MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 6

Ethyl chloride is converted to butane by :

0%
Kolbes synthesis
0%
Williamson's synthesis
0%
Wurtz reaction
0%
Gattermann reaction

Explanation

Williamson's synthesis: Sodium ethoxide reacts with alkyl halide and ether forms as product.

$RONa+RX\rightarrow ROR +NaX$

Wurtz reaction:

$RX+2Na+RX\rightarrow R-R+2NaX$

It is clear from above reactions Option C is correct.

Chloroethane is reacted with alcoholic potassium hydroxide. The product formed is:

0%
$C_{2}H_{6}O$
0%
$C_{2}H_{6}$
0%
$C_{2}H_{4}$
0%
$C_{2}H_{4}O$

Reaction of ethyl chloride with sodium leads to (in the presence of dry ether):

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

Explanation

The above reaction is Wurtz Reaction where alkyl halides react with sodium in presence of dry ether to form higher hydrocarbon.

$2C_2H_5-Cl\xrightarrow{2Na}C_4H_{10}+2NaCl$

Option C is correct.

$1$ -Chlorobutane on reaction with alcoholic potash gives:

0%
$1$ -butene
0%
$1$ -butanol
0%
$2$ -butene
0%
$2$ -butanol

Explanation

Alc.

$KOH$ is a dehydrohalogenation agent, it reacts with an alkyl halide and alkene forms as a product.

Reaction of process:

$CH_3CH_2CH_2CH_2Cl\xrightarrow {alc.KOH} CH_3CH_2CH=CH_2$

Option A is correct.

Which of the following is not true for

$S_{N}1$ reactions ?

0%
These occur through a single step.
0%
These are favoured by polar solvents.
0%
$3^{o}$ alkyl halides generally react through this mechanism.
0%
Concentration of nucleophile does not affect the rate of such reactions.

Explanation

The mechanism above shows a

$S_N1$ reaction.

The first step involves the slow loss of leaving group

$L$ giving a carbocation. In the second step, attack by nucleophile

$(Nu^-)$ occurs on the carbon centre of carbocation.

$S_N1$ mechanism proceeds through carbocation formation and so, it is a two-step process and like the stability of carbocation increases, reaction favours

$S_N1$ mechanism. This is because the formation of the carbocation is a slow and rate-determining step.

Thus, the greater is the stability of carbocation, greater will be the possibility of its formation and hence, greater will be the rate.

$3^o$ carbocations are more stable than others due to the electron donating effect of three alkyl groups. Polar solvents also increases the stability of carbocation by surrounding it through negative ends. Concentration of nucleophile does not affect the rate of such reactions.

Hence, the statement that the

$S_N1$ reactions occur through a single step is incorrect.

$\therefore$ Option A is the correct answer.

$C_6H_5-\!\!\!\!\!\underset{H}{\underset{|}{\overset{\,\,\,\,\,\,CH_3}{\overset{|}{C}}}}\!\!\!\!\!-Br+H_2O \to HO -\!\!\!\!\!\underset{H}{\underset{|}{\overset{\,\,\,\,\,\,CH_3}{\overset{|}{C}}}}\!\!\!\!\! -C_6H_5+HBr$

The above reaction result in 98% racemization, the reaction mainly follows:

0%
$S_{N}1$ mechanism
0%
$S_{N}2$ mechanism
0%
$E^{1}$ mechanism
0%
$E^{2}$ mechanism

Explanation

An alkyl halide with chiral carbon forms a racemic mixture through

$SN^1$ mechanism.

$SN^2$ mechanism changes the configuration of a compound due to formation of the transition state.

$SN^1$ mechanism proceeds through carbocation formation so it's a two-step process and as stability of carbocation increases, reaction favours

$SN^1$ mechanism.

$3^0$ carbocation are more stable than others and also polar solvent increases the stability of carbocation so they both favours

$SN^1$ mechanism.

