10th Class Chemistry Chapter Hydrocarbons solved exercise

Class 10 Chemistry Notes

Chapter: Hydrocarbons (Solved Exercise)

Multiple Choice Questions:

Which molecule contains a C-C double bond?

Ethane
Ethene
Ethyne
Ethyl alcohol

Solution:

The hydrocarbons with at least one C-C double bond are known as alkenes. Ethene, the second member of the alkene family with a C-C double bond, is also present. Therefore, the answer is (ethene)

2. Which product is obtained when chloromethane (or methyl chloride) is reduced?

Ethan
Ethene
Methane
Ethyne

Solution:

When methyl chloride is reduced then methane is produced as a product. Therefore, the answer is (Methane).

3. Which reacts explosively with methane?

F₂
Cl₂
Br₂
I₂

Solution:

Florine reacts explosively with methane. Therefore the answer is (F₂).

4. By dehydration we mean, removal of

Hydrogen
Water
Halogen
Hydrogen halide

Solution:

Hydra means water and de means removal.
So, dehydration means removal of water.
Therefore, the answer is (water).

5. Ethene and ethyne can be differentiated by

Hydrogenation
Bromine water
Strong alkaline aqueous solution of KMnO₄
Hydro halogenation

Solution:

Bromine water can discriminate between ethene and ethyne.
Ethene is used to treat bromine water, which prevents the brown coloration from dissipating.
However, bromine water loses its brown coloration when it is treated with ethyne.
Therefore, the answer is (bromine water)

6. Which is used for dehydrohalogenation?

Br₂water
Conc. H₂SO₄
Al₂O₃
Alcoholic KOH

Solution:

For dehydrohalogenation alcoholic KOH is used. Therefore, the answer is (alcoholic KOH)

Dehydrohalogenation of alkyl halide:

7. Which substance reacts with KMnO₄ to produce oxalic acid?

Ethane
Ethene
Ethyne
Ethyl alcohol

Solution:

Ethyne on reaction with KMnO₄ produces oxalic acid.
Therefore, the answer is (Ethyne).

8. The reduction of alkyl halides takes place in the presence of

Al₂O₃ at 350 ^oC
Conc. H₂SO₄ at 170 ^oC
Zn + Dust
Zn + HCl

Solution:

Reduction of alkyl halide takes place in the presence of Zn + HCl. Therefore, the answer is (Zn + HCl).

9. Which produces an alkane?

Combustion
Hydration
Dehydration
Hydrogenation

Solution:

Hydrogenation (addition of hydrogen to alkene or alkyne) leads to the production of alkanes. Therefore the answer is (Hydrogenation).

10. Does not react with aqueous solution of bromine

C₂H₆
C₂H₄
C₂H₂
C₃H₆

Solution:

Reaction of bromine with hydrocarbons is an addition reaction.
Alkanes are the saturated hydrocarbons only single covalent bonds are present in them.
Therefore, no addition reaction takes place in alkanes.
C₂H₆ is ethane, the second member of the alkane family, and because of its saturated state, it does not react with bromine water, means no addition reaction possible in it.
Therefore, the answer is (C₂H₆).

Answer the following questions:

Give three examples of saturated and un-saturated hydrocarbons.

Answer: Saturated hydrocarbons:

Three examples of saturated hydrocarbons are:

CH₃-CH₂-CH₂-CH₃(Propane)
CH₃-CH₂-CH₂-CH₂-CH₃ (Butane)
CH₃-CH₂-CH₂-CH₂-CH₂-CH₃ (Pentane)

Un-saturated hydrocarbons:

Three examples of unsaturated hydrocarbons are:

CH₃-CH₂-CH=CH₂ (Butene)
CH₃-CH₂-CH=CH-CH₂-CH₃ (3-Hexene)
CH₂=CH₂ (Ethene)

2. Draw electron dot and cross structure for ethene.

Answer: Electron dot and cross structure for ethene:

Electron dot and cross structure for ethene is illustrated below:

3. Draw structural formulas of an alkane, an alkene and an alkyne containing five carbon atoms.

Answer:

Structural formula of alkane having five C atoms:

Structural formula of alkene having five C atoms:

Structural formula of alkyne having five C atoms:

4. How can you differentiate ethane from ethene?

Answer:

Ethane are hydrocarbons have only single covalent bonds between C atoms.
While ethene are the hydrocarbons have at least one double covalent bond between any of two C atoms.

For example:

5. What do you mean by dehydration reaction? Give one example.

Answer: Dehydration refers to the removal of water.

