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Chapter 34: Problem 761

\(\beta\) -Ionone is an important chemical in the synthesis of vitamin A. It is prepared by the reaction of pseudo-ionone with sulfuric acid, \(\mathrm{d}, \ell-\alpha\) -ionone being produced in the same reaction. Suggest a mechanism for the acid-catalyzed conversion of pseudo-ionone to \(\alpha\) - and \(\beta\) -ionone. Why might you anticipate that the cyclization produces more \(\beta\) - than \(\alpha\) -ionone?

Short Answer

Expert verified
The acid-catalyzed conversion of pseudo-ionone to α- and β-ionone involves protonation of the carbonyl oxygen by sulfuric acid, followed by C=C bond formation through a cyclization reaction, and ultimately deprotonation to yield α- and β-ionone. The preference for β-ionone formation over α-ionone is attributed to sterics and stability of intermediates: the transition state leading to β-ionone is more stable and lower in energy than the transition state leading to α-ionone, and the overall ring strain and destabilizing interactions in the α-ionone product are higher.

Step by step solution

01

Understand the Structure of Pseudo-Ionone, α-Ionone, and β-Ionone

Before proposing a mechanism for the conversion of pseudo-ionone to α- and β-ionone, we need to have a clear understanding of the structure of these molecules. Please sketch the structures of pseudo-ionone, α-ionone, and β-ionone or use molecular modeling software to visualize them.
02

Identify Potential Acid-Base Reaction Sites

Since the reaction is acid-catalyzed, it will likely involve a proton transfer step. Examine the pseudo-ionone molecule and identify potential sites where protonation might create a stable intermediate that can undergo further transformation. The carbonyl oxygen is the most likely site for protonation, due to its high electronegativity and lone pair of electrons.
03

Propose the Protonation Step

After identifying the carbonyl oxygen as the most likely protonation site, the next step is to propose the protonation of the carbonyl group by sulfuric acid, which is a strong acid. This will result in the formation of a carbonyl oxygen bearing a positive charge. Draw the resonance structures to better understand the effect of protonation on the pseudo-ionone molecule.
04

C=C Bond Formation

The protonation of the carbonyl group will lead to increased electron deficiency, which can be stabilized by the formation of a new C=C bond, along with the elimination of a water molecule. This step involves a cyclization reaction in which the double bond in the side chain attacks the carbonyl carbon. Carefully choose the most likely carbohydrate for the double bond to attack and provide a sketch of the product after the elimination of the water molecule.
05

Deprotonation to Yield α- and β-Ionone

After the C=C bond formation, a new C=C bond will be formed in the ring, and the carbonyl will be restored – leading to α-ionone and/or β-ionone. Deprotonation can happen at two locations, leading to a mixture of two products. Sketch the deprotonation process and show the formation of both α- and β-ionone.
06

Explain the Preference for β-Ionone Formation Over α-Ionone

To explain why the reaction favors the formation of β-ionone over α-ionone, consider factors such as sterics and stability of intermediates. The steric factors, such as overall ring strain and destabilizing interactions, will contribute to the preference for β-ionone formation. Additionally, the transition state leading to β-ionone is more stable and lower in energy than the transition state leading to α-ionone. A brief explanation of these factors should suffice.

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