Class 11 Chemistry – Unit 15: Organic Chemistry (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Unit 15 – Organic Chemistry strictly according to the Federal Board (FBISE) syllabus. The unit introduces students to fundamental concepts of organic compounds, nomenclature, and functional groups.
Major topics include introduction to organic chemistry, classification of organic compounds, hybridization, functional groups, isomerism, homologous series, and basic reactions of alkanes, alkenes, and alkynes. Reaction mechanisms, tables, and important examples are included for clarity and exam focus.
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1. Catenation
Definition: The ability of carbon atoms to form long chains and rings. The word comes from the Latin "catena," meaning chain.
Importance in Organic Chemistry: Catenation is the primary reason for the vast number of organic compounds. Factors include:
- Carbon's ability to form single, double, and triple bonds.
- Kinetic stability of organic compounds.
- The strength of carbon-carbon bonds due to carbon's small atomic size.
2. Representation of Organic Molecules
Organic molecules are represented in several ways depending on the level of detail required:
A. Molecular Formula
Shows the actual whole number ratio and type of atoms of different elements in a molecule.
- Ethanol: $C_{2}H_{6}O$
- Glucose: $C_{6}H_{12}O_{6}$
B. Empirical Formula
The simplest whole-number ratio of atoms of each element in a compound.
- Benzene: Molecular Formula = $C_{6}H_{6} \rightarrow$ Empirical Formula = $CH$
- Ethanoic Acid: Molecular Formula = $C_{2}H_{4}O_{2} \rightarrow$ Empirical Formula = $CH_{2}O$
Note: Some compounds like sucrose ($C_{12}H_{22}O_{11}$) have the same molecular and empirical formulas.
C. Structural Formula
Shows the arrangement of atoms and the types of functional groups present. There are three main types, including:
- Condensed Structural Formula: Shows the relative positions of atoms without showing single covalent bonds. Carbon atoms are shown individually with attached hydrogens. Branches and methylene groups are shown in brackets.Example: Acetic Acid is $CH_{3}COOH$.
3. Key Calculations
The relationship between molecular and empirical formulas is defined by the factor $n$:
$$n = \frac{\text{molecular mass}}{\text{empirical formula mass}}$$
$$\text{Molecular Formula} = n \times (\text{Empirical Formula})$$
Review Questions & Answers
Q1: Define Catenation and why it is significant for carbon.
A: Catenation is the linking of atoms of the same element into chains or rings. It is significant for carbon because carbon's small size allows it to form very strong and stable $C-C$ bonds, enabling the existence of millions of organic compounds.
Q2: Calculate the molecular formula for a compound with an empirical formula of $CH$ and a molecular mass of $78\text{ amu}$.
A:1. Empirical formula mass of $CH = 12 + 1 = 13\text{ amu}$.2. Find $n$: $n = 78 / 13 = 6$.3. Molecular formula: $6 \times (CH) = C_{6}H_{6}$.
Q3: Why are structural formulas often preferred over molecular formulas in organic chemistry?
A: Molecular formulas do not show the arrangement of atoms or functional groups. Different compounds (isomers) can have the same molecular formula but different properties; structural formulas allow us to distinguish between them by showing the bonding patterns.
Q4: What is the first step in determining an empirical formula from percentage composition?
A: The first step is to calculate the number of moles of each element by dividing the percentage by mass of each element by its atomic mass ($n = \text{mass} / \text{molar mass}$).
These notes summarize the representation of organic molecules, homologous series, and functional groups based on the provided material.
1. Types of Chemical Formulae
- Full Structural (2D Displayed) Formula: Shows all atoms and all bonds in a molecule. It represents the molecule in two dimensions ($x$ and $y$ planes).
- Skeletal Formula: The most simplified representation. Carbon and hydrogen atoms are not explicitly shown. Vertices and ends of lines represent carbon atoms, and hydrogen atoms are assumed to be present to satisfy carbon's valency of 4.
- Stereochemical (3D) Formula: Shows the arrangement of atoms in 3D space ($x, y, z$ planes).
- Straight line: Bond in the plane of the paper.
- Wedge (solid): Bond coming out of the plane toward the viewer.
- Dashed line/Hatched wedge: Bond going into the plane away from the viewer.
