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Organic Chemistry: Basic Principles and Techniques - Isomerism (Structural and Stereo)

Grade 11ICSEChemistry

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

🔑Concepts

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Isomerism is the phenomenon where two or more organic compounds possess the same molecular formula but exhibit different physical and chemical properties due to the difference in the arrangement of atoms.

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Structural Isomerism arises due to differences in the connectivity of atoms. It includes: Chain Isomerism (different carbon skeletons), Position Isomerism (different positions of functional groups/substituents, e.g., 1−propanol1-propanol and 2−propanol2-propanol), Functional Isomerism (different functional groups, e.g., C2H6OC_2H_6O as CH3CH2OHCH_3CH_2OH and CH3OCH3CH_3OCH_3), Metamerism (unequal distribution of alkyl groups around a polyvalent functional group like −O−-O- or −S−-S-), and Tautomerism (dynamic equilibrium between two structural isomers, typically Keto-Enol form).

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Stereoisomerism occurs when isomers have the same connectivity but differ in the spatial arrangement of atoms. It is divided into Geometrical and Optical isomerism.

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Geometrical Isomerism (Cis-Trans) occurs due to restricted rotation around C=CC=C bonds. In the Cis form, identical groups are on the same side; in the Trans form, they are on opposite sides. For example, cis−but−2−enecis-but-2-ene and trans−but−2−enetrans-but-2-ene.

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Optical Isomerism occurs in molecules containing an asymmetric or Chiral Carbon (a carbon bonded to four different groups). Such molecules are non-superimposable on their mirror images (Enantiomers) and rotate the plane of polarized light.

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Chirality is the property of non-superimposability. A molecule with a plane of symmetry is Achiral and optically inactive (Meso compounds).

📐Formulae

Degree of Unsaturation (DoU)=C+1−H2−X2+N2Degree\ of\ Unsaturation\ (DoU) = C + 1 - \frac{H}{2} - \frac{X}{2} + \frac{N}{2}

Specific Rotation [α]λT=αl×cSpecific\ Rotation\ [\alpha]_{\lambda}^{T} = \frac{\alpha}{l \times c}

Number of Geometrical Isomers=2n (where n is the number of double bonds, for unsymmetrical molecules)Number\ of\ Geometrical\ Isomers = 2^n \text{ (where } n \text{ is the number of double bonds, for unsymmetrical molecules)}

Number of Optical Isomers=2n (where n is the number of chiral centers)Number\ of\ Optical\ Isomers = 2^n \text{ (where } n \text{ is the number of chiral centers)}

💡Examples

Problem 1:

Identify the type of isomerism between CH3CH2CH2OHCH_3CH_2CH_2OH and CH3CH(OH)CH3CH_3CH(OH)CH_3.

Solution:

Position Isomerism.

Explanation:

Both compounds share the molecular formula C3H8OC_3H_8O. In 1−propanol1-propanol, the −OH-OH group is at C1C_1, while in 2−propanol2-propanol, it is at C2C_2. Since the carbon chain is the same but the position of the functional group differs, they are position isomers.

Problem 2:

Calculate the Degree of Unsaturation for the molecular formula C4H6C_4H_6.

Solution:

DoU=4+1−62=5−3=2DoU = 4 + 1 - \frac{6}{2} = 5 - 3 = 2

Explanation:

A DoUDoU of 22 suggests the molecule could have two double bonds, one triple bond, one ring and one double bond, or two rings.

Problem 3:

Draw the Geometrical isomers of 1,2−dichloroethene1,2-dichloroethene.

Solution:

Cis-isomer: both ClCl atoms on the same side of the C=CC=C bond. Trans-isomer: ClCl atoms on opposite sides.

Explanation:

Because rotation is restricted around the double bond, the relative positions of the ClCl atoms are fixed, leading to distinct physical properties like different dipole moments (ciscis has a higher dipole moment than transtrans).

Problem 4:

Determine if 2−chlorobutane2-chlorobutane is optically active.

Solution:

Yes, it is optically active.

Explanation:

The second carbon (C2C_2) in CH3−C∗H(Cl)−CH2CH3CH_3-\overset{*}{C}H(Cl)-CH_2CH_3 is bonded to four different groups: −H-H, −Cl-Cl, −CH3-CH_3, and −C2H5-C_2H_5. This makes it a chiral center, allowing for the existence of dd and ll enantiomers.

Isomerism (Structural and Stereo) Revision - Class 11 Chemistry ICSE