Haloalkanes and Haloarenes Test

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Haloalkanes and Haloarenes Test - 62

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Weekly Quiz Competition

Question 1
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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?

A
Reactions $$(I)$$ and $$(II)$$ show primary kinetic isotope effect, whereas reactions $$(III)$$ and $$(IV)$$ show $${2}^{o}$$ kinetic isotope effect

B
Reactions $$(I)$$ and $$(II)$$ show $${2}^{o}$$ kinetic isotope effect, whereas reactions $$(III)$$ and $$(IV)$$ show $${1}^{o}$$ kinetic isotope effect

C
All reactions show $${1}^{o}$$ kinetic isotope effect

D
All reactions show $${2}^{o}$$ kinetic isotope effect

Solution

$$(I)$$ 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.
Question 2
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In the given reactions, the rate of reaction of $$(I)$$ and $$(II)$$ are same. Both reactions proceed by which mechanism?

A
$$E_1$$

B
$$E_2$$

C
$$E_1cb$$

D
Anti-elimination

Solution

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.
Question 3
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Which of the given mentioned positions in the given compound is more reactive towards electrophilic substitutions?

A
$$3$$

B
$$2$$

C
$$5$$

D
$$6$$

Solution

Due to resonance, $$C-3$$ acquires a negative charge and electrophilic substitution reaction at $$C-3$$ takes place.
Question 4
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In the reaction $${CH}_{3}C\equiv \overset { \oplus }{ CNa } +{ \left( { CH }_{ 3 } \right) }_{ 2 }CHCl\longrightarrow $$ the product formed is:

A
$$4$$-Methyl-$$2$$-pentyne only

B
Propyne

C
Propyne and propylene

D
Mixture of propene, propyne, and $$4$$-methyl-$$2$$-pentyne

Solution

Hence the answer is $$(D)$$
Question 5
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Compounds $$(B)$$ and $$(C)$$ respectively are :

A

B

C

D

Solution

Hence, the answer (c).
Question 6
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Which of the following synthesis could not be done without involving blocking position on the ring?

A

B

C

D

Solution

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.
Question 7
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In the given reactions, the rate of reaction of $$(I)$$ is faster than that of $$(II)$$. By which mechanism do both reactions proceed?

A
E1

B
E2

C
E1cB

D
$$\alpha$$-Elimination

Solution

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.
Question 8
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Choose the incorrect reaction.

A
$$2{C}_{2}{H}_{5}I+2Na\xrightarrow [ ]{ { \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }O } {C}_{4}{H}_{10}+2NaI$$

B
$$2{C}_{2}{H}_{5}Br+Zn \xrightarrow [ ]{ EtOH } { \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }Zn+{Br}_{2}$$

C
$$2{C}_{2}{H}_{5}I+{Na}_{2}S \longrightarrow { \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }S+2NaI$$

D
$$2{C}_{2}{H}_{5}Br+NaI \xrightarrow [ ]{ { \left( { CH }_{ 3 } \right) }_{ 2 }C=O } {C}_{2}{H}_{5}I+NaBr$$

E
All are correct reations

Solution

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.
Question 9
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Consider the following anions.
When attached to $$sp^{3-}$$ hybridized carbon, their leaving group ability in nucleophilic substitution reactions decreases in the order.

A
$$\;I>II>III>IV$$

B
$$\;I>II>IV>III$$

C
$$\;IV>I>II>III$$

D
$$\;IV>III>II>I$$

Solution

When attached to $$sp^3$$−hybridized carbon, their leaving group ability in nucleophilic substitution reactions decreases in the order $$\displaystyle I>II>IV>III $$

$$I$$ is best leaving group due to electron withdrawing $$(-I)$$ effect of trifluoromethyl group and delocalization of negative charge over $$3\: O$$ atoms.

$$III$$ is worst leaving group because negative charge is delocalized over $$C$$ atoms.

The stabilization effect of resonance for the anion in which negative charge is delocalized over $$O$$ atoms is more pronounced over the delocalization in which the negative charge is delocalized over $$C$$ atoms.

II is better leaving group than $$IV$$ because in $$II$$, the extent of delocalisation of negative charge is much more than the extent of delocalization in $$IV$$.
Question 10
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The yield of chlorobenzene obtained by reaction of phenols with $$P{Cl}_{5}$$ is less, due to the formation of :

A
o-chlorophenol

B
p-chlorophenol

C
phosphorus oxychloride

D
triphenyl phosphate

Solution

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.