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Organic Chemistry: Basic Principles and Techniques - Inductive, Electromeric, and Resonance Effects

Grade 11ICSEChemistry

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

🔑Concepts

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Inductive Effect (II-effect): This is a permanent effect involving the displacement of σ\sigma-electrons along a saturated carbon chain when an electron-withdrawing or electron-donating group is attached. It is distance-dependent and diminishes rapidly after three carbon atoms.

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−I-I Effect (Electron-Withdrawing): Groups that pull electron density away from the carbon chain, such as −NO2-NO_2, −CN-CN, −COOH-COOH, and halogens (−F,−Cl,−Br,−I-F, -Cl, -Br, -I). They increase the acidity of carboxylic acids.

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+I+I Effect (Electron-Donating): Groups that push electron density toward the carbon chain, primarily alkyl groups like −CH3-CH_3, −C2H5-C_2H_5, and −(CH3)3C-(CH_3)_3C. They stabilize carbocations and increase the basicity of amines.

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Electromeric Effect (EE-effect): A temporary effect involving the complete transfer of a shared pair of π\pi-electrons to one of the atoms joined by a multiple bond in the presence of an attacking reagent. It is categorized into +E+E (electrons move toward the reagent) and −E-E (electrons move away from the reagent).

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Resonance Effect (RR or MM effect): The permanent polarity produced in a molecule by the interaction between two π\pi-bonds or a π\pi-bond and a lone pair of electrons. +R+R effect (e.g., −OH-OH, −NH2-NH_2) increases electron density in the ring, while −R-R effect (e.g., −NO2-NO_2, −CHO-CHO) decreases it.

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Stability of Carbocations: Stabilized by +I+I and +R+R effects. The general order of stability for alkyl carbocations is: 3∘>2∘>1∘>methyl3^\circ > 2^\circ > 1^\circ > \text{methyl}.

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Stability of Carbanions: Stabilized by −I-I and −R-R effects. The general order of stability is: methyl>1∘>2∘>3∘\text{methyl} > 1^\circ > 2^\circ > 3^\circ.

📐Formulae

Acid Strength∝−I effect∝1+I effect\text{Acid Strength} \propto -I \text{ effect} \propto \frac{1}{+I \text{ effect}}

Base Strength∝+I effect∝1−I effect\text{Base Strength} \propto +I \text{ effect} \propto \frac{1}{-I \text{ effect}}

Stability of Carbocation∝+I,+R effects\text{Stability of Carbocation} \propto +I, +R \text{ effects}

Stability of Carbanion∝−I,−R effects\text{Stability of Carbanion} \propto -I, -R \text{ effects}

💡Examples

Problem 1:

Arrange the following carboxylic acids in increasing order of their acidic strength: CH3COOHCH_3COOH, ClCH2COOHClCH_2COOH, Cl2CHCOOHCl_2CHCOOH, and Cl3CCOOHCl_3CCOOH.

Solution:

CH3COOH<ClCH2COOH<Cl2CHCOOH<Cl3CCOOHCH_3COOH < ClCH_2COOH < Cl_2CHCOOH < Cl_3CCOOH

Explanation:

Chlorine exerts a −I-I effect (electron-withdrawing). As the number of chlorine atoms increases, the electron density is pulled away from the O−HO-H bond more effectively, facilitating the release of the H+H^+ ion and stabilizing the resulting carboxylate ion.

Problem 2:

Between Phenol and Ethanol, which is more acidic and why?

Solution:

Phenol is more acidic than Ethanol.

Explanation:

In phenol, the conjugate base (phenoxide ion, C6H5O−C_6H_5O^-) is stabilized by the resonance effect (delocalization of the negative charge over the benzene ring). In ethanol, the ethoxide ion (C2H5O−C_2H_5O^-) is destabilized by the +I+I effect of the ethyl group, which increases the negative charge on oxygen.

Problem 3:

Explain the stability of the following carbocations: (CH3)3C+(CH_3)_3C^+, (CH3)2CH+(CH_3)_2CH^+, and CH3CH2+CH_3CH_2^+.

Solution:

(CH3)3C+>(CH3)2CH+>CH3CH2+(CH_3)_3C^+ > (CH_3)_2CH^+ > CH_3CH_2^+

Explanation:

The tertiary carbocation (CH3)3C+(CH_3)_3C^+ is the most stable because it has three methyl groups providing +I+I effect (electron donation) and 99 hyperconjugative HH-atoms to neutralize the positive charge. The stability decreases as the number of alkyl groups decreases.

Inductive, Electromeric, and Resonance Effects Revision - Class 11 Chemistry ICSE