Class 10 Chemistry Chapter Hydrocarbons important questions

Class 10 Chemistry

Chapter Hydrocarbons Important Questions

Answer the following questions:

1. Define hydrocarbons.

Answer: Definition of Hydrocarbons:

Only carbon and hydrogen atoms containing organic compounds are called hydrocarbons. Hydrocarbons are of three types namely; Alkanes, Alkenes and Alkynes. These three hydrocarbons can be differentiated from each other on the basis of the type of covalent bond present between their carbon atoms. However, we can say that;

• Since only carbon and hydrogen atoms are found in organic substances, hydrocarbons are known as such.
• They could have a ring structure or an open chain structure.

2. Set apart saturated from unsaturated hydrocarbons.

Answer: Hydrocarbons are classified into two classes.

1. Saturated hydrocarbons
2. Un-saturated hydrocarbons

Saturated Hydrocarbons:

• Saturated hydrocarbons, are hydrocarbons in which all of the carbon atoms are connected by a single covalent bond, either in a straight chain structure or a ring shape.
• There won’t be any double or triple covalent bonds in saturated hydrocarbons, only single covalent bonds
• For example, alkanes

Unsaturated Hydrocarbons:

• Unsaturated hydrocarbons, are hydrocarbons in which at least one double or triple covalent bond is present between any of two carbon atoms.
• For example, Alkenes have at least one double bond between any two carbon atoms and alkynes have at least one triple bond between any two carbon atoms.

1. Specify three different types of hydrocarbons. Give some examples.

Answer: Hydrocarbons are of three types namely; Alkanes, Alkenes and Alkynes. These three hydrocarbons can be differentiated from each other on the basis of type of covalent bond present between their carbon atoms.

Explanation of alkane, alkene and alkyne:

Alkanes:

• Alkanes, also known as saturated hydrocarbons, are hydrocarbons in which all of the carbon atoms are connected by a single covalent bond, either in a straight chain structure or a ring shape.
• There won’t be any double or triple covalent bonds in saturated hydrocarbons, only single covalent bonds. 
• For example,    CH₄,      CH₃-CH₃,    CH₃-CH₂-CH₂-CH₃        etc.

• The general formula for alkanes are C_nH_2n+2
• n is the number of carbon atoms n = 1, 2, 3, 4………..
• If n = 1 then according to general formula, C₁H_{2 x 1 + 2} = C₁H₂₊₂ = C₁H₄ = CH₄ (Methane), same for carbon 2, 3, and so on.

Alkenes:

• Alkenes, also known as “unsaturated hydrocarbons,” are hydrocarbons in which at least two carbon atoms are connected together by a double covalent bond.
• For example,   CH₂=CH₂,    CH₃-CH=CH-CH₃      etc.
• The general formula for alkenes are C_nH_2n
•  n is the number of carbon atoms n = 1, 2, 3, 4………..
• If n = 2 then according to general formula, C₂H_{2 x 2} = C₂H₄ (Ethene), same for carbon 3, 4, 5 and so on.

Alkynes:

• Alkynes, as well known as “unsaturated hydrocarbons” are hydrocarbons in which at least two carbon atoms are linked together by a triple covalent bond.
• For example,

• The general formula for alkynes is C_nH_2n-2
• n is the number of carbon atoms n = 1, 2, 3, ……………….
• If n = 2 then according to general formula, C₂H_{2 x 2 – 2} = C₂H_4-2 = C₂H₂ (Ethyne), same for carbon 3, 4, 5, and so on.

4.List three saturated and unsaturated hydrocarbon examples.

Answer: Saturated hydrocarbons:

Three examples of saturated hydrocarbons are:

1. CH₃-CH₂-CH₂-CH₃ (Propane)
2. CH₃-CH₂-CH₂-CH₂-CH₃ (Butane)
3. CH₃-CH₂-CH₂-CH₂-CH₂-CH₃ (Pentane)

Un-saturated hydrocarbons:

Three examples of unsaturated hydrocarbons are:

1. CH₃-CH₂-CH=CH₂ (Butene)
2. CH₃-CH₂-CH=CH-CH₂-CH₃ (3-Hexene)
3. CH₂=CH₂ (Ethene)

5.For ethene, depict the electron dot and cross structure.

