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CBSE Questions for Class 12 Engineering Chemistry Haloalkanes And Haloarenes Quiz 11 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Haloalkanes And Haloarenes
Quiz 11
$$A + Na/\Delta \rightarrow$$ no gaseous product
$$A + Br_2/FeBr_3 \rightarrow C_7H_7OBr$$ (two isomers).
Based on the above reactions, identify A from the given options.
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Explanation
If phenolic $$OH$$ group was present, treatment with sodium would have liberated hydrogen gas. Since no gaseous product is obtained, phenolic $$OH$$ group is absent.
The alkene limonene has following structure, (Ref. image). Which product results from the reaction of Limonene and $$1$$ molar equivalent chlorine water?
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Explanation
Chlorine water $$HOCl$$ reacts as $$Cl^+OH^-$$. $$Cl^+$$ being strong electrophile attacks on electron rich $$\pi$$ bond.
Action of alcoholic $$AgNO_3$$ on chlorobenzene is similar to the action on:
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allyl chloride
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vinyl chloride
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isopropyl chloride
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benzyl chloride
Explanation
In chlorobenzene, the chlorine atom is directly attached to the unsaturated carbon atom.
In vinyl chloride also, the chlorine atom is directly attached to the unsaturated carbon atom.
Hence, the action of alcoholic $$AgNO_3$$ on chlorobenzene is similar to the action on vinyl chloride.
Both vinyl chloride and chlorobenzene give no precipitate with alcoholic $$AgNO_3$$ because both have chlorine atoms that are not reactive.
Option B is correct.
Correct statement (s) about reaction & related graph is (are):
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P & Q are nucleophilic substitution products by $$S_{N}2$$ reaction mechanism
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only R & S are elimination products by $$E2$$ reaction mechanism
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P is nucleophilic substitution product by $$S_{N}2$$ reaction mechanism
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Q, R, S are elimination products by $$E_2$$ reaction mechanism
Explanation
$$2^0$$ halide give both $$S_N2$$ and elimination reaction.
Here nucleophile is $$OH^-$$ ion which is not a bulky group so it will prefer $$S_N2$$ and $$P$$ is $$S_N2$$ product.
and $$Q,R,S$$ are elimination products by $$E_2$$ reaction mechanism.
Which of the following is fast debrominated?
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IV
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II
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III
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I
Explanation
The debromination of compound $$(IV)$$ is fast as it forms benzene which is aromatic and highly stable.
$$Ph-CH=CH_{2}+BrCCl_{3}\xrightarrow{peroxide} $$
Product is :
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Explanation
In presence of peroxide, anti-markonikoff product forms, so Br attached at benylic carbon.
and as product is optically active so it forms two product.
Which alkyl halide undergo $$E_{2}$$ elimination?
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Explanation
$$A$$ and $$B$$ will give elimination reaction according to -
In anti elimination, the base attacks the $$\beta $$- hydrogen on the opposite side of the leaving group.
It has been experimentally determind the $$E_2$$ elimination occurs through an anti mechanism.
In which of the following cases, configuration about chiral $$C^*$$ is retained?
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Explanation
A: Still $$^*C$$ retained chirality.
B: Lost chirality by $$2^{nd}$$ Methyl group attached.
C: Lost chirality same as in the above reaction.
D: chirality retained.
An aromatic compound $$\displaystyle 'A' (C_{7}H_{6}Cl_{2}$$), gives $$AgCl$$ on boiling with alcoholic $$AgNO_3$$ solution, and yields $$\displaystyle C_{7}H_{7}OCl$$ on treatment with sodium hydroxide. 'A' on oxidation gives a mono chlorobenzoic acid which affords only one mononitro derivative. The compound A is :
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Explanation
$$SN^1$$ mechanism proceeds through carbocation formation, as stability of carbocation increases, reaction favours $$SN^1$$ mechanism.Benylic carbocation is most stable due to more resonating structures and allylic carbocations are more stable than aliphatic carbocation due to resonance so order of stability of carbocations is benzyll > allyl > $$3^0 > 2^0 > 1^0$$.
$$3^0$$ carbocation are more stable than others or stability order of carbocations is $$3^0 > 2^0 > 1^0$$ so order of alkyl halides towards $$SN^1$$ reaction is $$3^0 > 2^0 > 1^0$$.
So, order of reactivity for $$SN^1$$ mechanism is
benzyll > allyl > $$3^0 > 2^0 > 1^0$$.
Here, given compound is $$Cl-C_6H_4-CH_2-Cl$$, in presence of $$AgNO_3$$ its benylic chloride reacts and AgCl forms white ppt.
In presence of NaOH, benzylic chlorine has been replaced by OH- group and $$Cl-C_6H_4-CH_2-OH$$ forms as product, which forms acid on oxidation.
