Lahore Board Chemistry

Equilibria Class 9th Chemistry – Chapter 6 Solved Exercises (2025 Punjab Boards)

by

his web page provides detailed solutions and explanations for the Equilibria chapter (Chapter 6) of Class 9th Chemistry, designed according to the 2025 Punjab Board syllabus. The content includes solved exercises, short and long answer questions, and additional tips to help students prepare for exams. Specifically crafted for Lahore Board and other Punjab Boards, it ensures clarity and understanding with simplified explanations.

Solutions to the given MCQs:

(i) What will happen if the rates of forward and reverse reactions are very high?

  • Options:
    1. The equilibrium point will reach very soon.
    2. The equilibrium point will reach very late.
    3. The reaction will not attain the state of dynamic equilibrium.
    4. The reaction will be practically irreversible.
  • Correct Answer: (a) The equilibrium point will reach very soon.
  • Explanation:
    When the rates of the forward and reverse reactions are very high, equilibrium is established faster as the reaction quickly balances the forward and reverse processes.

(ii) Predict which components of the atmosphere react in the presence of lightning.

  • Options:
    1. N2 and H2O
    2. O2 and H2O
    3. CO2 and O2
    4. N2 and O2
  • Correct Answer: (d) N2 and O2
  • Explanation:
    Lightning provides energy for the nitrogen and oxygen in the atmosphere to react, forming nitrogen oxides (NO and NO2).

(iii) An inorganic chemist places one mole of PCl5 in container A and one mole each of Cl2 and PCl3 in container B. Both containers are sealed and heated to the same temperature to reach equilibrium. Guess the composition of mixtures in both containers.

  • Options:
    1. Both containers will have the same composition of mixtures.
    2. Container A will have more concentration of PCl5 than B.
    3. Container A will have less concentration of PCl5 than B.
    4. Both containers will have zero concentration of its reactants.
  • Correct Answer: (a) Both containers will have the same composition of mixtures.
  • Explanation:
    Regardless of the initial concentrations, equilibrium is defined by the same equilibrium constant at a given temperature. Therefore, both containers will reach the same composition of mixtures.

(iv) Choose the conditions to produce the maximum amount of lime (CaO) in the decomposition of lime (CaCO3).

  • Options:
    1. Heating at high temperature in a closed vessel.
    2. Heating at high temperature in an open vessel.
    3. Cooling in a closed vessel.
    4. Cooling in an open vessel.
  • Correct Answer: (b) Heating at high temperature in an open vessel.
  • Explanation:
    Decomposition of CaCO3 into CaO and CO2 is favored by removing CO2 gas (a product) continuously. This is best achieved in an open system.

(v) What condition should be met for the reversible reaction to achieve the state of equilibrium?

  • Options:
    1. All the reactants should be converted into products.
    2. 50% of the reactants should be converted into products.
    3. The concentration of all the reactants and products should become equal.
    4. The rate of the forward and reverse reactions should become equal.
  • Correct Answer: (d) The rate of the forward and reverse reactions should become equal.
  • Explanation:
    At equilibrium, the forward and reverse reaction rates are equal, leading to constant concentrations of reactants and products (though not necessarily equal in value).

(vi) Why does gas start coming out when you open a can of fizzy drink?

  • Options:
    1. Because the solubility of the gas increases.
    2. Because the gas is insoluble in water.
    3. Because the gas is dissolved under pressure and comes out when the pressure is decreased.
    4. Because the solubility of the gas decreases at high pressure.
  • Correct Answer: (c) Because the gas is dissolved under pressure and comes out when the pressure is decreased.
  • Explanation:
    Carbon dioxide is dissolved in the drink under high pressure. When the can is opened, the pressure decreases, causing the gas to come out of the solution.

(vii) The following reaction is performed in an open vessel: CaCO3(s)↔CaO(s)+CO2(g). How will the equilibrium be affected if you increase the pressure?

  • Options:
    1. The forward reaction will be favored.
    2. The backward reaction will be favored.
    3. No effect on the backward reaction.
    4. No effect on forward or backward reaction.
  • Correct Answer: (b) The backward reaction will be favored.
  • Explanation:
    Increasing pressure favors the reaction that decreases the number of gaseous molecules. In this case, the backward reaction reduces CO2, which is a gas.

(viii) When will a reaction become a reversible one?

  • Options:
    1. If the activation energy of the forward reaction is comparable to that of the backward reaction.
    2. If the activation energy of the forward reaction is higher than that of the backward reaction.
    3. If the activation energy of the forward reaction is lower than that of the backward reaction.
    4. If the enthalpy change of both the reactions is zero.
  • Correct Answer: (a) If the activation energy of the forward reaction is comparable to that of the backward reaction.
  • Explanation:
    For a reaction to be reversible, both the forward and backward reactions must occur at comparable rates, which requires similar activation energies.

