenol$ .

The reaction is-

Final Answer: When phenol is treated with excess bromine water, it gives $2,4,6 tribromophenol$ .

Correct option (D).

Arrange the following in the increasing order of their ability as a leaving group:

$CF_3SO_3^-,\,CH_3SO_3^-\,and\,CH_3COO^-$

0%
$CH_3SO_3^-$ > $CF_3SO_3^-$ > $CH_3COO^-$
0%
$CF_3SO_3^-$ > $CH_3SO_3^-$ > $CH_3COO^-$
0%
$CF_3SO_3^-$ > $CH_3COO^-$ > $CH_3SO_3^-$
0%
$CF_3SO_3^-$ < $CH_3SO_3^-$ < $CH_3COO^-$

Explanation

The increasing order of their ability as a leaving group is

$CF_3SO_3^-$ >

$CH_3SO_3^-$ >

$CH_3COO^-$ .
The electronegative

$F$ atoms show

$-I$ effect and remove the electron density and stabilise the negative charge. Hence, the leaving group ability of

$CF_3SO_3^-$ is maximum. Acetate ion has the least ability as a leaving group.

Which of the following products will be obtained when neopentyl alcohol is treated with conc.

$HCl$ in presence of

$Zn{Cl}_{2}?$

0%
t-Butyl chloride
0%
Isobutylene
0%
t-Pentyl chloride
0%
Neo pentyl chloride

Explanation

When neopentyl alcohol is treated with conc.

$HCl$ in presence of

$ZnCl_2,$ t-pentyl chloride will be final product.
In this reaction, less stable primary carbocation rearranges to more stable tert carbocation through methyl shift. This gives rearranged product.
The overall reaction is the replacement of

$-OH$ group with chloride.

Which of the following order is correct for the decreasing reactivity of ring monobromination of the following compounds?

0%
$I > II > III > IV$
0%
$II > III > IV > I$
0%
$I > III > II > IV$
0%
$III > I > II > IV$

Explanation

Ring monobromination is electrophilic aromatic substitution reaction. Higher is the electron density on aromatic ring, higher will be the reactivity towards ring monobromination. Alkyl substituents increases the electron density by

$+I$ effect. Hence, the reactivity of toluene (I) is maximum.

Substituents such as

$-COOH$ group and

$-NO_2$ group are electron withdrawing groups. hence, they decrease the electron denstiy of the ring.

Hence, nitrobenzene has lowers reactivity towards ring monobromination.

Hence, the correct order for the decreasing reactivity of ring monobromination is

$I$ (toluene) >

$III$ (benzene) >

$II$ (benzoic acid)>

$IV$ (nitrobenzene)

Which of the following compounds will give

$S_{N}1$ and

$S_{N}2$ reactions with considerable rate?
I.

$\;C_6H_5-CH_2-Br$
II.

$\;CH_2=CH-CH_2-Br$
III.

$\,CH_3-CH(Br)CH_3$
Select the correct answer from the codes below below.

0%
$\;I,\,II\,and\,III$
0%
$\;I,\,II\,and\,IV$
0%
$\;II,\,III\,and\,IV$
0%
$\;I,\,III\,and\,IV$

Explanation

The following compounds will give

$S_{N}1$ and

$S_{N}2$ reactions with a considerable rate.

$I.\;C_6H_5-CH_2-Br$

$II.\;CH_2=CH-CH_2-Br$

$III.\,CH_3-CH(Br)CH_3$

I and II are primary alkyl halides. Hence, they do react by

$S_N2$ mechanism. However, the carbocation formed is stabilized by resonance with either C=C double bond or benzene ring. Hence, they also react by

$S_N1$ mechanism.

The compound III is a secondary alkyl bromide. It reacts by

$S_N1$ and

$S_N2$ mechanism.

However, the compound IV will undergo

$S_N1$ reaction faster than

$S_N2$ reaction as it is sterically hindered and the approach of the nucleophile from the side opposite to that of leaving group is difficult.

