Question

Five alcohols (A) \(-(\mathrm{E})\) are given below (A) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{OH}\) (B) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{3} \mathrm{H}_{7}\) (C) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{C}_{3} \mathrm{H}_{7}\) (D) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (E) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) The ease of dehydration in the decreasing order will be (a) \(\mathrm{D}>\mathrm{C}>\mathrm{E}>\mathrm{B}>\mathrm{A}\) (b) \(C>D>B>A>E\) (c) \(\mathrm{D}>\mathrm{C}>\mathrm{A}>\mathrm{B}>\mathrm{E}\) (d) \(\mathrm{D}>\mathrm{E}>\mathrm{B}>\mathrm{A}>\mathrm{C}\)

Short answer

Question: Arrange the given alcohols in the decreasing order of ease of dehydration: (A) CH3(CH2)4CH2OH, (B) C2H5CH(OH)C3H7, (C) (CH3)2C(OH)C3H7, (D) (CH3)2C(OH)CH(CH3)2, and (E) C2H5CH(OH)CH(CH3)2. Answer: The correct order of ease of dehydration is (a): D>C>E>B>A.

Step-by-step solution

Find the structure of alcohol (A)

The structure of alcohol A is given by \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2}\mathrm{OH}\)

Find the structure of alcohol (B)

The structure of alcohol B is given by \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{C}_{3}\mathrm{H}_{7}\)

Find the structure of alcohol (C)

The structure of alcohol C is given by \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH}) \mathrm{C}_{3}\mathrm{H}_{7}\)

Find the structure of alcohol (D)

The structure of alcohol D is given by \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH})\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\)

Find the structure of alcohol (E)

The structure of alcohol E is given by \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH})\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\)

Determine the carbocation stability for each alcohol

Based on the structures, we can find the carbocation formed during the dehydration as: (A): Primary carbocation (B): Secondary carbocation (C): Tertiary carbocation (D): Tertiary carbocation (with 3 alkyl groups) (E): Secondary carbocation (with 2 alkyl groups)

Order of carbocation stability

The order of carbocation stability in general is: Tertiary > Secondary > Primary. Thus, we can give the order as: (D) > (C) > (E) > (B) > (A)

Determine the correct order of ease of dehydration

Based on our analysis of carbocation stability, the ease of dehydration in decreasing order is: \(\mathrm{D}>\mathrm{C}>\mathrm{E}>\mathrm{B}>\mathrm{A}\) Therefore, the correct answer is (a): \(\mathrm{D}>\mathrm{C}>\mathrm{E}>\mathrm{B}>\mathrm{A}\).

Key concepts

Dehydration of Alcohols

Dehydration of alcohols is a common process in organic chemistry where an alcohol is transformed into an alkene. This usually involves the removal of a water molecule (hence the term "dehydration"). The reaction typically requires an acid catalyst, such as sulfuric acid or phosphoric acid, and heat to proceed.

In the dehydration reaction, the hydroxyl group (-OH) of the alcohol is protonated, leading to the formation of water. Then, a carbocation intermediate is formed by the departure of water. This intermediate is crucial, as its stability greatly influences the ease and speed of the dehydration process.

The carbocation's stability is the deciding factor. More stable carbocations form more readily and lead to faster dehydration reactions. Thus, tertiary alcohols dehydrate more easily than secondary, and secondary more easily than primary. This order stems from the increasing stability of the carbocations due to the electron-donating effects of alkyl groups.

Organic Chemistry Problems

Dehydration of alcohols is one of many classic organic chemistry problems that involve functional groups and their reactivity. Understanding these problems requires knowledge of several key concepts, including reaction mechanisms, stability of intermediates, and the effects of substituents.

Often, students are faced with problems requiring them to predict reaction products, such as the major alkene formed via dehydration. Determining which alkene is more stable involves understanding Markovnikov's rule and the various possible shifts and rearrangements that can occur with intermediates.

These problems help students develop strategic problem-solving skills that cover a range of scenarios in organic chemistry. Mastery comes from practicing and recognizing patterns of reactivity and stability that apply broadly, including in competitive exams.

Functional Groups Analysis

In functional groups analysis, alcohols serve as a fundamental class of compounds in organic chemistry. The presence of the hydroxyl group (-OH) confers distinct chemical properties to these molecules, contributing to their reactivity and transformation in reactions like dehydration.

Analyzing functional groups involves identifying the specific group present and understanding its effect on the molecule's overall chemical behavior. For alcohols, the position of the hydroxyl group determines the alcohol classification – primary, secondary, or tertiary. This classification is crucial because it directly impacts the resulting carbocation's stability during a reaction.

For example, tertiary alcohols, which have their hydroxyl group attached to a carbon that is bonded to three other carbons, often undergo dehydration reactions more readily because they form more stable carbocations. This knowledge is integral to predicting and explaining the outcomes of transformations in organic reactions.

IIT JEE Chemistry Preparation

For students preparing for the IIT JEE Chemistry exam, mastering concepts like the dehydration of alcohols is essential. This competitive exam tests a deep understanding of chemistry principles, including reaction mechanisms and the stability of intermediates.

In the context of the IIT JEE, students should focus on comprehending how different factors influence carbocation stability and how this affects the ease of alcohol dehydration. Practice with problems involving various alcohols and their structures helps students anticipate which reactions occur more readily.

To do well in the IIT JEE, it's also important that students integrate knowledge from other chemistry areas, such as acid-base chemistry and thermodynamics, as these domains often intersect in organic reactions. Participating in mock tests and solving previous years' exam papers will enhance the ability to tackle such problems effectively under exam conditions.