MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 8

Which is the correct increasing order of bond dissociation enthalpy of

$C-X$ bond?

0%
$C-F , C-Cl , C-Br , C-I$
0%
$C-I , C-F , C-Cl , C-Br$
0%
$C-Cl , C-F , C-Br , C-I$
0%
$C-I , C-Br , C-Cl , C-F$

Explanation

Due to high bond length C-I has less bond strength and has less bond dissociation enthalpy. Because of short bond length C-F has high bond dissociation enthalpy so that increasing order of bond dissociation enthalpies is

$C-I<C-Br<C-Cl<C-F$ .

$CH_3 - CH_2 - Br \xrightarrow[\Delta]{Alco. KOH} B \xrightarrow{HBr} C\xrightarrow{Na/ether}D$ the compound D is

0%
ethane
0%
propane
0%
n-butane
0%
n-pentane

Explanation

The compound D is

$\displaystyle n-butane$ .

$\displaystyle CH_3 - CH_2 - Br \xrightarrow[\Delta]{Alco. KOH} \underset {B}{CH_2 = CH_2 } \xrightarrow{HBr} \underset {C}{CH_3 - CH_2 -Br} \xrightarrow{Na/ether}\underset {D}{CH_3 - CH_2 -CH_2-CH_3}$

When

$\displaystyle ethyl \: bromide$ is heated with

$\displaystyle alc. KOH$ , it undergoes dehydrohalogenation to form

$\displaystyle ethylene$ . Addition of a molecule of

$\displaystyle HBr$ to

$\displaystyle C=C$ double bond gives

$\displaystyle ethyl \: bromide$ . In presence of

$\displaystyle Na/ether$ , two molecules of

$\displaystyle ethyl \: bromide$ couple to form

$\displaystyle n-butane$ . In this reaction,

$\displaystyle C-C$ single bond is formed.

The reaction of

$1-$ bromo

$-3-$ chlorocyclobutane with metallic sodium in dioxane gives:

Which of the following is allylic halide?

0%
(1-Bromoethyl) benzene
0%
Benzyl chloride
0%
1-Bromo benzene
0%
3-Chloro cyclohex-1-ene

Explanation

$\bf{Explanation \ for \ Correct \ Option}$

$\bullet$ Allylic halides are the halides in which the halogen atom is bonded to

$sp^3-$ hybridized carbon atom next to carbon-carbon double bond

$(C=C)$ .

$\bullet$ General chemical structure of allylic halides:

$CH_2=CH−CH_2−X$

Here, X is any halogen atom.

So, according to the rule, only in 3-chloro cyclohex-1-ene halide

$(Cl)$ is attached with

$sp^3$ carbon which is next to carbon-carbon double bond.

$\bf{Explanation \ for \ Incorrect \ Option}$

$\bullet$

$(1-Bromoethyl)benzene$ and

$Benzyl \ Chloride$ are benzylic halides while

$1-Bromo benzene$ is Aryl halide.

$\bf{Conclusion-}$ Hence, the correct option is D.

Which of the following cycloalkane gives open chain compound when it reacts with bromine?

0%
Cyclopropane
0%
Cyclopentane
0%
Cyclohexane
0%
Cyclo-octane

Explanation

Cyclo propane with

$Br_{2}$ .

The ring is broken because cyclopropane suffers badly from ring strain. The bond angle in the ring are

$60^{\circ}$ rather than the normal value of about

$109.5^{\circ}$ when the carbon makes four single bonds.

$C_6H_6+Cl_2(excess)\xrightarrow[dark,\, cold]{Anhy. Alcl_3}P$

Product,

$P$ is:

0%
${ C }_{ 6 }{ H }_{ 5 }Cl$
0%
${ C }_{ 6 }{ H }_{ 4 }{ Cl }_{ 2 }$
0%
${ C }_{ 6 }{ H }_{ 6 }{ Cl }_{ 6 }$
0%
${ C }_{ 6 }{ Cl }_{ 6 }$

Explanation

Benzene on treatment with excess of chlorine in the presence of anhydrous

$Al{ Cl }_{ 3 }$ in dark yields hexachlorobenzene.