Also, the rate of

$SN^1$ mechanism depends on nucleophile concentration only so as conc. of nucleophile increases, reaction favours

$SN^1$ mechanism.

Reaction of Chlorobenzene with

$CH_3Cl$ in presence of

$AlCl_{3}$ gives:

0%
Toulene
0%
m-chloro toulene
0%
Only o-chloro toulene
0%
Mixture of o- and p-chlorotoulene

Explanation

$Cl^-$ group attached to benzene is ortho, para directing so it forms

$o-,p-$ chloro toluene.
This is a Fridel Craft's alkylation reaction.

The correct match is :

	List-I		List-II
A	$C_2H_5Cl$	1	Williamson synthesis
B	R-X+Mg+dry ether	2	Wurtz reaction
C	$C_2H_5Cl+C_2H_5ONa$	3	Anaesthetic
D	Na+dry ether	4	Antiseptic
		5	Grignard reagent

0%
$A-3, B-5, C-1, D-2$
0%
$A-5, B-3, C-1, D-2$
0%
$A-3, B-4, C-1, D-2$
0%
$A-3, B-5, C-1, D-4$

Explanation

Wurtz reaction:

$R-X+2Na+R-X\xrightarrow{dry\ ether} R-R+2NaX$

Grignard reaction:

$R-X+Mg\xrightarrow {dry\ ether} R-Mg-X$

Williamson's synthesis:

$C_2H_5Cl+C_2H_5ONa\rightarrow C_2H_5OC_2H_5$

$C_2H_5Cl$ , ethyl chloride, is a local anesthetic. It was commonly used in the production of tetraethyllead (TEL), which is an additive for gasoline.

The IUPAC name is:

$H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{Cl}{\overset{|}{C}}}}-Cl$

0%
$1,2$ - dichoroethane
0%
$2,2$ - dichloroethane
0%
$1,1$ - Dichloroethane
0%
dichloroethane

Which of the following statement is not correct?

0%
Chlorobenzene is more reactive than benzene towards electrophilic substitution
0%
$C- Cl$ bond in chlorobenzene is less polar than $CH Cl_{3}$
0%
Chlorobenzene is less reactive than $CH Cl_{3}$
0%
In chlorobenzene further substitution takes place at ortho and para positions

Explanation

$Chlorine$ shows two effects

$-I$ &

$+M$ .

Due to

$the - I$ effect of

$Cl$ group, it reduces the electron density of benzene so chlorobenzene is less reactive than benzene for electrophilic substitution reaction.

But due to the

$+M$ effect, it is ortho, para directing, and in chlorobenzene further substitution takes place at ortho and para positions.

Option A is correct.

Which of the following molecules is expected to rotate the plane of plane-polarised light?

Which of these is o, p-directing and deactivating group?

0%
$-F$
0%
$-Cl$
0%
$-Br$
0%
All of these

Explanation

$X-$ (halogen group) has two effects

$-I$ &

$+M$ .

Due to the

$-I$ group, it reduces the electron density of benzene so halobenzene is less reactive than benzene for electrophilic substitution reaction.

But due to the

$+M$ effect, it is ortho, para directing (resonance effect predominates over inductive effect) and in halobenzene, further substitution takes place at ortho and para positions.

The compound

$X$ is:

0%
o-methylchlorobenzene
0%
p-methylchlorobenzene
0%
m-methylchlorobenzene
0%
both $A$ & $B$

Explanation

$Cl-$ group attached to benzene is ortho, para directing so it forms

$o-,p-$ chloro toluene.

Reaction is as shown.

The reaction of toluene with

$Cl_{2}$ in presence of

$Fe$ , the product(s) formed is/are:

0%
o- and p-chlorotoluene
0%
benzyl chloride
0%
m-chlorotoluene
0%
only p-chlorotoluene

Explanation

$-CH_3$ group on benzene is ortho and para directing.

In presence of

$Cl^-$ , it forms ortho (minor) and para (major) chlorotoluene as products.

Option A is correct.