(ix) Is a reversible reaction useful for preparing compounds on a large scale?

  • Options:
    1. No.
    2. Yes.
    3. They are useful only when equilibrium lies far to the right side.
    4. They are useful only when equilibrium lies far to the left side.
  • Correct Answer: (c) They are useful only when equilibrium lies far to the right side.
  • Explanation:
    To maximize product yield in a reversible reaction, the equilibrium must favor product formation, meaning it must lie far to the right.

(x) What will happen to the concentrations of the products if a reversible reaction at equilibrium is not disturbed?

  • Options:
    1. They will remain constant.
    2. They will keep on increasing.
    3. They will keep on decreasing.
    4. They will first increase and then decrease.
  • Correct Answer: (a) They will remain constant.
  • Explanation:
    At equilibrium, the rates of the forward and reverse reactions are equal, so the concentrations of products and reactants remain unchanged.

Answers to the Questions:

2. Questions for Short Answers

(i) How is dynamic equilibrium different from static equilibrium?

  • Answer:
    In dynamic equilibrium, the forward and backward reactions occur at the same rate, maintaining constant concentrations of reactants and products. In static equilibrium, there is no movement or reaction, and the system is at rest.

(ii) How will the following reversible reaction be affected if its temperature is increased?
Reaction: 2H2O(l)→Electricity 2H2(g)+O2(g)

  • Answer:
    Increasing temperature will favor the forward reaction because the decomposition of water into hydrogen and oxygen is an endothermic process, which absorbs heat.

(iii) How can you get the maximum yield in a reversible reaction?

  • Answer:
    Maximum yield can be obtained by:
    • Adjusting temperature and pressure to favor the desired reaction.
    • Removing the product as it forms to shift equilibrium toward the product side.

(iv) How can you decrease the time to attain the position of equilibrium in a reversible reaction?

  • Answer:
    • Increase the concentration of reactants.
    • Use a catalyst to speed up the forward and reverse reactions.
    • Raise the temperature to increase reaction rates (if temperature is favorable).

(v) What is the effect of increasing pressure on the following reaction?
Reaction: N2+O2↔2NO

  • Answer:
    There is no change because the number of moles of gas on both sides of the reaction is the same. Pressure changes do not affect equilibrium in such cases.

3. Constructed Response Questions

(i) Why are some reactions irreversible while others are reversible?

  • Answer:
    Irreversible reactions go to completion because their products are stable and cannot revert to reactants (e.g., combustion). Reversible reactions occur when products can react back to form the reactants, maintaining equilibrium.

(ii) Why are combustion reactions generally irreversible?

  • Answer:
    Combustion reactions are highly exothermic, releasing large amounts of energy. The products (e.g., CO2 and H2O) are stable and cannot revert to the reactants under normal conditions.

(iii) Can you make an irreversible reaction reversible and vice versa?

  • Answer:
    Some irreversible reactions can become reversible under specific conditions, such as high pressure or low temperature. However, most irreversible reactions (e.g., combustion) cannot be reversed due to product stability.

(iv) How do you know if a reaction is reversible or irreversible?

  • Answer:
    A reaction is reversible if:
    • The products can convert back into reactants.
    • It reaches equilibrium.
      A reaction is irreversible if it goes to completion without equilibrium.

(v) Do the phase changes in water (solid to liquid, liquid to vapor) reverse or irreversible?

  • Answer:
    Phase changes in water are reversible because the process can go back and forth under appropriate temperature and pressure conditions.

4. Descriptive Questions

(i) How can you drive the reversible reaction at equilibrium?

  • Answer:
    • In the forward direction: Increase the concentration of reactants, remove products, or adjust temperature/pressure to favor the forward reaction.
    • In the backward direction: Increase the concentration of products or decrease reactants.

(ii) Explain how the forward and backward reactions change when the system approaches equilibrium.

  • Answer:
    Initially, the forward reaction rate is high, and the backward rate is low. As products form, the backward reaction rate increases. At equilibrium, both reaction rates become equal.

(iii) Describe the effect of a catalyst on the reversible reaction.

  • Answer:
    A catalyst speeds up both the forward and backward reactions equally, helping the system reach equilibrium faster without changing the equilibrium position.

(iv) How can a reversible reaction be forced to go to completion?

  • Answer:
    • Continuously remove the product.
    • Adjust temperature and pressure to favor the desired reaction completely.
    • Use excess reactants.

(v) How does a change in temperature affect the reaction at equilibrium?

  • Answer:
    • For an exothermic reaction: Increasing temperature shifts equilibrium to the reactants.
    • For an endothermic reaction: Increasing temperature shifts equilibrium to the products.

5. Investigate

(i) Study the effect of heat on hydrated CuSO4. Why does this salt look colored, and why does it lose color upon heating?

