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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

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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.

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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 (120∘120^\circ), and spsp leads to linear geometry (180∘180^\circ).

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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.

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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 C−CC-C bonds and vertices represent Carbon atoms).

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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).

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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).

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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)}

CnH2n−2 (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 CH3−C≡NCH_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.