$C_{4}H_{9}Cl$ with

$Na/ether$ gives

$2,5$ -dimethylhexane, then the compound,

$C_{4}H_{9}Cl$ is:

0%
n-butylchloride
0%
sec-butylchloride
0%
iso-butylchloride
0%
tert-butylchloride

Explanation

The reaction of Isobutyl chloride

$(C_{4}H_{9}Cl)$ with

$Na/Ether$ gives

$2,5$ -dimethylhexane. This is a condensation reaction in which two molecules of sodium chloride are obtained for every molecule of

$2,5$ -dimethylhexane.

Option C is correct.

In the following pair of compounds, which compound is more nucleophilic in a polar-protic solvent?

${ CH }_{ 3 }Cl$ or

${ CH }_{ 3 }OH$

0%
${ CH }_{ 3 }Cl$
0%
${ CH }_{ 3 }OH$
0%
${ CH }_{ 3 }Cl = { CH }_{ 3 }OH$
0%
None of these

Explanation

Hydrogen bonding takes place in the polar-protic solvent when solute compound includes one of FON (Flourine, Oxygen and Nitrogen) atom.

So, in case of

$CH_3OH$ hydrogen bonding will take place between O and H atom, which decreases its nucleophilicity. Polar protic-solvent consists of

$H^+$ ions which can form H-bond with methanol and shield the solute molecule from donating electron thus decreases its nucleophilicity. However, there is no H-bonding present as such in case of methyl chloride

$(CH_3OH)$ .

Therefore,

$CH_3Cl$ is more nucleophilic than

$CH_3OH$ .
Hence, the correct option is A.

For the given pair of

${{ S }_{ N }}1$ reactions, the ______ reaction is faster.

0%
first
0%
second
0%
both progress at the same rate
0%
cannot be determined

Explanation

The given pair of

${ S }_{ N }1$ reactions, the first reaction in

${ S }_{ N }1$ reaction is faster because, after heterolysis of

$C-Br$ bond, the primary carbocation formed undergoes rearrangement via hydride shift which leads to the formation of a more stable tertiary carbocation.

Which of the following relation is correct for recativity?

${ CH }_{ 3 }COO^{ - }$ or

${ CH }_{ 3 }{ CH }_{ 2 }S^{ - }$ with

${ CH }_{ 3 }I$ in ethanol.

0%
${ CH }_{ 3 }{ CH }_{ 2 }S^{ - } > { CH }_{ 3 }COO^{ - }$
0%
${ CH }_{ 3 }{ CH }_{ 2 }S^{ - } < { CH }_{ 3 }COO^{ - }$
0%
${ CH }_{ 3 }{ CH }_{ 2 }S^{ - } = { CH }_{ 3 }COO^{ - }$
0%
None of these

Explanation

Higher electronegativity means lower nucleophilicity because the role of a nucleophile is to share e-. If the atom is more electronegative it is less willing to share its e- and wants to hold onto them. In the polar protic solvent like ethanol

$CH_3COO^-$ participating in Hydrogen bonding, the nucleophile is considerably less reactive than

$CH_3CH_2S^-$ as it interacts less to hydrogen ion. Thus the reaction will be faster for

$CH_3CH_2S^-$ than

$CH_3COO^-$ .

Which of the given pair of

${ S }_{ N }1$ reaction would you expect to proceed faster? Explain your answer.

0%
${3}^{o}$ -alcohol > ${2}^{o}$ -alcohol
0%
${3}^{o}$ -alcohol < ${2}^{o}$ -alcohol
0%
${3}^{o}$ -alcohol = ${2}^{o}$ -alcohol
0%
None of these

Explanation

Option A is the correct answer as tertiary alcohol reacts faster than secondary alcohol which in turn reacts faster than primary alcohol towards

$S_N1$ reaction.

Which of the following chlorides is highly reactive in

${{ S }_{ N }}1$ reaction, even solvolysed in cold water?

