1st PUC Chemistry Question Bank Chapter 13 Hydrocarbons

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Karnataka 1st PUC Chemistry Question Bank Chapter 13 Hydrocarbons

Question 1.
What are hydrocrabons?
Answer:
The organic compounds containing only carbon and hydrogen are called hydrocarbons.

Question 2.
Give important applications of hydrocarbons.
Answer:

• Hydrocarbons play a key role in our daily life.
• LPG and CNG used as fuels
• Hydrocarbons are used in the manufacture of polymers like polythene, polystyrene etc.
• Higher hydrocarbons are used as solvents for paints.
• They are used as the starting materials for many dyes and drugs.

Question 3.
Expand LPG.
Answer:
Liquified petroleum gas.

Question 4.
Expand CNG.
Answer:
Compressed natural gas.

Question 5.
Expand LNG.
Answer:
Liquified natural gas.

Question 6.
How do you obtain petrol, diesel and kerosene oils from petroleum?
Answer:
By the fractional distillation of petroleum.

Question 7.
How do you obtain LNG?
Answer:
By liquefaction of natural gas.

Question 8.
Name the method from which coal gas is obtained from coal.
Answer:
Destructive distillation.

Classification

Question 1.
What are saturated hydrocarbons?
Answer:
Hydrocarbons containing carbon – carbon and carbon-hydrogen single bonds are called saturated hydrocarbons. Example: Alkanes.

Question 2.
What are unsaturated hydrocarbons?
Answer:
Hydrocarbons containing carbon – carbon multiple bonds (double bonds or triple bonds of both) are called unsaturated hydrocarbons. Examples: Alkenes and alkynes

Question 3.
Give the classification of hydrocarbons.
Answer:
Depending on the structure,
hydrocarbons are classified as follows;

Alkanes

Question 1.
What are alkanes (paraffins)? Write the general formula of alkanes.
Answer:
Alkanes are the saturated aliphatic hydrocarbons containing only carbon-carbon single bond. They are represented by the general formula C_nH_2n+2, where n = 1, 2, 3, ….

Question 2.
Name the first member of alkane series.
Answer:
Methane

Question 3.
Alkanes are inert under normal conditions. Why?
Answer:
Because alkanes do not react with acids, bases and other reagents under normal conditions. Hence they were earlier known as paraffins (little affinity).

Question 4.
Write the name, formula and structure of first ten members of alkane series.
Answer:

Question 5.
Give the IUPAC and common names of following alkanes.
Answer:

Question 6.
Write structures of different chain isomers of alkanes corresponding to the formula C₆H₁₄ (hexane). Also write their IUPAC names.
Answer:
Hexane exists in five isomeric forms

Question 7.
Name one chain isomer of n-pentane.
Answer:
2-methyl butane.

Question 8.
What are primary (1°), secondary (2°), tertiary (3°) and quaternary (4°) carbon atoms?
Answer:
(i) Carbon attached to no other carbon atom (Example: CH₄) or to only one carbon atom (example: ethane) is called primary carbon atom.
Examples:

Note: Terminal carbon atoms are always primary.

(ii) Carbon atom attached to two carbon atoms is known as secondary carbon atom.
Example:

(iii) Tertiary carbon is attached to three carbon atoms.
Example:

(iv) Quaternary carbon (neo) is attached to four carbon atoms.

Question 9.

Answer:

Question 10.
Give two main natural sources of alkanes.
Answer:
Petroleum and natural gas.

Question 11.
How do you prepare alkanes from alkenes and alkynes by hydrogenation?
Answer:
Alkanes are prepared by the catalytic hydrogenation of alkenes and alkynes in the presence of catalyst like platinum, palladium, or nickel. The catalyst Ni is used in finely divided form at 200°C and relatively higher pressures. The Pt or Pd catalyse the reaction at room temperature.
Examples:

Question 12.
What is hydrogenation?
Answer:
The process of addition of dihydrogen gas to alkenes or alkynes in the presence of finely divided catalyst like Pt, Pd or Ni to form alkanes is called hydrogenation.

Question 13.
Name the catalyst used in hydrogenation of unsaturated hydrocarbons to alkanes.
Answer:
Pt or Ni or Pd.

Question 14.
How do you prepare alkanes from alkyl halides? Explain with examples.
Answer:
Alkanes are prepared by the reduction of alkyl halides with zinc and dilute hydrochloric acid.
Examples:

Question 15.
What is the reducing agent used in the reduction of alkyl halides to alkanes?
Answer:
Zn and dil HCl.

