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Chapter 6: Problem 116

Describe dehydration of alcohols to obtain an alkene.

Short Answer

Expert verified
Dehydration of alcohols is an elimination reaction where a water molecule is removed from an alcohol molecule, resulting in the formation of an alkene. This process requires a strong acid catalyst, such as sulfuric acid or phosphoric acid, and high temperature (140 to 180 °C) to promote the reaction. The reaction mechanism can follow either E1 (common with secondary and tertiary alcohols) or E2 (common with primary alcohols) pathways, involving protonation of the hydroxyl group and subsequent elimination of water to form the alkene. An example is the dehydration of ethanol to ethene using sulfuric acid as a catalyst at 180 °C.

Step by step solution

01

Understanding the dehydration reaction

Dehydration of alcohols is an elimination reaction in which a water molecule is removed from an alcohol molecule, leading to the formation of an alkene. This process typically requires a strong acid catalyst and high temperature.
02

Write down the general chemical equation

The general chemical equation for the dehydration of alcohol to form an alkene can be represented as follows: Alcohol + strong acid catalyst \(\xrightarrow[]{heat}\) Alkene + Water
03

Explain the role of acid catalyst

A strong acid catalyst, such as sulfuric acid (\(H_2SO_4\)) or phosphoric acid (\(H_3PO_4\)), is required to protonate the hydroxyl group (-OH) in alcohol, enabling the elimination of water molecule. The protonation of the hydroxyl group leads to a good leaving group, due to its positive charge, ensuring a smooth reaction.
04

Mechanism of the reaction

The dehydration of alcohols follows the E1 or E2 mechanism, depending on the type of alcohol (primary, secondary, or tertiary) and reaction conditions. In both mechanisms, the first step involves protonation of the hydroxyl group. The difference lies in the order of the remaining steps: E1 Mechanism (common with secondary and tertiary alcohols): 1. Formation of a carbocation by the leaving group, water. 2. Elimination of a proton from the carbocation by a base to form the alkene. E2 Mechanism (common with primary alcohols): 1. Simultaneous removal of a proton from the adjacent carbon by a strong base and leaving of the water molecule 2. Formation of the alkene through rearrangement.
05

Describe the reaction conditions

The temperature plays a crucial role in this reaction. Generally, the dehydration of alcohols requires high temperatures (140 to 180 °C) to shift the equilibrium towards the formation of alkene. If the temperature is not sufficiently high, the reaction might not proceed to completion.
06

Example of dehydration reaction

As an example, let's consider the dehydration of ethanol (ethyl alcohol) to form ethene (ethylene): Ethanol + \(H_2SO_4\) \(\xrightarrow[]{180 ^{\circ}C}\) Ethene + Water \[CH_3CH_2OH + H_2SO_4 \xrightarrow[]{180 ^{\circ}C} CH_2=CH_2 + H_2O\] In this reaction, sulfuric acid (\(H_2SO_4\)) acts as the acid catalyst, and the reaction occurs at 180°C. The hydroxyl group in the ethanol molecule is protonated, followed by the elimination of water, resulting in the formation of ethene and water.

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Most popular questions from this chapter

Indicate how you would synthesize each of the following compounds from any one of the given organic starting materials and inorganic reagents. Specify reagents and the reaction conditions. Starting materials: propylene, isobutylene.

Write the chemical structures for each compound listed. (a) 1 -Hexene (b) 3 -Methyl-1-butane (c) 2, 4-Hexadiene (d) 1-Iodo-2-methy1-2-pentene (e) 2-Chloro-3-methy1-2-hexane (f) 6,6 -Dibromo-5-methy1-5-ethy1-2,3-heptadiene

The trans alkenes are generally more stable than the cis alkenes. Give two examples of unsaturated systems where you would expect the cis form to be more stable and explain the reason for your choice.

What product is formed when \(\mathrm{X}_{2}(\mathrm{X}=\mathrm{Br}, \mathrm{Cl})\) is added to alkenes in the dark? Explain the mechanism of the reaction.

When isopropyl bromide is treated with sodium ethoxide in ethanol, propylene and ethyl isopropyl ether are formed in a \(3: 1\) ratio. If the hexadeuteroisopropy1 bromide, \(\mathrm{CD}_{3} \mathrm{CHBrCD}_{3}\) is used, \(\mathrm{CD}_{3} \mathrm{CH}=\mathrm{CD}_{2}\) and \(\left(\mathrm{CD}_{3}\right)_{2} \mathrm{CHOC}_{2} \mathrm{H}_{5}\) are formed in a ratio of \(1: 2\). Explain.
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