Answer: Electron dot and cross structure for ethene:

Electron dot and cross structure for ethene is illustrated below:

6. Generate the structural formulas for a five-carbon alkane, alkene, and alkyne.

Answer: Structural formula of an alkane having five C atoms:

Structural formula of alkene having five C atoms:

Structural formula of alkyne having five C atoms:

7. What distinguishes 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:

8. By “dehydration reaction,” what do you mean? Give one illustration.

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

9. How do you change?

• 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.

1. By reaction with halogens:
2. By reaction with KMnO₄:

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

2. 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.

10. To demonstrate how an alkane is created from an alkene and an alkyne, write a chemical equation.

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-dichloroethene.
• In second step, repeat the same reaction.
• React 1,2-dichloroethene with Cl₂ molecule in the same way and convert it into 1,1,2,2-tetrachloroethane.

11.Generate a chemical equation to illustrate how an alcohol is dehydrated and how alkyl halides are dehydrohalogenated to produce ethene.

Answer:

Chemical equations for synthesis of Ethene:

1. 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.

1. 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.

12. To prove how 1,2-dihalide and a tetra halide are dehalogenated to produce ethyne, generate a chemical equation.

Answer:

Methods for synthesis of alkynes:

1. 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

2. 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

13. Construct chemical equations that illustrate how KMnO₄ reacts 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.

14. State a few industrial applications for ethene and ethyne.

Answer: Industrial uses of ethene:

1. 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:

1. 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.

15. Describe the need for a standardized procedure for naming chemical substances.

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.

16. Generate an electron cross and dot 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:

17. Design chemical equations for producing propene from

Answer:

18. Note the structural formulas for the compounds that result from the reaction between 1-butene and

• H₂ / Ni
• Dilute alkaline aqueous KMnO₄ solution
• Bromine water
• Chlorine

Answer:

19. Find A, B, C, and D in the subsequent reactions.

Answer:

20. Lay out a plan to transform ethene into ethyne?

Answer:

21. Two flammable liquid hydrocarbons are handed to you. The first is an alkene, while the second is an alkane. Develop a test to determine 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.

22. How many different outcomes are there when chlorine and ethane interact? Draw each of them.

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

23. Make a distinction 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.

24. Indicate the characteristic of alkanes.

Answer: Alkanes:

Properties of alkanes:

• Alkanes are saturated hydrocarbons
• 1st four members of alkane family e.g. methane, ethane, propane and butane are colorless and odorless gases
• The next two members of alkane family e.g. pentane and hexane are colorless and odorless liquids
• And remaining members next to hexane e.g. heptane, octane, Nonane and Decane and so on are colorless and odorless solids
• Alkanes are non-polar in nature
• Their non-polar nature is due to very less electronegativity difference between C and H atom or due to absence of electronegative atom
• Density of alkanes is less than water
• Due to non-polar nature alkanes are insoluble in water according to like dissolve like rule, polar substances dissolve in polar solvents and non-polar substances dissolve in non-polar solvents
• Water is polar solvent. Therefore alkanes (non-polar) are insoluble in water
• Alkanes are soluble in non-polar solvents e.g. ether and benzene etc.
• Due to non-polar nature, alkanes are good solvents and used in useful reactions
• They do not react with ionic compounds
• Hexane is the member of alkane family and used for oil extraction from cotton seed, soya beans and corn etc.
• Due to saturated nature addition reactions are not possible in alkanes due to strong sigma bonds whose breakage is not easy. Therefore, only substitution reactions are possible in it.