Which of the following is the correct order for strength of

$C - X$ bond ?

0%
$CH_3F < CH_3Cl < CH_3Br < CH_3I$
0%
$CH_3F > CH_3Cl >CH_3Br >CH_3I$
0%
$CH_3I>CH_3F > CH_3Cl > CH_3Br$
0%
$CH_3Cl > CH_3Br > CH_3F >CH_3I$

Explanation

As the size of halogen atom increases, the

$C-X$ bond length increases and hence, the

$C-X$ bond strength decreases.
Since, the order of size of halogen atom is

$F < Cl < Br < I$
Thus, the order of C-X bond strength is

$C-F > C-Cl > C-Br > C- I$

On mixing alkane with chlorine and irradiating with ultraviolet light, it forms only one monochloro alkane. The alkane is:

0%
propane
0%
pentane
0%
isopentane
0%
neopentane

Explanation

Neo-pentane

$H_3C - \overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}- CH_3$
Neo-pentane contains all hydrogen atoms equivalent, so it will give only one monochloro derivative on halogenation.

Which of the following cannot undergo

$E_2$ reaction?

0%
0%
0%
0%
None of these

Explanation

${ E }_{ 2 }$ elimination requires the presence of hydrogen at

$\alpha$ carbon of the substance.

Hence,

$\alpha -H$ is present as

${ 3 }^{ 0 }$ carbon

Therefore does not undergo

${ E }_{ 2 }$ elimination reaction.

The correct order of reactivity of the halides,
(I) ethyl chloride; (II) isopropyl chloride and (III) benzyl chloride in

$S_N 1$ reaction is:

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

Explanation

For $S_N 1$ reactions, the greater the stability of carbocations formed, faster will be the rate of reaction. Thus, the order of reactivity of halides is $\text{benzyl chloride > isopropyl chloride > ethyl chloride}$ .

(Due to greater stabilisation of benzylic carbocation intermediate by resonance, it shows higher reactivity in

$S_N 1$ reactions than other aliphatic halides).

The products formed in the reaction is(are):

0%
ethane, propane
0%
butane
0%
phrpane
0%
ethane, propane, butane

The major product is:

Explanation

Hence, option

$A$ is correct.

The treatment of

${R}_{3}MgX$ with ethyl formate leads to the formation of:

0%
$RCHO$
0%
${ CH }_{ 3 }{ CH }_{ 2 }OR$
0%
$RCHOH{ C }_{ 2 }{ H }_{ 5 }$
0%
${ C }_{ 2 }{ H }_{ 5 }OH\quad$

Explanation

Ethyl formate is written as $HCOO{C_2}{H_5}$ . The reaction is as follows

$HCOO{C_2}{H_5} + RMgX \to RCHO + Mg(O{C_2}{H_5})X$

Hence the correct answer is option A

The compound

$C_7 H_8 \xrightarrow { 3Cl_2/\triangle } X\xrightarrow { Br_2/Fe } Y\xrightarrow { Zn/HCl } Z$
The compound

$Z$ is :

0%
m-bromotoluene
0%
p-bromotoluene
0%
o-bromotoluene
0%
2, 4, 6-trichlorotoluene

Explanation

The above reaction is the preparation of aniline from chlorobenzene. In

$KNH_{2} - {NH_{2}}^{-}$ forms and substitutes on Benzene.
1175757_712220_ans_6bb183b3ac3040bebb26afaa6b82602a.png

Arrange the following alkyl chlorides in order of decreasing reactivity in a

$S_N1$ reaction.
(I) isopropyl bromide
(II) propyl bromide
(III) tert-butyl bromide
(IV) methyl bromide

0%
(III) > (I) > (II) > (IV)
0%
(I) > (III) > (IV) > (II)
0%
(IV) > (III) > (II) > (I)
0%
(I) > (II) > (III) > (IV)

Explanation

tret+-buty 1 bromide > isomeroply 1 bromide > propyle bromide > methyl bromide

i.e.