Dehydration of alcohol:

In this process, alcohol is passed over heated Al₂O₃ between 340-450 ^oC to evaporate water from it.
This reaction produces an alkene and a water molecule as its byproduct.
In this reaction, the catalyst is Al₂O₃.
Instead of Al₂O₃ (alumina), we can use H₃PO₄ (phosphoric acid), P₄O₁₀ (phosphorus pentoxide), and concentrated Sulphuric acid.

Note:

Never forget that during hydrogenation, a hydroxyl atom will be removed from one carbon atom, followed by the removal of a hydrogen atom from a carbon atom next to the hydroxyl bounded carbon atom.
A single (same) carbon atom should never have its hydroxyl and hydrogen atoms removed.

Example 1:

Mechanism of reaction is given below in which;

Mechanism of a reaction is illustrated below in which you can see that how Al₂O₃ react with ethanol and form ethene from it.

Example 2:

Mechanism of reaction is given below in which;

Mechanism of a reaction is illustrated below in which you can see that how H₂SO₄ reacts with ethanol and form ethene from it.

3. How can you convert:

Ethene into ethane
Methane into carbon tetrachloride
Ethene into glycol
Ethyl chloride into ethane
Ethyl bromide into ethene

Answer: Ethene into ethane:

Ethene can be converted to ethane by the following two reactions.

By reaction with halogens:
By reaction with KMnO₄:

By reaction with halogens:

Double covalent bond is present between any two carbon atoms in alkenes.
In this double covalent bond, one bond is sigma bond (strong bond) and another bond is pi bond (weak bond).
Therefore, due to presence of this weak pi bond, alkenes perform addition reaction.
Pi bond is very weak covalent bond and can be easily break down.
By breakage of this old pi bond alkene can easily convert to alkane by formation of two new sigma bonds with halogen atoms.

Reaction with chlorine:

Balanced chemical reaction and mechanism is given below in which we can see that,

In this reaction as the name indicates (reaction with halogens), halogens will reacts with alkene.
Here, pi bond in alkene will break by homo-lytically and both carbon atoms will take back their shared electrons.
In this way each C atom will convert to alkyl free radicals by gaining one unpaired electron.
Similarly, in Cl₂ molecule homolytic fission will occur and two chlorine free radicals will produced.
These two chlorine free radicals will react with two alkyl free radicals (unpaired electrons present at two carbon atoms in alkene) and will lead to the formation of 1,2-dichloromethane.

Reaction with Bromine:

Reaction and its mechanism is illustrated below. This reaction and its mechanism is same as for mentioned above reaction of alkene with chlorine. Just use Br₂ molecule instead of chlorine.

By reaction with KMnO₄:

This reaction is also known as Bayer’s test.
This test is used for identification of presence of double bond in a substance.
In this reaction KMnO₄ react with alkene in the presence of water molecules and produces ethylene glycol by addition of two hydroxyl groups to alkene.
Ethylene glycol is used as an anti-freeze agent.

Reaction and its mechanism is given below in which we can see that how alkene reacts with KMnO₄ and form ethylene glycol.

4. Write a chemical equation to show the preparation of an alkane from an alkene and an alkyne.

Answer:

Alkynes by addition of halogens converted into alkenes and then alkenes again on hydrogenation converted into alkane. The reaction and its mechanism is illustrated below, in which we demonstrate that:

Triple covalent bond is present between any two carbon atoms in alkynes.
In this triple covalent bond, one bond is sigma bond (strong bond) and another two bonds are pi bonds (weak bonds).
Therefore, due to presence of these weak pi bonds, alkynes perform addition reactions.
Pi bond is very weak covalent bond and can be easily break down.
By fission of this old pi bond alkyne can easily convert to alkane through two step reaction by formation of four new sigma bonds with halogen atoms.

Reaction with chlorine:

Reaction and mechanism is given below in which we can see that,

In this reaction as the name indicates (reaction with halogens), halogens will react with alkyne.
Here in first step, one pi bond in alkyne will break homoliticaly and both carbon atoms will take back their shared electrons.
In this way both C atoms involved in this pi bond will convert to alkyl free radicals by gaining unpaired electron.
Similarly, in Cl₂ molecule homolytic fission will occur and two chlorine free radicals will produced.
These two chlorine free radicals will react with alkyl free radicals (two unpaired electrons present on two carbon atoms in alkene) and will lead to the formation of 1,2-dichloromethene.
In second step, repeat the same reaction.
React 1,2-dichloromethene with Cl₂ molecule in the same way and convert it into 1,1,2,2-tatrachloromethane.

5. Write a chemical equation to show the preparation of ethene from dehydration of an alcohol and dehydrohalogenation of alkyl halides.

