Haloalkanes and Haloarenes Test

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Haloalkanes and Haloarenes Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

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

A

B

C

D
mixture of (B) and (C)

Solution

The reaction of toluene with chlorine in the presence of iron and in the absence of light yields a mixture of $$1$$-chloro-$$2$$-methylbenzene and $$1$$-chloro-$$4$$-methylbenzene.
Methyl group is activating and ortho-para directing towards electrophilic aromatic substitution.
Question 2
1 / -0

What will be the product of following reaction

$$\overset{OH}{|}$$ $$\overset{OH}{|}$$
$$CH_3-C-C-CH_3\xrightarrow[]{H_2SO_4}Q$$
$$\underset{H}{|}$$ $$\underset{H}{|}$$

Q is ___________.

A
$$CH_3-C-CH_2-CH_3$$
$$\; \; \; \; \;\; \; \; \; \;\; \; \underset{O}{||}$$

B
$$CH_3-C=CH-CH_3$$
$$\; \; \; \; \;\; \; \; \; \; \underset{OH}{|}$$

C
Both

D
None

Solution

Dehydration reaction takes place with the hydride shift. The product is $$2-Butanone$$.
Question 3
1 / -0

The product in the following reaction is:
$$Ph - Cl + {Fe}/{{Br}_{2}} \longrightarrow $$ Product

A
o-bromo-chloro benzene

B
p-bromo-chloro benzene

C
both of the above

D
2,4,6-tribromo chloro benzene

Solution

The product in the following reaction is 2,4,6-tribromo chloro benzene.
Question 4
1 / -0

The least reactive chlorine is present in:

A
Methyl chloride

B
Allyl chloride

C
Ethyl chloride

D
Vinyl chloride

Solution

The least reactive chloride is vinyl chloride due to partial double bond character and unstable carbocation formation.

Vinyl carbocation is least stable.
Question 5
1 / -0

Which of the following compound is optically active?

A
$$CH_3 CH_2 COOH$$

B
$$CH_3 CHOH COOH$$

C
$$HOOC.CH_2.COOH$$

D
$$CH_3.CO.COOH$$

Solution

Chiral compounds / chiral molecules are the one's which generally show optical activity and optical isomers are mirror images called enantiomers.

Chiral compounds are those, that have asymmetric centre.

Option: $$B$$
Question 6
1 / -0

Which of the following compounds will given racemic mixture on nucleophilic substitution by $$O{H}^{\ominus}$$ ion?
(a) $$C{ H }_{ 3 }-\underset { \overset { | }{ { C }_{ 2 }{ H }_{ 5 } } }{ CH } -Br$$
(b) $$C{ H }_{ 3 }-\overset { \underset { | }{ Br } }{ \underset { \overset { | }{ { C }_{ 2 }{ H }_{ 5 } } }{ C } } -C{ H }_{ 3 }$$
(c) $$C{ H }_{ 3 }-\underset { \overset { | }{ { C }_{ 2 }{ H }_{ 5 } } }{ CH } -C{ H }_{ 2 }Br$$

A
$$(a)$$

B
$$(a), (b), (c)$$

C
$$(b), (c)$$

D
$$(a), (c)$$

Solution

$$C : $$ $$1^{\circ}$$ Alkyl halide can directly forms substitution product without inversion.
$$ B : $$ $$3^{\circ}$$ Alkyl halide allows $$OH^{-}$$ to attack only from one side d after leaving of halide $$OH^{-}$$ attacks due steric hinderance.
$$ A : $$ $$2^{\circ}$$ Alkyl halide allows $$OH^{-}$$ to attack while leaving of $$Br^{-}$$ hence inversion of molecule occurs forms racemic mixture.
Question 7
1 / -0

Most stable carbocation formed from $${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Br, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }CBr, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 2 }CHBr$$ and $${ C }_{ 6 }{ H }_{ 5 }C{ H }_{ 2 }Br$$ would be:

A
$${ C }_{ 6 }{ H }_{ 5 }\overset { \oplus }{ C } { H }_{ 2 }$$

B
$${ \left( C{ H }_{ 3 } \right) }_{ 3 }\overset { \oplus }{ C } $$

C
$${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }\overset { \oplus }{ C } $$

D
$${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 2 }\overset { \oplus }{ C } H$$

Solution

Most stable carbocation formed will be $${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }\overset { \oplus }{ C }$$ obtained from the hydrolysis of $${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Br, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }CBr, { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 2 }CHBr$$.

This is because triphenyl methyl carbocation $${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }C^+$$ is highly stable due to delocalization of positive charge on three phenyl groups.

$${ \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }CBr \rightarrow { \left( { C }_{ 6 }{ H }_{ 5 } \right) }_{ 3 }C^+$$
Question 8
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Alkyl halide can be converted into alkene by:

A
nucleophilic substitution reaction

B
elimination reaction

C
both nucleophilic substitution and elimination reaction

D
rearrangement

Solution

$$R- CH_2 - CH_2 - X + KOH_{(alc)} \xrightarrow{\Delta} \underset{alkene}{RCH = CH_2 + }KX + H_2O$$
The $$H_2O$$ and KX molecules are eliminated during reaction, so we can say alkene is formed from alkyl halide by elimination reaction.
Question 9
1 / -0

$$S{_N}{1}$$ reaction on optically active substrates mainly gives:

A
retention in configuration

B
inversion in configuration

C
racemic product

D
no product

Solution

since the nucleophile attacks the carbocation only after the leaving group has departed, in $${S_N}^1$$, there is no need for back-side attack. The carbocation and its substituents are all in the same plane (Figure 1), meaning that the nucleophile can attack from either side. As a result, both enantiomers are formed in a $${S_N}^1$$ reaction, leading to a racemic mixture of both enantiomers.

Hence, the correct option is $$\text{C}$$
Question 10
1 / -0

Consider following three halides:
(I) $$C{H}_{3}-C{H}_{2}-Cl$$ ;
(II) $$C{H}_{2}=CH-Cl$$ ;
(III) $${C}_{6}{H}_{5}Cl$$
Arrange $$C-Cl$$ bond length of these compounds in decreasing order.

A
$$I > II > III$$

B
$$I > III > II$$

C
$$III > II > I$$

D
$$II > III > I$$

Solution

I > II > III because