MCQ Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 9

Which compound is a possible product from addition of

$Br_2$ to 1- butene?

Explanation

As refer to the above image, Bromination results in the formation of

$1,2-dibromobutane$ .
1171443_989550_ans_e17303c33ac24a319e93fcce44547ad8.png

$CH_3 - \overset{CH_3}{\overset{|}{C}}H - CH_3 \xrightarrow{Cl_2/hv} \ product$

Major product in the above reaction is :

0%
$CH_3 - \overset{CH_3}{\overset{|}{C}}H - CH_2 - Cl$
0%
$CH_3 -\underset {Cl} {\overset{CH_3}{\overset{|}{\underset{|}{C}}}} - CH_3$
0%
$CH_3 - CH_2 - CH_2 - Cl$
0%
$CH_3 - \underset{Cl} {\underset{|}C}H - CH_2 - CH_3$

$\underset{a}{CH_3 }\, - \,\underset{b}{CH_2 }\, - \,\underset{c} {CH_2} \, - \,\underset{d} CH_2 \, - \, F$
Arrange the hydrogens

$a, b, c, d,$ in decreasing order of their reactivities towards chlorination:

0%
$a > b > c > d$
0%
$b > c > d > a$
0%
$b > c > a > d$
0%
$c > b > a > d$

Explanation

Among all of them hydrogen 'b' will be most leachive towards chlorination as it is

$2^o$ hydrogen and it will get less electronegativity of fluorine atom and 'd' will be least stable as it is adjacent to the fluorine atom and does not have

$3^o-2^o$ hydrogen.

Taking into account the stability of various cycloalkanes and carbocations, as well as the rules governing mechanisms of carbocation rearrangements What is the most likely product of this reaction ?

The IUPAC name of

$(CH_3)_3C. CH_2CH_2Cl$ is:

0%
$2, 2$ - Dimethyl - $4$ -chloro butane
0%
$1$ - Chloro - $3, 3$ - dimethylbutane
0%
$4$ - Chloro - $2, 2$ - dimethylbutane
0%
none of these

Explanation

Parent chain consist of

$4-$ carbonation and it is also saturated.So,butane is parent chain.

$3rd$ carbon atom two methyl substituent are attached.

So its name is

$1-$ chloro

$-3,3-$ dimethyl butane.

1069404_996512_ans_cfbd5b2da6b04de19c26c135e24b4628.png

Identify the product.

Explanation

$KSH$ is potassium hydrosulfide. It is a colourless salt & consists of the cation

$K^+$ and the bisulphide anion

$SH^-$ . It is the product of the half neutralization of hydrogen sulphide with potassium hydroxide. The compound is used in the synthesis of some organosulphur compound. It is prepared by neutralizing aqueous

$KOH$ with

$H_2S$ .

1119613_989653_ans_54e4be9476b04f838231f79968bd5219.png

Product (B) of the reaction is :

Compound A, which is the degradation product of the antibiotic vermiculine has following structure?

$\underset{C_{11}H_{14}O_4}{(A)} \, \xrightarrow[pd/c]{H_2} \, \underset{C_{11}H_{18}O_4}{(B)} \, \xleftarrow{(CH_3)_2S} \, \xleftarrow[CH_2Cl_2]{O_3} \, \underset{C_{11}H_{18}O_2}{(C)} \, .$ Unknown C is:

0%
0%
0%
0%
None of these

Product (A) is:

Explanation

Allylic carbocation is more stable.So,

$Br^-$ at allylic position will get substituted by

$SN^1$ mechanism.
1175216_991838_ans_4c2e04582a014d1983241286f26d4002.png

In the given pair of alcohol, in which pair second alcohol is more reactive then first towards hydrogen bromide?