$(III) > (I) > (II) > (IV)$

Reason:

Since we know that in

$SN^{\prime}$ reaction first is

$10\ ss$ of a leaving group to give a carbocation, then nuclophile is attack

So carbocation must be stable

Carbocation Stability increases with increasing substitution of carbon

i.e. tertiary > secondry >> primary

The best yield of given product can be obtained by using which set of reactants

$X$ and

$Y$ respectively

0%
$PhLi+$ Neopentyl chloride
0%
$PhMgBr+$ Neopentyl bromide
0%
$t-Bu+MgBr+$ Benzyl bromide
0%
Benzylchloride $+$ t-Butyl chloride

Explanation

Given: In the given reaction the catalyst provided is Ether.

Solution: Usually, metallic reactions (Grignard's reaction) take place in the presence of ether. Since Grignard's Reagents are highly reactive compounds, they react with any compound present so as to obtain a proton in the presence of di-ethyl ether.

In the options provided, since the B option has a Grignard' Reagent present which is

$PhMgBr$ , it will react with Neopentyl Bromide

$(CH_3)_3C-CH_2Br$ and form the given product as shown in the above picture.

So, the Correct option : B

$CH_3.CH_2O\overset{CH_3}{\overset{|}{C}}H.CH_2.CH_2CH_2Cl$

The correct IUPAC name of the given compound is:

0%
2-ethoxy-5-chloropentane
0%
1-chloro-4-ethoxy-4-methylbutane
0%
1-chloro-4-ethoxypentane
0%
ethyl-1-chloropentylether

Explanation

$CH_3-CH_2-O-\overset{5CH_3}{\overset{|}{\underset{4}{CH}}}-\underset{3}{CH_2}-\underset{2}{CH_2}-\underset{1}{CH_2}-Cl$

$1-chloro-4-Ethoxypentane$

Parent chain is pentane

The product formed in the reaction is:

0%
$(BrCH_{2})_{3}CCH_{2}CH_{2}C(CH_{2}Br)_{3}$
0%
0%
0%

Which of the following will be most reactive for

$S_N1$ reaction?

0%
$(CH_3)_3C-Cl$
0%
$(CH_3)_2CH-Cl$
0%
$(CH_3)_2N-CH_2Cl$
0%
$(CH_3)_3C-Br$

Which does not have conjugate system?

0%
$CH_{2} = CHCl$
0%
$CH_{2} = CHCHO$
0%
$CH_{3}CH = CH_{2}$
0%

During the course of an

$S_N1$ reaction, the intermediate species formed is:

0%
a carbocation
0%
a free radical
0%
a carbanion
0%
an intermediate complex

Which one of the following is best catalyst for the reaction shown below ?

${ CH }_{ 3 }({ CH }_{ 2 }{ ) }_{ 8 }{ CH }_{ 2 }{ Br }\overset { KCN }{ \underset { benzene }{ \longrightarrow } } { CH }_{ 3 }({ CH }_{ 2 }{ ) }_{ 8 }{ CH }_{ 2 }CN$

Which of the following compounds is optically active?

Explanation

Refer to image 01

Optically inactive due to centre of symmetry.

Refer to image 02

Both the wedges (are on the same side of plane

$\rightarrow$ Optically active

1020791_874584_ans_fbfaecd70ef6460ba07f497bfbb32ec1.png

Among the following, the optically inactive compound is :

Explanation

a) Here aminos are optically active all the attached

$3$ groups are different, thus making it available for pyramidal inversion. (Ref. Image [a])

b) Same in the case of phosphines but the rate of inversion is lesser than that of aminos. (Ref. Image [b])

c) Allenes exhibit optical activity, if these is no condition like

$a = b$ or

$e = d$ (Ref. Image [c])

d) Chiral carbon

$\rightarrow$ optically active. (Ref. Image [d])

1132419_878567_ans_e9c0d6aa846b4db4914e855c0713660d.png

Specific rotation is the rotation produced by

$1$ g of material in

$1$ ml of liquid in a

$1$ dm tube. Which of the following molecules have largest value of specific rotation? (Hint : Molecular mass increasing).