So, answer is option A.
$$CH_3-\!\!\overset{\,\,CH_3}{\overset{|}{\underset{\,\,\,CH_3}{\underset{|}{C}}}\!\!\!-}\,CH=CH_2\xrightarrow{dil.H_2SO_4}\,\,\xrightarrow[\Delta]{conc.H_3PO_4}\,\,\xrightarrow[(ii)\, Zn / H_2O]{(i)\, O_3}$$
Final product is / are:
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$$\underset{CH_3}{\underset{|}{C}=}\,O$$
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$$HCHO + CH_3-\overset{O}{\overset{||}{C}}-\underset{CH_3}{\underset{|}{C}H}-CH_3$$
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$$CH_3CHO +$$
$$CH_3-\overset{O}{\overset{||}{C}}-CH_3$$
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$$CH_{3}CHO+HCHO$$
Explanation
$$CH_3-\!\!\overset{\,\,CH_3}{\overset{|}{\underset{\,\,\,CH_3}{\underset{|}{C}}}\!\!\!-}\,CH=CH_2\xrightarrow{dil.H_2SO_4}CH_3-\!\!\overset{\,\,CH_3}{\overset{|}{\underset{\,\,\,CH_3}{\underset{|}{C}}}\!\!\!-}\,\underset{OH}{\underset{|}{C}H}-CH_3 \xrightarrow[\Delta]{conc.H_3PO_4} \,\,CH_3-\underset{CH_3}{\underset{|}{C}=}\,\underset{CH_3}{\underset{|}{C}-}\,CH_3$$
$$\xrightarrow[(ii)\, Zn / H_2O]{(i)\, O_3}CH_3-\underset{CH_3}{\underset{|}{C}=}\,O$$
Here first $$2^0$$ alcohol forms as a product which forms symmetrical alkene as secondary product.
Then this alkene in presence of ozone forms ketone as a product.
Suppose the double bonded methylene group of $$1$$-bromo-$$2$$-propene is labelled with $$_{ }^{ 13 }{ C },{ H }_{ 2 }{ }^{ 13 }{ C }=CH-{ CH }_{ 2 }Br$$. What product will be formed by methanolysis of this substrate?
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$${ H }_{ 2 }{ }^{ 13 }{ C }=CH-{ CH }_{ 2 }-O{ CH }_{ 3 }$$ and $${ H }_{ 2 }C=CH-_{ }^{ 13 }{ { CH } }-O{ CH }_{ 3 }$$
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$${ H }_{ 2 }C=_{ }^{ 13 }{ { CH } }-{ CH }_{ 2 }-O{ CH }_{ 3 }$$
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$${ H }_{ 2 }{ }^{ 13 }{ { C } }=CH-{ CH }_{ 2 }-O{ CH }_{ 3 }$$
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$${ H }_{ 2 }C=CH-_{ }^{ 13 }{ { CH }_{ 2 } }-O{ CH }_{ 3 }$$
Explanation
Two products are obtained. In the first product, the bromine atom is replaced by the methoxy group. In the second product, the Br atom is replaced by the methoxy group and the double bond is also rearranged.
$$^{13}CH_2=CHCH_2Br \xrightarrow [-HBr] {CH_2OH} ^{13}CH_2=CHCH_2OCH_3+CH_2=CH^{13}CH_2OCH_3$$
Select correct statement about the product (P) of the reaction:
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P is optically inactive due to internal compensation
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P is optically inactive due to the presence of plane of symmetry in the molecule
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The structure of P can have three optical isomers possible.
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P can have four possible optical isomers.
Explanation
Because of polar mechanism, the addition takes place in trans form with the formation of cyclocarboration P is $$C{ H }_{ 3 }-CHBr-CHBr-C{ H }_{ 3 }$$. In this case of meso compounds with two chiral centres, the configuration of compound either R,S or S,R, meso compounds with such internal compensation are optically inactive as rotation by one chiral center is compensated by second chiral center.
In an Aprotic solvent, which relative ordering best describes the nucleophilicity of the halide ions?
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Iodide> Bromide> Chloride>> Flouride
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Iodide> Bromide> Chloride> Flouride
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Iodide< Bromide< Chloride< Flouride
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Iodide< Bromide> Chloride> Flouride
Explanation
In an aprotic solvent, the increasing ordering for the nucleophilicity of the halide ions
is iodide< bromide< chloride< flouride .
A polar aprotic solvent does not hydrogen bond to nucleophiles to a significant extent, meaning that the nucleophiles have greater freedom in solution. Under these conditions, nucleophilicity correlates well with basicity and fluoride ion, being the most unstable of the halide ions, reacts fastest with electrophiles.