0%
0%
0%
$CH_3-\underset{\underset{OH}{|}}{CH}-CH_2-CH_3\, and \, CH_3-CH_2-\underset{\underset{CH_3}{|}}{CH}-CH_2-OH$
0%
$CH_3-\underset{\underset{OH}{|}}{CH}-CH_2-CH_3\, and \, (CH_3)_2-\underset{\underset{OH}{|}}{C}-CH_2-CH_3$

Explanation

$CH_{ 3 }-\overset { \oplus }{ \underset { \underset { 2^{ 0 } }{ \downarrow } }{ CH } }-CH_2-CH_3<(CH_3)_2 \underset { \underset { \underset { 3^ 0 }{ \downarrow } }{\oplus } }{ C } -CH_2-CH_3$

Alcohols are more reactive towards

$HBr$ , if the formed carbocation is highly stable.

$3^o cation >2^o cation > 1^o cation$ [stability]

(Refer to Image)

$CH_3$

$3^o$ cation is more stable than

$2^o$ ,

$(CH_3)_2\underset {|}{C}-CH_2-CH_3$ is more stable than

$CH_3-\underset {|}{CH}-\overset {|}{CH_2}$ .

$OH$

Which of the following solvents is most suitable for

$S_N1$ reaction?

0%
Polar protic
0%
Polar aprotic
0%
Non-polar
0%
All of these

$E_2$ elimination some compounds follow Hofmann's rule which means:

0%
the double bond goes to the most substituted carbon
0%
the compound is resistant to elimination
0%
no double bond is formed
0%
the double bond mainly goes to the less substituted carbon

Compare rate of reaction

$Ag^+NO^-_3$ or rate of

$S_N1$ reaction.
The correct choice is:

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

Explanation

Reaction with

$AgNO_3$ will give precipitate of AgI only in that case where stable carbocation get form

so order will be ii.>iii.>i

1019542_1035507_ans_57ff2002818b48f087bb50759d650153.png

State true or false.

Nature of C-X bond in haloarene is Non-Polar.

0%
True
0%
False

$2$ -Methyl propane on monochlorination under photochemical condition give:

0%
$2$ -chloro- $2$ -methylpropane as major product
0%
$(1:1)$ mixture of $1$ -chloro- $2$ -methylpropane and $2$ -chloro- $2$ -methylpropane
0%
$1$ -chloro- $2$ -methylpropane as a major product
0%
$(1:9)$ mixture of $1$ -chloro- $2$ -methylpropane

The addition of HBr is easiest with:

0%
$CH_2 = CH- Cl$
0%
$Cl - CH = CH - Cl$
0%
$CH_3 - CH = CH_2$
0%
$(CH_3)_2C = CH_2$

Explanation

Addition of

$H-X$ (Where

$x$ is halogen) on alkenes is

electrophilic addition reaction

The rate of reaction depends on the stability of intermediate

carbocation.

Thus, that carbocation is more stable for cempound (2)

So, correct option is (D)

1999813_1044182_ans_4a11653b201f4b82ae7e7cae32855b5e.PNG

For the reaction, the major product is formed by:

0%
An E $1$ reaction
0%
An E $2$ reaction
0%
$S_N1$ reaction
0%
$S_N2$ reaction

Explanation

Since,

$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}$ is a moderate base, so there is no possibility

$E_{1}$ reaction and in

$E_{1 C B}$ , there takes place the carbanion

formation. Hex, there is no anion stabilising group, so

$E_{1} C B$ is not possible.

E. mechanism is possible because of presence of

$\mathrm{H}$ -atom anti

to the leaving group

$(B r)$ .

So, option(B) is correct.

1999680_1036903_ans_2feb32f23bd54708ac899743ae134ca5.PNG

Which of the following reactions is

$S_N1$ type ?

0%
Only (i)
0%
Only (ii)
0%
Both (i) and (ii)
0%
None of these

Explanation

Since in (ii) reaction configuration is retained so it undergoes by sN' mechanism as in

$SN^2$ inversion of configuration occur.

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is:

0%
$CH_3 - CH_3$
0%
$CH_2 = CH_2$
0%
$CH \equiv CH$
0%
$CH_4$

Explanation

$R-H+{ Br }_{ 2 }\quad \underrightarrow { Substitution } \quad R-Br+HBr$

$2R-Br+2Na\quad \underrightarrow { wurtz\quad reaction } \quad R-R+2{ Na }^{ + }{ Br }^{ - }$

In substitution reaction alkanes produces alkyl bromides and in Wurtz reaction two alkyl halides are reacted with sodium

$(Na)$ metal to form higher alkane.