Answer:

Chemical equations for synthesis of Ethene:

By dehydration of alcohol:

Dehydration refers to the removal of water.
In this process, alcohol is passed over heated Al₂O₃ between 340-450 ^oC to evaporate water from it.
This reaction produces an alkene and a water molecule as its byproduct.
In this reaction, the catalyst is Al₂O₃.
Instead of Al₂O₃ (alumina), we can use H₃PO₄ (phosphoric acid), P₄O₁₀ (phosphorus pentoxide), and concentrated Sulphuric acid.

Note:

Never forget that during hydrogenation, a hydroxyl atom will be removed from one carbon atom, followed by the removal of a hydrogen atom from a carbon atom next to the hydroxyl bounded carbon atom.
A single (same) carbon atom should never have its hydroxyl and hydrogen atoms removed.

Example 1:

Mechanism of reaction is given below in which;

Mechanism of a reaction is illustrated below in which you can see that how Al₂O₃ react with ethanol and form ethene from it.

Example 2:

Mechanism of reaction is given below in which;

Mechanism of a reaction is illustrated below in which you can see that how H₂SO₄ reacts with ethanol and form ethene from it.

By dehydrohalogenation of alkyl halides:

In this chemical process, the hydrogen halide from the alkyl halide is removed.
Dehydrohalogenation is the reaction’s scientific name, which stands for the elimination of the hydrogen and halide group (de means removal).
Here, alcohol-infused KOH is used to perform the removal.
Always remember that hydrogen and halogen both will be removed from two adjacent carbon atoms not from a single carbon atom.
From one C atom halide group will be removed and from another adjacent C atom hydrogen atom will be removed.

Mechanism of reaction is given below in which we can see that;

In a reaction, alcoholic KOH is utilized as a base (OH^–)
This base will extract proton from alkyl halide
When base will extract proton of 1^st carbon atom then, the sigma bond between carbon and hydrogen atom will shift between CH₂-CH₂, resulting in the formation of pi bond between these two carbon atoms.
Alkene is created when a proton is removed and a sigma bond is moved.
The halide ion will react with the K⁺ of alcoholic KOH to generate KCl, whereas the proton will react with the hydroxyl group OH^– of alcoholic KOH to form H₂O.

6. Write a chemical equation to show the preparation of ethyne from dehalogenation of 1,2-dihalide and a tetra halide.

Answer:

Methods for synthesis of alkynes:

By dehydrohalogenation of vicinal di-halides:

According to the method’s name, dehydrohalogenation, the vicinal di-halide will be freed of its hydrogen and halide ions.

What is vicinal di-halide, exactly?

A compound in which two halogen atoms are attached with two covalently bounded carbon atoms is called vicinal di-halide.
In this method, Hydrogen and halogen atom both will not be removed from one single C atom.
Removal will be like, hydrogen atom will be removed from one carbon atom and halogen atom will be removed from next adjacent carbon atom.

Mechanism of a reaction is given below in which we can see that,

Step 1:

Alcoholic KOH is used in a reaction which act as a base OH^–
This base will extract proton from 1^st carbon of vicinal di-halide
When base will extract proton of 1^st carbon atom then, sigma bond between carbon and hydrogen atom will shifts between CH-CH, resulting in the formation of pi bond between these two carbon atoms.
Removal of proton and shifting of sigma bond leads to the formation of vinyl chloride
The proton will react with hydroxyl group OH^– of alcoholic KOH and will form H₂O
The halide will react with K⁺of alcoholic KOH and will form KCl

Step 2:

Alcoholic KOH is used in a reaction which act as a base OH^–
This base will extract proton from 2^nd carbon of vinyl chloride
When base will extract proton of 2^nd carbon atom then, sigma bond between carbon and hydrogen atom will shifts between CH=CH, resulting in the formation of pi bond between these two carbon atoms.
Removal of proton and shifting of sigma bond leads to the formation of alkyne
The proton will react with hydroxyl group OH^– of alcoholic KOH and will form H₂O
The halide will react with K⁺of alcoholic KOH and will form KCl.

By dehalogenation of tetra-halides:

In this method as the name indicates (dehalogenation) means halogen atom will be removed from tetra-halides. Now what is a Tetra-halide?

A compound in which four halogen atoms are attached with two covalently bounded carbon atoms is called tetra-halide.
In this method, in one step two halogen atoms will be removed at a time.
But remember two halogen atoms will not be removed from a single C atom.
The removal will be like; one halogen atom will be removed from one C atom and second halogen atom will be removed from next adjacent C atom.
Same for the next two halogen atoms.