What is the best term to describe the rearrangement that causes the conversion of $$3$$-bromo-$$2,2$$-dimethylbutane to $$2,3$$-dimethyl-$$2$$-butanol?
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$$1,2$$-methyl shift
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$$1,3$$- hydride shift
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$$1,3$$-methyl shift
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$$1,2$$- hydride shift
Explanation
$$1,2$$-methyl shift is the best term to describe the rearrangement that causes the conversion of $$3$$-bromo-$$2,2$$-dimethylbutane to $$2,3$$-dimethyl-$$2$$-butanol.
In the following pair of compounds, which of the following relation is correct for nucleophilicity in a polar-protic solvent?
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$${ CH }_{ 3 }S{ CH }_{ 3 } >{ CH }_{ 3 }O{ CH }_{ 3 }$$
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$${ CH }_{ 3 }S{ CH }_{ 3 } <{ CH }_{ 3 }O{ CH }_{ 3 }$$
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$${ CH }_{ 3 }S{ CH }_{ 3 } ={ CH }_{ 3 }O{ CH }_{ 3 }$$
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None of these
Explanation
In a polar-protic solvent, the order of nucleophilicity is $$\displaystyle { CH }_{ 3 }S{ CH }_{ 3 } >{ CH }_{ 3 }O{ CH }_{ 3 } $$
as $$O$$ is more electronegative than $$S$$. Hence, the lone pair of electrons is more strongly held by $$O$$ than by $$S$$.
Hence, it cannot be easily donated by $$O$$.
Hence, dimethyl ether is less nucleophilic than dimethyl sulphide.
Which of the given pair of $${ S }_{ N }1$$ reaction would you expect to proceed faster?
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$$HBr > NaBr $$
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$$HBr < NaBr $$
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$$HBr = NaBr $$
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None of these
Explanation
The reaction with HBr is faster than the reaction with NaBr. HBr protonates the oxygen atom of tert butyl alohol and helps in the formation of tert butyl carbocation (by loss of molecule of water) which is the rate determining step.
Which alkene is formed as the major product in the $$E_2$$ reaction of the given compound?
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$$3$$-methylcyclonex-($$1$$)-ene
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$$1$$-deutero-$$3$$-methylcyclohex-($$1$$)-ene
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$$1$$-methyl-$$3$$-deuterocyclohex-(1)ene
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both A and B
Explanation
$$3$$-methylcyclohex-$$1$$-ene is formed as the major product in the $$E_2$$ reaction of the given compound.
The most stable carbocation formed on debromination is :
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Explanation
In option C the formed carbocation is of $$3^o$$,in conjugation with double bond hence due to resonance and hyperconjugation it is most stable.
Explain the relative rate of $$E2$$ reaction of the given compounds.
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$$ III < I < II $$
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$$ I < II < III $$
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$$ III < II < I $$
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None of these
Explanation
The relative rate of $$E_2$$ reaction of the given compounds is $$III<II<I$$.
In the elimination reaction, a molecule of $$HCl$$ is eliminated. Elimination will be fastest when $$H$$ atom and $$Cl$$ atom are trans to each other. In compound $$I$$, both $$H$$ atoms are trans to $$Cl$$. Hence, the rate of elimination for compound $$I$$ is maximum.
In compound $$III$$, both $$H$$ atoms are cis to $$Cl$$. Hence, the rate of elimination for compound $$III$$ is minimum.
In compound $$II$$, only one $$H$$ atom is trans to $$Cl$$. Hence, the rate of elimination for compound $$II$$ is intermediate.
Hence option C is correct
$$CH{Cl}_{3}$$ does not give white ppt. with $$Ag{NO}_{3}$$ because it:
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is a covalent compound
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does not give $$Cl$$ ions in solution
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is not dissociated in water
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none of the above
Explanation
$$CH{Cl}_{3}$$ does not give white ppt. with $$Ag{NO}_{3}$$ because it:
(A) is a covalent compound
(B) does not give $$Cl$$ ions in solution
(C) is not dissociated in water
On the other hand, ionic compounds such as $$NaCl$$ dissociate in water to give sodium ions and chloride ions. The chloride ions thus obtained react with silver ions to form white ppt of $$AgCl$$.
Which of the following will give $$S_N$$ reaction?
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$$R-Br$$
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$$R-{N}_{3}$$
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$$R-\overset { \oplus }{ { OH }_{ 2 } } $$
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All
Explanation
As all of these compounds contains one stable nucleophile so they all will give $$S_N$$ reaction.
Compound $$(A)$$ on reaction with $${NH}_{2}{NH}_{2}+OH$$ gives:
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Which of the following gives $${S_N}{1}$$ reaction?