The IUPAC name of

$(CH_3)_2CH-CH_2-CH_2Br$ is ?

0%
1-bromopentane
0%
1-bromo-3-methylbutane
0%
2-methyl-4-bromobutane
0%
2-methyl-3-bromopropane

Explanation

4 3 2 1

$CH_{3}-\underset {\underset{\displaystyle {CH_{3}} }{\displaystyle |}}{C}H–CH_{2}–CH_{2}Br$

Its IUPAC name is 1-bromo-3-methylbutane.

The correct option is B.

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

$\beta$ - elimination reaction with alcoholic KOH.

${ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }Br\quad { CH }_{ 3 }-\underset { \underset { { \displaystyle \,\,\,\,CH }_{ 3 } }{ | } }{ CH } -{ CH }_{ 2 }Br\quad { CH }_{ 3 }-{ CH }_{ 2 }-\underset { \underset {\displaystyle Br }{ | } }{ CH } -{ CH }_{ 3 }\quad$
(i) (ii) (iii)

0%
$(ii) > (iii) > (i)$
0%
$(iii) > (ii) > (i)$
0%
$(i) > (ii) > (iii)$
0%
$(ii) > (i) > (iii)$

Explanation

Rate of dehydrohalogenation in alkyl halide is in the order

$\Rightarrow \; 3°>2°>1°$ .

Thus, correct order is

$(iii)>(ii)>(i)$ .

The IUPAC name of tertiary butyl chloride is ?

0%
2-chloro-2-methylpropane
0%
3-chlorobutane
0%
4-chlorobutane
0%
1, 2-chloro-3-methylpropane.

Explanation

Tertiary butyl chloride is an organochloride with formula

$(CH_3)_3CCl$ . The longest continuous chain of carbons is 3. It is propane. The remaining groups are chloro and methyl. Both are at position 2. Chlorine comes before methyl in the alphabet, so the IUPAC name of tertiary butyl chloride is 2-chloro-2-methylpropane.

The reaction of Sodium methoxide with

${R}_{2}CHX$ is:

0%
${SN}^{1}$
0%
${SN}^{2}$
0%
Elimination
0%
Substitution

Explanation

Sodium methoxide

$\Rightarrow { CH }_{ 3 }\overset { \left( - \right) }{ O } \overset { + }{ Na }$

As the corboncation formed is stable by inductive effect and hyper conjugation. It undergoes

${ SN }^{ 1 }$ reaction.

${ SN }^{ 1 }$ reaction means the formed carboncation should be stable.

1034931_1075134_ans_fbf8733f07064141bbd296851145f071.jpg

The reaction proceeds with 98% racemisation. The reaction may follow:

0%
$S_N1$ mechanism
0%
$S_N2$ mechanism
0%
E1 mechanism
0%
E2 mechanism

Explanation

The mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer. Such mixture is racemic modification. The process of conversion of enantiomer into a racemic mixture is called as racemisation. In case of optically active alkyl halides

$S_N1$ reactions are accompanied by racemisation. This is because the carbocation formed in slow step is planar.

Choose the correct answer among the alternatives given :Bromination of methane in presence of sunlight is a:

0%
nucleophilic substitution
0%
free radical substitution
0%
electrophilic substitution
0%
nucleophilic addition

Explanation

The reaction between bromine and methane happens in the presence of sunlight.

$CH_4 + Br_2 \to CH_3 + H Br$ . The product obtained is bromomethane. One of the hydrogen atom in the methane is replaced by a bromine atom. So this is substitution reaction. During the chain reaction, a new species is generated at the end. The overall process is called as free radical substitution.

Which of the following is not correctly matched with its IUPAC name?

0%
$CHF_2CBrClF$ : 1-Bromo-1-chloro-1, 2, 2 -trifluoroethane
0%
$(CCl_3)_3CCl$ : 2-(Trichloromethyl)-1,1,1,2,3,3, 3 -heptachioropropane
0%
$CH_3C(p-ClC_6H_4)_2CH(Br)CH_3$ : 2-Bromo-3,3-bis(4-chlorophenyl) butane
0%
$o-BrC_6H_4CH(CH_3)CH_2CH_3$ : 2-Bromo-1-methylpropylbenzene

Explanation

IUPAC system is one of the systematic method of naming organic compound. The numerals 1,2; 1,3 and 1,4 are used in IUPAC system. The option D is not correctly matched with its IUPAC name because the correct IUPAC name for the structure

$o - BrC_6H_4CH(CH_3)CH_2CH_3$ is 1 Bromo-2-1-methlypropylbenzene.

