Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 42

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Question 1
1 / -0

On mixing alkane with chlorine and irradiating with ultraviolet light, it forms only one monochloro alkane. The alkane is:

A
propane

B
pentane

C
isopentane

D
neopentane

Solution

Neo-pentane
$$H_3C - \overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}- CH_3$$
Neo-pentane contains all hydrogen atoms equivalent, so it will give only one monochloro derivative on halogenation.
Question 2
1 / -0

Which of the following cannot undergo $$E_2$$ reaction?

A

B

C

D
None of these

Solution

$${ E }_{ 2 }$$ elimination requires the presence of hydrogen at $$\alpha $$ carbon of the substance.
Hence,
no $$\alpha -H$$ is present as $${ 3 }^{ 0 }$$ carbon
Therefore does not undergo $${ E }_{ 2 }$$ elimination reaction.
Question 3
1 / -0

The correct order of reactivity of the halides,
(I) ethyl chloride; (II) isopropyl chloride and (III) benzyl chloride in $$S_N 1 $$ reaction is:

A
$$I > II > III$$

B
$$III > II > I$$

C
$$II > I > III$$

D
$$I > III > II$$
Solution
- For $$S_N 1 $$ reactions, the greater the stability of carbocations formed, faster will be the rate of reaction. Thus, the order of reactivity of halides is $$\text{benzyl chloride > isopropyl chloride > ethyl chloride}$$.
(Due to greater stabilisation of benzylic carbocation intermediate by resonance, it shows higher reactivity in $$S_N 1$$ reactions than other aliphatic halides).
Question 4
1 / -0

The major product is:

A

B

C

D

Solution

Hence, option $$A$$ is correct.
Question 5
1 / -0

The product formed in the reaction is:

A
$$(BrCH_{2})_{3}CCH_{2}CH_{2}C(CH_{2}Br)_{3}$$

B

C

D

Solution

It is intramolecular Wurtzs reaction
Question 6
1 / -0

The treatment of $${R}_{3}MgX$$ with ethyl formate leads to the formation of:

A
$$RCHO$$

B
$${ CH }_{ 3 }{ CH }_{ 2 }OR$$

C
$$RCHOH{ C }_{ 2 }{ H }_{ 5 }$$

D
$${ C }_{ 2 }{ H }_{ 5 }OH\quad $$

Solution

Ethyl formate is written as $$HCOO{C_2}{H_5}$$ . The reaction is as follows
$$HCOO{C_2}{H_5} + RMgX \to RCHO + Mg(O{C_2}{H_5})X$$
Hence the correct answer is option A
Question 7
1 / -0

The compound
$$ C_7 H_8 \xrightarrow { 3Cl_2/\triangle } X\xrightarrow { Br_2/Fe } Y\xrightarrow { Zn/HCl } Z$$
The compound $$Z$$ is :

A
m-bromotoluene

B
p-bromotoluene

C
o-bromotoluene

D
2, 4, 6-trichlorotoluene

Solution

In the first step, Toluene undergoes side-chain halogenation to give Benzotrichloride which is compound A. In the second step, benzotrichloride reacts with bromine in presence of Fe to give m-bromobenzotrichloride, which is compound B. And in the final step compound, B undergoes reduction with Zn and HCl to give the final product C as m-bromotoluene.
Question 8
1 / -0

During the course of an $$S_N1$$ reaction, the intermediate species formed is:

A
a carbocation

B
a free radical

C
a carbanion

D
an intermediate complex

Solution

Solution
When nucleophilic substitution takes place bond between C and X dissociates.
Now there are only 3 bonds on carbon And Carbon bears a positive Charge.
This is called carbocation.
The correct option is A
Question 9
1 / -0

Arrange the following alkyl chlorides in order of decreasing reactivity in a $$S_N1$$ reaction.
(I) isopropyl bromide
(II) propyl bromide
(III) tert-butyl bromide
(IV) methyl bromide

A
(III) > (I) > (II) > (IV)

B
(I) > (III) > (IV) > (II)

C
(IV) > (III) > (II) > (I)

D
(I) > (II) > (III) > (IV)

Solution

tret+-buty 1 bromide > isomeroply 1 bromide > propyle bromide > methyl bromide
i.e. $$ (III) > (I) > (II) > (IV)$$
Reason:
Since we know that in $$SN^{\prime}$$ reaction first is $$10\ ss$$ of a leaving group to give a carbocation, then nuclophile is attack
So carbocation must be stable
Carbocation Stability increases with increasing substitution of carbon
i.e. tertiary > secondry >> primary
Question 10
1 / -0

In which of the following can $$C-Cl$$ bond be easily broken?

A
Chlorobenzene

B
Vinyl chloride

C
Allyl chloride

D
p-Chlorotoluene

Solution

$$C-Cl$$ bond is easily broken in allyl chloride
Reason:- It is form stable carbocation as an intermediate.