What will be the product of the following the reaction

$\underset{|}{CH_3}$

$CH_3-C-CH_2-Br\xrightarrow[]{NaOH} P$ ,

$\overset{|}{CH_3}$

P is ________.

0%
$CH_3-C\! \! \! =CH-CH_3$
$\; \; \; \; \; \; \; \; \; \overset{|}{CH_3}$
0%
$CH_3-CH-CH=CH_2$
$\; \; \; \; \; \; \; \; \; \overset{|}{CH_3}$
0%
Both of these
0%
None of these

cis-2-butene

$\underset{CCl_4}{\xrightarrow{Br_2}}$

$P$

$(P)$ is:

0%
0%
0%
0%
Both $A$ and $B$

Explanation

cis-2-butene undergo anti-addition of bromine to give a mixture of products

$A$ and

$B$ respectively.

Find the major product of the following reaction.

0%
0%
0%
0%
none

Explanation

The compound undergoes pyrolytic elimination with heat. The pyrolysis of

$N,N-dimethyl-2-phenylcyclohexylamine-N-oxide$ shows how conformational effects and the stability of the transition state affect product composition for cyclic substrates.

In the trans isomer, there are two

$cis-\beta-hydrogens$ that can be eliminated. The major product is the alkene that is in conjugation with the phenyl ring, presumably due to the stabilizing effect on the transition state.

In the cis isomer, there is only one

$cis-\beta-hydrogen$ that can be eliminated, giving the nonconjugated regioisomer as the major product.

An organic compound

$A(C_4H_9Cl)$ on reaction with

$Na/diethyl \hspace{1mm}ether$ gives a hydrocarbon which on monochlorination gives only one chloro derivative.

$A$ is:

0%
t-butyl chloride
0%
sec-butyl chloride
0%
isobutyl chloride
0%
n-butyl chloride

A mixture of two organic compounds was treated with sodium metal in ether solution. Isobutane was obtained as a product. The two chlorine compounds are:

0%
methyl chloride and propyl chloride
0%
methyl chloride and ethyl chloride
0%
isopropyl chloride and Methyl chloride
0%
isopropyl chloride and ethyl chloride

Explanation

Wurtz reaction:

$RX + 2 Na + RX \xrightarrow[ether]{pure \, \& \, dry} R - R + 2 NaX$

$CH_3Cl + 2 Na + Cl \, CH_3 \xrightarrow[ether]{pure \, \& \, dry} CH_3CH_3 + 2NaCl$

$2CH_3 - \underset{CH_3}{\underset{|}{CH}} - Cl + 2Na \xrightarrow[ether]{pure \, \& \, dry} CH_3 \underset{CH_3}{\underset{|}{CH}} - \underset{CH_3}{\underset{|}{CH}} CH_3 + 2NaCl$

$RX + 2 Na + XR' \xrightarrow[ether]{pure \, \& \, dry}$ Mixture of

$R' - R' + R' - R + R - R$ (mixture of three)

$CH_3 Cl + 2Na + Cl \, C_2H_5 \xrightarrow[ether]{pure \, \& \, dry} CH_3CH_3 + C_3H_8 + C_4H_{10}$

Product formed in the reaction is__________.

Explanation

The pyrolytic decomposition of esters is an example of a thermal syn elimination. Esters containing

$\beta-hydrogens$ can eliminate a carboxylic acid through a 6-membered transition state, resulting in an alkene.

Under identical conditions, solvolysis of which of the following substrates would lead to maximum racemization?

Explanation

The solvolysis of the compound

$D$ will result in maximum racemization.

The methoxy group is electron releasing group and increases the electron density.

Hence, the carbocation intermediate is stabilised and easily formed.

Option D is correct.

For the following reaction which of the following statement is/are correct?

