Question

Many years before the Hofmann degradation of optically active a-phenylpropionamide was studied, the following observations were made: when the cyclopentane derivative, \(\mathrm{I}\), in which the \(-\mathrm{COOH}\) and \(-\mathrm{CONH}_{2}\) groups are cis to each other, was treated with hypobromite, compound II was obtained; compound II could be converted by heat into the amide III (called a lactam). What do these results show about the mechanism of the rearrangement? (Use models.)

Short answer

The mechanism of the rearrangement involves two main reactions: Hofmann rearrangement and intramolecular rearrangement. In the first reaction, when Compound I is treated with hypobromite, it favors Hofmann rearrangement, converting the -CONH2 group into an isocyanate group to form Compound II. The steric effects and adjacent -COOH group play a crucial role in this reaction. In the second reaction, heating Compound II causes the isocyanate group to rearrange into a lactam (intramolecular amide), forming Compound III. Intramolecular rearrangement occurs due to the strained ring structure of cyclopentane and is favored for stability reasons.

Step-by-step solution

Visualize the given compounds

First, let's get a clear understanding of the three compounds mentioned in the problem: Compound I: A cis-cyclopentane derivative - both the -COOH and -CONH2 groups are on the same side of the cyclopentane ring. Compound II: Created when Compound I is treated with hypobromite (OBr-). Compound III: An amide (lactam) - formed when Compound II is heated.

Analyze the reactions

Now let's examine the reactions in detail: Reaction 1: Formation of Compound II When a cyclopentane derivative (Compound I) is treated with hypobromite (OBr-), a Hofmann rearrangement occurs. In this reaction, the amide (-CONH2) group gets converted to an isocyanate group (-N=C=O) with the help of the hypobromite. This transformation results in the formation of Compound II. Reaction 2: Formation of Compound III When Compound II is heated, the isocyanate group rearranges itself to form a lactam, also known as an intramolecular amide. This rearrangement results in the formation of Compound III (amide III).

Determine the mechanism of rearrangement

We can discern the following clues about the mechanism of these rearrangements: Reaction 1: The treatment of Compound I with hypobromite favors the Hofmann rearrangement due to the relatively favorable formation of an isocyanate. The steric effect caused by the adjacent -COOH group is likely to favor the -CONH2 group's transformation over other possible transformations. Reaction 2: When Compound II is heated, the isocyanate group prefers to rearrange into an intramolecular amide (lactam) due to the strained ring structure of cyclopentane and the need to maximize stability. The intramolecular rearrangement, in this case, leads to greater stability compared to any other possible rearrangement. In conclusion, these results indicate that the mechanism of the rearrangement occurs due to the steric effects and the preference for stable intermediates and products during the reaction. Utilizing models to represent different chemical structures and mechanisms can help us better understand these complex transformations.

Key concepts

Optically Active Compounds

Optically active compounds are those that can rotate plane-polarized light. This property is due to the asymmetric nature of their molecular structure. Such compounds usually contain chiral centers, which are carbon atoms bonded to four different groups. In the context of our problem, optically active a-phenylpropionamide was analyzed before its Hofmann degradation.

Understand that each molecule's chirality is key to its optical activity. This chirality means the molecules are not superimposable on their mirror images, much like our left and right hands.

• Chiral molecules have specific "handedness," which leads to dextrorotatory (right-rotating) and levorotatory (left-rotating) isomers.
• Molecules with non-superimposable mirror images are known as enantiomers.

The study of optically active compounds aids in understanding how their structure affects reactions like the Hofmann rearrangement.

Mechanism of Rearrangement

The Hofmann rearrangement mechanism reveals the transformation of primary amides into isocyanates when treated with a halogen and a base. In our specific cyclopentane example, we observe this change when the amide group in Compound I reacts with hypobromite.

As a result, the amide ( -CONH2 ) group turns into an isocyanate ( -N=C=O ) group, forming Compound II. The mechanism involves several critical steps:

• Base-induced halogenation at the amide nitrogen.
• Rearrangement is facilitated by intramolecular shifts that lead to the formation of an isocyanate intermediate.
• Ultimately, the intermediate is stabilized due to distinctive effects and interactions within the molecular framework.

This rearrangement highlights the importance of steric and electronic factors, which influence reaction pathways in organic molecules.

Stability of Intermediates

The stability of intermediates plays a crucial role in dictating the course and outcome of chemical reactions. In the context of the Hofmann rearrangement, we are particularly concerned with the isocyanate intermediate formed in the rearrangement of cyclopentane derivatives.

Once the isocyanate group is formed after treating Compound I with hypobromite, this intermediate is crucial for the progression into Compound II. A stable intermediate determines whether the rearrangement proceeds smoothly.

• For Compound II to form reliably, the stability of this intermediate is essential.
• Factors such as ring strain, electronic effects, and steric hindrance can significantly affect this stability.

Understanding intermediate stability helps in foretelling the viability and speed of reactions, allowing chemists to control processes more effectively.

Intramolecular Reactions

Intramolecular reactions occur within a single molecule, rather than between separate moleculs . In the Hofmann rearrangement, after Compound II formation, heating leads to an intramolecular process resulting in Compound III - a lactam.

The reaction pathway shows the transformation of the isocyanate group into an intramolecular amide, stabilizing the system due to the cyclopentane’s unique structure. This efficiently produces Compound III or the lactam.

• Intramolecular reactions often provide higher reaction rates and selectivity due to closer proximity of reactive sites.
• For strained rings like cyclopentane, intramolecular processes can release tension, leading to more stable structures.

These reactions demonstrate intricate molecular dynamics and illustrate the importance of ring strain and molecular conformations in organic chemistry.