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Chapter 10: Problem 47

Take a proton away and simultaneously break a bond to give stable molecules or ions. Simultaneous proton transfer to solvent and lass of \(\mathrm{H}_{2} \mathrm{O}\) gives the carbon-carbon double bond of the terminal alkene.

Short Answer

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Question: Describe the process of deprotonation and elimination reaction in forming a carbon-carbon double bond, using 2-methyl-2-butanol as an example. Answer: In the case of 2-methyl-2-butanol, the process involves removing an acidic proton from the alpha carbon of the molecule by a solvent such as NaOH, resulting in a carbocation intermediate and a negatively charged conjugate base (OH-). Then, the oxygen atom of the hydroxyl group (-OH) on the third carbon takes one of the hydrogen atoms attached to the second carbon from the carbocation, resulting in the elimination of water (H2O) and the formation of a double bond between the second and third carbon atoms. This process leads to the formation of the terminal alkene, 2-methyl-2-butene, with a carbon-carbon double bond.

Step by step solution

01

Identify the starting molecule

We are not given a specific starting molecule, but based on the description, it seems like we are dealing with a molecule that has an acidic proton and a hydroxyl group (-OH) that can be eliminated to form a carbon-carbon double bond. A suitable example for this scenario would be 2-methyl-2-butanol.
02

Deprotonation

When transferring a proton from the molecule to a solvent, we remove an acidic proton from the alpha carbon of the molecule. In our example of 2-methyl-2-butanol, the acidic proton is on the alpha carbon (second carbon) adjacent to the tertiary alcohol. The solvent, usually a base, will remove this proton, forming a carbocation intermediate and leaving a negatively charged conjugate base of the solvent. In the case of using a base like \(\mathrm{NaOH}\), we would have a \(\mathrm{Na}^{+}\) and an \(\mathrm{OH}^{-}\) ion.
03

Loss of water

The next step in this process is the elimination of water from the molecule. To do this, the oxygen atom of the hydroxyl group (-OH) on the third carbon will take one of the hydrogen atoms attached to the second carbon from the carbocation. This results in the formation of a \(\mathrm{H}_{2}\mathrm{O}\) molecule, and a double bond is formed between the second and third carbon atoms.
04

Formation of terminal alkene

Following the removal of the proton and elimination of water, a carbon-carbon double bond is formed. In our example, this reaction leads to the formation of 2-methyl-2-butene, a terminal alkene with a double bond between the second and third carbon atoms.
05

Summary

In this exercise, we have analyzed a chemical reaction where a proton is taken away, and a bond is broken simultaneously, leading to the formation of stable molecules or ions. We considered the example of 2-methyl-2-butanol and described the process of deprotonation and elimination of water, resulting in the formation of the terminal alkene 2-methyl-2-butene with a carbon-carbon double bond.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Deprotonation Reaction
In organic chemistry, a deprotonation reaction is one where an acidic hydrogen atom is removed from a molecule, typically by the action of a base. This action not only strips away the hydrogen ion (H+) but also contributes to the stability of the resulting structure by forming a conjugate base.

Usually, the base used is stronger than the conjugate base of the acid. This difference in strength drives the reaction forward. In our textbook example, the molecule 2-methyl-2-butanol undergoes deprotonation facilitated by a strong base, such as NaOH. The deprotonation step is critical as it sets the stage for the formation of a carbocation intermediate, an essential species that gives rise to further reaction pathways leading to the creation of alkenes.
Carbocation Intermediate
A carbocation intermediate is a highly reactive species that contains a carbon atom bearing a positive charge. Formation of carbocations is a common motif in many organic reactions, especially those involving alkenes, such as the E1 or E2 elimination reactions.

These intermediates are typically formed during reactions by the loss of a leaving group, which provides the necessary positive charge on the carbon atom. The stability of carbocations depends on the number of alkyl groups attached to the positively charged carbon, with tertiary carbocations being the most stable. In our scenario, the carbocation forms as a transient species after the deprotonation of 2-methyl-2-butanol, providing a cornerstone for the subsequent steps of the reaction.
Loss of Water
The loss of water, or dehydration reaction, occurs in many organic mechanisms and is crucial in converting alcohols to alkenes. Typically, a hydrogen atom from the alcohol's neighboring carbon will bond with the hydroxyl group's oxygen, leading to the release of water and the formation of a double bond.

In our textbook example with 2-methyl-2-butanol, after deprotonation and formation of the carbocation, the -OH group attached to the carbon adjacent to the charged carbon acts as a leaving group, resulting in the expulsion of a water molecule. The removal of water is a critical step that allows the carbon atoms to form a double bond, leading us towards the synthesis of the desired alkene.
Carbon-Carbon Double Bond
The carbon-carbon double bond is a pivotal feature of alkenes, defined by the sharing of four electrons between two carbon atoms, instead of the usual two-electron single bond. This double bond is formed through the elimination of atoms or groups from adjacent carbon atoms, known as an elimination reaction.

In our context, the elimination reaction involves the deprotonation of an alcohol, followed by the loss of water to form this highly important double bond. This bond imparts characteristics such as geometric isomerism and a lower saturation point to alkenes, making them suitable for a variety of chemical reactions, such as polymerization and hydrohalogenation.
Terminal Alkene
A terminal alkene refers to an alkene with the double bond located at the end of the carbon chain, adjacent to a primary carbon atom. These alkenes are typically more reactive in certain reactions, like hydroboration-oxidation, because the terminal position presents a less steric hindrance and often has distinctive synthetic value.

In the exercise, the deprotonation, the formation of the carbocation intermediate, and the subsequent loss of water build upon one another to yield a terminal alkene. In our discussed example, this terminal alkene is 2-methyl-2-butene, showcasing the carbon-carbon double bond at the second position, hence, the terminal location in its carbon chain.

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Most popular questions from this chapter

Make a new bond between a nucleophile and an electrophile. Reaction of the \(3^{*}\) carbocation (an electrophile) with chloride ion (a nucleophile) gives the so haloalkane.

Thiols are much more acidic than alcohols.

According to the principle of microscopic reversibility, the sequence of transition states and reartive intermediates (that is, the mechanism) for any reversible reaction must be the sime, but in reverse order, for the reverse reaction as for the forward reaction.

Primary alcohols with extensive \(\beta\)-hranching give large amounts of a product erined from rearrangement. For example, treatment of 2,2 -dimethyl-1 propanol neopentyl alcohol) wìth HBr gives a rearranged product almost exclasively. \(10.5\) Cornersion of Alcohols to Haloakanes and sulfons CC(C)(C)CO CCC(C)(C)Br 2.2-Dimethyy-1-propanol 2-Bromo-2-tuetlyylbutane (a product of rearrangement)

Thiols are named in the same manner as aleohols, but the suffix e of the parent alkane is retained and thiol added. \- In compounds containing other functional groups of higher precedence, the presence of \(-\$ M\) is indicated by the prefix werapto.
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