0%
Both $X$ and $Y$ are five membered cyclic ring
0%
Reaction (i) follows $E2$ path where as reaction (ii) follows $E1$ path
0%
Reaction (ii) has carbocation as an intermediate
0%
Carbocation rearrangemnet takes place in reaction (i)

Explanation

(A)

$X$ is five membered cyclic ring and

$Y$ is four membered cyclic ring.
(B) reaction (i) follows

$E1$ path where as reaction (ii) follows

$E2$ path.
(C) reaction (ii) does not have a carbocation as an intermediate.
(D) carbocation rearrangement takes place in reaction (i)
Due to 1,2 alkyl shift, 4 member ring expands to 5 member ring. This relieves ring strain.
Hence, only statement (D) is correct.

Which of the following applies in the reaction?

${ CH }_{ 3 }CH(Br){ CH }_{ 2 }{ CH }_{ 3 }\xrightarrow [ ]{ Alc.KOH }$ ?

(I)

${ CH }_{ 3 }CH=CH-{ CH }_{ 3 }$ (major product)
(II)

${ CH }_{ 2 }=CH{ CH }_{ 2 }{ CH }_{ 3 }$ (minor product)

0%
Hofmann's rule
0%
Saytzeff's rule
0%
Kharasch effect
0%
Markovnikov’s rule

Explanation

This reaction is governed by Saytzeff's rule. According to this rule, the elimination of

$\beta$ -hydrogen atom takes place from the carbon having the less number of H-atoms or in other words, a stable alkene is formed (More substituted alkene is more stable)

Alkyl halides and alcohols easily undergo nucleophilic substitution either through

$S_N1$ or

$S_N2$ mechanism. The relative case of these two processes depends upon the nature of the substrate (alkyl group) as well as leaving group, nature of nucleophile and also upon the nature of the solvent.

$S_N1$ mechanism involves the formation of carbocation as intermediate while

$S_N2$ mechanism involves the formation of a transition pentavalent state.

$S_N1$ is the main mechanism in

$3^0$ alkyl halides and alcohols, while

$S_N2$ mechanism is the path adopted by most of the

$1^o$ alkly halides;

$2^o$ alkyl halides may follow

$S_N1$ as well as

$S_N2$ .

What happens when

$1^o ROH$ react with

$HX$ ?

0%
The $1^o$ carbocations are unstable and hence are not formed.
0%
The carbocations are unstable and hence undergo rearrangement.
0%
The $1^o$ alcohol forms protonated alcohol and then follow $S_N2$ path.
0%
$X^-$ makes an $S_N2$ attack at $\alpha - C$ in $1^oROH$ .

Explanation

$1^o$ carbocations are not formed as they are unstable.

$1^o$ alcohols first form protonated alcohols and then halide ions attack it forcing

$H_2O$ to leave in an

$S_N2$ path.

$\xrightarrow{1eqv.\, of Br_2 / Fe}$ A. Compound A is:

Explanation

The position of Bromine depends on substituting group not on the reactant As the

$C$ atom acquires positive charge due to electronegative chlorine atoms. Hence, the group becomes deactivating and meta-directing, that is it will have negative inductive effect on the ring more than the positive mesomerio effect.

Correct option:

$A$

1998033_305135_ans_196b966de3a44c4b80edf8dd17e8356e.jpg

Which one of the following compounds will be most readily dehydrated?

Explanation

$\bf{Hint}$ : Dehydration depends on the stability of carbocation formed.

$\bf{Correct Answer:}$ Option

$C$

$\bf{Explanation \ for \ correct \ option:}$

Easy dehydration of compound depends on the following parameters.

Resonance stabilized carbocation formation after dehydration.
$+M$ and $+I$ effect of an adjacent group of atoms.
The number of $\alpha$ hydrogen attached with carbocation.

Compound

$A$ is the one that is most readily dehydrated because the

$H$ that is removed from the compound is the most acidic one as it is next to the carbonyl group.

So, the positive charge formed will be more stabilized by the removal of acidic

$H$ that is why compound

$A$ is the most readily dehydrated.

Moreover, the carbocation formed in the case of option

$C$ is most stable because it is far away from the carbonyl carbon and the carbocations formed in option

$A$ and

$B$ are least stable because there is

$+ve$

$+ve$ charge repulsion between the carbonyl carbon and carbocation.