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$$MeBr$$
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All
Explanation
It is $${3}^{o}$$ $$RX$$ and forms stable $$Ph{Me}_{2}{ C }^{ \oplus }$$ ion due to resonance and $$+I$$ effect of methyl groups.
$$2^o$$ and $$ 1^o$$ halide follows $$S_N2$$ mechanism.
For the reaction given, which substrate will give maximum racemisation?
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Explanation
$${Ph}_{3}{ C }^{ \oplus }$$ with EDG is more stable and favours $${SN}^{1}$$ reaction and gives maximum racemisation.
$$\beta$$-Elimination or anti-elimination reaction is carried out with base $$({ B }^{ \overset { .. }{ \ominus } })$$ as shown.
The following bases are used.
$$(I) \overset { \ominus \quad }{ OH } $$ $$(II) R{ O }^{ \ominus }$$ $$(III) RCO{ O }^{ \ominus }$$ $$(IV) \overset { \ominus }{ C } N$$ $$(V) N{ O }_{ 3 }^{ \ominus }$$
The decreasing order of reactivity for the above elimination is :
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$$(II)>(I)>(IV)>(III)>(V)$$
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$$(V)>(III)>(IV)>(I)>(II)$$
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$$(II)>(I)>(III)>(IV)>(V)$$
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$$(I)>(II)>(III)>(IV)>(V)$$
Explanation
The reagent should be a strong bronsted base.
Acidic order: $$H{NO}_{3}> RCOOH> HCN> {H}_{2}O> ROH$$
Basic order : $${NO}_{3}^{\ominus }< RCO{O}^{\ominus }< {CN}^{\ominus }< \overset { \ominus }{ OH } < {RO}^ { \ominus }$$
Decreasing order of basicities and hence $$\beta$$-elimination is : $$(II)> (I)> (IV)> (V)$$.
Fugacity power of which group will be maximum?
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$${ CH }_{ 3 }-O-\overset { O\\ \parallel }{ \underset { \parallel \\ O }{ S } } -{ C }_{ 2 }{ H }_{ 5 }$$
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$${ CH }_{ 3 }-O-\overset { O\\ \parallel }{ \underset { \parallel \\ O }{ S } } -{ CH }_{ 3 }$$
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$${ CH }_{ 3 }-O-\overset { O\\ \parallel }{ C } -{ CH }_{ 3 }$$
Explanation
Stronger the acid or weaker the base, better is the leaving group. In (c), due to $$-I$$ effect of three $$F$$-atoms, it makes stronger acid.
Therefore, the order of leaving group:$$(c)>(b)>(a)>(d)$$
In Dow's process for the manufacture of phenol, $$PhCl$$ is fused with $$NaOH$$ at elevated temperature under pressure.
$$PhCl\xrightarrow [ 623K,300atm ]{ NaOH } \left[ Intermediate \right] \xrightarrow [ ]{ { H }_{ 2 }O } \underset { (A)+(B+C)\\ side\ product }{ Phenol } $$
Which of the following statements are correct?
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Phenol is formed via the formation of benzyne intermediate
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p-phenyl phenol is also formed as a by-product
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Diphenylether is also formed as a by-product
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Biphenylene is also formed as a by-product
Explanation
PhCl+NaOH given phenol as the major product
After the formation of benzyne, $$OH^− $$nucleophilic attack takes place. the reaction mixture also contains some phenoxide ions.
So Diphenyl ether is a possible side product.
Third by product is shown in the figure.
Which of the following give $${S_N}{1}$$ reaction?
Select the correct answer.
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$$(I),(II)$$ and $$(III)$$
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$$(I)$$ and $$(II)$$
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$$(II),(III)$$ and $$(IV)$$
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$$(I),(III)$$ and $$(IV)$$
Explanation
Stability of carbocation:Benzyl $${ C }^{ \oplus }>$$ Allyl $${ C }^{ \oplus }> {3}^{o}{ C }^{ \oplus }(I)>(II)$$
$$(IV)$$ gives $${Me}_{3}-\overset { \oplus }{ { CH }_{ 2 } } $$(neopentyl $${ C }^{ \oplus }$$) (sterically hendered). It undergoes neither $${SN}^{1}$$ nor $${SN}^{2}$$.
In the given reactions, the rate of reaction of $$(I)$$ and $$(II)$$ are same. Both reactions proceed by which mechanism?
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$$E_1$$
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$$E_2$$
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$$E_1cb$$
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Anti-elimination
Explanation
Following the $$\beta $$-elimination reaction mechanism we can determine the reaction pathway.
In E1 mechanism, cabocation forms heterolytic cleavage of C-Br bond. This is a slow and rate determining step. Followed by fast step of proton abstraction from adjacent carbon resulting in alkene formation. Thus only halide affects the rate of reaction
In E2 mechanism, both $$Br$$ and $$D/H$$ leaves simultaneously, therefore both halide and D/H affects the rate.