Choose the correct option from the following:

0%
In the electrophilic substitution of toluene with $Br_2$ , iron (III) bromide acts as a Lewis acid.
0%
In the reaction of toluene with $Cl_2/FeCl_3$ , ortho and para isomers are easily separated.
0%
Similar reaction with iodine is reversible in nature
0%
All of these

A compound X with molecular formula

$C_7H_8$ is treated with

$Cl_2$ in presence of

$FeCl_3$ . Which of the following compounds are formed during the reaction?

Which of the following is the most reactive towards nucleophilic substitution reaction?

0%
$ClCH_2-CH=CH_2$
0%
$CH_2=CH-Cl$
0%
$CH_3CH=CH-Cl$
0%
$C_6H_5Cl$

The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ________.

0%
0%
0%
0%
mixture of (b) and (c)

Explanation

$-CH_3$ is ortho & para directing group so

$Cl$ will come at p-position.But major product is para-chloro toluene.The reaction is electrophilic aromatic substitution reaction.
1032548_939598_ans_c2426e54a44a41a084a81e5791272f09.PNG

1032548_939598_ans_c2426e54a44a41a084a81e5791272f09.PNG

Choose the correct answer among the alternatives given :

In the following pairs of halogen compounds , which compound undergoes faster

$S_N1$ reaction?

Explanation

$S_N1$ reaction proceeds via formation of carbocation. In option B, the alkyl halide (1) is

$3^{\circ}$ while (2) is

$2^{\circ}$ .

Therefore greater the stability of the carbocation, faster is the rate of

$S_N1$ reaction.

Therefore option B, the compound pair of 2-chloro-2 methylpropane and 2- chloroheptane is the correct option.

Reaction of

$C_6H_5CH_2Br$ with aqueous sodium hydroxide follows ___________.

0%
$S_N1$ mechanism
0%
$S_N2$ mechanism
0%
Any of the above two depending upon the temperature of reaction
0%
Saytzeff rule

Explanation

In aqueous media

$SN^1$ mechanism follow.In this reaction stable carbocation get form as intermediate.
1174084_939677_ans_026a24d409564dc19696c12c5a4754cb.PNG

1174084_939677_ans_026a24d409564dc19696c12c5a4754cb.PNG

Chlorobenzene is formed by reaction of chlorine with benzene in the presence of

$A1C1_3$ .Which of the following species attacks the benzene ring in this reaction?

0%
$C1^-$
0%
$Cl^+$
0%
$AlCl_3$
0%
$[AlCl_4]^-$

Which is the correct IUPAC name for,

$CH_3-{\underset{C_2H_5}{\underset{|}{CH}}}-CH_2-Br$ :

0%
1-Bromo-2-ethylpropane.
0%
1-Bromo-2-ethyl-2-methylethane.
0%
1-Bromo-2-methylbutane.
0%
2-Methyl-1-bromobutane.

Explanation

Rules for IUPAC Name:

Numbering should start at the substituent carbon and longest chain should be covered.

$1-Bromo-2-methylbutane$

1032525_939586_ans_bf71af14bf7f40008cb3d9a3603a55f4.gif

For the given Question (1,2,3) consider the following reaction.
Which halogen will give the best yield of a single monohalogenation product ?

0%
$F_2$
0%
$Cl_2$
0%
$Br_2$
0%
$I_2$

Explanation

For monohalogenation product we prefer bromination as compared to flourination or chlorination to cleave the

$H-C$ bond requires

$397Kj/mol$ and

$H-Br$ is formed which has bond dissociation energy of

$366Kj/mol$ . From Bond energy value we can say that bromination is more selective.