Since after Wurtz reaction, the gaseous hydrocarbon has less than

$4$ carbon atoms. So,

$'R'$ should contain only

$1$ carbon atom. So

$A$ is

${ CH }_{ 4 }$ .

State true or false.

The nature of C-X bond in

$CH_3-CH_2-CH_2-X$ is non-polar.

0%
True
0%
False

Explanation

Nature of C-X bond in

$CH_3-CH_2-CH_2-X$ is polar due to electronegativity difference

$CH_3CHI_2 \xrightarrow{KCN}\, \xrightarrow[\Delta]{H_2O}$ ?;
Here the end product would be :

0%
2 - cyanopropionic acid
0%
ethane - 1, 1 - dicarboxylic acid
0%
2 - methyl ethanoic acid
0%
propionic acid

Explanation

$CN^{-}$ will substitute the iodine,
Hydrolysis of cyanide results into $COOH$ group,
Heat will eliminate one $COOH$ group from the structure.

1015901_1034579_ans_4fa887d840fa4b978d2298c273a031e8.PNG

What are possible product obtained in the following reaction?

$(CH_3)_3-C-Br\overset{80\% EtOH}{\underset{20\% H_2O}{\rightarrow}}$ .

0%
$(CH_3)_2=CH_2$ via E $1$
0%
$(CH_3)_3-C-OH$ via $S_N1$
0%
$(CH_3)_3C-O-Et$ via $S_N1$
0%
$(CH_3)_2=CH_2$ via $E2$

$CH_{ 3 }CH_{ 2 }Cl\quad \xrightarrow [ ]{ NaCN } \times \xrightarrow [ ]{ Ni/H_{ 2 } } Y$
Y in the above reacting sequence is:

0%
$CH_3CH_2CH_2NHCOCH_3$
0%
$CH_3CH_2CH_2NH_2$
0%
$CH_3CH_2CH_2CONHCH_3$
0%
$CH_3CH_2CH_2CONHCOCH_3$

Explanation

$Ni/H_2$ helps in reduction -CN reduces to

$-CN_2NH_2$
1157646_1073097_ans_82036b7b63244541af661c510b107d9e.PNG

The correct statement for

$\alpha$ elimination is?

0%
It forms cyclic compounds
0%
It forms carbine or substituted carbine
0%
Two atoms are removed from $\alpha$ and $\beta$ positions
0%
In $CHCl_{3};\alpha$ elimination is not possible

In a set reactions p-nitrotoluene yielded a product

$E$ . The product

$E$ would be:

0%
0%
0%
0%
none of these

Explanation

$Br_2/FeBr_3$ gives mono bromo product.
1380041_1117564_ans_3f3aacb9790e4b4fbc9e3c3b887cf498.jpg

$CH_3CH=CH_2 \xrightarrow {Cl-I} P$ ,

$P$ is:

0%
$CH_3-CHI-CH_2Cl$
0%
$CH_3CHCl-CH_2I$
0%
$CH_3-ICHCl-CH_2$
0%
$CH_3-CH_2-ICHCl$

Explanation

$CH_3-CH=CH_2 \xrightarrow {I^+- \ Cl^-} CH_3-CHCl-CH_2-I$

The reaction is electrophilic addition and the product obtained is called Markonikov product.

Reagent which can convert an alkyl amine into alkyl chloride.

0%
Tilden reagent (NOCI)
0%
Lucas reagent
0%
Hisnberg's reagent
0%
None

Explanation

Alkylamine + Tilden's Reagent

$\rightarrow$ Alkyl chloride.

$R-{ NH }_{ 2 }+NoCl\rightarrow R-Cl+NO-{ NH }_{ 2 }$

$\hookrightarrow$ Tiden's Regent.

Which of the following reaction will yield,

$2,2$ - dibromopropare?

0%
$CH_{3}-C\equiv CH+2HBr$
0%
$CH_{3}-CH=CHBr + 2HBr$
0%
$CH\equiv CH+2HBr$
0%
$CH_{3}-CH=CH_{2}+HBr$

Which of the following has the lowest barrier to rotation about the indicated bond ?