0%
0%
0%
0%
None of the above

Explanation

Chlorocycloheptane is more reactive in

$S_{N}1$ reaction.

As water is a polar protic solvent which has more number of highly polarised hydrogen atoms which react with hydrogen attached to an electronegative atom resulting in cleavage of

$Cl$ .

As for size increases, the basicity decreases and nucleophilicity increases which makes the reaction to occur more reactively.

The correct option: A

Statement: In the above reaction, first compound in the product side is the major product.

State whether the given statement is true or false

0%
True
0%
False

Explanation

It is an example of SN

$^1$ reaction

${SN}'$ reaction

$3^{\circ}>2^{\circ}>1^{\circ}$ is the order of reactivity.

$[ I ]$ formed from

$2^{\circ}$ carbocation

$[ II ]$ formed from

$3^{\circ}$ carbocation which will be major.

Hence, the given statement is

$false$

Which of the following statements/reactions are correct?

0%
$S_N2/E2$ ratio is higher with $RS^{\circleddash}$ than for those with $RO^{\circleddash}$
0%
$S_N2/E2$ ratio is higher for $1^o RX$ and least for $3^oRX$
0%
$Me_3C-Br+KCN\longrightarrow Me_3C-CN$
0%
None of the above

Explanation

The statements (A) and (B) are correct.
(A)

$\displaystyle S_N2/E2$ ratio is higher with

$\displaystyle RS^-$ than for those with

$\displaystyle RO^-$ . With stronger nucleophiles,

$\displaystyle S_N2$ reaction is favored. With stronger bases,

$\displaystyle E2$ reaction is favored.

(B)

$\displaystyle S_N2/E2$ ratio is higher for

$\displaystyle 1^o$ RX and least for

$\displaystyle 3^o$ RX.
Primary alkyl haldies (

$\displaystyle 1^o$ RX) prefer

$\displaystyle SN^2$ over

$\displaystyle SN^1$ .
Tertiary alkyl haldies (

$\displaystyle 3^o$ RX) prefer

$\displaystyle SN^1$ over

$\displaystyle SN^2$ .

(C) The reaction

$\displaystyle Me_3C-Br+KCN\longrightarrow Me_3C-CN$ will not occur as

$\displaystyle Me_3C-Br$ is a tertiary alkyl halide. Eliminiation will be favouerd our substitution.

$\displaystyle Me_3C-Br\xrightarrow [-HBr]{KCN} Me_2C=CH_2$

The alkyl halides required to prepare

$(CH_3)_2CH-CH_2-CH_2-CH_3$ by Wurtz reaction are:

$\underset { Specific\quad rotaion=+{ 50 }^{ o } }{ Ph-CH(OH){ CH }_{ 3 } } \xrightarrow [ ]{ SO{ Cl }_{ 2 } } \quad \underset { |\\ Cl }{ PhCl } -{ CH }_{ 3 }$
Which of the following mechanisms does the reaction proceed?

0%
${S_N}i$
0%
${S_N}^{2}$
0%
${S_N}^{1}$
0%
$E_2$

Explanation

The reaction proceeds through

$SN^1$ mechanism. The

$-OH$ group of optically active substrate is replaced with

$-Cl$ atom to form racemic mixture as the specific rotation of substrate is mentioned but that of product is not mentioned.

Assertion:

$E1cB$ reaction is favoured by stabilisation of carbanion and poor leaving group
Reason: The reaction is kinetically of the second-order and unimolecular

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

$E1cB$ reaction is a second-order and unimolecular reaction. The reaction proceeds in two steps. In the first step, the base abstracts an acidic hydrogen atom from the beta carbon atom to form the carbanion transition state, which is stabilized by resonance. In the second step, the lone pair of electrons on the anion moves to the neighbouring atom forming a pi bond with elimination of leaving group. The rate law for this reaction can be given as

$Rate\quad =k\left[ concentration\ of\ substrate \right] \left[ concentration\ of\ base \right]$

Hence, greater the stability of carbanion intermediate faster will be reaction. Hence, both assertion and reason are correct.