2. Functional Groups
A functional group is an atom or group of atoms attached to carbon chains/rings that determines the characteristic chemical properties of an organic compound.
- Reactive Part: The functional group (e.g., $-OH$ for alcohols, $-COOH$ for carboxylic acids).
- Unreactive Part: The hydrocarbon network (carbon chain).
3. Homologous Series
A series of organic compounds with the same functional group but different lengths of carbon chains. Members of a series share several features:
- They share the same functional group.
- Each member differs from the next by a methylene group ($-CH_2-$).
- They follow a General Formula (e.g., Alkanes: $C_nH_{2n+2}$, Alkenes: $C_nH_{2n}$, Alcohols: $C_nH_{2n+1}OH$).
- They have similar chemical properties.
- They show a gradual change in physical properties (like boiling point) as the carbon chain length increases.
- They have similar methods of preparation.
Concept Review: Questions and Answers
Q1: Define a functional group and provide two examples.
A: A functional group is an atom or group of atoms that determines the chemical properties of a molecule. Examples include the hydroxyl group ($-OH$) found in alcohols and the carboxyl group ($-COOH$) found in carboxylic acids.
Q2: What are the key differences between a displayed formula and a skeletal formula?
A: A displayed formula shows every atom and every bond explicitly. In contrast, a skeletal formula uses lines where vertices represent carbon atoms; carbon and hydrogen labels are omitted (unless the hydrogen is part of a functional group like $-OH$).
Q3: How do members of a homologous series differ from one another structurally?
A: Successive members of a homologous series differ by exactly one methylene group, which is written as $-CH_2-$.
Q4: In stereochemical formulae, what does a solid wedge represent?
A: A solid wedge represents a chemical bond that is projecting out of the plane of the paper toward the viewer.
Q5: What is the general formula for a mono-functional alcohol?
A: The general formula is $C_nH_{2n+1}OH$.
1. Hydrocarbons Overview
Hydrocarbons are organic compounds composed entirely of carbon and hydrogen. Carbon atoms form the backbone (skeleton) of the molecule, while hydrogen atoms act as the "skin."
- Alkanes: The simplest organic compounds. They lack functional groups and are relatively unreactive.
- Reactivity: Alkanes are unreactive because C-C and C-H bonds are very strong and non-polar, making them resistant to attack by polar reagents like acids or bases.
2. Saturated vs. Unsaturated Hydrocarbons
| Feature | Saturated Hydrocarbons | Unsaturated Hydrocarbons |
|---|---|---|
| Bonding | Only single covalent bonds. | Contain one or more double or triple bonds. |
| Examples | Alkanes (e.g., ethane, butane), Cycloalkanes. | Alkenes (double bonds), Alkynes (triple bonds). |
| Capacity | Carbon has reached maximum combining capacity (4 atoms). | More atoms can be added to the carbon atoms. |
| Reactions | Substitution (typically). | Addition reactions across multiple bonds. |
3. Alkyl Groups ($R-$)
An alkyl group is formed by removing one hydrogen atom from an alkane. They are named by replacing the suffix -ane with -yl.
- Methane $\rightarrow$ Methyl ($-CH_{3}$)
- Ethane $\rightarrow$ Ethyl ($-CH_{2}CH_{3}$)
- Propane can form n-propyl (terminal H removed) or isopropyl (middle H removed).
Concept Assessment Questions & Answers
Q1: Provide the general formulae and functional groups for the following homologous series:
- Alcohols: $R-OH$; General Formula: $C_{n}H_{2n+1}OH$
- Aldehydes: $R-CHO$; Functional group contains a carbonyl bonded to at least one H.
- Ketones: $R-CO-R'$; Carbonyl group bonded to two carbons.
- Carboxylic Acids: $R-COOH$; Carboxyl group.
- Ethers: $R-O-R'$; Oxygen atom connected to two alkyl groups.
- Esters: $R-COO-R'$; Formed from an acid and an alcohol.
- Amines: $R-NH_{2}$; Contains nitrogen.
- Nitriles: $R-C \equiv N$; Cyano group.
Q2: Why are alkanes considered the least reactive organic compounds?