  • Answer:
    Hydrated CuSO4CuSO_4 (blue in color) loses its water molecules on heating, forming anhydrous CuSO4 (white). The color is due to the presence of water of crystallization, which is lost upon heating.

(ii) Synthesis of ammonia gas is very important industrially because it is used in the preparation of urea fertilizer. Explain the conditions you will use to get the maximum yield of ammonia.

  • Answer:
    • Use high pressure to favor the forward reaction (fewer moles of gas).
    • Use moderate temperature (450°C) to balance rate and yield (since the reaction is exothermic).
    • Use an iron catalyst to speed up the reaction.

Aliphatic hydrocarbons Solved Exercise

by

Unlock the key to mastering aliphatic hydrocarbons with this comprehensive guide to solved exercises. Covering topics such as alkanes, alkenes, and alkynes, this resource is aligned with the latest syllabus for Lahore Board, Federal Board, and other educational boards. Understand the structure, nomenclature, reactions, and properties of aliphatic hydrocarbons through step-by-step solutions to textbook exercises, conceptual questions, MCQs, and past paper problems. Designed to simplify learning, this content is ideal for students preparing for their chemistry exams

Q4. Write the structural formula for each of the following compounds:

i) 2-Methylpropane:
Structure:
CH3-CH(CH3)-CH3

ii) 3-Ethylpentane:
Structure:
CH₃-CH₂-CH(CH₂CH₃)-CH₂-CH₃

iii) 2,2,3,4-Tetramethylpentane:
Structure:
CH₃-C(CH₃)-C(CH₃)(CH₃)-CH(CH₃)-CH₃

iv) Neopentane (2,2-Dimethylpropane):
Structure:
(CH₃)₄C

v) 4-Ethyl-3,4-Dimethylheptane:
Structure:
CH₃-CH₂-CH(CH₃)-CH(CH₂CH₃)-CH₂-CH₃

vi) 4-iso-Propylheptane:
Structure:
CH₃-CH₂-CH(CH₂CH(CH₃)₂)-CH₂-CH₂-CH₃

vii) 2,2-Dimethylbutane:
Structure:
(CH₃)₃C-CH₂CH₃

Q5. Write down the names of the following compounds according to the IUPAC system:

i)
Structure: CH₃-CH₂-CH₂-CH₂-CH(CH₃)-CH₃
Name: 3-Methylhexane

ii)
Structure: (CH₃)₃C-CH₂-C(CH₃)₃
Name: 2,2,3,3-Tetramethylbutane

iii)
Structure: CH₃-C(CH₃)-CH(CH₃)-CH₂CH₃
Name: 2,3-Dimethylpentane

iv)
Structure: (CH₃)₃-C-CH₂CH₃
Name: 2,2-Dimethylbutane

v)
Structure: CH₃CH₂C(CH₂CH₃)-CH(CH₂CH₃)-CH₂CH₃
Name: 3-Ethyl-4-propylhexane

vi)
Structure: (C₆H₅)₃CH
Name: Triphenylmethane

vii)
Structure: CH₃-CH(CH₃)-CH₂-CH(CH₃)-CH₂-CH₃
Name: 2,4-Dimethylhexane

viii)
Structure: (CH₃)₂CH-CH₂-CH(CH₃)-CH₃
Name: 3-Methylhexane

Q6. What are the rules for naming alkanes? Explain with suitable examples.

The IUPAC rules for naming alkanes are:

  1. Identify the longest continuous chain of carbon atoms. This chain determines the base name of the alkane (e.g., “pentane” for five carbon atoms).
  • Example: In CH₃-CH₂-CH₂-CH₃, the longest chain has four carbon atoms, so it’s called “butane.”
  1. Number the chain from the end closest to the first substituent (branching group).
  • Example: In 2-Methylbutane, the chain is numbered from the side closest to the methyl group.
  1. Name the substituents (side chains) attached to the main chain, giving them the lowest possible numbers.
  • Example: In CH₃CH₂CH(CH₃)-CH₃, the substituent (CH₃) is on the second carbon, so it is named “2-methylbutane.”
  1. If there are multiple identical substituents, use prefixes like “di-“, “tri-“, etc., and number them.
  • Example: In 2,3-Dimethylbutane, two methyl groups are attached at positions 2 and 3.
  1. Arrange the substituents alphabetically when writing the complete name.
  • Example: In 3-Ethyl-2,3-Dimethylpentane, “ethyl” comes before “methyl” alphabetically.

Q7. (a) Write down the structural formulas for all the isomeric hexanes and name them according to the IUPAC system.