Question 16.
Give the preparation of alkanes by Wurtz reaction.
Answer:
Higher alkanes are prepared by warming alkyl halides with sodium metal in dry ether. This reaction is called Wurtz reaction.
Example 1.

Example 2.

Question 17.
What will happen if two different alkyl halides are taken in Wurtz reaction?
Answer:
Mixture of alkanes are formed.

Question 18.
What is decarboxylation? How do you prepare methane from sodium acetate by decarboxylation? Write the equation.
Answer:
The removal of CO₂ from molecules having – COOH group by using soda lime is called decarboxylation.
Example: Methane is formed when sodium acetate is heated strongly in the presence of soda lime.

Question 19.
What is soda lime?
Answer:
Soda lime is a mixture of sodium hydroxide and calcium oxide. (NaOH + CaO)

Question 20.
Sodium salt of which acid will be needed for the preparation of propane? Write the chemical equation of the reaction.
Answer:
Butanoic acid

Question 21.
Describe the preparation of alkanes by Kolbe’s electrolytic method.
Answer:
Alkanes are prepared by the electrolysis of concentrated aqueous solution of sodium or potassium salt of carboxylic acids.

Example:

The reaction path of the above reaction is as follows:
CH₃COO ^– Na⁺ → CH₃COO^– + Na⁺

Question 22.
Name the product formed at the cathode during electrolysis of aqueous solution of sodium or potassium salts of carboxylic acids by Kolbe’s electrolytic method.
Answer:
Hydrogen gas.

Question 23.
What type of alkanes can be prepared by Kolbe’s electrolytic method?
Answer:
Alkanes containing even number of carbon atoms.

Question 24.
Name the product formed at the anode when aqueous solution of sodium acetate is electrolysed.
Answer:
Ethane

Question 25.
Alkanes are almost non polar molecules. Why?
Answer:
Because C – C and C – H bonds are covalent in nature and due to very little electronegativity difference between carbon and hydrogen.

Question 26.
Alkanes are not soluble in water. Why?
Answer:
Because they do not form hydrogen bond with water.

Question 27.
Boiling point of alkanes increases with increase in atomic mass. Why?
Answer:
Due to increase in van der Waals forces.

Question 28.
Boiling point of isomeric alkanes decreases with increasing branching. Why?
Answer:
Because branching decreases the surface for contact, which results in decrease in van der Waals forces.

Question 29.
What is halogenation of alkanes? Under which condition halogenation of alkanes takes place?
Answer:
The replacement of one or more hydrogen atoms of alkanes by halogen atoms is known as halogenation. Halogenation of alkanes takes place either at higher temperatures (573 – 773K) or in the presence of diffused sun light.

Question 30.
Explain chlorination of methane.
Answer:
When methane is treated with chlorine in presence of diffused sunlight, the hydrogen atoms of methane are successively replaced by chlorine atoms to form chloromethane, dichloromethane, trichloromethane and tetrachloromethane.

Note:

• Fluorination of alkanes takes place violently and results in rupture of C – C bond. Therefore fluorination is controlled by diluting F₂ with nitrogen.
• Iodination is very slow and reversible. Hence iodination is carried out in the presence of HIO₃ which reacts with HI liberating I₂ and thus prevents the reverse reaction.
• The reactivity of different hydrogen atoms of alkanes towards substitution reaction is in the order:
  tertiary hydrogen > secondary hydrogen > primary hydrogen > CH₄
• The order of reactivity of halogens is in the order: fluorine > chlorine > bromine > iodine

Question 31.
Describe the mechanism of chlorination of methane.
Answer:
The chlorination of. methane is a free radical chain reaction. The mechanism of this reaction involves following steps.
1. Chain initiation: Chlorine molecule undergoes homolytic fission in the presence of diffused sun light to form chlorine free radicals.

2. Chain propagation: The chlorine free radical abstracts a hydrogen atom from a methane molecule to form methyl free radical.

The methyl radical abstracts a chlorine atom from another Cl₂ molecule forming chloromethane and a new chlorine free radical.

These two reactions take place several times and a chain reaction occurs.

3. Chain termination: Chain reaction stops when two radicals combine to form a stable covalent molecule.

Question 32.
Explain complete oxidation or combustion of alkanes with examples.
Answer:
Alkanes burnt in air or oxygen with non luminous flame to form carbon dioxide and water with evolution of heat. Therefore alkanes are used as fuels.