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

Explanation

lowest barsier to rotation indicates

easier to rotate ( activation energy is less)

So option

$B$ is corrcet

1996315_1103464_ans_50f91a000c754cf1b03ec8132432f488.PNG

Which of the following react with

$Cl_2$ and

$Br_2$ at room temperature and in the absence of diffused sunlight to produce dihalogen derivatives?

0%
Cyclobutane
0%
Cyclopentane
0%
Cyclohexane
0%
All of these

Explanation

Cyclo butane on reaction with

${ Cl }_{ 2 }$ at 298K produce dihalogen derivatives mechanism.

1174544_1082122_ans_155f5c796e78428882aa16b325f5ba79.jpg

Identify the products

$A$ and

$B$ .

0%
0%
0%
0%
None of these

Which of the following hydrocarbons is not formed when Wurtz reaction takes place between ethyl iodide and propyl iodide?

0%
$Butane$
0%
$Propane$
0%
$Pentane$
0%
$Hexane$

The racemic mixture of Alanine (

$CH_8-CH-COOH$ ) can be resolved by

$NH_2$
1) (+)-2-Butanol
2) (l)-2-Chlorobutanoic acid
3) ()-2-Butanol
4)

$(dil$ mix)-2-Chlorobutanoic acid

0%
1&2 only
0%
1&3 only
0%
2&4only
0%
3&4only

${ CH }_{ 2 }={ CH }_{ 2 }+{ Br }_{ 2 }\xrightarrow { NaCl } \quad Products$ . Which of the following product is not formed in the above reaction?

0%
$Br{ CH }_{ 2 }{ CH }_{ 2 }Br$
0%
$Cl{ CH }_{ 2 }{ CH }_{ 2 }Cl$
0%
$Br{ CH }_{ 2 }{ CH }_{ 2 }Cl$
0%
All are formed

Explanation

Solution:- (B)

$Cl-C{H}_{2}-C{H}_{2}-Cl$

$C{H}_{2} = C{H}_{2} + {Br}_{2} \xrightarrow{NaCl} {C{H}_{2}}^{+}-C{H}_{2}-Br$

The

$\pi$ -electrons will attack an electrophile forming the carbocation.

Now the carbocation will be attacked by a nucleophile.

Hence the product will be a mixture of

$Br-C{H}_{2}-C{H}_{2}-Br$ and

$Br-C{H}_{2}-C{H}_{2}-Cl$ .

Which of the following statements regarding the

$S_N1$ reaction shown by alkyl halide is NOT CORRECT?

0%
The added nucleophile plays no kinetic role in $S_N1$ reaction.
0%
The $S_N1$ reaction involves the inversion of configuration of the optically active substrate
0%
The $S_N1$ reaction on the chiral starting material ends up with racemization of the product.
0%
The more stable the carbocation intermediate the faster the $S_N1$ reaction

$CH_3-CH_3+Cl_2\xrightarrow{hv}$ ?

0%
$C_2H_5Cl$
0%
$Cl_2$
0%
$CH_2(Cl)-CH_2(Cl)$
0%
All of the above

Explanation

The reaction given in the question is free radical halogenation reaction of alkanes in the presence of sunlight. The free radical halogenation in the given conditions gives the major product as

${C_2}{H_5}Cl$ .

The whole reaction will be:

$C{H_3} - C{H_3}(g) + C{l_2} \to {C_2}{H_5}Cl(g) + HCl(g)$

Hence, the correct option will be option A

The product Y is:

Explanation

In the given reaction first chlorination of the 4-methyl benzene hydrogen sulphate takes place where all the hydrogen atoms of the methyl group are replaced by

$Cl$ group and given compound

$X$ .

Wkhen this compound

$x$ reacts with

$Na_2CO_3$ it forms compound

$Y$ which is given in option B.

1192269_1117922_ans_fb29063c2cc544cc82dedca001b3a02b.png

Above reaction is an example of 1-4 -elimination .Predict the product.

