Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Choose the correct option from the following:

A
In the electrophilic substitution of toluene with $Br_2$ , iron (III) bromide acts as a Lewis acid.

B
In the reaction of toluene with $Cl_2/FeCl_3$ , ortho and para isomers are easily separated.

C
Similar reaction with iodine is reversible in nature

D
All of these

Solution

Chlorides of bromides can be easily prepared by electrophilic substitution with chlorine and bromine in the presence of Lewis acid catalysts like iron. The ortho and para isomers can be easily separated due to large difference in their melting points. Reactions with iodine are reversible in nature in presence of an oxidizing agent. So all teh options are correct.
Question 2
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

$-CH_3$ is ortho & para directing group so $Cl$ will come at p-position.But major product is para-chloro toluene.The reaction is electrophilic aromatic substitution reaction.
Question 3
1 / -0

Choose the correct answer among the alternatives given :

In the following pairs of halogen compounds , which compound undergoes faster $S_N1$ reaction?

A

B

C

D

Solution

$S_N1$ reaction proceeds via formation of carbocation. In option B, the alkyl halide (1) is $3^{\circ}$ while (2) is $2^{\circ}$ .

Therefore greater the stability of the carbocation, faster is the rate of $S_N1$ reaction.

Therefore option B, the compound pair of 2-chloro-2 methylpropane and 2- chloroheptane is the correct option.
Question 4
1 / -0

An alkyl chloride produces a single alkene on reaction with sodium ethoxide and ethanol. The alkene further undergoes hydrogenation to yield 2- methyl butane. Identify the alkyl chloride from amongst the following.

A
$ClCH_2CH(CH_3)CH_3$

B
$ClCH_2CH_2CH_2CH_3$

C
$ClCH_2CH(CH_3)CH_2CH_3$

D
$CH_3C(Cl)(CH_3)CH_2CH_3$

Solution

$Cl-CH_2-{\underset{CH_3}{\underset{|}{CH}}}-CH_2CH_3\xrightarrow[EtoH]{NaoEt}CH_2={\underset{CH_3}{\underset{|}{C}}}-CH_2CH_3\xrightarrow[H_2]{catalyst}CH_3={\underset{CH_3}{\underset{|}{C}}}-CH_2CH_3$
2methylbutane
Question 5
1 / -0

Chlorobenzene on treatment with sodium in dry ether gives diphenyl. The name of the reaction is ?

A
fittig reaction

B
Wurtz.Fittg reaction

C
Sandmeyer reaction

D
Gattermann reaction

Solution

It is called as Fittig reaction, two aryl halides are coupled in presence of sodium metal in dry ether to furnish biarlys. So from the above image we can see that chlorobenzene on treatment with sodium in dry ether gives diphenyl.
Question 6
1 / -0

Chlorobenzene is formed by reaction of chlorine with benzene in the presence of $A1C1_3$ .Which of the following species attacks the benzene ring in this reaction?

A
$C1^-$

B
$Cl^+$

C
$AlCl_3$

D
$[AlCl_4]^-$

Solution
Question 7
1 / -0

Which is the correct IUPAC name for,

$CH_3-{\underset{C_2H_5}{\underset{|}{CH}}}-CH_2-Br$ :

A
1-Bromo-2-ethylpropane.

B
1-Bromo-2-ethyl-2-methylethane.

C
1-Bromo-2-methylbutane.

D
2-Methyl-1-bromobutane.

Solution

Rules for IUPAC Name:
Numbering should start at the substituent carbon and longest chain should be covered.
$1-Bromo-2-methylbutane$
Question 8
1 / -0

Choose the correct answer among the alternatives given :
The end product of (Q) in the following sequence of reaction is:

A

B

C

D

Solution

Option C is correct.
Question 9
1 / -0

Chloromethane on treatment with excess of ammonia yields mainly :

A
N, N-dimethylmethanamine

B
N-methylmethanamine $(CH_3-NH-CH_3)$

C
methanamine $(CH_3NH_2)$

D
mixture containing all these in equal proportion.

Solution

This reaction of chloromethane results in formation of methanamine after reacting with $NH_3$
Question 10
1 / -0

Alkyl halides react with metallic sodium in dry ether producing?

A
Alkanes with same number of carbon atoms

B
Alkanes with double number of carbon atoms

C
Alkenes with triple number of carbon atom

D
Alkenes with same number of carbon atom

Solution

$RX+2Na+X-R\xrightarrow {dry ether} R-R+2Na-X$
alkyl halide
alkane sodium halide