Question

Which compound(s) shows faster rate of solvolysis than compound $$ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl} $$ (A) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{Cl}\) (B) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl}\) (C) \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl}\) (D) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}\)

Short answer

The compounds (A), (B), (C), and (D) have faster solvolysis rates than compound CH3-CH2-CH2-CH2-Cl due to their more stable carbocations.

Step-by-step solution

Identify the original compound

First, let's identify the initial compound: \[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Cl} \]

Form carbocations from the original and comparison compounds

Next, we will form carbocations by removing the chlorine atom from the original compound and each of the comparison compounds: Original compound: \[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^+ \] (A) \[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}^+ \] (B) \[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^+ \] (C) \[ \mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^+ \] (D) \[ \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}^+=\mathrm{CH} \]

Compare the stability of carbocations

Now, let's compare the stability of the carbocations formed. Carbocation stability follows the order: Allylic and benzylic > 3° (tertiary) > 2° (secondary) > 1° (primary) > methyl 1. Original carbocation is a primary carbocation. 2. Carbocation (A) is a secondary carbocation adjacent to an electron-donating oxygen atom, making it more stable than the original. 3. Carbocation (B) possesses two electron-donating groups: a secondary carbocation and an oxygen atom, making it more stable than the original. 4. Carbocation (C) is a primary carbocation, but the presence of an electron-donating oxygen atom makes it more stable than the original. 5. Carbocation (D) is an allylic carbocation, which is stabilized by resonance, making it more stable than the original.

Determine the compounds with faster solvolysis rates

Based on the stability of the carbocations, we can conclude that the following compounds have faster solvolysis rates than compound CH3-CH2-CH2-CH2-Cl: (A) CH3-CH2-CH2-O-CH2-Cl (B) CH3-CH2-O-CH2-CH2-Cl (C) CH3-O-CH2-CH2-CH2-CH2-Cl (D) CH3-CH2-CH2-CH=CH-Cl

Key concepts

Carbocation Stability

Understanding carbocation stability is crucial in predicting the outcome of many organic reactions, including solvolysis. A carbocation is a positively charged carbon atom with three bonds instead of its usual four, making it very reactive.

The stability of carbocations depends on several factors:

• Alkyl groups adjacent to the positively charged carbon can donate electron density through inductive effect, stabilizing the carbocation. This is why tertiary carbocations (carbons with three alkyl groups) are more stable than secondary, which are more stable than primary.
• Resonance stabilization can occur when a carbocation is adjacent to a double bond or aromatic ring, allowing for the positive charge to be delocalized over multiple atoms.
• Adjacent atoms or groups that can donate electron density, like oxygen in ethers, can also stabilize carbocations via hyperconjugation or through inductive effects.

Carbocations are intermediates in many reactions. Their stability plays a pivotal role in determining the rate and pathway of reaction, as displayed in the given problem where stability directly influences the solvolysis rate.

Organic Reaction Mechanisms

In the context of organic chemistry, the mechanism by which a reaction occurs is the step-by-step sequence of elementary reactions leading to the product formation. Grasping these mechanisms allows chemists to predict the behavior of organic molecules in a variety of chemical reactions.

For example, solvolysis is a type of substitution reaction wherein the solvent acts as the nucleophile and replaces a leaving group. The route by which solvolysis occurs involves several stages:

• Formation of a carbocation intermediate upon departure of the leaving group.
• Nucleophilic attack by the solvent on the carbocation, leading to the product.

This mechanism demonstrates why carbocation stability is so important: the more stable the carbocation, the faster the nucleophilic attack, and thus, the faster the overall reaction rate. By comparing stability, like in the textbook problem, you can predict which compound undergoes faster solvolysis.

JEE Organic Chemistry

JEE, the Joint Entrance Examination, is an engineering entrance assessment conducted in India and is known for its challenging and comprehensive test of concepts, including organic chemistry. Students preparing for JEE need to have a deep understanding of organic reaction mechanisms and the stability of intermediates like carbocations.

JEE organic chemistry often tests knowledge through problems that examine reaction rates, product prediction, and mechanism proposals. In the case of the solvolysis problem, understanding the nuanced effect of substituents on the reaction center is crucial for success. For example, recognizing the stabilizing effects of electron-donating groups and resonance structures on carbocations can help in solving such problems. Therefore, it's imperative for students to not only remember these concepts but also to apply them strategically to answer questions accurately and efficiently.