Federal Board syllabus

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 ]

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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.