A: They are unreactive due to two main reasons:
- The $C-C$ and $C-H$ bonds are very strong (high bond energy, e.g., $C-C$ is $346\text{ kJ/mol}$), making them hard to break.
- The bonds are non-polar, so they do not attract polar reagents like acids, bases, or oxidizing agents.
Q3: Distinguish between "n-propyl" and "isopropyl" groups.
A: Both are derived from propane ($C_{3}H_{8}$). The n-propyl group is formed by removing a hydrogen from a terminal (end) carbon, resulting in a straight chain: $-CH_{2}CH_{2}CH_{3}$. The isopropyl group is formed by removing a hydrogen from the central carbon: $-CH(CH_{3})_{2}$.
Q4: Why do unsaturated hydrocarbons undergo addition reactions?
A: Because they contain double or triple bonds, the carbon atoms are not "full." Each double-bonded carbon is only attached to three atoms, leaving the "ability" to bond with one more atom by breaking the pi-bond. This allows new molecules to be added directly across the multiple bonds.
Organic Chemistry: Nomenclature and Structures
Concept Assessment Exercise 15.3: Alkanes
Topic: Systematic naming of branched-chain alkanes and cycloalkanes.
- i. $C(CH_3)_4$: 2,2-dimethylpropane
- ii. $CH_3CH_2CH(C_2H_5)CH_3$: 3-methylpentane
- iii. $(CH_3)_3CH$: 2-methylpropane
- iv. $CH_3(CH_2)_4C(CH_3)_3$: 2,2-dimethylheptane
- v. $(CH_3)_3C(CH_2)_3C(C_2H_5)_3$: 3,3-diethyl-6,6-dimethylnonane
- vi. $C(C_2H_5)_4$: 3,3-diethylpentane
- vii. Cyclopentane structure: 1-ethyl-3-ethyl-3-methylcyclopentane (or 1,3-diethyl-3-methylcyclopentane)
Concept Assessment Exercise 15.4: Unsaturated Hydrocarbons
Task 1: Draw Skeletal Formulae
- i. but-1-ene: 4-carbon chain with a double bond at carbon 1.
- ii. penta-1,2-diene: 5-carbon chain with double bonds at carbons 1 and 2.
- iii. 5-methylhepta-1,3-diyne: 7-carbon chain, triple bonds at 1 and 3, methyl group at 5.
- iv. hex-1-en-4-yne: 6-carbon chain, double bond at 1, triple bond at 4.
Task 2: Give Systematic Names
- i. $CH_2=CH(CH_2)_3C \equiv CCH_3$: hept-1-en-5-yne
- ii. $CH_2=CH(CH_2)_4CH(CH_3)C \equiv CH$: 7-methyloct-1-en-8-yne
- iii. $CH_3CH=C=CH_2$: buta-1,2-diene
- iv. $(CH_3)CH=C(CH_3)CH(CH_3)_2$: 3,4-dimethylpent-2-ene
Concept Assessment Exercise 15.5: Alcohols and Halides
- i. $(CH_3)_2CHCH(Br)CH_2CH_3$: 3-bromo-2-methylpentane
- ii. $CH_3CH(OH)CH_2CH_3$: butan-2-ol
- iii. $CH_3CH(OH)CH(CH_3)CH_2CH_3$: 3-methylpentan-2-ol
- iv. $CH_3CH(CH_3)CH_2CH(OH)CH_3$: 4-methylpentan-2-ol
- v. $CH_3CH=C(OH)CH_3$: but-2-en-2-ol
- vi. $CH_3CH(OH)CH(Br)CH_3$: 3-bromobutan-2-ol
Concept Assessment Exercise 15.6: Carbonyl Compounds
- i. $HCHO$: methanal
- ii. $(CH_3)_2CHCHO$: 2-methylpropanal
- iii. $C_6H_5COCH_3$: acetophenone (or 1-phenylethanone)
- iv. $CH_3COCH_2CH_3$: butanone
- v. $(CH_3)_2CHCH_2COCH_2CH_3$: 5-methylhexan-3-one
Concept Assessment Exercise 15.7: Nitrogen and Oxygen Compounds
Task 1: Names
- i. $CH_3OCH_2CH_3$: methoxyethane
- ii. $(C_2H_5)_2CHNH_2$: pentan-3-amine
- iii. $(CH_3)_3N$: N,N-dimethylmethanamine
Task 2: 2D Displayed Formulae
Concept Assessment Exercise 15.8: Acids and Esters
Task 1: Names
- i. $(CH_3)_2CH(CH_2)_3COOH$: 5-methylhexanoic acid
- ii. $(CH_2COOH)_2$: butanedioic acid
- iii. $(COOH)_2$: ethanedioic acid
- iv. $CH_3COOCH_2CH_3$: ethyl ethanoate
- v. $HCOOCH_3$: methyl methanoate
- vi. $CH_3CH(Cl)CH_2COOH$: 3-chlorobutanoic acid
1. Fundamental Terminology
Organic reactions involve the movement of electrons and the breaking/forming of covalent bonds. Understanding the reagents and how bonds break is essential.
