When warmed in dilute sulfuric acid, 1-phenyl-1,2-propanediol undergoes dehydration and rearrangement to give 2-phenylpropanal. (a) Propose a mechanism for this example of a pinacol rearrangement (Section 10.7). (b) Account for the fact that 2-phenylpropanal is formed rather than its constitutional isomer, 1-phenyl-1-propanone.

Short Answer

Expert verified
Answer: The pinacol rearrangement mechanism involves three main steps: 1) Protonation of the 1,2-diol, 2) Rearrangement of the cation intermediate, and 3) Deprotonation of the rearranged cation to form the carbonyl compound. In the case of 1-phenyl-1,2-propanediol, the phenyl group migration during the rearrangement step leads to a more stable carbocation due to resonance with the phenyl ring. Thus, the formation of 2-phenylpropanal is favored over 1-phenyl-1-propanone.

Step by step solution

01

Understand the Pinacol Rearrangement Mechanism

The pinacol rearrangement is an organic reaction that involves the acid-catalyzed rearrangement of a 1,2-diol (such as 1-phenyl-1,2-propanediol) to give a carbonyl compound (in this case, 2-phenylpropanal). The general mechanism involves three main steps: 1. Protonation of the 1,2-diol by the acidic conditions 2. Rearrangement of the cation intermediate 3. Deprotonation of the rearranged cation to give the final carbonyl compound
02

Apply the Pinacol Rearrangement Mechanism to This Example

Now let's apply this general mechanism to the specific case of 1-phenyl-1,2-propanediol: 1. Protonation: The oxygen atom on the 1-phenyl-1,2-propanediol molecule gets protonated by the available sulfuric acid, forming an oxonium ion. 2. Rearrangement: A 1,2-migration occurs where the bond between the carbon and the substituent (phenyl group) migrates to the adjacent carbocation carbon, creating a new positive charge on the phenyl-substituted carbon. 3. Deprotonation: Finally, a water molecule deprotonates the protonated carbonyl carbon, resulting in the formation of the final pinacol rearrangement product, 2-phenylpropanal.
03

Explain the Formation of 2-Phenylpropanal over 1-Phenyl-1-Propanone

During the rearrangement step of the mechanism, the migration of the phenyl group leads to the formation of a new carbocation, which is stabilized by resonance with the phenyl ring. This rearrangement is preferred because the resulting carbocation is more stable compared to the alternative migration of the methyl group. If the methyl group had migrated, the carbocation would not have the stabilization effect provided by resonance with the phenyl ring, and therefore, the resulting carbocation would be less stable. Thus, based on the stability of the intermediate carbocations, the formation of 2-phenylpropanal is favored over the formation of its constitutional isomer, 1-phenyl-1-propanone.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Following is a synthesis for toremifene, a nonsteroidal estrogen antagonist whose structure is closely related to that of tamoxifen. (a) This synthesis makes use of two blocking groups, the benzyl (Bn) group and the tetrahydropyranyl (THP) group. Draw a structural formula of each group, and describe the experimental conditions under which it is attached and removed. (b) Discuss the chemical logic behind the use of each blocking group in this synthesis. (c) Propose a mechanism for the conversion of D to \(E\). (d) Propose a mechanism for the conversion of \(F\) to toremifene. (e) Is toremifene chiral? If so, which of the possible stereoisomers are formed in this synthesis?

Account for the fact that water-insoluble carboxylic acids ( \(\mathrm{p} K_{\mathrm{a}} 4-5\) ) dissolve in \(10 \%\) aqueous sodium bicarbonate \((\mathrm{pH} 8.5)\) with the evolution of a gas but that waterinsoluble phenols ( \(\mathrm{p} K_{\mathrm{a}} 9.5-10.5\) ) do not dissolve in \(10 \%\) sodium bicarbonate.

Compound \(\mathrm{A}\left(\mathrm{C}_{9} \mathrm{H}_{12}\right)\) shows prominent peaks in its mass spectrum at \(\mathrm{m} / \mathrm{z} 120\) and 105. Compound \(\mathrm{B}\) (also \(\mathrm{C}_{9} \mathrm{H}_{12}\) ) shows prominent peaks at \(m / z 120\) and 91 . On vigorous oxidation with chromic acid, both compounds give benzoic acid. From this information, deduce the structural formulas of compounds A and B.

In certain clinical situations, there is need for an injectable \(\beta\)-blocker with a short biological half-life. The clue to development of such a drug was taken from the structure of atenolol, whose corresponding carboxylic acid (the product of hydrolysis of its amide) has no \(\beta\)-blocking activity. Substitution of an ester for the amide group and lengthening the carbon side chain by one methylene group resulted in esmolol. Its ester group is hydrolyzed quite rapidly to a carboxyl group by serum esterases under physiological conditions. This hydrolysis product has no \(\beta\)-blocking activity. Propose a synthesis for esmolol from 4-hydroxycinnamic acid, epichlorohydrin, and isopropylamine. (a) Propose a synthesis for esmolol from 4-hydroxycinnamic acid, epichlorohydrin, and isopropylamine. (b) Is esmolol chiral? If so, which of the possible stereoisomers are formed in this synthesis?

Arrange these compounds in order of increasing acidity: 2,4-dichlorophenol, phenol, cyclohexanol.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free