In E1cb mechanism, first step of proton abstraction resulting in carbanion formation and second step of halide/ leaving group cleavage. Since it is two step mechanism, therefore isotope effect will be present.
Anti-elimination and E2 are same
Therefore, option A is correct.
Which of the following statements are correct about the following reactions?
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Reaction $$1$$ proceeds by $${S_N}{2}$$ and reaction $$2$$ by $${S_N}{1}$$ mechanism
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Reaction $$1$$ proceeds by $${S_N}{1}$$ and reaction $$2$$ by $${S_N}{2}$$ mechanism
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(B)$$\Rightarrow$$
(-) $$Ph-\overset{OH}{\overset{|}CH}-CH=CH-Me$$
(C)$$\Rightarrow$$
$$(\pm)$$
$$ Ph-CH=CH-\underset{OH}{\underset {|}CH}-Me$$
The products $$(B)$$ and $$(C)$$ are respectively
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(B)$$\Rightarrow$$
(-) $$Ph-\overset{OH}{\overset{|}CH}-CH=CH-Me$$
(C)$$\Rightarrow$$
$$(\pm)$$
$$Ph-\underset{OH}{\underset{|}CH}-CH=CH-Me$$
The products $$(B)$$ and $$(C)$$ are, respectively
Explanation
For Reaction 1 :
5N of $$NaOH$$ is added in $$Ph -CH-CH=CH-Me$$ thus it proceeds through $$S_N2 $$ mechanism
|
$$OCOR$$
Thus it produces a (-) product =
$$Ph-\overset{OH}{\overset{|}CH}-CH=CH-Me \ \ \ \ \ =(B)$$
For Reaction 2 :
Addition of dip. $$NaOH$$ makes it follow a $$S_N1$$ mechanism.
Thus it forms 2 products
$$(\pm)$$
$$ Ph-CH=CH-\underset{OH}{\underset {|}CH}-Me \ \ \ = (C)$$
Which of the following side chain reaction/s can be used to reduce the activity groups such as $$(OH)$$ or $$(-{NH}_{2})$$?
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Benzoylation
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Acetylation
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Tosylation
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Sulphonation
Explanation
Benzoylation (introducing benzoyl group $$PhCO-$$), acetylation (introducing acetyl group $$CH_3CO-$$) and tosylation (introducing tosyl group $$p-CH_3C_6H_4SO_2-$$) can be used to reduce the activity groups such as ($$OH$$) or ($$−NH_2$$).
Consider the following reactions:
$$I. {Me}_{3}C-Br \xrightarrow [ { SN }^{ 1 } ]{ { H }_{ 2 }O+NaBr } Products$$
$$II. {Ph}_{3}C-Br \xrightarrow [ { SN }^{ 1 } ]{ { H }_{ 2 }O+NaBr } Products$$
Which of the following statements are correct about the above reactions?
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The products in reactions $$(I)$$ and $$(II)$$ are mixture of $$({Me}_{3}-OH+{Me}_{3}Br)$$ and $$({Ph}_{3}C-OH+{Ph}_{3}C-Br)$$, respectively
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The product in $$(I)$$ is $$({Me}_{3}C-OH)$$ and in $$(II)$$ is $$({Ph}_{3}C-OH+{Ph}_{3}C-Br)$$
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The product in $$(I)$$ is $$({Me}_{3}C-OH)$$ and in $$(II)$$ is $$({Ph}_{3}C-OH)$$
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$${Ph}_{3}{ C }^{ \oplus }$$ is more stable than $${Me}_{3}{ C }^{ \oplus }$$
Explanation
Tertiary alkyl halides in the presence of a mixture of $$NaBr$$ and water undergo hydrolysis in which halogen is replaced with hydroxide group. The reaction proceeds via $$S_N1$$ mechanism.
The product in $$(I)$$ is $$({Me}_{3}C-OH)$$ and in $$(II)$$ is $$({Ph}_{3}C-OH+{Ph}_{3}C-Br)$$
$${Ph}_{3}{ C }^{ \oplus }$$ is more stable than $${Me}_{3}{ C }^{ \oplus }$$. Hence, in reaction $$(1)$$ only one product is formed and in reaction $$(2)$$ both the products are formed.
Compounds $$(B)$$ and $$(C)$$ respectively are :
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Explanation
Hence, the answer (c).
Which of the following synthesis could not be done without involving blocking position on the ring?
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Explanation
In $$(c)$$ phenol is $$o-$$ and $$p$$-directing, so both the products would be obtained. Thus, to obtain ortho-product exclusively, p-position has to be blocked first, and then $$Br$$ introduced in ortho-position, followed by the removal of group from p-position.