Chloromethane on treatment with excess of ammonia yields mainly :

0%
N, N-dimethylmethanamine
0%
N-methylmethanamine $(CH_3-NH-CH_3)$
0%
methanamine $(CH_3NH_2)$
0%
mixture containing all these in equal proportion.

Explanation

This reaction of chloromethane results in formation of methanamine after reacting with

$NH_3$

1156965_939631_ans_d2b2a897322b4eb49cad03b80a96286a.PNG

Alkyl halides react with metallic sodium in dry ether producing?

0%
Alkanes with same number of carbon atoms
0%
Alkanes with double number of carbon atoms
0%
Alkenes with triple number of carbon atom
0%
Alkenes with same number of carbon atom

Explanation

$RX+2Na+X-R\xrightarrow {dry ether} R-R+2Na-X$
alkyl halide
alkane sodium halide

What should be the correct IUPAC name for diethylbromomethane?

0%
1 - Bromo- 1, 1 - diethylmethane
0%
3 - Bromopentane
0%
1 - Bromo-1 - ethylpropane
0%
1 - Bromopentane

Explanation

The correct IUPAC name for diethylbromomethane is 3 - Bromopentane. It has molecular formula

$C_5H_{11}Br$ .

Therefore

$B$ is the correct option.

Which of the following reactants is suitable for preparation of methane and ethane by using one step only?

0%
$H_2C = CH_2$
0%
$CH_3OH$
0%
$CH_3-Br$
0%
$CH_3-CH_2-OH$

Explanation

$2CH_3-Br+Na \longrightarrow \underset {Ethane}{CH_3-CH_3}+2NaBr$

Mechansim:-

The above reaction is Wurtz reaction., in which alkane are produced from alkyl halide in presence of metal.

Thus, this above method can be used to prepare Ethane.

$CH_3-Br \xrightarrow []{LiAlH_4}CH_4+Br^-$

Mechanism:-

$LiAlH_4$ produce

$H^{\circleddash}$ ion which acts as nucleophile and undergoes nucleophilic substitution gives alkane.

$Li^+$

$Al\overset {-}{H_4}$

$LiAlH_4$ reduce alkyl halide to alkane.

Thus the above reactant can give methane and ethane in

$1$ step.

Additional of

$Br_2$ to trans-2- butene would give a product which is :

0%
achiral
0%
racemic
0%
meso
0%
optically active

Rank in the order of increasing acidic strength:

0%
$A < B < C$
0%
$A < C < B$
0%
$B < A < C$
0%
$B < C < A$

Explanation

The greater

$I$ effect is favorable or greater acidic strength. The

$-I$ effect of aldehyde group is greater than the ester group. Due to this

$C$ will be most acidic. In

$A$ and

$B$ ,

$B$ will be less acidic as it has

$-I$ effect sort of localized due to ester group being on a single carbon. In

$A$ it is more delocalized is not effective on more by adjacent carbons.

Hence, the required order is-

$B<A<C$ .

Match the column I with column II and with column III.

0%
a - q - w; b - s - x; c - p - y; d - q - z
0%
a - b - x; q - s - z; c - p - w; d - q - x
0%
a - q - x; b - s - z; c - p - w; d - q - x
0%
a - s - x; b - p - z; c - q - w; d - q - x

Explanation

$(a)$ As refer to the IMAGE $01\rightarrow$ Due to geometrical Isomerism, it has structure at $C-2$

$(b)$ As refer to the IMAGE $02\rightarrow$ Possible structure has $4$ isomers and $5$ stereoisomers.

$(c)$ As refer to the IMAGE $03\rightarrow$ Only at terminal carbons same for any substitution.

$(d)$ As refer to the IMAGE $04\rightarrow$ Structure are $3$ stereoisomers.

1100143_988890_ans_1096a82729c54e9093348cfbeebcaef5.png

$\underrightarrow{AgNO_3} (A)$ ;
Which statement is incorrect in respect of the above reaction?

0%
Product is aromatic
0%
product has high dipole moment
0%
Product has less resonance energy
0%
Product is soluble in polar-solvent

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CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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