Class 11 Chemistry – Unit 16: Hydrocarbons (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Unit 16 – Hydrocarbons strictly according to the Federal Board (FBISE) syllabus. The unit focuses on structure, classification, properties, and reactions of hydrocarbons.
Major topics include alkanes, alkenes, alkynes, aromatic hydrocarbons, isomerism, physical and chemical properties, combustion, substitution and addition reactions, and industrial applications. Important reactions, mechanisms, and exam-focused points are highlighted for effective learning.
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16.1 Hydrocarbons: Study Notes
Hydrocarbons are organic compounds containing only hydrogen and carbon atoms. They are the fundamental building blocks of organic chemistry.
Classification of Hydrocarbons
- Aliphatic Hydrocarbons: From the Greek "fatty." These include open-chain (acyclic) or closed-chain (cyclic) compounds that behave like open chains.
- Aromatic Hydrocarbons: From the Greek "fragrant." These compounds contain at least one benzene ring or have properties resembling benzene.
1. Open Chain (Acyclic) Hydrocarbons
These have carbon skeletons forming straight or branched chains with terminal carbon atoms.
A. Saturated Hydrocarbons (Alkanes/Paraffins)
- Contain only single covalent bonds between carbon atoms.
- Carbon atoms are bonded to four other atoms; no more atoms can be added.
- Hybridization: Each carbon is $sp^3$ hybridized.
- Examples: Ethane ($C_2H_6$), Butane, 2-methylpropane.
B. Unsaturated Hydrocarbons
Contain at least one multiple bond (double or triple).
- Alkenes (Olefins): Contain at least one carbon-carbon double bond ($C=C$). Carbon atoms in the double bond are $sp^2$ hybridized.
Examples: Ethene ($H_2C=CH_2$), 4-methylpent-2-ene. - Alkynes (Acetylenes): Contain at least one carbon-carbon triple bond ($C \equiv C$). Carbon atoms in the triple bond are $sp$ hybridized.
Examples: Ethyne ($CH \equiv CH$), 3-methylpent-1-yne.
2. Closed Chain (Cyclic) Hydrocarbons
Carbon atoms form a ring; there are no terminal carbon atoms.
A. Alicyclic Hydrocarbons
- Rings of carbon atoms that show properties similar to aliphatic (open-chain) hydrocarbons.
- Examples: Cyclobutane, Cyclopentene, Cyclohexane.
B. Aromatic Hydrocarbons
- Contain at least one benzene ring (a hexagonal ring with alternating single and double bonds).
- Hybridization: Each carbon in the benzene ring is $sp^2$ hybridized.
- Examples: Benzene, Ethylbenzene, 1-methylethylbenzene.
Concept Assessment Exercise 16.1: Questions & Answers
i. Differentiate between saturated and unsaturated compounds. Give examples.
Answer:
| Feature | Saturated Compounds | Unsaturated Compounds |
|---|---|---|
| Bond Type | Only single covalent bonds. | At least one double or triple bond. |
| Reactivity | Less reactive (no more atoms can be added). | More reactive (atoms can be added across multiple bonds). |
| Hybridization | $sp^3$ hybridized carbons. | $sp^2$ or $sp$ hybridized carbons. |
| Examples | Methane ($CH_4$), Ethane ($C_2H_6$). | Ethene ($C_2H_4$), Ethyne ($C_2H_2$). |
ii. Define aliphatic and aromatic compounds. Give two examples of each.
Answer:
- Aliphatic Compounds: Organic compounds with carbon atoms joined in straight chains, branched chains, or non-aromatic rings.
Examples: Propane, Cyclohexane. - Aromatic Compounds: Cyclic organic compounds that contain at least one benzene ring ($C_6H_6$) or exhibit similar chemical stability.
Examples: Benzene, Naphthalene (or Ethylbenzene).
iii. What type of hybridization is shown by each carbon of hexa-1,2-diene?
Answer:
The structure of hexa-1,2-diene is: $H_2C=C=CH-CH_2-CH_3$ (Correction based on the image "5-methylhexa-1,2-diene" example, but for a standard hexa-1,2-diene):
- C1 ($H_2C=$): $sp^2$ hybridized (double bonded to one C).
- C2 ($=C=$): $sp$ hybridized (double bonded to two different carbons).
- C3 ($=CH-$): $sp^2$ hybridized (double bonded to one C).
- C4, C5, C6 ($-CH_2-$ and $-CH_3$): $sp^3$ hybridized (only single bonds).
16.2 Reactivity of Alkanes
Alkanes are known as paraffins (Latin: parum = little, affinis = reactivity). They are generally inert under normal conditions due to two primary factors:
- Non-polar Bonds: The electronegativity of Carbon (2.5) and Hydrogen (2.1) is nearly equal. This results in a symmetrical distribution of electrons, making the molecules non-reactive toward acids, bases, or oxidizing/reducing agents.