In which of the following reactions, the correct major product formed?
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Explanation
$$(A)$$ The reaction is Friedel craft's alkylation reaction. The isopropyl group enters ortho to the$$ -OH$$ group.
$$(B)$$ The reaction is Friedel craft's acylation reaction. The acetyl group enters ortho to the methyl group.
$$(C)$$ The $$H$$ atoms of the aromatic ring are replaced with $$Br$$ atoms to form tribromo derivative. Bromine enters at ortho and para position to amino group. This reaction occurs in presence of polar solvent such as water.
$$(D)$$ In presence of non polar solvent such as $$CS_2$$, the bromination of phenol gives ortho and para bromo phenol in the ratio $$1: 4$$.
Consider the given reactions:
Which statement(s) is/are wrong?
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The product by path $$(I)$$ is $$Me-CH={CH}_{2}(I)$$
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The product by path $$(II)$$ is $$Me-CH(OEt)Me(II)$$
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The products are mixture of $$(I)$$ and $$(II)$$ by both paths
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Path $$(I)$$ proceeds via E2 mechanism, while path $$II$$ proceeds via
$${S_N}{1}$$
mechanism
Explanation
Statement (C) is wrong
i. In path $$(I)$$, $$Et{ O }^{ \ominus }$$ is a strong base and with $${2}^{o}$$ $$RX$$ groups. The $$E_2$$ product predominates over the $${S_N}{2}$$ product to give $$(Me-CH={CH}_{2})$$
ii. In path $$(II)$$, $$EtOH$$ is a weak base, but a better nucleophile. so $${S_N}{1}$$ reaction is favoured to give $$MeCH(OEt)Me$$
Product $$(C)$$ is:
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None of the above
Explanation
Bromobenzene reacts with methyl bromide in presence of anhydrous aluminium chloride to form a mixture of o-bromotoluene and p-bromotoluene. Aryl halides undergo electrophilic aromatic substitution reaction in the ortho and para positions. Thus, halogens are ortho-para directing groups due to +M and + E effects. Halogens are deactivating due to strong -I effect.
I. $$CH_3CH_2I\xrightarrow[E2]{EtO^-} CH_2=CH_2+EtOH+I$$
II. $$D_3C-CH_2I\xrightarrow[E2]{EtO^-}D_2C=CH_2+EtOD+I$$
III. $$Me_3C-I\xrightarrow[SN^1 \ and\ E1]{EtO^-} Me_3C-OEt+(Me)_2C=CH_2$$
IV. $$(CD_3)_3C-I\xrightarrow[SN^1 and\ E1]{EtO^-}(CD_3)_3C-OEt+(CD_3)_2C=CD_2$$
Consider the given reactions.
Which of the following statement(s) is/are correct?
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Reactions $$(I)$$ and $$(II)$$ show primary kinetic isotope effect, whereas reactions $$(III)$$ and $$(IV)$$ show $${2}^{o}$$ kinetic isotope effect
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Reactions $$(I)$$ and $$(II)$$ show $${2}^{o}$$ kinetic isotope effect, whereas reactions $$(III)$$ and $$(IV)$$ show $${1}^{o}$$ kinetic isotope effect
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All reactions show $${1}^{o}$$ kinetic isotope effect
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All reactions show $${2}^{o}$$ kinetic isotope effect
Explanation
$$(I)$$ and $$(II)$$ are $$E2$$ reactions in which the R.D.S is breaking of $$(C-H)$$ or $$(C-D)$$ bond. Thus, $$E2$$ elimination of $$(I)$$ is faster than that of $$(II)$$ since, $$(C-H)$$ bond is weaker than $$(C-D)$$ bond. Hence, $$(I)$$ and $$(II)$$ show $${1}^{o}$$ kinetic isotope effect.
$$(III)$$ and $$(IV)$$ are either $$E1$$ or $${S_N}{1}$$ reactions, which involve the formation of same intermediate $${Me}_{3}{ C }^{ \oplus }$$ or $${({CD}_{3})}_{3}{ C }^{ \oplus }$$ in R.D.S. This step does not involve any $$(C-H)$$ or $$(C-D)$$ bond breaking, so H/D effect is not $${1}^{o}$$ but rather a small $${2}^{o}$$ isotope effect, where $${K}_{H}/{K}_{D}=0.7/1.5$$.
Reaction $$(III)$$ is faster than $$(IV)$$, since, $${({CD}_{3})}_{3}{ C }^{ \oplus }$$ is not as stable as $${Me}_{3}{ C }^{ \oplus }$$ because, $${CD}_{3}$$ is not as good as $$\overline { e } $$-donator as $${CH}_{3}$$. Moreover $$(C-D)$$ is not a good hyperconjugative participant as $$(C-H)$$
Option A is correct.