  1. n-Hexane:
    CH₃-(CH₂)₄-CH₃
  2. 2-Methylpentane:
    CH₃-CH(CH₃)-CH₂-CH₂-CH₃
  3. 3-Methylpentane:
    CH₃-CH₂-CH(CH₃)-CH₂-CH₃
  4. 2,2-Dimethylbutane:
    (CH₃)₂C-CH₂-CH₃
  5. 2,3-Dimethylbutane:
    CH₃-CH(CH₃)-CH(CH₃)-CH₃
  6. 3-Ethylpentane:
    CH₃-CH₂-CH(CH₂CH₃)-CH₂-CH₃

(b) The following names are incorrect. Give the correct IUPAC names:

i) 4-Methylpentane
Correct name: 2-Methylpentane

ii) 2-Methyl-3-Ethylbutane
Correct name: 3-Methylpentane

iii) 3,5,5-Trimethylhexane
Correct name: 2,2,4-Trimethylpentane

Q8. (a) Explain why alkenes are less reactive than alkynes? What is the effect of branching on the melting point of alkanes?

  1. Reactivity of Alkenes vs. Alkynes:
  • Alkenes have a double bond (C=C) consisting of one sigma (σ) bond and one pi (π) bond.
  • Alkynes have a triple bond (C≡C), consisting of one sigma (σ) bond and two pi (π) bonds. The presence of two pi bonds in alkynes makes them more reactive than alkenes because pi bonds are weaker and more easily broken in chemical reactions.
  1. Effect of Branching on Melting Point of Alkanes:
  • Branching decreases the surface area available for van der Waals forces, leading to a lower melting point. Linear alkanes can pack more tightly, while branched alkanes are less dense, resulting in a lower melting point.

(b) Three different alkenes yield 2-methylbutane when they are hydrogenated in the presence of a metal catalyst. Give their structures and write equations for the reactions involved.

  1. 2-Methyl-1-butene:
    Structure: CH₃-CH=CH-CH₃
    Equation: CH₃-CH=CH-CH₃ + H₂ → CH₃CH₂CH₂CH₃
  2. 2-Methyl-2-butene:
    Structure: CH₂=C(CH₃)-CH₃
    Equation: CH₂=C(CH₃)-CH₃ + H₂ → CH₃CH₂CH₂CH₃
  3. 3-Methyl-1-butene:
    Structure: CH₂=CH-CH(CH₃)-CH₃
    Equation: CH₂=CH-CH(CH₃)-CH₃ + H₂ → CH₃CH₂CH₂CH₃

Q9. (a) Outline the methods available for the preparation of alkanes.

  1. Hydrogenation of Alkenes and Alkynes:
  • Alkanes can be prepared by the catalytic hydrogenation of alkenes and alkynes using a metal catalyst such as nickel (Ni), palladium (Pd), or platinum (Pt).
  • Example:
    CH2=CH2 + H2 → Ni CH3-CH3
  1. Wurtz Reaction:
  • Alkanes can be formed by the coupling of alkyl halides in the presence of sodium metal in dry ether.
  • Example:
    2CH3Cl + 2Na → CH3-CH3 + 2NaCl
  1. Decarboxylation of Carboxylic Acids (Kolbe’s Electrolysis):
  • Alkanes are produced when sodium salts of carboxylic acids undergo electrolysis.
  • Example:
    CH3COONa + NaOH → CaO, heat CH4 + Na2CO3
  1. Reduction of Alkyl Halides:
  • Alkyl halides can be reduced to alkanes by treatment with zinc and hydrochloric acid.
  • Example:
    CH3Cl + H2 → Zn, HCl CH4 + HCl

(b) How will you bring about the following conversions?

i) Methane to Ethane:

  • Wurtz reaction:
    2CH3Cl + 2Na → dry ether C2H6 + 2NaCl

ii) Ethane to Methane:

  • Cracking of Ethane
    C2H6 → high temp CH4 + C

iii) Acetic Acid to Ethane:

  • Decarboxylation of acetic acid:
    CH3COONa + NaOH → CaO, heat C2H6 + Na2CO3

iv) Methane to Nitromethane:

  • Nitration of Methane (Halogenation):
    CH4 + HNO3 → CH3NO2 + H2O

Q10. (a) What is meant by octane number? Why does a high-octane fuel have a less tendency to knock in an automobile engine?

  1. Octane Number:
  • The octane number of a fuel is a measure of its ability to resist knocking during combustion. It is based on a scale where iso-octane (2,2,4-trimethylpentane) is given a rating of 100 (least knocking), and n-heptane is given a rating of 0 (most knocking). Fuels with higher octane numbers burn more smoothly in engines.
  1. Why High-Octane Fuels Have Less Knocking:
  • High-octane fuels are composed of branched-chain hydrocarbons, which burn more evenly. This reduces the chances of pre-ignition or “knocking” (where the air-fuel mixture combusts prematurely). This smooth combustion is essential for modern engines to operate efficiently without damage.

(b) Explain the free radical mechanism for the reaction of chlorine with methane in the presence of sunlight.