A. Types of Reagents
- Free Radical: An atom or group with an unpaired electron (e.g., $Cl^{\cdot}$, $CH_3^{\cdot}$). They are highly unstable and formed via homolytic fission.
- Electrophile: An "electron-loving" species that is electron-deficient. It often carries a positive charge or an empty orbital (e.g., $CH_3^{+}$).
- Nucleophile: A "nucleus-loving" species rich in electrons. It carries a lone pair or a negative charge (e.g., $OH^{-}$).
B. Bond Breakage (Fission)
- Homolytic Fission: Equal splitting of a covalent bond where each atom retains one electron, forming free radicals.
Example: $Cl_2 \rightarrow 2Cl^{\cdot}$ - Heterolytic Fission: Unequal breakage where one atom takes both electrons, resulting in ions (cation and anion).
Example: $CH_3Cl \rightarrow CH_3^{+} + Cl^{-}$
2. Types of Organic Chemical Reactions
| Reaction Type | Description | Example Equation |
|---|---|---|
| Free Radical Substitution | A halogen radical replaces hydrogen in alkanes. | $CH_4 + Cl_2 \xrightarrow{hv} CH_3Cl + HCl$ |
| Electrophilic Addition | Electrophile attacks a double/triple bond, converting it to a single bond. | $CH_3CH=CH_2 + HBr \rightarrow CH_3CH(Br)CH_3$ |
| Elimination | Removal of atoms from adjacent carbons to form a double bond. | $CH_3CH_2OH \xrightarrow{H_2SO_4} CH_2=CH_2 + H_2O$ |
| Nucleophilic Substitution | A nucleophile replaces an atom or group in a molecule. | $CH_3CH_2Br + NaOH \rightarrow CH_3CH_2OH + NaCl$ |
| Nucleophilic Addition | Nucleophile adds to a carbonyl group ($C=O$). | $CH_3CHO + HCN \xrightarrow{NaCN/HCl} CH_3CH(CN)OH$ |
| Hydrolysis | Water molecule breaks an organic compound, often catalyzed by acid/alkali. | $CH_3COOCH_2CH_3 + H_2O \xrightarrow{H_2SO_4} CH_3COOH + CH_3CH_2OH$ |
| Condensation | Two molecules combine to form a larger one, usually losing a small molecule like $H_2O$. | $CH_3COOH + CH_3CH_2OH \xrightarrow{H_2SO_4} CH_3COOCH_2CH_3 + H_2O$ |
3. Oxidation and Reduction
- Oxidation: Gain of oxygen, loss of hydrogen, or increase in carbon-oxygen bonds.
Example: $CH_3OH + 3[O] \xrightarrow{K_2Cr_2O_7/H_2SO_4} 2HCOOH$ - Reduction: Loss of oxygen, gain of hydrogen, or decrease in carbon-oxygen bonds.
Example: $CH_3CHO + 2[H] \xrightarrow{LiAlH_4} CH_3CH_2OH$
4. Questions and Answers
Q1: What is the primary difference between homolytic and heterolytic fission?
A: In homolytic fission, the bond breaks equally and each atom takes one electron to form radicals; it usually occurs when the electronegativity difference is zero. In heterolytic fission, the bond breaks unequally, and one atom takes both electrons to form ions; it occurs between atoms with a significant electronegativity difference.
Q2: Why are Alkenes susceptible to Electrophilic Addition reactions?