Choose the incorrect reaction.
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$$2{C}_{2}{H}_{5}I+2Na\xrightarrow [ ]{ { \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }O } {C}_{4}{H}_{10}+2NaI$$
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$$2{C}_{2}{H}_{5}Br+Zn \xrightarrow [ ]{ EtOH } { \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }Zn+{Br}_{2}$$
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$$2{C}_{2}{H}_{5}I+{Na}_{2}S \longrightarrow { \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }S+2NaI$$
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$$2{C}_{2}{H}_{5}Br+NaI \xrightarrow [ ]{ { \left( { CH }_{ 3 } \right) }_{ 2 }C=O } {C}_{2}{H}_{5}I+NaBr$$
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All are correct reations
Explanation
A. The reaction given in option A is correct. The preparation of higher alkanes from lower alkyl halides using sodium metal is called as Wurtz reaction. When two molecules of ethyl iodide are treated with sodium metal in the presence of diethyl ether as solvent, n-butane is formed.
B. The reaction given in option B is incorrect. When sodium metal in Wurtz reaction is replaced by zinc metal, then that reaction is called as Frankland reaction and it also gives a higher alkane as the product. Hence the product of the reaction given in option B should be n-butane.
C. Williamson synthesis of alkyl sulfides is carried out by treating alkyl halides with sodium sulfide. The reaction given in option C is correct.
D. When alkyl bromide is treated with NaI and acetone, then halogen exchange reaction takes place to give alkyl iodide as the product. This is called as Finkelstein reaction. Hence the reaction given in option D is also correct.
Which of the given mentioned positions in the given compound is more reactive towards electrophilic substitutions?
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$$3$$
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$$2$$
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$$5$$
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$$6$$
Explanation
Due to resonance, $$C-3$$ acquires a negative charge and electrophilic substitution reaction at $$C-3$$ takes place.
In the reaction $${CH}_{3}C\equiv \overset { \oplus }{ CNa } +{ \left( { CH }_{ 3 } \right) }_{ 2 }CHCl\longrightarrow $$ the product formed is:
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$$4$$-Methyl-$$2$$-pentyne only
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Propyne
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Propyne and propylene
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Mixture of propene, propyne, and $$4$$-methyl-$$2$$-pentyne
Explanation
Hence the answer is $$(D)$$
The yield of chlorobenzene obtained by reaction of phenols with $$P{Cl}_{5}$$ is less, due to the formation of :
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o-chlorophenol
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p-chlorophenol
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phosphorus oxychloride
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triphenyl phosphate
Explanation
The $$(-OH)$$ group on phenol unlike that of alcohol, is difficult to replace by a halogen, e.g., halogen acids have no action, and $$P{X}_{3}$$ yields only phosphorous esters.
Phenol reacts with $$P{Cl}_{5}$$ or $$P{Br}_{5}$$, when the $$(-OH)$$ group of phenol is replaced by a halogen atom. The yield of chloro or bromo benzene is small the main product is triphenyl phosphate $${(PhO)}_{3}P=O$$ or $${Ph}_{3}{PO}_{4}$$.
Phenol further reacts with $$PO{Cl}_{3}$$ to give $${Ph}_{3}{PO}_{4}$$.
Option D is correct.
Which of the following sequence of reactions (reagents) can be used for the conversion of $$Ph{CH}_{2}{CH}_{3}$$ into $$PhCH={CH}_{2}$$?
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$$SO{Cl}_{2};\ {H}_{2}O$$
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$${SO}_{2}{Cl}_{2};\ alc. KOH$$
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$${Cl}_{2}/hv;\ {H}_{2}O$$
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$$SO{Cl}_{2}; \ alc. KOH$$
Explanation
Ethyl benzene can be converted to styrene by using sulphuryl chloride followed by alc. KOH. Sulphuryl chloride is used for the allylic substitution and alc. KOH is used for dehydrochlorination.
In the given reactions, the rate of reaction of $$(I)$$ is faster than that of $$(II)$$. By which mechanism do both reactions proceed?
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E1
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E2
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E1cB
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$$\alpha$$-Elimination
Explanation
Since, the rates of two reactions are different, the breaking of C-H and C-D bonds are involved in the slow step (or rate determining effect).
This indicates $$E_2$$ mechanism in which the rate determining step involves cleavage of C-H (or C-D) and C-Br bond.
Option B is correct.