The chlorination of methane proceeds via a free radical chain reaction with the following steps:

  1. Initiation Step:
  • UV light causes the dissociation of chlorine molecules into free radicals.
    Cl2 → UV light 2Cl .
  1. Propagation Step:
  • The chlorine radical abstracts a hydrogen atom from methane, forming hydrogen chloride and a methyl radical.
    CH4 + Cl → CH3^. + HCl ]
  • The methyl radical reacts with another chlorine molecule to form chloromethane and regenerate a chlorine radical.
    CH3. + Cl2 →CH3Cl + Cl.
  1. Termination Step:
  • The chain reaction is terminated when two radicals combine to form a stable molecule.
    Cl + CH3. →CH3Cl ]
    Cl. + Cl. →Cl2 ]

Q11. (a) Write structural formulas for each of the following compounds:

i) Isobutylene (2-methylpropene):
Structure:
CH2=C(CH3)-CH3

ii) 2,3,4,4-Tetramethyl-2-pentene:
Structure:
CH2=C(CH3)-CH(CH3)-CH2-CH3

iii) 2,5-Heptadiene:
Structure:
CH2=CH-CH2-CH2-CH=CH2

iv) 4,5-Dimethyl-2-hexene:
Structure:
CH3-CH=CH-CH(CH3)-CH(CH3)-CH3

v) Vinylacetylene (1-buten-3-yne):
Structure:
CH2=CH-C ≡ CH

vi) 1,3-Pentadiene:
Structure:
CH2=CH-CH=CH-CH3

vii) 1-Butyne:
Structure:
CH≡ CH-CH2-CH3

viii) 3-n-Propyl-1,4-pentadiene:
Structure:
CH2=CH-CH2-CH2-CH=CH-CH2-CH2-CH3

ix) Vinyl bromide:
Structure:
CH2=CH-Br

x) But-1-en-3-yne:
Structure:
CH2=CH-C≡CH

xi) 4-Methyl-2-pentyne:
Structure:
CH3-CH2-C≡CH-CH3

xii) Isopentane (2-methylbutane):
Structure:
CH3-CH(CH3)-CH2-CH3

(b) Name the following compounds by IUPAC system:

i) 3-Methylhex-2-ene:
Structure:
CH3-CH=CH-CH2-CH(CH3)-CH3

ii) 2-Methyl-1-propene:
Structure:
CH2=C(CH3)-CH3

iii) 1-Heptyne:
Structure:
C≡CH-(CH2)4-CH3

Q. 12
(a) Methods for Preparation of Alkenes:

  1. Dehydration of alcohols: Alcohols are heated with a strong acid like sulfuric acid (H₂SO₄) to remove water and form an alkene.
  2. Dehydrohalogenation of alkyl halides: Alkyl halides react with a strong base such as KOH in ethanol, which removes hydrogen and halogen atoms to form an alkene.
  3. Catalytic cracking: Larger hydrocarbons are broken down into smaller alkenes through thermal or catalytic cracking.

To establish that ethylene contains a double bond, use Bromine Water Test: Ethylene decolorizes bromine water, indicating the presence of a double bond.

(b) Structure formulas of alkenes formed by dehydrohalogenation:

  1. 1-Chloropentane:
    CH2=CH-CH2-CH2-CH3 (Pent-1-ene)
  2. 2-Chloro-3-methylbutane:
    CH2=CH-CH3-CH3 (3-methylbut-1-ene)
  3. 1-Chloro-2,2-dimethylpropane: No reaction, as no hydrogen is available on adjacent carbon for elimination.

Q. 13
(a) Preparation of Propene:

  1. From 1-propanol (CH₃–CH₂–CH₂–OH):
  2. From propyne (CH≡C–CH₃):
  3. From isopropyl chloride (CH₃–CHCl–CH₃):

(b) Skeletal formula of all possible alkenes (C₄H₈):

  1. But-1-ene: CH₂=CH-CH₂-CH₃
  2. But-2-ene: CH₃-CH=CH-CH₃
  3. 2-methylprop-1-ene: CH₂=C-CH₃-CH₃

Q. 14
(a) Conversion of ethene to ethyl alcohol:


(Ethene reacts with water in the presence of an acid catalyst to form ethanol.)

(b) Reactions for preparation of:

  1. 1,2-Dibromoethane:
    CH2=CH2 + Br2 —–>CH2Br-CH2Br
    ]
  2. Ethyne:
    CHBr=CHBr —–>HC≡CH + ZnBr2
    ]
  3. Ethane:
    CH2=CH2 + H2 ——>Ni CH3-CH3
    ]
  4. Ethylene glycol:

(c) Conversions:

  1. 1-Butene to 1-Butyne:
  2. 1-Propanol to propene:

Q. 15
Reaction scheme:

(Ethane is first cracked to ethene, which then undergoes further dehydrogenation to form ethyne.)