0%
$C{ H }_{ 3 }-CH=C=C{ H }_{ 2 }$
0%
$C{ H }_{ 3 }-C=C=C{ H }_{ 3 }$
0%
$C{ H }_{ 3 }-C{ H }_{ 2 }=C=C{ H }$
0%
${ H }_{ 2 }C=CH-CH=C{ H }_{ 2 }$

Here decreasing order of the rate of Bromination of the following compounds is:
I.

${ Ph }\overset { \oplus }{ NM } e_{ 3 }$
II.

$Ph{ CH }_{ 2 }\overset { \oplus }{ N } Me_{ 3 }$
III

$PhMe$
IV.

$PhNM{ e }_{ 2 }$

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

Alkyl halides can be obtained by all methods excepts?

0%
$CH_{3}-CH_{2}-+HCl\overset{znci_{2}}{\rightarrow}$
0%
$CH_{2}=CH_CH_{3}-+HBr\rightarrow$
0%
both a and b
0%
none of these

Which of the following can undergo nucleophilic substitution reactions?

0%
$CH_3-C\equiv C-H$
0%
$CH_3-CH_2-CH_3$
0%
$CH_3-CH=CH_2$
0%
$CH_3-CH_2-Br$

Explanation

$C{H_3} - C{H_2} - Br$

This reaction is applied for alkyl halide.

Bromide ion is a very good leaving group.

Hence option (D) is correct.

Which is more acidic

0%
$CHCl_3$
0%
$CHI_3$
0%
$CHBr_3$
0%
$CH(CN)_3$

Explanation

Hint: The acidic strength of a compound is directly proportional to the stability of conjugate base.

Step 1: Compare stability of conjugate base

$CHCl_3$ : It break down in solution as-

$CHC{l_3} \to {H^ + } + CC{l_3}^ -$

Here, the conjugate base is stabilized by the back bonding of carbanion with vacant d-orbitals of $Cl.$

$CHBr_3$ : It break down in solution as-

$CHB{r_3} \to {H^ + } + CB{r_3}^ -$

Here, the conjugate base $CBr{_3}^ -$ is stabilized by the back bonding of carbanion with vacant d-orbitals of $Br.$

$CHI_3$ : It break down in solution as-

$CH{I_3} \to {H^ + } + C{I_3}^ -$

Here, the conjugate base $CI{_3}^ -$ is stabilized by the back bonding of carbanion with vacant d-orbitals of $I.$

$CH(CN)_3$ : It breaks down in solution as-

$CH{(CN)_3} \to {H^ + } + C{(CN)_3}^ -$

Here, the conjugate base is stabilized by the $-M$ effect of $CN^-$ .

The mesomeric effect dominates on back bonding. So $CH(CN)_3$ has the most stable conjugate base.

Also $Cl>Br>I$ ( Extent of back bonding. )

So $CH{(CN)_3} > CHC{l_3} > CHB{r_3} > CH{I_3}$ ( Order of stability of conjugate base. )

Step 2: Acidic order

$CH{(CN)_3} > CHC{l_3} > CHB{r_3} > CH{I_3}$

Final Step: Correct option (D) $CH(CN)_3$ .

In which of the following reaction, the major product alkene is formed by

$E_1CB$ mechanism?

0%
$CH_{3} - CH_{2} - CH_{2} - CH_{2} - Br \underset {\triangle}{\xrightarrow {alc. KOH}}$
0%
$CH_{3} - CH_{2} - \underset {\underset {Br}{|}}{CH} - CH_{3} \underset {\triangle}{\xrightarrow {alc.KOH}}$
0%
$CH_{3} - CH_{2} - \underset {\underset {F}{|}}{CH} - CH_{3} \underset {\triangle}{\xrightarrow {alc.KOH}}$
0%
$CH_{3}F \underset {\triangle}{\xrightarrow {alc.KOH}}$

Identify the major product

Consider the above reaction :

The major products formed in this reaction are :

0%
0%
0%
0%
No reaction

Which among the following compounds is used to decaffeinate coffee?