- Strength of Sigma ($\sigma$) Bonds: Alkanes only contain $\sigma$ bonds. These are very strong because the bonding electrons are held tightly between the nuclei, making them difficult to break for chemical reactions.
16.2.1 Free Radical Substitution of Ethane
Alkanes undergo substitution reactions where an atom (usually hydrogen) is replaced by another atom or group. The halogenation of ethane occurs in three distinct stages:
1. Initiation
Ultraviolet (UV) light provides the energy to break the chlorine-chlorine bond via homolytic fission, creating two highly reactive chlorine free radicals.
$$Cl-Cl \xrightarrow{h\nu} Cl^\bullet + Cl^\bullet$$
2. Propagation
This is a chain reaction consisting of two main steps:
- The chlorine radical attacks ethane, removing a hydrogen atom to form $HCl$ and an ethyl free radical ($CH_3CH_2^\bullet$).
- The ethyl radical then attacks a new chlorine molecule to form chloroethane ($CH_3CH_2Cl$) and generates a new chlorine radical to continue the cycle.
Note: If chlorine is in excess, substitution continues until hexachloroethane ($C_2Cl_6$) is formed.
3. Termination
The reaction ends when any two free radicals collide and combine to form a stable molecule. Examples include:
- $Cl^\bullet + Cl^\bullet \rightarrow Cl_2$
- $CH_3CH_2^\bullet + Cl^\bullet \rightarrow CH_3CH_2Cl$ (Chloroethane)
- $CH_3CH_2^\bullet + CH_3CH_2^\bullet \rightarrow CH_3CH_2CH_2CH_3$ (Butane)
Self-Assessment: Questions & Answers
| Question | Answer |
|---|---|
| Why are alkanes called "paraffins"? | The name comes from Latin words meaning "little affinity" or "little reactivity," describing their chemically inert nature. |
| What is the electronegativity difference in a C-H bond? | The difference is $2.5 - 2.1 = 0.4$, which is considered negligible, making the bond non-polar. |
| What type of fission occurs during initiation? | Homolytic fission occurs, where the bond breaks evenly and each atom retains one electron to become a radical. |
| Why is UV light necessary for this reaction? | UV light provides the specific energy required to break the covalent bond in the chlorine molecule ($Cl_2$) to start the reaction. |
| How can butane be formed during the chlorination of ethane? | During the termination step, two ethyl radicals ($C_2H_5^\bullet$) can collide and bond together to form a four-carbon chain ($C_4H_{10}$). |
16.3 Shapes of Ethane and Cyclopropane
1. Shape of Ethane Molecules ($C_2H_6$)
- Ethane consists of two carbon atoms bonded by a single covalent bond, with three hydrogen atoms bonded to each carbon.
- All single covalent bonds in ethane are sigma ($\sigma$) bonds, formed by the end-on overlap of atomic orbitals.
- Each carbon atom undergoes $sp^3$ hybridization.
- The molecule is non-planar because the bond angles are $109.5^\circ$, resulting in a tetrahedral shape around each carbon atom.
2. Shape of Cyclopropane Molecules ($C_3H_6$)
- Cyclopropane features three carbon atoms bonded in a ring, with two hydrogens attached to each carbon.
- Like ethane, each carbon atom shows $sp^3$ hybridization with tetrahedral orbital arrangement.
- The molecule is non-planar as the bonds are directed in three dimensions.
- The internal C-C-C bond angle is $60^\circ$, while H-C-H and C-C-H angles remain $109.5^\circ$.
16.4 Preparation of Alkenes
Alkenes can be synthesized through elimination reactions, where atoms are removed from adjacent carbons to form a double bond.
1. Dehydration of Alcohols
- This involves the removal of a water molecule ($H_2O$) from an alcohol.
- Catalysts: Passing ethanol vapors over hot aluminium oxide ($Al_2O_3$) powder, or using concentrated sulphuric acid ($H_2SO_4$) or phosphoric acid ($H_3PO_4$).
- Reaction: $CH_3-CH_2-OH \xrightarrow{Al_2O_3, \text{Heat}} H_2C=CH_2 + H_2O$
2. Dehydrohalogenation of Halogenoalkanes
- This involves the removal of a hydrogen halide molecule (like $HBr$ or $HCl$).
- Conditions: Heating a halogenoalkane with ethanolic sodium hydroxide ($NaOH$).
