Problem 31 - Basic Principles of Organic Chemistry

Question

Which compound in each of the following pairs would lose chloride ion more readily and form a carbonium ion? Explain. and C=CCCCl C=CCCl C=CC(Cl)C=C and \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Cl} \quad(21 . \mathrm{E} .2)\) C=CC=CCCl

Step-by-Step Solution

Answer

The compound \( \text{C=CCCCl} \) and \( \text{C=CC(Cl)C=C} \) lose chloride ion more readily.

Step 1: Understand the Concept

To determine which compound will more readily form a carbonium ion, we need to consider how stable the resulting carbocation will be after losing a chloride ion. Carbocations are more stable when they are allylic or benzylic due to resonance stabilization, or more substituted due to hyperconjugation.

Step 2: Analyze the First Pair

The first pair involves: 1. \( \text{C=CCCCl} \) and 2. \( \text{C=CCCl} \). The compound \( \text{C=CCCCl} \), upon losing a chloride ion, forms an allylic carbocation, which is resonance-stabilized. In contrast, \( \text{C=CCCl} \) forms a primary carbocation that is not resonance-stabilized. Therefore, \( \text{C=CCCCl} \) will lose its chloride ion more readily.

Step 3: Analyze the Second Pair

The second pair involves: 1. \( \text{C=CC(Cl)C=C} \) and 2. \( \text{CH}_2=CH-CH=CH-CH_2Cl \). In \( \text{C=CC(Cl)C=C} \), losing a chloride forms an allylic carbocation between two double bonds, leading to significant resonance stability. \( \text{CH}_2=CH-CH=CH-CH_2Cl \) also forms a primary allylic carbocation but with fewer resonance structures. Therefore, \( \text{C=CC(Cl)C=C} \) would more readily lose a chloride ion as it forms a more stabilized carbocation.

Key Concepts

Resonance Stabilization

Resonance stabilization is a key concept in determining the stability of carbocations, which are positively charged carbon atoms formed during certain chemical reactions. When a carbocation is resonance stabilized, its positive charge is delocalized across multiple atoms in the molecule through the overlapping of π orbitals. This delocalization reduces the electron deficiency on the carbocation, making it more stable.

Here's how it works:

• The positive charge in a carbocation is not confined to a single carbon atom but is spread out over several adjacent atoms.
• This spreading out, or resonance, happens through the movement of electrons in double bonds or lone pairs, allowing multiple resonance structures to exist.
• Each resonance structure contributes to the overall stability, as no single atom bears the entire positive charge.

As a result, compounds that can form resonance-stabilized carbocations, such as those with allylic positions, tend to lose leaving groups like chloride ions more easily, because the resulting cation is highly stable.

Allylic Carbocation

An allylic carbocation forms when a positive charge resides on a carbon atom that is adjacent to a double bond, known as an allylic position. This position allows the carbocation to undergo significant resonance stabilization.

For example:

• In the compound C=CCCCl, when the chloride ion leaves, the positive charge that develops on the carbon atom next to the double bond is allylic.
• The presence of the double bond enables the delocalization of the positive charge, thus increasing the stability of the carbocation.
• The ability for the charge to resonate over multiple structures offers additional stabilization, making the carbocation formation more favorable.

This increased stability is why allylic carbocations form readily and why compounds like C=CCCCl can lose chloride ions, forming allylic carbocations more easily than similar compounds lacking this resonance potential.

Hyperconjugation

Hyperconjugation is another stabilizing interaction that helps carbocations become more stable. It involves the donation of electron density from adjacent σ bonds (such as C-H bonds) into the empty p-orbital of the positively charged carbon atom.

Here's how it aids stability:

• This overlap of orbitals helps disperse the positive charge over several bonds, rather than just one atom.
• The more hyperconjugative structures available, the greater the electron donation, and thus the more stable the carbocation.
• Hyperconjugation is particularly effective in carbocations that are tertiary because they have more adjacent C-H bonds compared to primary or secondary carbocations.

Although less significant than resonance, hyperconjugation still contributes meaningfully to carbocation stability making compounds more likely to lose a leaving group and form stable carbocations.