Question

Marked similarities exist between the mechanism of nitrous acid deamination of \(\beta\)-aminoalcohols and the pinacol rearrangement. Following are examples of each. (a) Analyze the mechanism of each rearrangement and list their similarities. (b) Why does the first reaction, but not the second, give ring expansion? (c) Suggest a \(\beta\)-aminoalcohol that would give cyclohexanecarbaldehyde as a product.

Short answer

Also, suggest a β-aminoalcohol that would yield cyclohexanecarbaldehyde as a product. Answer: Both nitrous acid deamination of β-aminoalcohols and pinacol rearrangement mechanisms involve the formation of carbonyl groups, are intramolecular reactions, and involve the migration of groups within the molecule. Ring expansion occurs in the deamination reaction due to the formation of a carbocation that can rearrange to form a more stable carbocation whereas, in the pinacol rearrangement, ring expansion does not occur as the migration of the hydride/methyl group is accompanied by simultaneous creation of a carbonyl group without involving a carbocation that would allow ring expansion. To obtain cyclohexanecarbaldehyde as a product, the starting β-aminoalcohol should be 1-amino-2-cyclopentanol.

Step-by-step solution

Understand the Two Mechanisms

The deamination of β-aminoalcohols using nitrous acid is a process where the amine group (NH\(_2\)) is removed from the molecule and the carbon that bears the amine group becomes a carbonyl group (C=O). The pinacol rearrangement is a reaction that shifts the position of the carbonyl group within the molecule, typically accompanied by the migration of a methyl or hydride (H) group.

Analyze the Mechanism of Each

Let's discuss each mechanism separately: (a) Deamination of β-aminoalcohol: 1. β-Aminoalcohol reacts with nitrous acid to form a diazonium intermediate. 2. Loss of molecular nitrogen (\(N_2\)) from the diazonium intermediate results in the formation of a carbocation. 3. A neighboring hydroxyl group (OH) donates its electrons to the carbocation to form a carbonyl group (C=O). Pinacol rearrangement: 1. A proton acid catalyst protonates the carbonyl oxygen in the pinacol molecule. 2. A neighboring hydride or methyl group migrates to the carbonyl carbon, with simultaneous loss of a water molecule (H\(_2\)O) from the molecule. 3. Final deprotonation leads to formation of the rearranged carbonyl compound.

Identify Similarities

Both reactions share the following similarities: 1. Both mechanisms involve the formation of carbonyl groups. 2. Both rearrangements are intramolecular reactions, meaning the reaction takes place within a single molecule. 3. Both reactions involve the migration of groups within the molecule (i.e., the formation of a carbocation and the migration of the hydride/methyl group).

Explain Ring Expansion in Deamination

(b) The first reaction (deamination of β-aminoalcohols) involves the loss of an amine group and the formation of a carbocation, which can rearrange to form a more stable carbocation. This rearrangement can result in ring expansion if the newly formed carbocation is more stable than the original one. The ring expansion does not occur in pinacol rearrangement because the migration of the hydride/methyl group is accompanied by simultaneous creation of a carbonyl group and does not involve the formation of a carbocation that would allow ring expansion.

Suggest a β-Aminoalcohol for the Given Product

(c) To obtain cyclohexanecarbaldehyde as a product from the deamination reaction, we need a β-aminoalcohol that has a five-membered ring and an amine group on the β-carbon. Thus, the starting β-aminoalcohol should be 1-amino-2-cyclopentanol.