Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 82

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Weekly Quiz Competition

Question 1
1 / -0

The $$IUPAC$$ name of $$n-$$butyl chloride is:

A
$$1-$$chlorobutane

B
$$n-$$chlorobutane

C
tre-butylchloride

D
$$2-$$methylbutane
Question 2
1 / -0

Dehydrohalogenation of an alkyl halide is

A
An addition reaction

B
A substitution reaction

C
An elimination reaction

D
An oxidation reaction

Solution

Alkyl halide gives alkene on Dehydrohalogenation, reaction takes place in presence of alc. $$KOH.$$This reaction is called an elimination reaction.
$$CH_3CH_2Br + Alc. KOH \rightarrow CH_2 = CH_2 + KBr + H_2O$$

There are two types of elimination reactions.
(a) $$E_1 \rightarrow$$ Unimolecular elimination
(b) $$E_2 \rightarrow$$ Bimolecular elimination
Question 3
1 / -0

Optically active compound is

A
$$3-$$chloropentane

B
$$2-$$chlorobutane

C
$$2-$$chloropropane

D
None of these

Solution

$$2-$$ chloro butane will be optically active as it has 1 chiral center.
Question 4
1 / -0

What would be the product formed when 1-Bromo-3-chlorocyclobutane reacts with two equivalents of metallic sodium in ether?

A

B

C

D

Solution
Question 5
1 / -0

The final product formed when ethyl bromide is treated with an excess of alcoholic KOH is

A
Ethylene

B
Ethane

C
Ethyne

D
Vinyl bromide

Solution

$$CH_{3}-CH_{2}-Br + KOH\rightarrow \underset{\text{Ethylene}}{CH_{2}=CH_{2}} + KBr + H_{2}O$$

This reaction is known as dehydrohalogenation.
Question 6
1 / -0

A mixture of ethyl iodide and n-propyi iodide is subjected to Wurtz reaction. The hydrocarbon that will not be formed is

A
n-butane

B
n-propane

C
n-pentane

D
n-hexane

Solution

Ethyl iodide and n-propyl iodide are allowed to undergo Wurtz reaction and they form butane and hexane as self-addition products and pentane as cross addition product. ∴ So,$$n- propane$$ will not be formed.
Question 7
1 / -0

The reaction , $$CH_3Br + Na \rightarrow Product ,$$ is called

A
Perkin reaction

B
Levit reaction

C
Wurtz reaction

D
Aldol condensation

Solution

It is a common method to prepare alkanes. Methane cannot be prepared by Wurtz reaction.
$$CH_3Br + 2Na + BrCH_3 \rightarrow \underset{ethane}{C_2H_6} + 2NaBr$$

Option C is correct.
Question 8
1 / -0

Action hydrogen chloride on $$CH_3 - \underset{CH_3}{\underset{|}{C}} = CH_2$$ and $$CH \equiv CH$$ will predominantly give the compounds, respectively

A
$$CH_3 - \underset{CH_3}{\underset{|}{C}}H = CH_2Cl$$ and $$CH_2Cl - CH_2Cl$$

B
$$CH_3 - \underset{CH_3}{\underset{|}{C}}Cl - CH_3$$ and $$CH_3 - CHCl_2$$

C
$$CH_3 - \underset{CH_3}{\underset{|}{C}}H - CH_2Cl$$ and $$CH_3 - CHCl_2$$

D
$$CH_3 - \underset{CH_3}{\underset{|}{C}}H = CH_3$$ and $$CH_2Cl - CH_2Cl$$

Solution

$$CH_3 - \underset{CH_3}{\underset{|}{C}} = CH_2 + HCl \rightarrow CH_3 - \overset{Cl}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}} - CH_3$$

$$CH \equiv CH+HCl\to CH_2=CHCl+HCl\to CH_3-CHCl_2$$

Markownikoff's rule is followed.
Question 9
1 / -0

Which one of the following compounds cannot be prepared by Wurtz reaction?

A
$$CH_4$$

B
$$C_2H_6$$

C
$$C_3H_8$$

D
$$C_4H_{10}$$

Solution

It is not possible to prepare $$CH_4$$ by Wurtz reaction. By Wurtz reaction we get minimum 2 C product.

Example: $$2CH_3-X+2Na \xrightarrow{Dry\ ether} H_3C-CH_3+2 NaX$$
Question 10
1 / -0

The reaction of an aromatic halogen compound with an alkyl halides in presence of sodium and ether is called

A
Wurtz reaction

B
Sandmeyer's reaction

C
Wurtz-fittig reaction

D
Kolbe reaction

Solution

It is Wurtz fittig reaction
$$C_6H_5Br + CH_3Br \xrightarrow[Ether]{Na} C_6H_5CH_3 + 2NaBr$$