Mechanism of a reaction is given below in which we can see that;

Two Zn metal atoms are used in a reaction which help in the abstraction of halogen atom
This Zn metal atom will abstract two halogen atoms from 1^st and 2^nd carbons of tetra-halide
In first step, Zn metal atom will attack on halogen atom bounded to 1^st carbon atom.
As a result, sigma bond between C and Cl will shift between C-C atoms.
This bond shifting leads to the removal of halogen atom from 2^nd C atom.
Removal of halogen atom from 2^nd C atom leads to the formation of 1,2-Dichlororthane and ZnCl₂
In second step, second Zn metal will attack on halogen atom bounded to 2^nd C atom.
As result, sigma bond between C and Cl will shift between C-C atoms.
This bond shifting leads to the removal of halogen atom from 1^st C atom.
Removal of halogen atom from 1^st C atom leads to the formation of alkyne and ZnCl₂

7. Write chemical equations showing reaction of KMnO4 with ethene and ethyne.

Answer: Reaction of KMnO4 with ethene:

This reaction is also known as Bayer’s test. This test is used for identification of presence of double bond in a substance.
In this reaction KMnO₄ react with alkene in the presence of water molecules and produces ethylene glycol by addition of two hydroxyl groups to alkene. Ethylene glycol is used as an anti-freeze agent.

Reaction and its mechanism is given below in which we can see that how alkene reacts with KMnO₄ and form ethylene glycol.

Reaction of KMnO4 with ethyne:

This reaction is used for identification of presence of triple bond in a substance.
In first step alkaline KMnO₄ react with alkyne in the presence of water molecules and produces tetrahydroxy ethane by addition of four hydroxyl groups to alkyne.
Then in second step, by removal of two water molecules tetrahydroxy ethane convert into glyoxal.
This glyoxal on oxidation convert into oxalic acid.

Reaction and its mechanism is illustrated below in which we can see that how alkyne reacts with KMnO₄ and form oxalic acid.

8. List some industrial uses of ethene and ethyne.

Answer: Industrial uses of ethene:

Synthesis of polymers:

Ethene is used in the manufacturing of polythene.

For artificial ripening of food:

Artificial food ripening is accomplished using ethene.

As an antifreeze-agent:

Used as an anti-freeze agent (ethylene glycol)

For welding and cutting of metals:

Metals are chopped and joined together using an oxy-ethylene flame.

Industrial uses of ethyne:

Synthesis of polymers:

Acetylene is utilized as a starting ingredient in the manufacture of polymers like nylon, rubber, and polyvinyl chloride (PVC).

For welding and cutting of metals:

Metals are chopped and joined together using an oxy-acetylene flame.

For artificial ripening of food:

Artificial food ripening is accomplished using acetylene.

9. Explain why a systematic method of naming chemical compounds is necessary.

Answer: There are countless numbers of organic compounds. Systematic naming of organic compounds is required in order to comprehend, identify, and categorize these chemicals. Therefore, some guidelines for naming organic compounds were developed by an international union of pure and applied chemistry. Each organic compound has a unique name under this system.

10. Draw electron dot and cross structure for

Propane
Propyne
Propene

Answer:

Electron dot and cross structure for propane:

Electron dot and cross structure for propyne:

Electron dot and cross structure for propene:

Think – Tank

11. Write chemical equations for the preparation of propene from

Answer:

12. Write down structural formulas for the products which are formed when 1-butene is reacted with

H₂ / Ni
Dilute alkaline aqueous KMnO₄ solution
Bromine water
Chlorine

Answer:

13. Identify A, B, C, and D in the following reactions.

Answer:

14. Construct a scheme to convert ethene into ethyne?

Answer:

15. You are given two flammable liquid hydrocarbons. One of them is alkene and the other is an alkane. Design an experiment to find out which is which.

Answer:

Take two test tubes.
In one test tube take sample of one flammable liquid.
In another test tube take sample of another flammable liquid.
Add bromine water in each test tube.
Observe the color of the samples placed in both test tubes.
When the bromine water disappear then it is alkene.
While when no color change takes place then it is alkane.

16. How many possible products are there when chlorine reacts with ethane? Sketch them all.

Answer: The possible products formed by reaction of ethane with chlorine are chloro-ethane and hydrochloric acid.

17. Differentiate between ethene and ethyne.

Answer:

Two C atoms are present in both ethene and ethyne, which are both unsaturated hydrocarbons.
The primary distinction between them is that at least one double covalent connection between any two C atoms exists in ethene
Since ethyne contains at least one triple covalent connection between any two C atoms.
However, in laboratory bromine water can discriminate between ethene and ethyne.
Ethene is used to treat bromine water, which prevents the brown coloration from dissipating.
However, bromine water loses its brown coloration when it is treated with ethyne.

Numerical problems:

Self-assessment exercise 12.1:

Write structural formula for