Example 1: When methane is burnt in air, carbon dioxide and. water are formed with the liberation of heat.
CH₄ + 2O₂ → CO₂ + 2H₂O ; ΔH = -890 kJ mol^-1

Example 2: When butane is burnt in air, carbon dioxide and water are formed with the liberation of heat.
C₄H₁₀ + \(\frac { 13 }{ 2 }\)O₂ → 4CO₂ + 5H₂O ΔH = -2875.84 kJ mol^-1

Question 33.
Explain controlled oxidation of alkanes under different experimental conditions.
Answer:
Alkanes are on heating with a limited supply of air at high pressure and in the presence of suitable catalyst give a variety of oxidation products.
Examples
1. Conversion of methane to methanol
When methane is oxidized with oxygen in presence of copper catalyst at 523K under 100 atm pressures, methyl alcohol is formed.

2. Conversion of methane to formaldehyde
When methane is heated with oxygen in presence of Mo₂O₃ (molybdenum (III) oxide) catalyst, formaldehyde is formed.

3. Oxidation of alkanes in the presence of manganese acetate catalyst to give fatty acids.

4. Alkanes containing tertiary carbon are oxidised to tertiary alcohols by KMnO₄.

Question 34.
Name the products formed during incomplete combustion of alkanes. Give the uses of products formed.
Answer:
Incomplete oxidation of alkanes in limited supply of air gives carbon black and carbon monoxide. Carbon black is used in the manufacture of printer ink, black pigments and as filters.
2CH₄ + 3O₂ → 2CO + 4H₂O

Question 35.
Explain isomerisation of alkanes with an example.
Answer:
The process of conversion straight chain alkanes on heating with AlCl₃ + HCl at about 200°C and 35 atm pressure into branched chain alkanes is called isomerisation.

Question 36.
Explain aromatization of n-alkanes with an example.
Answer:
The process of conversion of aliphatic compound into aromatic compound is known as aromatisation. Alkanes having six to 10 carbon atoms are converted into benzene and its homologues at high pressure and temperature in presence of catalyst.

Example: When n-hexane is heated in the presence of oxides of vanadium, molybdenum or chromium supported over alumina, benzene is formed.

Question 37.
What is cracking? What is the role of cracking in petroleum industry?
Answer:
The phenomenon of conversion of bigger hydrocarbons into smaller hydrocarbons by means of heating is called cracking.
For example;

Cracking plays an important role in petroleum industry to convert higher alkanes into lower alkanes. For example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of platinum, palladium or nickel catalyst gives a mixture of heptane and pentene.

Question 38.
Explain the reaction of methane with steam.
Answer:
Methane reacts with super heated steam in the presence of nickel catalyst at 1273 K to form carbon monoxide and hydrogen gas.

Question 39.
What are conformations or conformers or rotamers?
Answer:
The different arrangements of the atoms in space that results from the rotation of groups about C-C bond are called conformers or conformations or rotomers.

Question 40.
Rotation around the C-C single bonds in alkanes is not completely free. Why?
Answer:
Due to weak repulsive interaction between the adjacent bonds in alkanes.

Question 41.
What is torsional strain?
Answer:
The repulsive interaction between the electron clouds which affects the stability of a conformation is called torsional strain.

Question 42.
What are eclipsed and staggered conformations?
Answer:
Eclipsed conformation: The conformation in which hydrogen atoms attached to two carbons are overlap together is called eclipsed conformation.
Staggered conformation: The conformation in which hydrogen atoms attached two carbons are as far apart as possible is called staggered conformation.
Any other intermediate conformation is called skewed conformation.
Chemists have adopted two ways of representing conformational isomers.
1. Sawhorse representation
2. Newman projections

Question 43.
Write the Sawhorse representations of ethane.
Answer:
The saw horse eclipsed and staggered conformations of ethane are as follows:

Question 44.
Write the Newman projections of ethane.
Answer:
The Newman projections of eclipsed and staggered conformations of ethane are as follows:

Question 45.
Among staggered conformation and eclipsed conformation of ethane, which form is more stable?
Answer:
Staggered conformation is more stable than eclipsed conformation by about 12.55 kJ mol^-1.

Question 46.
What is dihedral angle or torsion angle?
Answer:
The angle of rotation about C – C bond due to torsional strain is called dihedral angle or torsional angle.

Question 47.
Among staggered and eclipsed form ethane, which form has maximum torsional strain.
Answer:
Eclipsed form.

Alkenes

Question 1.
What are alkenes? Write the general formula of alkenes.
Answer:
Alkenes are unsaturated aliphatic hydrocarbons containing carbon-carbon double bond. They are represented by the general formula C_nH_2n where n = 2, 3, 4, 5,….

