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Organic Chemistry – Some Basic Principles and Techniques - General Introduction

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Carbon exhibits tetravalency and the unique property of catenation (self-linking), which allows it to form a vast variety of stable chains and rings. The ground state electronic configuration of carbon is 1s22s22p21s^2 2s^2 2p^2.

Hybridization in organic compounds explains the geometry of molecules: sp3sp^3 hybridization leads to tetrahedral geometry (bond angle 109.5\approx 109.5^\circ), sp2sp^2 leads to trigonal planar geometry (120120^\circ), and spsp leads to linear geometry (180180^\circ).

A single covalent bond is always a σ\sigma (sigma) bond. A double bond consists of one σ\sigma and one π\pi (pi) bond. A triple bond consists of one σ\sigma and two π\pi bonds.

Organic structures can be represented in multiple ways: Complete structural formulas (showing all bonds), Condensed structural formulas (omitting some or all bonds, e.g., CH3CH2OHCH_3CH_2OH), and Bond-line structural formulas (where lines represent CCC-C bonds and vertices represent Carbon atoms).

Organic compounds are broadly classified into Acyclic (open chain) and Cyclic (closed chain) compounds. Cyclic compounds are further divided into Alicyclic and Aromatic compounds (like Benzene C6H6C_6H_6).

A functional group is an atom or a group of atoms joined in a specific manner which determines the chemical properties of the organic compound (e.g., OH-OH for alcohols, CHO-CHO for aldehydes).

A Homologous series is a group of organic compounds having the same functional group and similar chemical properties, where successive members differ by a CH2-CH_2- unit and 1414 units of molecular mass.

📐Formulae

CnH2n+2 (General formula for Alkanes)C_nH_{2n+2} \text{ (General formula for Alkanes)}

CnH2n (General formula for Alkenes)C_nH_{2n} \text{ (General formula for Alkenes)}

CnH2n2 (General formula for Alkynes)C_nH_{2n-2} \text{ (General formula for Alkynes)}

CnH2n+1X (General formula for Haloalkanes)C_nH_{2n+1}X \text{ (General formula for Haloalkanes)}

Bond Order=Total number of bonds between two atomsTotal number of resonating structures\text{Bond Order} = \frac{\text{Total number of bonds between two atoms}}{\text{Total number of resonating structures}}

💡Examples

Problem 1:

Determine the number of σ\sigma and π\pi bonds in the molecule CH2=C=CH2CH_2=C=CH_2 (Propadiene).

Solution:

In CH2=C=CH2CH_2=C=CH_2, the first carbon is bonded to two HH (2 σ\sigma) and double bonded to the middle CC (1 σ\sigma, 1 π\pi). The middle CC is double bonded to the third CC (1 σ\sigma, 1 π\pi). The third CC is bonded to two HH (2 σ\sigma). Total σ=2+1+1+2=6\sigma = 2+1+1+2 = 6; Total π=1+1=2\pi = 1+1 = 2.

Explanation:

Every single bond is 1σ1 \sigma. Every double bond contains 1σ1 \sigma and 1π1 \pi bond.

Problem 2:

Identify the hybridization of each carbon atom in CH3CNCH_3-C \equiv N.

Solution:

Carbon 1 (in CH3CH_3) is attached to four atoms via single bonds, so it is sp3sp^3 hybridized. Carbon 2 (in CN-CN) is attached to one carbon via a single bond and one nitrogen via a triple bond, so it is spsp hybridized.

Explanation:

Hybridization is determined by the number of σ\sigma bonds and lone pairs. 4 σ\sigma bonds = sp3sp^3, 3 σ\sigma bonds = sp2sp^2, 2 σ\sigma bonds = spsp.

Problem 3:

Draw the bond-line structure for 22-Bromobutane.

Solution:

A zig-zag line consisting of four vertices (representing 4 Carbons). A branch with the symbol BrBr is attached to the second vertex.

Explanation:

In bond-line notation, Carbon atoms are not shown; they are assumed at corners and ends. Hydrogen atoms attached to carbons are also omitted.

General Introduction - Revision Notes & Key Formulas | CBSE Class 11 Chemistry