Assertion:

$1^o$ allylic halides are more reactive than

$1^oRX$ in

$S_N1$ reaction
Reason: Allylic carbocation intermediate is stabilised by resonance

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

Reactivity of

$S_N1$ reaction is compared by the stability of its carbocation intermediate.

$1^o$ allylic halides are more reactive than

$1^o$

$RX$ in

$S_N1$ reaction due to the formation of the allylic carbocation.

Allylic carbocation intermediate is stabilised by resonance.

Hence correct option is A

Which of the following aromatic compound is least reactive towards electrophilic substitutions?

Explanation

Pyridine is least reactive towards ES because pyridine is resonance stabilised as shown in first figure.
Because of withdrawal of

${ e }^-$ 's from the ring by

$N$ atom, the ring is deactivated, thereby resembling the benzene ring in nitrobenzene. So, it is least reactive towards ES reaction.

$(b)$ and

$(d)$ are resonance structures so ES reaction takes place easily.
ES reaction in pyrrole, thiophene, and furan takes place at

$C-2$ or

$C-5$ positions because there are three resonating structures when

${E}^{\oplus }$ attacks at

$C-2$ or

$C-5$ and two resonating structures when

${E}^{\oplus }$ attacks at

$C-3$ .

The first steps of

${SN}^{1}$ and

${SN}^{2}$ reactions are, respectively:

0%
Both exothermic
0%
Both endothermic
0%
Endothermic and exothermic
0%
Exothermic and endothermic

Explanation

${SN}_{2}$ , there is only one step, and it is exothermic. But in

${SN}_{1}$ there are two steps first step; is endothermic and second step is exothermic.

Which one of the following compounds gives

${S_N}{1}, {S_N}{2}$ and

${S_N}{2'}$ mechanisms?

Explanation

Allyl halides undergo

${SN}^{1}, {SN}^{2}$ and

${SN}^{2'}$

Nitration of xylene gives only one mono-nitro derivative. Which xylene is it?

0%
ortho
0%
meta
0%
para
0%
Both o and p

During

$S_N1$ reactions, the leaving group leaves the molecule before the incoming group is attached to the molecule

0%
True
0%
False

Explanation

$S_N1$ , first the ionisation of

$RX$ takes place. which means the leaving group leaves the molecule before the attack of

$Nu^-$

Mechanism:

An example of a reaction taking place with an

$S_N1$ reaction mechanism is the hydrolysis of tert-butyl bromide with water forming tert-butanol:

By which mechanism does the above reaction proceed?

0%
$E_1$
0%
$E_2$
0%
$E_1cb$
0%
$\gamma$ -Elimination

Explanation

If the reaction proceeds by

$E_1$ mechanism, then the breaking of

$C-H$ and

$C-D$ bonds is not involved in the slow step (or rate determining effect). Based on this both products

$\displaystyle CH_2=CH-CD_3$ and

$\displaystyle CH_3-CH=CD_2$ are expected. However only

$\displaystyle CH_2=CH-CD_3$ is obtained. This indicates that breaking of

$C-H$ bond (which is faster than breaking of

$C-D$ bond) is involved in the mechanism along-with breaking of

$C-Br$ bond. This indicates

$E_2$ mechanism for elimination.

Which of the following represents Westrosol?

0%
$CH{Cl}_{3}$
0%
${CH}_{2}{Cl}_{2}$
0%
$CH{Cl}_{2}{CH}_{2}Cl$
0%
$C{Cl}_{2}=CHCl$

Explanation

When westron vapours are passed over heated

$Ba{Cl}_{2}$ or lime, westrol (trichloro ethylene) is obtained.