Question 2.
Alkenes are also called olefines. Why?
Answer:
Alkenes are also called OLEFINS because they form oily liquids on reaction with chlorine or bromine.

Question 3.
Give the formula, name and structure of first nine members of alkenes.
Answer:

Question 4.
Explain the structure of alkenes.
Answer:
1. In alkenes, the carbon atom of the double bond undergoes sp² hybridisation. Hence they are having triangular planar geometry with the three sp² hybrid orbitals directed towards three comers of an equilateral triangle with a bond angle of 120°.
Example: Structure of ethene

2. Alkenes are more reactive than alkanes due to the presence of weaker pi bond. The reactions of alkenes are essentially the reactions of pi bond.

3. The bond length of carbon-carbon double bond in alkene (134 pm) is less than the bond length of carbon carbon single bond (154 pm) in alkanes.

Question 5.
Write the structural isomerism in propene.
Answer:
Alkenes higher than propene show structural isomerism. The structural isomers of the formula C₄H₈ are

The structure I and III, II and III are chain isomers whereas structure I and II are position isomers.

Question 6.
Write structures and IUPAC names of different structural isomers of alkenes corresponding to C₅H₁₀.
Answer:

Question 7.
Explain geometrical isomerism (cis-trans isomerism) in alkenes.
Answer:
Two compounds having the same molecular formula and same structural formula but differs in the spacial arrangement of atoms or groups across the double bond are called geometrical isomers. The phenomenon is called geometrical isomerism. It is due to the restricted rotation across the carbon-carbon double bond.
There are two forms of geometrical isomers;

• cis isomers
• trans isomers

In cis isomer, the identical groups are on the same side of the double bond.
In trans isomers, the identical groups are on the opposite side of the double bond.
For example, 2-butene exhibit cis trans isomerism as shown below :

Question 8.
Discuss the preparation of alkenes from alkynes.
Answer:
Alkenes can be obtained by partial reduction of alkynes with calculated amount of hydrogen gas in presence of Lindlar’s catalyst (palladium on charcoal partially deactivated with BaSO₄ or quinoline). This catalyst prevents further hydrogenation. Alkenes obtained by this method are having cis geometry.

However, alkynes on reduction with sodium in liquid ammonia form trans alkenes.

Question 9.
Explain the preparation of alkenes by dehydrohalogenation of a alkyl halides.
Answer:
Alkenes are obtained by heating alkylhalides with alcoholic potash (Alcoholic KOH) or Sodamide (NaNH₂).
This reaction is called dehydrohalogenation. This reaction is also known as β-elimination since hydrogen atom is eliminated from β-carbon atom.

where X = Cl, Br, I

Example 1. Ethene is prepared by heating bromoethane with alcoholic potash.

Example 2: Propene is obtained by heating n-propyl chloride with alcoholic potash.

Note: The ease of dehydrohalogenation of alkyl halides is in the order: tertiary > secondary > primary alkyl halides.
The ease of dehydrohalogenation of alkyl halides for different halides is in the order : iodide > bromide > chloride.

Question 10.
Explain the preparation of alkenes from vicinal dihalides with an example.
Answer:
Alkenes are obtained by treating vicinal dihalidcs with zinc metal. This reaction is called dehalogenation.

Question 11.
Describe the preparation of alkenes by dehydration of alcohols.
Answer:
Alkenes are obtained by heating alcohols with concentrated sulphuric acid.

Example: Ethene is obtained by heating ethyl alcohol with concentrated sulphuric acid at 170°C.

Note: The ease of dehydration of alcohols is in the order: tertiary > secondary > primary alcohols.

Question 12.
Give the addition reaction of alkenes with
(i) hydrogen
(ii) halogens.
Answer:
(i) Addition of hydrogen (Sabatier-Senderens reduction)
Addition of hydrogen to alkenes in the presence of platinum or nickel catalyst to form alkanes. The reaction is called hydrogenation. This reaction is exothermic reaction.

(ii) Addition of halogens: Alkenes react with halogens to form dihaloalkanes in inert solvent like carbon tetra chloride. The order of reactivity is chlorine > bromine > iodine.

Note: Bromination reaction is used as a test for unsaturation.

Question 13.
Explain the addition of HBr to symmetrical alkene with equation.
Answer:
HBr adds to symmetrical alkenes (alkenes containing similar groups on eitherside of double bond) to form alkyl bromides.

Question 14.
State Morkovnikov’s rule. Illustrate with an example.
Answer:
The rule states that “negative part of the Hydrogen halide (addendum) gets attached to the unsaturated carbon atom of the alkene containing less number of hydrogen atoms.”
Example: Addition of HBr to propene takes place according to Markovnikov’s rule.