- Reaction: $CH_3-CH_2-Br + NaOH \text{ (ethanolic)} \xrightarrow{\text{Heat}} H_2C=CH_2 + H_2O + NaBr$
Self-Assessment: Questions & Answers
| Question | Answer |
|---|---|
| What is the hybridization and bond angle in ethane? | The carbon atoms are $sp^3$ hybridized with bond angles of $109.5^\circ$. |
| Why is the C-C-C angle in cyclopropane $60^\circ$ instead of $109.5^\circ$? | Because the three carbon atoms are forced into a triangular ring structure. |
| Define a sigma ($\sigma$) bond as per the text. | A covalent bond formed by the end-on overlap of atomic orbitals. |
| What catalyst is used for the dehydration of ethanol? | Hot aluminium oxide ($Al_2O_3$) powder or concentrated acids like $H_2SO_4$. |
| How do you prepare but-1-ene from butan-1-ol? | By dehydration: Pass butan-1-ol vapors over hot $Al_2O_3$ to remove $H_2O$ and form the double bond. |
Notes: Structure and Reactivity of Alkenes
1. Structure of Ethene ($C_2H_4$)
- Bonding: Each carbon atom in ethene is $sp^2$ hybridized, resulting in a trigonal planar geometry with bond angles of approximately $120^\circ$.
- Sigma ($\sigma$) Bond: Formed by the head-on overlap of $sp^2$ hybridized orbitals. It is a strong bond.
- Pi ($\pi$) Bond: Formed by the parallel overlap of unhybridized $p$ orbitals. It is weaker than the sigma bond because the electron density is more exposed and less tightly held.
- Reactivity: Alkenes are more reactive than alkanes because the $\pi$ bond is easily broken, allowing for addition reactions.
2. Chemical Reactions of Alkenes
A. Hydrogenation
- The addition of $H_2$ to an alkene to form an alkane.
- Conditions: Requires a catalyst like nickel ($Ni$) or platinum ($Pt$) and heat.
- Application: Used industrially to prepare margarine from vegetable oils.
B. Hydrohalogenation (Addition of HX)
- Hydrogen halides (like $HBr$ or $HCl$) add across the double bond to form haloalkanes.
- Markovnikov’s Rule: In the addition of an unsymmetrical reagent to an unsymmetrical alkene, the hydrogen atom attaches to the carbon that already has the greater number of hydrogen atoms.
3. Carbocation Stability
The reaction mechanism involves the formation of a carbocation intermediate. The stability of these intermediates determines the major product:
- Stability Order: $Tertiary (3^\circ) > Secondary (2^\circ) > Primary (1^\circ)$
- Inductive Effect: Alkyl groups ($R$) are electron-donating. They push electron density toward the positive carbon, dispersing the charge and stabilizing the ion.
Concept Assessment Exercise 16.3: Questions & Answers
Question 1
What are the products of hydrochlorination of 2-methylbut-2-ene?
Answer:
- Reactant: $CH_3-C(CH_3)=CH-CH_3$
- Mechanism: According to Markovnikov's rule, the $H^+$ adds to the carbon with more hydrogens ($C3$), and the $Cl^-$ adds to the more substituted carbon ($C2$).
- Intermediate: A stable tertiary carbocation is formed at $C2$.
- Major Product: 2-chloro-2-methylbutane
Question 2
What are the products of hydrochlorination of 3,3-dimethylbut-1-ene?
Answer:
- Reactant: $CH_2=CH-C(CH_3)_3$
- Mechanism: The $H^+$ adds to $C1$ (the carbon with 2 hydrogens). This initially forms a secondary carbocation at $C2$.
- Rearrangement: Note that a 1,2-methyl shift may occur to form a more stable tertiary carbocation, but following standard Markovnikov addition:
- Major Product: 2-chloro-3,3-dimethylbutane (The $Cl$ attaches to $C2$).
1. Hydration of Alkenes
Hydration is the addition of water ($H_2O$) across the carbon-carbon double bond to produce an alcohol. This is a primary industrial method for preparing ethanol.
- Step 1: Ethene reacts with cold concentrated sulfuric acid ($H_2SO_4$) to form ethyl hydrogen sulphate.Reaction: $CH_2=CH_2 + H_2SO_4 \rightarrow H_3C-CH_2-O-SO_3H$
- Step 2: The mixture is heated with water, causing the ethyl hydrogen sulphate to decompose into ethanol.Reaction: $H_3C-CH_2-O-SO_3H + H_2O \xrightarrow{\text{Boil}} CH_3CH_2OH + H_2SO_4$
2. Halogenation
The addition of halogens (like $Br_2$ or $Cl_2$) to alkenes, typically carried out in an inert solvent like tetrachloromethane ($CCl_4$).
- Test for Unsaturation: Bromine water is reddish-brown. When added to an alkene, it decolors rapidly, serving as a test for the presence of a double bond.
- Mechanism:
- The $\pi$ electron cloud of ethene induces a dipole in the non-polar bromine molecule.
