Question

An alkyl bromide, \(\mathrm{C}_{5} \mathrm{H}_{11}\) Br reacted rapidly with water to give an alcohol \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\), but when the alkyl bromide was treated with sodium methoxide in an attempt to convert it to an ether, the only product was an alkene, \(\mathrm{C}_{5} \mathrm{H}_{10}\) Ozonolysis of this alkene gave two carbonyl compounds as products, a ketone \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\) and an aldehyde \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O} .\) What was the structure of the alkyl bromide?

Short answer

The structure of the alkyl bromide is 2-bromopentane (CH3CH(Br)CH2CH2CH3).

Step-by-step solution

Identify the functional groups in products

The products formed after the ozonolysis reaction are a ketone and an aldehyde. The ketone can be represented as: \(\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}\) with structure: CH3C(O)CH3 (propanone) The aldehyde can be represented as: \(\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}\) with structure: CH3CHO (ethanal)

Determine the alkene structure

Since ozonolysis cleaved the alkene into a ketone and an aldehyde, we can now infer that the alkene's structure is derived from the main chain of both compounds. Combining propanone and ethanal, we get the following alkene structure: CH3CH=CHCH2CH3 (pent-2-ene)

Determine the alcohol structure

Given that the alkyl bromide reacts with water to form an alcohol, and the alcohol has the molecular formula \(\mathrm{C}_5\mathrm{H}_{12}\mathrm{O}\), this alcohol must have the same carbon skeleton as the alkene formed in step 2, with an additional hydroxyl group. Thus, the alcohol must be pentan-2-ol (CH3CH(OH)CH2CH2CH3)

Determine the alkyl bromide structure

Knowing the alcohol structure, we can now determine the alkyl bromide structure by replacing the hydroxyl group in the alcohol with a bromide. This results in the following structure: CH3CH(Br)CH2CH2CH3 (2-bromopentane) The structure of the alkyl bromide is 2-bromopentane.

Key concepts

Understanding Ozonolysis

Ozonolysis is a critical reaction in organic chemistry, especially when it comes to identifying structures of unknown alkenes. In simpler terms, ozonolysis involves breaking the carbon-carbon double bond in alkenes using ozone, leading to the formation of carbonyl compounds, namely aldehydes and ketones.

Imagine our alkene like a paper link chain; when you introduce ozone, the links (or carbon atoms in this case) are cleaved at the double bond. Ozonolysis effectively slices through this bond and attaches an oxygen atom to each carbon that was originally part of the double bond. For a student grappling with organic structures, recognizing the original alkene from the ozonolysis products is like solving a puzzle by looking at the pieces provided—at the end of our exercise, we identified that our puzzle yielded propanone and ethanal, indicating a pent-2-ene precursor.

Alkene Formation via Nucleophilic Substitution

When crafting the narrative of alkene formation, students often witness a plot twist in the form of a reaction known as nucleophilic substitution. This process typically involves an exchange between a more nucleophilic atom or group (which donates an electron pair) and a leaving group in a molecule.

For the alkyl bromide in question, you'd expect the nucleophile, sodium methoxide in this case, to swap places with the bromide, leading to ether formation. However, the presence of bromide—which often serves as a good leaving group—instead results in the removal of a hydrogen atom from an adjacent carbon atom and the expulsion of bromide. This creates a double bond, thus an alkene is born. It may help to envision this as a tango dance where one partner gracefully exits the dance floor, allowing a new bond to form between the remaining dancers (carbon atoms).

The Mechanics of Nucleophilic Substitution

Nucleophilic substitution is not merely a reaction; it's a performance where electron-rich nucleophiles target electron-poor sites of a substrate and exchange places with leaving groups.

It's similar to a well-rehearsed play—each actor must know their role. The nucleophile is a molecule or ion ready to donate a pair of electrons, like the sodium methoxide in our session's exercise. The leaving group, typically a halide like bromine, must be ready to accept electrons and depart. Successful nucleophilic substitution hinges on this ebb and flow of electron exchange, often resulting in the creation of a new molecule—in our scenario, an alkene rather than the expected ether. Students should focus on understanding the traits of good leaving groups, the steric factors, and the electronic properties of the substrate to predict the outcomes of these intriguing reaction scenes.