Problem 184
Question
During dehydration of alcohols to alkenes by heating with concentration \(\mathrm{H}_{2} \mathrm{SO}_{4}\) the initiation step is \([\mathbf{2 0 0 3} \mid\) (a) protonation of alcohol molecule (b) formation of carbocation (c) elimination of water (d) formation of an ester
Step-by-Step Solution
Verified Answer
(a) Protonation of alcohol molecule is the initiation step.
1Step 1: Identify Goal of Dehydration
The process of dehydrating alcohols to form alkenes involves removing a water molecule from the alcohol. Therefore, the initial steps must facilitate the departure of the hydroxyl group (-OH) as water (H₂O).
2Step 2: Protonation of Alcohol
First, the alcohol interacts with concentrated sulfuric acid (H₂SO₄). The lone pair of electrons on the oxygen in the hydroxyl group of the alcohol attacks a proton (H⁺) from H₂SO₄, leading to the formation of an oxonium ion. This is the protonation of the alcohol molecule.
3Step 3: Consider Formation of Carbocation
Once the alcohol is protonated, it becomes a good leaving group in the form of water, which can be eliminated to form a carbocation. However, the formation of carbocation is not the initiation step; instead, it occurs after protonation.
4Step 4: Elimination Process
After protonation, the next step involves the removal of water, leaving behind a positively charged carbon atom (carbocation). This step is critical in forming the alkene in subsequent reactions, but is not the initiation step.
5Step 5: Conclusion on Initiation Step
Reviewing the options, the initiation step in the dehydration process is the protonation of the alcohol, as it alters the alcohol into a form that can proceed to become an alkene through further dehydration.
Key Concepts
ProtonationCarbocation FormationElimination of Water
Protonation
In the dehydration of alcohols, protonation is a crucial step that kickstarts the entire process. When an alcohol molecule encounters a strong acid, like sulfuric acid (\(\mathrm{H}_2\mathrm{SO}_4\)), it undergoes protonation. Protonation is where a proton (\(\mathrm{H}^+\)) from the acid attaches to the alcohol. This involves:
- The lone pair of electrons on the oxygen in the hydroxyl group (\(-\mathrm{OH}\)) of the alcohol attacking a proton.
- This forms an oxonium ion (\(\mathrm{R}-\mathrm{OH_2}^+\)), making the oxygen positively charged and turning the alcohol into a better leaving group.
Carbocation Formation
Following protonation, the next pivotal event in alcohol dehydration is carbocation formation. After the hydroxyl group in the alcohol is protonated, it becomes a water molecule that is primed to leave, or "exit," the rest of the structure:
- As the water molecule departs, it leaves behind a positively charged carbon atom, known as a carbocation.
- This carbocation is often stabilized by surrounding atoms or groups, which can influence its reactivity.
Elimination of Water
The elimination of water is the step in which the dehydrated structure takes shape. Once protonation has converted the hydroxyl group into a better leaving entity, the water molecule exits:
- This removal is known as the elimination step, and it is integral to forming the double bond characteristic of alkenes.
- After the water leaves, the remaining skeleton forms a pi bond between adjacent carbon atoms, resulting in the creation of an alkene.
Other exercises in this chapter
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