- The electrophilic bromine ($Br^{\delta+}$) attacks the alkene to form a cyclic bromonium ion.
- The nucleophilic bromide ion ($Br^-$) attacks from the opposite side, resulting in trans-1,2-dibromoethane.
3. Halohydration and Epoxidation
- Halohydration: Addition of hypohalous acid ($HOX$). When ethene reacts with $Br_2$ in water, it forms bromohydrin ($Br-CH_2-CH_2-OH$).
- Epoxidation: Alkenes react with peracids (e.g., peroxyacetic acid) to form epoxides (cyclic ethers). Ethene reacts to form epoxyethane.
Self-Assessment Questions
Q1: Why does bromine water decolorize when added to ethene?
A: The reddish-brown bromine molecule is consumed as it adds across the double bond of ethene to form colorless 1,2-dibromoethane.
Q2: What is the role of cold concentrated sulfuric acid in the industrial preparation of ethanol?
A: It acts as a reactant in the first step to create an intermediate (ethyl hydrogen sulphate), which can then be hydrolyzed by boiling with water to produce ethanol.
Q3: Define the intermediate formed during the halogenation mechanism.
A: The intermediate is a cyclic bromonium ion, a three-membered ring containing two carbon atoms and a positively charged bromine atom.
Q4: What are the products when ethene reacts with peroxyacetic acid?
A: The products are epoxyethane (an epoxide) and acetic acid ($CH_3COOH$).
1. Ozonolysis of Alkenes
Ozonolysis is a reaction where ozone ($O_3$) is bubbled into a cold solution ($-78^{\circ}\text{C}$) of an alkene in dichloromethane ($CH_2Cl_2$). It cleaves the carbon-carbon double bond to form aldehydes and/or ketones.
- Process: Alkene + $O_3$ $\rightarrow$ Molozonide (unstable) $\rightarrow$ Ozonide (stable intermediate).
- Reduction: When the ozonide is treated with $Zn$ and $H_2O$, it breaks down into carbonyl compounds.
2. Polymerization
The process where small molecules (monomers) react together to form large molecules (polymers). The word comes from Greek 'polu' (many) and 'meros' (parts).
- High-Pressure Method: Ethene polymerizes at $400^{\circ}\text{C}$ and $1000\text{ atm}$ with traces of $O_2$ to form polyethene.
- Ziegler-Natta Catalyst: A mixture of $Al(C_2H_5)_3$ and $TiCl_4$ produces high-quality, high-melting polyethene without branching.
$nCH_2=CH_2 \xrightarrow{400^{\circ}\text{C}, 1000\text{ atm}} \text{--}[CH_2-CH_2]\text{--}_n$
3. Conjugation in Alkenes
Conjugation occurs in alkenes with alternate double and single bonds. Unhybridized "$p$" orbitals on adjacent carbon atoms overlap to form a continuous bridge, allowing $\pi$ (pi) electrons to delocalize across the chain.
Example: Hexa-1,3,5-triene ($H_2C=CH-CH=CH-CH=CH_2$).
4. Redox Reactions of Organic Compounds
A. Oxidation Reaction
Defined as the addition of oxygen, loss of hydrogen, or increasing bonds to oxygen/electronegative elements.
- $CH_4 + \frac{1}{2}O_2 \rightarrow CH_3OH$ (Methane to Methanol)
- $CH_3OH + [O] \xrightarrow{K_2Cr_2O_7/H^+} HCHO + H_2O$ (Methanol to Methanal)
B. Reduction Reaction
Defined as the addition of hydrogen, loss of oxygen, or decreasing bonds to oxygen.
- $HCHO + 2[H] \xrightarrow{LiAlH_4} CH_3OH$ (Methanal to Methanol)
- Ethanamide $\xrightarrow{LiAlH_4}$ Ethanamine
Concept Assessment Exercises (Solved)
Exercise 16.4
Q: What are the products of ozonolysis of (i) propene and (ii) but-2-ene?
A: (i) Propene ($CH_3-CH=CH_2$) gives ethanal ($CH_3CHO$) and methanal ($HCHO$).
(ii) But-2-ene ($CH_3-CH=CH-CH_3$) gives two molecules of ethanal ($CH_3CHO$).
Q: Name the alkenes for the following products:
(i) Ethanal only: But-2-ene
(ii) Mixture of methanal and propanone: 2-methylprop-1-ene (Isobutylene)
Exercise 16.5
Q: Which of the following shows conjugation?
(i) ethene (No)
(ii) but-1-ene (No)
(iii) penta-1,2-diene (No - cumulative)
(iv) buta-1,2-diene (No)
(v) penta-1,3-diene (Yes - alternate double/single bonds)