Question 15.
Describe the mechanism of addition of HBr to propene.
Answer:
The mechanism of addition of HBr to propene involves the following steps.
1. Formation of electrophile, H⁺
HBr → H⁺ + Br^–

2. The electrophile, H⁺ which attacks the double bond to form carbocation.

The secondary carbocation (II) is more stable than primary carbocation (I).

3. The carbocation (II) is attacked by Br^– ion to form stable product 2-bromopropane.

Question 16.
What is peroxide effect or Kharash effect? Illustrate with an example.
Answer:
In the presence of peroxide, addition of HBr to propene (alkene) takes place against to the Markovnikov’s rule. This is called peroxide effect or Kharash effect or Antimarkovnikov’s rule.

Question 17.
Explain the mechanism of addition of HBr to propene in the presence of peroxide.
Answer:
The addition of HBr to propene in the presence of peroxide proceeds via free radical chain mechanism as follows:

4. Secondary free radical I (2°) is more stable than primary free radical II (1°). Hence the reaction proceeds via secondary free radical (I) resulting in the formation of n-propyl bromide.

Question 18.
Anti Morkownikov’s rule does not occur in case of HCl and HI in the presence of peroxide. Give reasons.
Answer:
This is due to
(i) H – Cl bond energy is more than H – Br bond energy. So that H – Cl bond is stronger. This does not form chlorine free radicals.
(ii) In case of HI, iodine free radicals are formed. They do not attack alkene instead they combine with each other to form iodine molecules.

Question 19.
Explain addition reaction of alkenes with concentrated sulphuric acid with an example.
Answer:
In accordance of Markovnikov’s rule, alkenes readily add concentrated sulphuric acid to form alkyl hydrogen sulphates.
For example,
Ethene reacts with concentrated sulphuric acid to give ethyl hydrogen sulphate.

Propene reacts with concentrated sulphuric acid to give Iso-propyl hydrogen sulphate.

Question 20.
Explain hydration of alkenes with an example.
Answer:
Water molecule adds to an alkene molecule across the double bond in the presence of dilute acids to give corresponding alcohol in accordance with the Markovnikov’s rule.
Example 1: Propene undergoes addition reaction with water in presence of concentrated sulphuric acid to give propan-2-ol.

Question 21.
Explain oxidation of alkenes with cold dilute aqueous KMnO₄.
Answer:
When alkenes are oxidised with cold, dilute aqueous KMnO₄ dihydroxy compounds (vicinal glycols) are formed. The KMnO₄ gets decolorised. This reaction is used as a test for unsaturation called Bayer’s test.
For example,
(i) Ethene gives ethane – 1, 2 – diol (ethylene glycol) when treated with cold alkaline KMnO₄ at 273 K.

(ii) Propene gives propane-1, 2-diol when treated with cold alkaline KMnO₄

Question 22.
How does alkenes undergo oxidation with acidified KMnO₄ or acidified K₂Cr₂O₇. Explain with example.
Answer:
Acidified KMnO₄ or acidified K₂Cr₂O₇ oxidises alkenes into ketones and / or acids depending upon the nature of the alkene and experimental conditions.
Example:

Question 23.
Explain ozonolysis reaction of alkenes with an example.
Answer:
The addition of ozone to alkenes to form ozonides and decomposition of ozonides with water and zinc to give carbonyl compounds is termed as ‘ozonolysis’.
Example 1: Propene reacts with ozone to give propeneozanide which on hydrolysis with zinc gives acetaldehyde and formaldehyde.

Example 2: 2-methyl propene reacts with ozone to form ozanide which on hydrolysis with zinc gives acetone.

Question 24.
What is polymerisation? How would you prepare polythene and polypropene by polymerization?
Answer:
Polymerisation is a process by which large number of simple molecules join together to form a giant molecule called polymer. The simple molecules are called monomers. When ethene is heated to 1000°C under 1000 atm pressure in presence of oxygen, polythene is formed.

Similarly, when propene is polymersied in the presence of catalyst, it forms polypropylene.

Question 25.
Give two uses of polymers.
Answer:
1. Polymers are used for the manufacture of plastic bags, squeeze bottles, refrigerator dishes, toys, pipes, radio and TV cabinets etc.
2. Polypropene is used for the manufacture of milk crates, plastic buckets and other moulded articles.