A: Alkenes contain a double bond ($\pi$-bond), which is a region of high electron density. This attracts electrophiles (electron-deficient species) which add across the double bond to form a saturated single bond.
Q3: Distinguish between Hydrolysis and Condensation reactions using the provided examples.
A: Hydrolysis is the breakdown of a molecule (like an ester) using water to form smaller products (acid and alcohol). Condensation is the opposite, where two molecules (acid and alcohol) combine to form a larger molecule (ester) while releasing a small molecule like water.
Q4: Identify the reagent type of $OH^{-}$ and justify its classification.
A: $OH^{-}$ is a nucleophile. This is because it possesses a negative charge and lone pairs of electrons, making it "nucleus-loving" and capable of donating electrons to an electron-deficient center (electrophile).
1. Free Radical Substitution of Alkanes
This reaction involves the replacement of hydrogen atoms in alkanes with halogen atoms (like chlorine) in the presence of ultraviolet (UV) light or sunlight. The process follows a three-step mechanism:
- Initiation: Homolytic fission of a halogen molecule into two free radicals using UV energy.Example: $$Cl-Cl \xrightarrow{h\nu} Cl\cdot + Cl\cdot$$
- Propagation: A continuous cycle where a radical reacts with a stable molecule to form a new radical and a new molecule.Step 1: $$CH_4 + Cl\cdot \rightarrow \cdot CH_3 + HCl$$Step 2: $$\cdot CH_3 + Cl_2 \rightarrow CH_3Cl + Cl\cdot$$
- Termination: Two free radicals combine to form a stable neutral molecule, ending the chain reaction.Example: $$\cdot CH_3 + \cdot CH_3 \rightarrow CH_3CH_3$$
Note: If chlorine is in excess, the reaction continues until all hydrogens are replaced, forming dichloromethane ($$CH_2Cl_2$$), trichloromethane (chloroform, $$CHCl_3$$), and finally tetrachloromethane ($$CCl_4$$).
2. Elimination Reactions
Elimination involves the removal of atoms or groups from adjacent carbon atoms to create an unsaturated molecule (alkene) and a small inorganic molecule (like water or $$HCl$$).
- Example: Dehydrohalogenation of bromoethane using a base ($$OH^-$$).
- The base attacks the acidic hydrogen, the $C-H$ bond electrons form a pi-bond, and the bromine leaves as a bromide ion.
- Result: $$CH_3CH_2Br + OH^- \rightarrow CH_2=CH_2 + H_2O + Br^-$$
3. Nucleophilic Substitution ($$S_N2$$)
A nucleophile (an electron-rich species) attacks the electrophilic carbon (the carbon attached to the halogen) and replaces the leaving group.
- Mechanism: It involves a transition state where the bond to the nucleophile is forming at the same time the bond to the halogen is breaking.
- Example: Formation of ethanol from bromoethane.
- $$CH_3CH_2Br + OH^- \rightarrow CH_3CH_2OH + Br^-$$
4. Electrophilic Addition
This reaction occurs in alkenes where an electrophile attacks the electron-rich double bond ($$\pi$$-bond), converting it into a single covalent bond.
- Example: Reaction of ethene with hydrogen bromide ($$HBr$$) to form bromoethane.
Review Questions & Answers
| Question | Answer |
|---|---|
| What condition is essential for the initiation step in the chlorination of methane? | The presence of ultraviolet (UV) light or sunlight is required to provide the energy for homolytic fission of the chlorine molecule. |
| Define "Homolytic Fission" in the context of these notes. | It is the breaking of a covalent bond where each atom takes one of the shared electrons, resulting in the formation of free radicals. |
| In the elimination reaction of bromoethane, what is the role of the $$OH^-$$ ion? | The $$OH^-$$ ion acts as a base, attacking the acidic hydrogen atom on the carbon adjacent to the halogen. |
| What is the chemical name for the product formed when all four hydrogen atoms of methane are replaced by chlorine? | The product is tetrachloromethane ($$CCl_4$$). |
| How does a nucleophilic substitution reaction differ from an addition reaction? | In substitution, an atom or group is replaced by another; in addition, a double bond is broken to allow new atoms to be added to the molecule without losing any existing atoms. |