25. List the properties of alkenes.

Answer: Alkenes:

Properties of alkenes:

• Alkenes are un-saturated hydrocarbons
• 1st three members of alkene family e.g. Ethene, propene and butene are gases
• The members of  alkene family next to butene e.g. pentene, hexene, heptene, octene, nonene, decene, undecene, dodecene, tridecene, tetradecene and pentadecene are liquids
• And remaining members next to pentadecene are solids
• Alkenes are non-polar in nature
• Their non-polar nature is due to very less electronegativity difference between C and H atom or due to absence of electronegative atom
• Density of alkenes is less than water
• Due to non-polar nature alkenes are insoluble in water according to like dissolve like rule, polar substances dissolve in polar solvents and non-polar substances dissolve in non-polar solvents
• Water is polar solvent. Therefore alkenes (non-polar) are insoluble in water
• Alkenes are soluble in non-polar solvents e.g. ether, alcohol and benzene etc.
• Due to non-polar nature, alkenes are good solvents and used in useful reactions
• They do not react with ionic compounds
• Due to un-saturation (presence of double covalent bond, one sigma and one pi bond) they perform addition reactions
• The pi bond is weak bond and can be easily break down. Therefore, alkenes undergo addition reaction easily by breaking old pi bond and can make new strong sigma bonds with other atoms

26. Outline the qualities of alkynes.

Answer: Alkynes:

Properties of alkynes:

• Alkynes are un-saturated hydrocarbons
• 1st three members of alkyne family e.g. Ethyne, propyne and butyne are gases
• The next eight members of  alkyne family e.g. pentyne, hexyne, heptyne, octyne, nonyne, decyne, undecyne, dodecyne are liquids
• And remaining members next to dodecyne are solids
• Ethyne, a first member of alkyne has garlic like smell
• Alkynes are non-polar in nature
• Their non-polar nature is due to very less electronegativity difference between C and H atom or due to absence of electronegative atom
• Density of alkynes is less than water
• Due to non-polar nature alkynes are insoluble in water according to like dissolve like rule, polar substances dissolve in polar solvents and non-polar substances dissolve in non-polar solvents
• Water is polar solvent. Therefore alkynes (non-polar) are insoluble in water
• Alkynes are soluble in non-polar solvents e.g. ether, alcohol and benzene etc.
• They do not react with ionic compounds
• Due to un-saturation (presence of triple covalent bond (one sigma and two pi bonds)) they perform addition reactions
• The pi bonds are weak bonds and can be easily broken down. Therefore, alkynes undergo addition reactions easily by breaking old pi bonds and can make new strong sigma bonds with other atoms.

27. Write the alkanes general formula. Utilize this formula to derive the first two alkanes.

Answer:

• General formula for an alkane is C_nH_2n+2
• n is the number of carbon atoms and is equal to n = 1, 2, 3………
• If n = 1 then C₁H_2×1+2 = C₁H₂₊₂ = CH₄ (methane)
• If n = 2 then C₂H_2×2+2 = C₂H₄₊₂ = C₂H₆ (Ethane)

28. Write the alkenes general formula. Utilize this formula to derive the first two alkenes.

Answer:

• General formula for an alkene is C_nH_2n
• n is the number of carbon atoms and is equal to n = 1, 2, 3………
• If n = 2 then C₂H_2×2 = C₂H₄ = C₂H₄ (Ethene)
• If n = 3 then C₃H_2×3  = C₃H₆ = C₃H₆ (Propene)

29. Write the alkynes general formula. Utilize this formula to derive the first two alkynes.

Answer:

• General formula for alkane is C_nH_2n-2
• n is the number of carbon atoms and is equal to n = 1, 2, 3………
• If n = 2 then C₂H_2×2-2 = C₂H_4-2 = C₂H₂ (Ethyne)
• If n = 3 then C₃H_2×3-2 = C₃H_6-2 = C₃H₄ (Propyne)

30. List commercial uses of halogenated hydrocarbons.

Answer: The following list of halogenated hydrocarbons’ commercial applications:

• Liquid solvents for grease, oils and other organic materials include dichloromethane, tri-chloromethane, and tetra-chloromethane.
• Anaesthesia is provided via chloroform.
• Methyl chloride, a gas at room temperature, is utilized in industry for a variety of things.

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