Alkynes

Question 1.
What are alkynes? Write their general formula.
Answer:
Alkynes are unsaturated aliphatic hydrocarbons containing carbon-carbon triple bond. They are represented by general formula C_nH_2n-2 where n = 2, 3,4, 5

Question 2.
Give the formula and structure of first nine alkynes.
Answer:

Question 3.
Which is the first member of alkyne series?
Answer:
Ethyne or Acetylene

Question 4.
Write structures of different isomers corresponding to the 5th member of alkyne series. Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different pairs of isomers?
Answer:
5th member of alkyne has the molecular formula C₆H₁₀. The possible isomers are:

Position and chain isomerism shown by different pairs.

Question 5.
Explain the structure of triple bond.
Answer:
In alkynes, the carbon atom of the triple bond under go sp – hybridization. The two sp-hybrid orbitals are diagonally opposite to each other resulting in linear structure with a bond angle of 180°.

Question 6.
How do you prepare acetylene from calcium carbide?
Answer:
Acetylene (ethyne) is prepared by treating calcium carbide with water.

Note: Industrially, calcium carbide is obtained by heating quick lime with coke. Quick lime can be obtained by heating lime stone.
CaCO₃ → CaO + CO₂
CaO + 3C → CaC₂ + CO

Question 7.
Give the preparation of acetylene (Ethyne) from 1,2-dibromethane.
Answer:

Question 8.
Explain the acidic nature of alkynes in terms of hybridization.
Answer:
Acidic nature can be explained on the basis of hybridisation in terminal alkynes. The triply bonded carbon is sp hybridised. In sp hybridisation, s character is more (50%) as compared to sp³ (25%) and sp² (33.3%) hybridisation. Due to this large s-character, the carbon atom is quite electronegative. Hence the sp hybridised carbon attract the shared electron pair of the C – H bond of 1-alkynes. Hence, the hydrogen atom of sp hybridized carbon atom can be easily replaced. Thus, hydrogen atoms attached to the triply bonded carbon atom in alkyne is acidic in nature.

Question 9.
Give any two reactions to acidic nature of acetylene (ethyne).
Answer:
The hydrogens of acetylene or hydrogen bonded to triple bonded carbon of any other terminal alkyne (1-alkynes) are acidic in nature. They can be easily replaced by metal atoms like Ag, Na, Cu to give corresponding metal acetylides. Thus, acetylene and terminal alkynes (alk-l-yne) are acidic.
Examples:

Question 10.
Alkynes and alkenes undergo electrophilic addition reaction. Why?
Answer:
Due to the presence of n bond.

Question 11.
How does dihydrogen adds to acetylene? Give equation.
Answer:
When ethyne is heated with H₂ in presence of Ni or Pt or Pd catalyst at 250°C, ethane is formed.

Question 12.

Answer:
Propane or CH₃CH₂CH₃

Question 13.
Give the chemical equation for the addition of bromine to propyne.
Answer:

Question 14.
Give a test show unsaturation of alkynes.
Answer:
Orange red colour of bromine in carbon tetrachloride solution is decolourised by alkynes. This test shows the unsaturation of alkynes.

Question 15.
How does halogen acids adds to alkynes? Explain with examples.
Answer:
Alkynes reacts with halogen acids according to the Markovnikov’s rule.

Example 1: Ethyne (acetylene) reacts with HBr to form bromoethene and finally 1,1 dibromoethane.

Example 2: Propyne reacts with HBr to forms 2,2-dibromopropane.

Note: The rate of addition of halogen acids follows the order, HI > HBr > HCl

Question 16.
Explain the addition reaction of water with alkynes.
Answer:
Alkynes are immiscible and do not react with water. But, when an alkyne is passed through dilute H₂SO₄ in presence mercuric sulphate at 333 K, addition of one molecule of water takes place to form carbonyl compound.

Example 1: When ethyne is passed through dilute H₂SO₄ in presence of mercuric sulphate at 333 K, ethanal is formed.

Example 2: When propene is passed through dilute H₂SO₄ in presence of mercuric sulphate at 333 K, propanone is formed.

Question 17.
Explain the polymerization reactions of acetylene.
Answer:
Acetylene undergo polymerization on heating in the presence of catalyst. The nature of products depends upon the conditions of polymerisation.
(i) Linear polymerization: Acetylene undergoes linear polymerisation to form polyacetylene or polyethylene under suitable conditions.

(ii) Cyclic polymerization: When ethyne (acetylene) is passed through a red hot iron tube, it polymerises to form benzene.

Aromatic Hydrocarbon

Question 1.
What are arenes?
Answer:
Aromatic hydrocarbons are known as arenes.