$\bf{Explanation \ for \ incorrect \ options:}$

Compound $A$ from a secondary carbocation which is not in conjugation with any group or atoms and directly attached to next carbon with electron-withdrawing group $CO$ , hence it will not go dehydration as fast as compound $C$ .
- Compound $B$ from a secondary carbocation which is not in conjugation with any group or atoms and directly attached to next carbon with electron-withdrawing group $CO$ , hence it will not go dehydration as fast as compound $C$ .
- Compound $D$ from a secondary carbocation which is not in conjugation with any group or atoms and directly attached to next carbon with electron-withdrawing group $CO$ , hence it will not go dehydration as fast as compound $C$ .

The major product is:

0%
0%
0%
0%
None of these

Which of the following solvent will give maximum yield for an alkyl halide undergoing

$S_N1$ mechanism?

0%
Water
0%
Ethanol
0%
Diethyl ether
0%
n-hexane

Explanation

The

$S_N1$ path is strongly favoured by the presence of polarising solvents which stabilizes both cation and anion effectively.

More polar is a solvent, easier is

$S_N1$ path. The polarity of the given solvents is water > ethanol > ethoxyethane > n-hexane.

Option A is correct.

Which of the following cannot undergo nucleophilic substitution under ordinary conditions?

0%
Chlorobenzene
0%
Tert-butyl chloride
0%
Iso-propyl chloride
0%
None of the above

In the above reactions, which of the following is/ are products?

0%
$1,2$ - Dichlorobenzene
0%
$1,3$ - Dichlorobenzene
0%
$1,4$ - Dichlorobenzene
0%
Both A and C

Explanation

$\begin{array}{l}\text { Cl}^{+}\text{ electrophile is formed during reaction of FeCl }_{3} \text { and } \mathrm{Cl}_{2} \text { . }\\\text { Now, in chlorobenzene chlorine increases the } \\\text { eleetron density at ortho and para position through } \\\text { resonance (as chlorine is ortho-para directing group) }\\\text{Hence, option D is correct.}\end{array}$

An organic compound which produces a bluish-green coloured flame on heating in presence of copper is:

0%
chlorobenzene
0%
benzaldehyde
0%
anlline
0%
benzoic acid

On treating a mixture of two alkyl halides with sodium metal in dry ether,

$2-$ methyl propane was obtained. The alkyl halides are:

0%
$2$ - chloropropane and chloromethane
0%
$2$ - chloropropane and chloroethane
0%
chloromethane and chloroethane
0%
chloromethane and $1$ - chloropropane

Haloarenes react with halogens in the presence of ferric salt as catalyst to give ortho and para dichloro benzene. Which of the following products predominates in ortho and para isomers in this case?

0%
Ortho-dichlorobenzene
0%
Para-dichlorobenzene
0%
Meta-dichlorobenzene
0%
A and C both

Which of the following is the best method for synthesis of 1-bromo-3-chlorobenzene?

Explanation

As,

$Br$ and

$Cl$ are o,p directing, they cannot be used directly for chlorination of bromobenzene. As

$NO_2$ is meta directing firstly

$Cl$ is placed over 2,4 position of nitrobenzene, then reduction is done using

$Zn/HCl$ to produce, 1,3 dichlorobenzene. In the last step one

$Cl$ is replaced by

$Br$ using

$Cu_2Br_2$

The reaction of 1-bromo-3-chlorocyclobutane with metallic sodium in dioxane under reflux condition gives:

In the following reaction, identify the product.