0%
Iodoform
0%
Carbon tetrachloride
0%
Methylene dichloride
0%
Chloroform

Explanation

The two most commonly used solvents for decaffeinating coffee today are methylene dichloride and ethyl acetate.

After soaking in these, the coffee beans are steamed, dried, and roasted. These processes ensure that no residual solvent remains in the beans. Soaking coffee beans in solvent removes caffeine as it dissolves in the solvent.

Hence, option

$C$ is correct.

Reactivity order for SN1
(i)

$CH_3-\underset{Cl}{\underset{|}{C}H}-CH_2-CH_3$
(ii)

$Ph-CH_3-\underset{Cl}{\underset{|}{C}H}-CH_3$
(iii)

$CH_3-\overset{\,\,\,Cl}{\overset{|}{\underset{\,\,\,\,\,\,OCH_3}{\underset{|}{C}}}}-CH_3$
(iv)

$CH_3-CH_2-CH_2-Cl$

0%
$i>ii>iii>iv$
0%
$ii>i>iii>iv$
0%
$iii>ii>i>iv$
0%
$iv>iii>ii>i$

Explanation

(i)

$CH_3-\overset{Cl}{\overset{|}{C}H}-CH_2-CH_3\rightarrow CH_3-\overset{\oplus}{C}H-CH_2-CH_3$
(ii)

$PH-CH_2-\underset{Cl}{\underset{|}{C}H}-CH_3\rightarrow PH-CH_2-\overset{\oplus}{C}H-CH_3$
(iii)

$CH_3-\overset{Cl}{\overset{|}{\underset{\,\,\,\,OCH_3}{\underset{|}{C}}}}-CH_3\rightarrow CH_3-\overset{\oplus}{\underset{\,\,\,\,\,\,\,\,\,OCH_3}{\underset{|}{C}}}H-CH_3$
(iv)

$CH_3-CH_2-CH_2-Cl\rightarrow CH_3-CH_2-CH_2^+$

$S_n^1$ reaction is favoured by:

0%
Bulky groups on the carbon atom attached to halogen atom
0%
Small groups on the carbon attached to halogen atom.
0%
None -polar solvents
0%
None of the above

Explanation

$S_{N^1}$ mechanism proceeds through carbocation formation and as stability of carbocation increases, it favours

$S_{N^1}$ mechanism and as we add bulky group to carbon attached with halogen, carbocation formed will become more stable due to

$+I$ effect of bulky alkyl group so it favours

$S_{N^1}$ mechanism.

$CH_3-\overset{CH_3}{\overset{|}{C}H}-CH_2-CH_2-\underset{O}{\underset{||}{C}}-CH_3 \xrightarrow[]{H_2/Pd}$ No. Of stereoisomerism.

0%
$2$
0%
$4$
0%
$8$
0%
$6$

Explanation

$CH_3-\underset{\,\,\,\,CH_3}{\underset{|}{C}}=CH-CH_2-\underset{O}{\underset{||}{C}}-CH_3 \xrightarrow []{H_2/Pd/C}CH_3-\underset{\,\,\,\,CH_3}{\underset{|}{C}}-CH_2-CH_2-\underset{OH}{\underset{|}{C}H}-CH_3$
Total number of stereo isomer =

$2$

n-Propyl chloride reacts with sodium metal in dry ether to give:

0%
$CH_{3} - CH_{2} - CH_{2} - CH_{2} - CH_{2} - CH_{3}$
0%
$CH_{3} - CH_{2} - CH_{3}$
0%
$CH_{3} - CH_{2} - CH_{2} - CH_{3}$
0%
$CH_{3} - CH_{2} - CH_{2} - CH_{2} - CH_{2} - CH_{2} - CH_{3}$

Explanation

Wurtz reaction

$n(CH_{3} CH_{2} CH_{2} Cl) \underset {dry\ ether}{\xrightarrow {Na}} CH_{3} - (CH_{2})_{4} - CH_{3}$ .

n-Propyl chloride reacts with sodium metal in dry ether to give n-hexane. The reaction is called a Wurtz reaction.

Hence, the correct option is

$A$

CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 9 - MCQExams.com

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CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 9 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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