Question 2.
What are benzenoids? Give examples.
Answer:
Aromatic compounds containing benzene ring are called benzenoids.
Examples:

Question 3.
What are nonbenzenoids?
Answer:
Aromatic compounds without benzene ring are called nonbenzenoids.
Example:

Question 4.
Name the scientist who isolated the benzene first.
Answer:
Michael Faraday.

Question 5.
Give a reaction to show the presence of three double bonds in benzene.
Answer:

This reaction indicates the presence of three double bonds.

Question 6.
How do you show that all the six carbon and six hydrogen atoms of benzene are identical?
Answer:
Benzene reacts with Cl₂ in the presence of FeCl₃ to give monosubstituted product. This shows that all the six hydrogen and carbon atoms are identical

Question 7.
Write the Kekule structure of benzene.
Answer:

Question 8.
Give two limitations of Kekule structure.
Answer:

1. Fails to explain stability of benzene.
2. Fails to explain the preference to substitution reactions than addition reactions.

Question 9.
Write the resonance structures and resonance hybrid structure of benzene.
Answer:

Question 10.
What type of hybridisation take place in each carbon atom of benzene?
Answer:
sp²

Question 11.
How many C – C, and C – H σ-bonds and n bonds are present in benzene?
Answer:
6 C – C σ, 6 C – H σ and 3 π bonds or Benzene contains 12 σ bonds and 3 – π bonds.
C – C, σ – bonds = 6
C-H, σ – bonds = 6
Total σ = 12
C = C, 3, π – bonds

Question 12.
Write the orbital structure of benzene.
Answer:

Question 13.
Give three necessary conditions to show aromaticity of organic compounds.
Answer:

• Planarity
• Complete delocalization of the n electrons in the ring
• Presence of (4n + 2) n electrons in the ring. Where n is an integer (n = 0, 1, 2, 3, )

Question 14.
State Huckel’s rule or (4n + 2)π rule. How do you decide the aromaticity of organic compounds by using this rule. Give some examples.
Answer:
This rule states that “cyclic, planar, conjugated systems containing a total of (4n + 2) delocalised π-electrons possess a high degree of unsaturation and behave as aromatic compounds.” where n = 0, 1, 2, 3
Examples:
(1) Benzene has 6 electrons due to 3 double bonds. Hence benzene is aromatic compound.
(2) Naphthalene (10π electrons due to ‘5’ double bonds). Hence it is aromatic.
Examples

Question 15.
Give any three methods for the preparation of benzene.
Answer:
(1) From ethyne: When ethyne is passed over heated iron tube at 873 K, benzene is formed

(2) From sodium benzoate: When sodium benzoate is heated with sodalime, decarboxylation takes place to form benzene.

(3) From pheno.1: When phenol is distilled with zinc dust, benzene is formed

Question 16.
What are common electrophilic substitution reactions of benzene?
Answer:

• Nitration
• Halogenation
• Sulphonation
• Friedel crafts alkylation and acylation

Question 17.
Explain nitration of benzene.
Answer:
Benzene on heating with nitrating mixture (cone. HNO₃ + conc. H₂SO₄) gives nitrobenzene. This reaction is known as nitration of benzene.

Question 18.
Explain the mechanism of nitration of benzene.
Answer:
It involves the following steps:

III Step: The carbocation loses a proton to HSO^–₄ (bisulphate ion) to give nitrobenzene.

Question 19.
Name the electrophile produced during the nitration of benzene.
Answer:

Question 20.
Explain halogenation of benzene with an example.
Answer:
Benzene on heating with a halogen in the presence of ferric chloride (halogen carrier) gives halobenzene.

Example:

Question 21.
Explain the mechanism of halogenation of benzene.
Answer:
It involves the following steps:
I Step: Formation of an electrophile, X⁺ (halonium ion).
The catalyst FeCl₃ reacts with halogen to give an electrophile, X⁺.

II Step: The electrophile attacks benzene molecule to give resonance stabilised carbocation.

III Step: The carbocation combines with FeCl₃X^– to give halobenzene.

Note: Learn the mechanism of chlorination of benzene by replacing X₂ by Cl₂ and X by Cl.

Question 22.
Name the electrophile produced during the chlorination of benzene.
Answer:
Cl⁺, chloronium ion.

Question 23.
Explain sulphonation of benzene.
Answer:
Benzene on heating with concentrated sulphuric acid gives benzene sulphonic acid. This is called sulphonation of benzene.

Question 24.
Explain the mechanism of sulphonation of benzene.
Answer:
It involves the following steps
I Step: Formation of neutral electrophile SO₃. (sulphur trioxide)
Two molecules of H₂SO₄ interact to give the neutral electrophile, sulphur trioxide.