Q. 16
Products from 1-butene (CH₂=CH–CH₂–CH₃):

  1. H₂ (Hydrogenation):
  2. Br₂ (Bromination):
  3. HBr (Hydrobromination):

Here are the answers to the new set of questions:

Q. 17
Identify each lettered product in the given reactions:

(i) Ethyl alcohol:

  • A: Ethene (CH₂=CH₂)
    (Dehydration of ethyl alcohol using concentrated H₂SO₄)
  • B: 1,2-Dibromoethane (CH₂Br–CH₂Br)
    (Addition of Br₂ in CCl₄ to ethene)
  • C: Ethylene glycol (CH₂OH–CH₂OH)
    (Oxidation with cold dilute KMnO₄)

(ii) Propene:

  • D: 1-Butene (CH₂=CH–CH₂–CH₃)
    (Dehydrohalogenation with alcoholic KOH)
  • E: 2-Butanone (CH₃CO–CH₂–CH₃)
    (Oxidation with alkaline KMnO₄)
  • F: Butanenitrile (CH₃CH₂CH₂CN)
    (Reaction with HCN)

Q. 18
After ozonolysis of the compound, acetaldehyde (CH₃CHO) was obtained, which suggests the structural formula of the compound was ethene (CH₂=CH₂).

Q. 19
(a) Markovnikov’s Rule Alcohol Products:

  • Propene: Propan-2-ol (CH₃CH(OH)CH₃)
  • 1-Butene: Butan-2-ol (CH₃CH₂CH(OH)CH₃)
  • 2-Butene: Butan-2-ol (same as for 1-butene)

(b) The most likely product from the addition of hydrogen iodide (HI) to 2-methyl-2-butene is 2-iodo-2-methylbutane. This follows Markovnikov’s rule, where the iodine attaches to the more substituted carbon.

Q. 20
Hydrocarbons are classified as:

  • Saturated hydrocarbons (alkanes): Contain only single bonds (e.g., ethane, propane).
  • Unsaturated hydrocarbons: Contain double or triple bonds (e.g., ethene, ethyne).

Characteristic reactions:

  • Saturated hydrocarbons: Undergo substitution reactions.
  • Unsaturated hydrocarbons: Undergo addition reactions.

Q. 21
(a) Preparation of Ethyne:

  • Calcium carbide (CaC₂) reacts with water to produce ethyne.

(b) Ethyne Reactions:

  • With Hydrogen: Ethyne (HC≡CH) reacts with hydrogen (H₂) to form ethane (CH₃CH₃).
  • With Halogen acid (e.g., HCl): Produces chloroethene.
  • With alkaline KMnO₄: Ethyne is oxidized to oxalic acid.
  • With 10% H₂SO₄ (in presence of HgSO₄): Forms acetaldehyde (CH₃CHO).
  • With ammoniacal cuprous chloride: Forms copper acetylide (Cu₂C₂).

(c) Importance of Ethene, Ethane, and Ethyne:

  • Ethene: Used in the production of polyethylene (plastics) and ethanol.
  • Ethane: Important fuel and feedstock for ethylene production.
  • Ethyne: Used in welding (as acetylene) and for organic synthesis.

Q. 22
To distinguish ethane, ethene, and ethyne:

  • Ethane: Does not react with bromine water.
  • Ethene: Decolorizes bromine water and reacts with KMnO₄ to give diols.
  • Ethyne: Forms a red precipitate with ammoniacal cuprous chloride and also decolorizes bromine water.

Q. 23
(a) Synthesis of compounds from ethyne:

  1. Ethene: Hydrogenate ethyne using a Lindlar’s catalyst.
  2. Ethanol: React ethene with water in the presence of an acid catalyst.

Q. 24
(a) Comparison of the Reactivity of Ethane, Ethene, and Ethyne:

  • Ethane (C₂H₆): It is a saturated hydrocarbon, so it undergoes substitution reactions (e.g., with halogens) but is less reactive than unsaturated hydrocarbons.
  • Ethene (C₂H₄): Ethene has a double bond, making it more reactive than ethane. It readily undergoes addition reactions, such as halogenation, hydration, and hydrogenation.
  • Ethyne (C₂H₂): Ethyne is even more reactive than ethene due to its triple bond. It can undergo addition reactions similar to ethene but can also react with certain metals (e.g., forming acetylides).

(b) Comparison of Physical Properties of Alkanes, Alkenes, and Alkynes:

  • Alkanes (e.g., ethane): Non-polar, low boiling and melting points, relatively inert, less dense than water, and generally insoluble in water but soluble in organic solvents.
  • Alkenes (e.g., ethene): Also non-polar but with slightly higher boiling points than corresponding alkanes due to the presence of a double bond, which introduces some polarity in the molecule.
  • Alkynes (e.g., ethyne): Non-polar and exhibit higher boiling points than both alkanes and alkenes. Due to the linear structure of the triple bond, alkynes have stronger intermolecular forces than alkanes and alkenes, leading to slightly higher melting and boiling points.