${Cl}_{2}CH-CH{Cl}_{2}\xrightarrow [ Ba{ Cl }_{ 2 } ]{ } {Cl}_{2}C=CHCl$ (Westrol)

The chemical reaction:

${ \left( { CH }_{ 3 } \right) }_{ 3 } CBr \xrightarrow [ { -H }_{ 2 }O,-KBr ]{ KOH(alc.) } { \left( { CH }_{ 3 } \right) }_{ 2 }C={CH}_{2}$ is an example of:

0%
nucleophilic substitution
0%
electrophilic substitution
0%
free radical substitution
0%
$\beta$ -elimination

${C}_{3}{H}_{7}Cl \xrightarrow [ ]{ KOH(alc.) } (A) \xrightarrow [ 770K ]{ { Cl }_{ 2 }(g) } (X)$ .

$(X)$ can be:

0%
vinyl chloirde
0%
allyl chloride
0%
ethyl chloride
0%
ethyl iodide

Explanation

$(A)\Rightarrow {CH}_{3}-CH={CH}_{2}$

$(X)\Rightarrow$ Allylic chlorination

$Cl-{CH}_{2}-CH={CH}_{2}$
Allyl chloride

Ethanol

$\xrightarrow [ ]{ { P }_{ 4 }/{ I }_{ 2 } } (X) \xrightarrow [ (ii)HBr ]{ (i)KOH(alc.) } (Y)$
In this sequence of reaction ,

$(Y)$ is:

0%
ethene
0%
bromoethane
0%
ethanol
0%
none of these

${C}_{6}{H}_{5}Cl \xrightarrow [ 625K, \ 300atm ]{ NaOH(aq.) }$

The product can be:

0%
benzal
0%
sodium benzoate
0%
benzol
0%
sodium phenate

Explanation

$PhCl\xrightarrow [ High\:temp.\: and\:pressure ]{ NaOH } \: \underset { Sodium\:Phenate }{ PhONa }$

Which of the following is called Westron?

0%
${CH}_{3}Cl$
0%
$CH{Cl}_{3}$
0%
$CH{Cl}_{2}.CH{Cl}_{2}$
0%
$C{Cl}_{2}=CHCl$

Explanation

$CH\equiv CH+2{Cl}_{2}\xrightarrow [ ]{ C{ Cl }_{ 4 } } \underset { Westron\: or\\ 1,1,2,2-Tetrachloro\: ethane\: or\\ Acetylene\: tetrachloride }{ C{ l }_{ 2 }-CH-CH-C{ l }_{ 2 } }$
Westron is used as industrial solvent for rubber, fats and varnishes. It has some insecticidal action.

$(X)$ on treatment with sodium hydroxide followed by the addition of silver nitrate gives white precipitate at room temperature which is soluble in

${NH}_{4}OH$ .

$(X)$ can be:

0%
chlorobenzene
0%
ethyl bromide
0%
benzyl chloride
0%
vinyl chloride

Explanation

Silver nitrate solution can be used to find out which halogen is present in a suspected halogenoalkane.

$NaOH$ substitutes the

$Cl^-$ from haloalkane.

In Chlorobenzene and Vinyl chloride, the

$Cl$ is in conjugation with benzene ring and does not give alcohol.

In Benzylchloride, the

$Cl$ is not directly linked to benzene and thus react as haloalkane and gives white precipitate of

$AgCl$ with

$AgNO_3$ which is soluble in

$NH_4OH$ .

Ethyl bromide gives yellow precipitate of

$AgBr$ which is insoluble in

$NH_4OH$ at room temperature.

Which of the following halides can yield ethane and also methane in a single step?

0%
${C}_{2}{H}_{5}Br$
0%
${CH}_{3}I$
0%
${ \left( { CH }_{ 3 } \right) }_{ 2 }CHBr$
0%
None

${CH}_{3}-CH(Cl)-{C}_{2}{H}_{5} \xrightarrow [ ]{ Alc.\quad KOH } {CH}_{3}-CH=CH-{CH}_{3}$
The above reaction proceeds via

$E_1cb$ mechanism?