II Step: The electrophile SO₃ attacks the molecule of benzene to give resonance stabilised carbocation.

III Step: The carbocation combines with HSO₄^– to give benzene sulphonate ion

IV Step: Benzene sulphonate ion on protonation gives benzene sulphonic acid

Question 25.
Name the electrophile produced during sulphonation of benzene.
Answer:
Sulphur trioxide or SO₃

Question 26.
Explain alkylation of benzene by Friedel – Crafts reaction with an example.
Answer:
Benzene on heating with alkyl halide in the presence of anhydrous aluminium chloride catalyst gives alkyl benzene. This reaction is called Friedel – Crafts reaction.

Benzene reacts with methyl chloride in presence of anhydrous AlCl₃ to form toluene.

Question 27.
Give the mechanism of alkylation of benzene by Friedel – Crafts reaction.
Answer:
It involves the following steps
I Step: Formation of an electrophile, R⁺
The catalyst AlCl₃ combines with the molecule of haloalkane (alkyl halide) to give the electrophile R⁺.

II Step: The electrophile R⁺ attacks the molecule of benzene to give resonance stabilised carbocation.

III Step: The carbocation combines with AlCl₃X to give alkyl benzene.

Question 28.
Explain Friedel-Crafts acylation of benzene.
Answer:
Benzene reacts with acid chloride in presence of anhydrous aluminium chloride to form aromatic ketone.

Example:
Benzene reacts with acetyl chloride (Ethanoyl chloride) in presence of anhydrous aluminium chloride to form aceto phenone.

Question 29.
Give the mechanism of Friedel Crafts acylation of benzene.
Answer:
The Mechanism of Friedal-craft’s acetylation involves the following steps.

ii. The electrophile attacks benzene to form resonance stabilised carbocation.

Step 3: The carbocation loses a proton to AlCl^–₄ to form acetophenone.

Question 30.
Explain addition of (a) hydrogen (b) chlorine to benzene.
Answer:
(a) Addition of hydrogen: In the presence of finely divided nickel, at 180° C, benzene undergoes hydrogenation to give cyclohexane.

(b) Addition of chlorine: Benzene undergoes addition reactions with chlorine in the presence of sunlight gives benzene hexachloride (BHC or gammaxane or indane), an insecticide and pesticide

Question 31.
Give the combustion reaction of benzene with air.
Answer:
When benzene is heated in air, it bums with sooty flame producing CO₂ and H₂O

Question 32.
What are 0- and p-directing groups? Give examples.
Answer:
The substituents or groups which directs the incoming group to ortho and para positions are called ortho and para directing groups.
Examples, – CH₃, – C₂H₅, – Cl, – OH, – Br, OCH₃, -NHR, -NH₂, -NHCOCH₃

Question 33.
Write the resonance structures of phenol. Why substitution takes place at o- and p- positions of phenol only?
Answer:

From the above resonating structure, it is known that the electron density is more in ortho and para positions. Hence the substitution takes place mainly at these positions.

Question 34.
Even though aryl halides are deactivates due to – I effect, substitution takes place only at o- and p-position only. Why?
Answer:
Due to resonance, the electron density on o- and p-position is greater than that at the m-position. Hence aryl halides are also o- and p-directing groups.

Question 35.
What are meta directing groups. Give examples.
Answer:
The substituents or groups which direct the incoming group to meta position are called meta directing groups.
Examples, – NO₂, – CN, -CHO, – COOH, – COR, -COOR, – SO₃H, etc.

Question 36.
Write the resonance structures of nitrobenzene. Why substitution takes place only at m-position in nitro benzene.
Answer:

In this case, the electron density on benzene ring decreases making further substitution difficult. Hence these groups are called deactivating groups. The electron density on m-position is comparatively greater than o- and p-positions. Therefore, electrophile attacks m-position to form meta substitution product.

Corcinogenicity and Toxicity

Question 1.
Explain carcinogenicity and toxicity of benzene and polynuclear compounds.
Answer:
Benzene and many of its poly nuclear compounds are toxic and corcinogenic (cancer producing property). The poly nuclear hydrocarbons are formed due to incomplete combustion of organic materials like tobacco, coal and petroleum. They enter into the human body and undergo various biochemical reactions and finally damage DNA and cause cancer.
Examples for corcinogens
1, 2-Benzanthracene, 3-methylcholanthrene 1, 2-Benzpyrene 1, 2, 5, 6 – Dibenzanthracene, etc.