Q. 25
Reactions of Propyne with the following reagents:

(a) AgNO₃/NH₄OH (Tollens’ reagent):
Propyne (HC≡C–CH₃) reacts with Tollens’ reagent to form a white precipitate of silver acetylide (AgC≡C–CH₃). This test is specific for terminal alkynes due to the acidic hydrogen attached to the terminal carbon.

(b) Cu₂Cl₂/NH₄OH:
Propyne reacts with ammoniacal cuprous chloride to form a red precipitate of copper acetylide (CuC≡C–CH₃). This is also a test for terminal alkynes.

(c) H₂O/H₂SO₄/HgSO₄ (Hydration):
Propyne undergoes hydration in the presence of H₂SO₄ and HgSO₄ to form a ketone, specifically propanone (acetone), via the addition of water across the triple bond.

Q. 26
A compound with the molecular formula C₄H₆ suggests an alkyne. When treated with hydrogen and a Ni catalyst, a new compound C₄H₁₀ is formed, indicating the complete hydrogenation of an alkyne to an alkane.

The reaction of C₄H₆ with ammoniacal silver nitrate forms a white precipitate, indicating a terminal alkyne. This points to but-1-yne (CH≡C–CH₂–CH₃) as the structure of the original compound. Upon hydrogenation, butane (C₄H₁₀) is formed.

Q. 21
(a) Identification of A and B:

  • Starting compound: 1-Propanol (CH₃CH₂CH₂OH)
  • Reaction with PCl₅ forms A: 1-chloropropane (CH₃CH₂CH₂Cl).
  • Reaction with sodium in ether leads to the formation of B: Hexane (CH₃(CH₂)₄CH₃) (via the Wurtz reaction).

(b) General Mechanism of Electrophilic Addition Reactions of Alkenes:

  1. Formation of carbocation (Electrophilic attack): The double bond in an alkene is electron-rich and attacks an electrophile (e.g., H⁺ from HCl), leading to the formation of a carbocation intermediate.
  2. Nucleophilic attack: The nucleophile (e.g., Cl⁻) then attacks the positively charged carbon atom (carbocation) to complete the addition, forming a saturated product.

Fundamental principles of Organic chemistry solved exercise

by

Master the core concepts of organic chemistry with this detailed guide to solved exercises from the ‘Fundamental Principles of Organic Chemistry’ chapter. This resource covers key topics such as bonding, hybridization, isomerism, functional groups, and reaction mechanisms. Aligned with the latest syllabus for Lahore Board, Federal Board, and other academic boards, it includes step-by-step solutions, solved MCQs, short questions, and conceptual problems to reinforce learning. Ideal for students aiming to excel in organic chemistry, this guide simplifies complex principles and enhances exam preparation.

Q4. How organic compounds are classified? Give a suitable example of each type.

Organic compounds are classified based on their structure, functional groups, and bonding into the following major categories:

  1. Acyclic or Open-Chain Compounds: These are compounds with straight or branched chains.
  • Example: Butane (C₄H₁₀)
  1. Cyclic Compounds: These compounds have atoms arranged in a ring structure.
  • Example: Cyclohexane (C₆H₁₂)
  1. Aromatic Compounds: Compounds containing one or more benzene rings (arenes).
  • Example: Benzene (C₆H₆)
  1. Heterocyclic Compounds: Cyclic compounds where one or more atoms in the ring are not carbon.
  • Example: Pyridine (C₅H₅N)

Q5. What are homocyclic and heterocyclic compounds? Give one example of each.

  • Homocyclic Compounds: Compounds whose rings are made up entirely of carbon atoms.
  • Example: Benzene (C₆H₆)
  • Heterocyclic Compounds: Compounds that contain at least one atom other than carbon in the ring structure.
  • Example: Pyridine (C₅H₅N) (contains nitrogen in the ring)

Q6. Write the structural formulas of the two possible isomers of C₄H₁₀.

The two isomers of C₄H₁₀ are:

  1. n-Butane (Straight-chain isomer):
    Structure: CH₃-CH₂-CH₂-CH₃
  2. Iso-Butane (Branched-chain isomer):
    Structure: (CH₃)₃CH

Q7. Why is ethene an important industrial chemical?

Ethene (ethylene) is crucial in the chemical industry because:

  1. It is used as a raw material for producing polymers such as polyethylene, the most widely used plastic.
  2. It is involved in the production of other chemicals such as ethanol, ethylene oxide, and ethylene glycol, which are used in manufacturing antifreeze, detergents, and solvents.
  3. Ethene is also used as a plant hormone to stimulate fruit ripening.