Bottles containing $$PhI$$ and $$Ph{CH}_{2}I$$ lost their original labels. They were labelled as $$(A)$$ and $$(B)$$ for testing. $$(A)$$ and $$(B)$$ were separately taken in test tubes and boiled with $$NaOH$$ solutions. The end solution in each tube was made acidic with dilutie $$H{NO}_{3}$$ and some $$Ag{NO}_{3}$$ solution was added. Substance $$(B)$$ gave a yellow precipitate. Which of the following statements is true for this experiment?
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Addition of $$H{NO}_{3}$$ was unnecessary
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$$(A)$$ was $$PhI$$
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$$(A)$$ was $$Ph{CH}_{2}I$$
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$$(B)$$ was $$PhI$$
Explanation
$$PhI+aq. NaOH\rightarrow No\ reaction$$
$$PhCH_2I+Aq. NaOH\rightarrow PhCH_2OH+I^{\ominus}\xrightarrow {AgNO_3}AgI(Yellow\ ppt.)$$
(A) was iodobenzene $$PhI$$ and (B) was benzyl iodide $$PhCH_2I$$.
Since (A) is aromatic iodide, it will not give free iodide ions. Hence, no precipitate will be obtained with silver nitrate.
(B) is alkyl iodide. It will give free iodide ions. Hence, yellow precipitate will be obtained with silver nitrate.
The product on monobromination of this compound is :
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0%
0%
0%
Explanation
It is electrophilic substitution, so electrophile must be attacked at o/p position due to higher electron density at these positions. Ring $$(A)$$ is activated by $$+R$$ effect of $$(-NH)$$ group, while ring $$(B)$$ has EW $$\left( -\overset { O\\ \parallel }{ C } - \right) $$ group.
So, SE reaction will take place at p-position of ring $$(A)$$ since o-position is blocked.
Which one of the following compound will give $$S_{N}1$$ reaction predominantly?
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0%
0%
0%
None of these
Explanation
The compound of option (A) $$\displaystyle H_5C_6-C(CH_3)_2Br$$ will give $$\displaystyle S_N1$$ reaction predominately.
It is a tertiary alkyl halide. Also the carbocation intermediate (formed by loss of bromide ion) is stabilized by resonance with the benzene ring.
$$Cl^-\;\;\;\;\;\;\;\;\;\;\;\;CH_3O^-\;\;\;\;\;\;\;\;\;\;\;\;CH_3S^-\;\;\;\;\;\;\;\;\;\;\;\;I^-$$
$$(I)\;\;\;\;\;\;\;\;\;\;\;\;(II)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(III)\;\;\;\;\;\;\;\;\;\;\;\;(IV)$$
The correct order of increasing leaving group capability of above anions:
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$$III < IV < II < I$$
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$$II < III < I < IV$$
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$$II < IV < III < I$$
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$$I < III < II < IV$$
Explanation
The correct order of increasing leaving group capability of the anions is
$$\displaystyle II (CH_3O^-) < III (CH_3S^-) < I (Cl^-) < IV (I^-) $$
Iodide ion is the best leaving group and methoxide ion is the worst leaving group.
The negative charge on large iodide atom is stabilized. In methoxide ion, the $$+I$$ (electron releasing) effect of methyl group intensifies the negative charge on $$O$$ atom and destablizes the ion.
A good leaving group should be
(a) electron-withdrawing to polarize the carbon
(b) stable once it has left (not a strong base)
(c) polarisable- to maintain partial bonding with the carbon in the transition state (both $$S_N1$$ and $$S_N2$$).
This bonding helps to stabilize the transition state and reduces the activation energy.
Among the following which is feasible?
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$$\;X^-+CH_3-CH_2-H\longrightarrow CH_3-CH_2-X+H^-$$
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$$\;X^-+CH_3-OH\longrightarrow CH_3-X+\bar{O}H$$
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$$\;{X}\bar+H_3C-\overset { \oplus }{ \underset { \underset { H }{ \mid } }{ O } } H\longrightarrow CH_{ 3 }-X+ H_{ 2 }O$$
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$$\;X^-+CH_3-CH_3\longrightarrow CH_3-X+\bar{C}H_3$$
Explanation
The reaction $$\displaystyle \;{X}\bar+H_3C-\overset { \oplus }{ \underset { \underset { H }{ \mid } }{ O } } H\longrightarrow CH_{ 3 }-X+ H_2O $$ is most feasible as water is the stable, neutral molecule and not a strong base.
Vinyl chloride does not give $$S_N$$ reaction but allyl chloride does. because ______________.
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in vinyl chloride, $$C\,-\,Cl$$ bond is stable due to resonance.
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in allyl chloride, $$C\,-\,Cl$$ bond is stable due to resonance.
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vinyl carbocation formed by removal of $$Cl^\ominus$$ is stable due to resonance.
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allyl carbocation formed by removal of $$Cl^\ominus$$ is stable due to resonance.
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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