0%
$\underset{\,\,\,OH}{\underset{|}{C}}\!\!\!H_2-\underset{CH_3}{\underset{|}{C}H}-C_2H_5$
0%
$C_2H_5-\underset{OH}{\underset{|}{C}H}-CH_2CH_3$
0%
$C_2H_5-CH_2-\!\!\underset{\,\,OH}{\underset{|}{C}}\!\!H - CH_3$
0%
$C_2H_5-CH = CH - CH_3$

Explanation

$\begin{array}{l}\text { Ethyl Magnesium Bromide undergoes S}_N {2} \text { type of reaction with the epoxy bond in the prescence } \\\text { of ether medium. Hence option (C) is correct. }\end{array}$

2002506_639705_ans_857248c5eaa44277872f67e9f83fe21b.PNG

Bromo ethane and isoproply chloride with metallic sodium in ether forms :

0%
Pentane
0%
2-methyl butane
0%
3-methyl butane
0%
2:3 dimethyl butane

$C_7H_8\overset{3Cl_2/ heat}{\rightarrow}A\overset{Fe/Br}{\rightarrow}B\overset{Zn/HCl}{\rightarrow}C$ . Here, the compound C is:

0%
$3$ -bromo $2,4,6$ -trichlorotoluene
0%
o-bromotoluene
0%
P-bromotoluene
0%
m-bromotoluene

Anisole can be prepared by the action of methyl iodide on sodium phenate. The reaction is called:

0%
Wurtz reaction
0%
Williamson's reaction
0%
Fittig's reaction
0%
Etard's reaction

Explanation

Anisole can be prepared by the action of methyl iodide on sodium phenate. The reaction is called Williamson's reaction.

$\displaystyle \underset { \displaystyle \text { phenol } }{ C_6H_5-OH} + NaOH \rightarrow \underset { \displaystyle \text { sodium phenoxide } }{ C_6H_5-ONa} \xrightarrow {\underset { \displaystyle \text { methyl iodide } }{ CH_3-I}} \underset { \displaystyle \text { anisole } }{ C_6H_5-O-CH_3}$

Williamson's reaction can be used for the preparation of simple as well as mixed ethers.

Wurtz reactions have low yield because of ___________

0%
the formation of multiple products.
0%
reaction is reversible
0%
reagent used are impure
0%
none of these

In electrophilic aromatic substitution reactions of chlorobenzene, the ortho/para-directing ability of chlorine is due to its _______.

0%
positive inductive effect $(+I)$
0%
negative inductive effect $(-I)$
0%
positive resonance effect $(+R)$
0%
negative resonance effect $(-R)$

Explanation

In electrophilic aromatic substitution reactions of chlorobenzene, the

$ortho /para$ ability of chlorine is due to its positive resonance effect (

$+R$ ).

The order of rate of

$S_N1$ in following iodo compounds is:

0%
I > II > III
0%
I > III > II
0%
III > II > I
0%
II > III > I

$A$ and

$B$ are:

0%
0%
0%
0%
none of these

Explanation

When dibromo alkaline is treated with alc.

$KOH$ followed by soda-amide and acidification, two molecules of

$HBr$ are lost and an alkyne is obtained. This alkyne is product (A).
This alkyne then reacts with sodamide followed by ethyl chloride to form product (B). H atom of terminal alkyne is replaced with ethyl group.

Which of the following reactions is not correctly matched?

0%
$2C_2H_5Br +2Na\xrightarrow{dry\ ether}C_4H_{10}+2NaBr$

Wurtz reaction
0%
$CH_3Br+AgF\rightarrow CH_3F +AgBr$

Etards reaction
0%
$C_6H_5Br +2Na+BrC_2H_5\xrightarrow{dry\ ether}C_6H_5C_2H_5+2NaBr$

Wurtz-fittig reaction
0%
$2C_6H_5Br +2Na\xrightarrow{dry\ ether}C_6H_5C_6H_5+2NaBr$

Fittig reaction

Explanation

Reaction in option

$B$ is not correctly matched. Etards reaction is an organic transformation in which alkyl group

$(-CH_3)$ bonded to an aromatic (benzene) ring undergoes oxidization and becomes an aldehyde functional group.

Also, the reaction given in option

$B$ is Swarts reaction. Here when methyl bromide reacts with silver fluoride, it decomposes and gives rise to methyl fluoride and silver bromide.