0%
It is second order and bumolecular
0%
It is first order and unimolecular
0%
It is first order an bimolecular
0%
It is second order and unimolecular

Explanation

The given reaction (dehydrochlorination of

$2$ -chlorobutane in presence of alcoholic

$KOH$ to form

$2$ -butene) proceeds via

$E_1cb$ mechanism.

$E_1$ = unimolecular elimination ;cB = conjugate base.
It is second order and unimolecular.

The reaction of toluene with chlorine in the presence of ferric chloride gives predominantly:

0%
Benzoyl chloride
0%
$m-chlorotoluene$
0%
Benzyl chloride
0%
$o-$ and $p-chlorotoluene$

Explanation

In presence of

$FeCl_3$ and in dark, electrophilic substitution on benzene ring by

$Cl^+$ on at ortho and para positions occurs to give o- and p-chlorotoluene.

Since the methyl group activates the ring at ortho and para position more than meta position, the electrophilic substitution occurs mostly at these positions.

Again, among these, the para isomer is the major product due to sterric reasons.

Hence the answer is (d)

For the reaction,

$C_2H_5OH+HX\overset{ZnX_2}{\longrightarrow} C_2H_5X$ , the order of reactivity is:

0%
$\;HI>HCl>HBr$
0%
$\;HI>HBr>HCl$
0%
$HCl>HBr>HI$
0%
$\;HBr>HI>HCl$

Explanation

Because the reaction follows

$S_N1$ mechanism and hence the halide formed is stable when it is of a larger size and size increases down the group.

$I>Br>Cl$

Tertiary butyl chloride undergoes

$S_{N}1$ reactions immediately .

0%
True
0%
False

Explanation

True,

Tertiary butyl chloride forms a very stable carbocation as an intermediate hence it undergoes

$S_N1$ reaction mechanism.

Allyl chloride is a .............. compound while vinyl chloride is inert towards nucleophilic substitution.

0%
reactive
0%
mildly reactive
0%
not reactive
0%
netural

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to:

0%
the formation of less stable carbonium ion
0%
resonance stabilisation
0%
longer carbon-halogen bond
0%
the inductive effect
0%
${sp}^{2}$ -hybridised carbon attached to the halogen

Explanation

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to the following reasons:
i. resonance stabilisation
ii.

${sp}^{2}$ -hybridised carbon attached to the halogen

$1,\:3$ - Dibromopropane reacts with metallic zinc to form:

0%
propene
0%
cyclopropane
0%
propane
0%
hexane

Vinyl chloride reacts with dilute

$NaOH$ to form vinyl alcohol.

0%
True
0%
False

Explanation

Vinyl chloride reacts with dil NaOH to form vinyl alcohol which tautomerizes to acetaldehyde.

$CH_2=CHCl + NaOH \xrightarrow {-NaCl} CH_2=CHOH \xrightarrow {tautomerization} CH_3CHO$
Hence, the given statemnt is false.

Secondary butyl chloride undergo alkaline hydrolysis in the polar solvent by ___________ mechanism.

0%
$\:S_{N}2$
0%
$\;S_{N}1$
0%
$\;S_{N}1$ and $S_{N}2$
0%
$\;E\ 2$

Explanation

In polar medium sec. alkyl halide undergo

$S_N1$ mechanism.

The carbocation intermediate formed during the

$S_N1$ mechanism is very stable in polar solvents.

So, option B is correct.

$1-phenyl-2-chloropropane$ on treating with alc.

$KOH$ gives mainly:

0%
$1$ -phenylpropene
0%
$2$ -phenylpropene
0%
$1$ -phenylpropane- $2$ -ol
0%
$1$ -phenylpropan- $1$ -ol

Explanation

$1$ −phenyl−

$2$ −chloropropane on treating with alc.

$KOH$ gives mainly

$1$ −phenylpropene. This is dehydrochlorination reaction.
The double bond in the product is in conjugation with benzene ring.

$1$ −phenyl−

$2$ −chloropropane.

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 6 - 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 6 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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