Q8. What is meant by a functional group? Name typical functional groups containing oxygen.

A functional group is a specific group of atoms within a molecule responsible for the characteristic chemical reactions of that molecule. Typical oxygen-containing functional groups include:

  1. Hydroxyl group (-OH): Found in alcohols (e.g., ethanol)
  2. Carbonyl group (C=O): Found in aldehydes and ketones (e.g., formaldehyde)
  3. Carboxyl group (-COOH): Found in carboxylic acids (e.g., acetic acid)
  4. Ether group (R-O-R’): Found in ethers (e.g., diethyl ether)

Q9. What is an organic compound? Explain the importance of Wöhler’s work in the development of organic chemistry.

An organic compound is a chemical compound containing carbon atoms, usually bonded to hydrogen, oxygen, and/or other elements. Organic compounds are the basis of life and include molecules such as carbohydrates, proteins, and fats.

Wöhler’s work was groundbreaking because he synthesized urea (an organic compound) from ammonium cyanate (an inorganic compound) in 1828. This demonstrated for the first time that organic compounds could be synthesized from inorganic substances, disproving the belief that organic compounds could only be produced by living organisms, leading to the rise of modern organic chemistry.

Q10. Write a short note on cracking of hydrocarbons.

Cracking is a process in which large hydrocarbon molecules (usually alkanes) are broken down into smaller, more useful molecules, often by applying heat and pressure. This process is crucial in the petroleum industry to convert long-chain hydrocarbons into gasoline, diesel, and other products. There are two main types of cracking:

  1. Thermal Cracking: High temperature and pressure are used to break the bonds.
  2. Catalytic Cracking: A catalyst is used to lower the temperature and pressure needed for the process.

Q11. Explain reforming of petroleum with the help of a suitable example.

Reforming is a chemical process used to convert low-octane hydrocarbons into high-octane gasoline components. This process improves the quality of gasoline by rearranging the molecular structure of hydrocarbons.

  • Example: In naphtha reforming, straight-chain alkanes are converted into branched-chain alkanes, cycloalkanes, and aromatic hydrocarbons. For instance, heptane (C₇H₁₆) can be converted into methylcyclohexane or toluene, which have higher octane ratings, improving fuel efficiency.

Q12. Describe important sources of organic compounds.

Important sources of organic compounds include:

  1. Petroleum: The largest source, used for producing fuels, plastics, and chemicals.
  2. Natural Gas: Contains methane and is used as a source for organic synthesis.
  3. Coal: A source of hydrocarbons, aromatic compounds, and various other organics.
  4. Plants and Animals: Provide carbohydrates, proteins, fats, and other biochemicals used in medicine, food, and textiles.

Q13. What is orbital hybridization? Explain sp³, sp², and sp modes of hybridization of carbon.

Orbital hybridization is the mixing of atomic orbitals in an atom to form new hybrid orbitals that influence molecular geometry and bonding properties.

  1. sp³ Hybridization: Involves the mixing of one s and three p orbitals. The geometry is tetrahedral with bond angles of 109.5°.
  • Example: Methane (CH₄)
  1. sp² Hybridization: Involves the mixing of one s and two p orbitals. The geometry is trigonal planar with bond angles of 120°.
  • Example: Ethene (C₂H₄)
  1. sp Hybridization: Involves the mixing of one s and one p orbital. The geometry is linear with bond angles of 180°.
  • Example: Ethyne (C₂H₂)

Q14. Explain the type of bonds and shapes of the following molecules using hybridization approach.

  • CH₃-CH₂-CH₂-CH₃ (Butane):
  • Hybridization: sp³ for each carbon atom
  • Shape: Tetrahedral around each carbon
  • CH=CH₂ (Ethene):
  • Hybridization: sp² for each carbon
  • Shape: Trigonal planar
  • CHCl (Chloromethane):
  • Hybridization: sp³ for the carbon
  • Shape: Tetrahedral around the carbon
  • HCHO (Formaldehyde):
  • Hybridization: sp² for carbon
  • Shape: Trigonal planar

Q15. Why is there no free rotation around a double bond and free rotation around a single bond? Discuss cis-trans isomerism.

In a double bond, one of the bonds is a pi bond (π) that restricts rotation because breaking this bond requires a significant amount of energy. This is unlike a single bond, which is a sigma bond (σ) that allows free rotation because of the symmetric overlap of orbitals along the bond axis.

Cis-trans isomerism occurs due to the restricted rotation around double bonds, resulting in different spatial arrangements of groups attached to the carbon atoms involved in the double bond. In cis-isomers, similar groups are on the same side of the double bond, while in trans-isomers, they are on opposite sides.

©2024. Everexams.com. All Rights Reserved.