The antiaromatic compounds are:

0%
$x,w$
0%
$y,z$
0%
$x,z$
0%
$y,w$

Explanation

$\begin{array}{l} x=2 \pi e^{-}(\text {aromatic }) \\ y=4 \pi e^{-}(\text {anti-aromatic }) \\ z=10 \pi e^{-}(\text {aromatic }) \\ \omega=12 \pi e^{-}(\operatorname{anti}-\text { aromatic }) \\ \text { Hence }(D) y, w \text { is correct option. } \end{array}$
2004098_874493_ans_6d50a959bb8848348edb2d95d31e755a.PNG

2004098_874493_ans_6d50a959bb8848348edb2d95d31e755a.PNG

Arrange the following alkyl halides in decreasing order of the rate of

$\beta$ -elimination reaction with alcoholic

$KOH$ .
(I)

$CH_{ 3 }-\overset { \overset { H }{ | } }{ \underset { \underset { CH_{ 3 } }{ | } }{ C } } -CH_{ 2 }Br$
(II)

$CH_3-CH_2-Br$
(III)

$CH_3-CH_2-CH_2-Br$

0%
$I > II > III$
0%
$III > II > I$
0%
$II > III > I$
0%
$I > III > II$

Explanation

More the number of

$\beta$ - substituents (alkyl groups), more stable alkene it will form on

$\beta$ - elimination and more will be the reactivity.

Hence, the decreasing order of the rate of

$\beta$ -elimination reaction with alcoholic

$KOH$ is

$\rightarrow$

$\underset{2 \; \beta-substituents}{I} > \underset{1 \; \beta-substituents}{III} > \underset{0 \; \beta-substituents}{II}$ .

The major product(s) obtained in the following reaction is/are.

$\xrightarrow [ 2Na ]{ dry\ ether }$

Which of alkyl halides exhibits complete racemisation in

$S_N1$ reaction?

0%
$(CH)_3CCl$
0%
$CH_3CH_2CH_2Cl$
0%
$CH_3CH_2Cl$
0%
$C_6H_5CH_2Cl$

Explanation

All the molecules undergoing

$S_N1$ reaction shows racemisation as there is formation of 2 types of product in equal amount. Out of all these 4 alkyl halides ,

$C_6H_5CH_2Cl$ i.e benzylic chloride has the strongest tendency to undergo

$S_N1$ here thus leading to formation of racemic mixture as it contains benzylic carbocation which is most stable.

Which of the following alkane cannot be prepared by Wurtz's reaction in good yield?

Match the column I with column II.

	Columns I		Columns II
A	Carbon tetrachloride	i	Paint remover
B	Methylene chloride	ii	Refrigerators and air conditioners
C	DDT	iii	Fire extinguisher
D	Freons	iv	Non biodegradable insectiside

0%
A-(ii), (B)-(iii), (C)-(i), (D)-(iv)
0%
A-(i), (B)-(ii), (C)-(iii), (D)-(iv)
0%
A-(iv), (B)-(iii), (C)-(ii), (D)-(i)
0%
A-(iii), (B)-(i), (C)-(iv), (D)-(ii)

What is the product of given reaction?

Which of the following order is/are correct for the solvolysis in

$50\%$ aqueous ethanol at

$44.6^oC$ ?

The major product obtained on treatment of

$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } ( \mathrm { F } ) \mathrm { CH } _ { 3 }$ with

$\mathrm { CH } _ { 3 } \mathrm { O } ^{- }/ \mathrm { CH } _ { 3 } \mathrm { OH }$ is

0%
$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } \left( \mathrm { OCH } _ { 3 } \right) \mathrm { CH } _ { 3 }$
0%
$\mathrm { CH } _ { 3 } \mathrm { CH } = \mathrm { CHCH } _ { 3 }$
0%
$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } = \mathrm { CH } _ { 2 }$
0%
$\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 } \mathrm { OCH } _ { 3 }$

Which of the following does not represent major product of that reaction?

Which of the following biphenyl compound is not optically active?

0%
0%
0%
0%
Both $(B)$ and $(C)$

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 12 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 12 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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