Both 1,2 -diols and 1,3 -diols can be protected by treatment with 2-methoxypropene according to the following reaction. (a) Propose a mechanism for the formation of this protected diol. (b) Suggest an experimental procedure by which this protecting group can be removed to regenerate the unprotected diol.

Short Answer

Expert verified
#Question# Propose a mechanism for the formation of the protected diol by treatment with 2-methoxypropene and suggest a procedure to remove the protecting group and regenerate the unprotected diol.

Step by step solution

01

Part (a): Mechanism of diol protection

1. Formation of carbocation In the first step, 2-methoxypropene undergoes the ionization process to generate a tertiary carbocation and hydroxide ion. This process occurs due to the presence of a strong acid catalyst in the reaction mixture. 2-methoxypropene ->[\(H^+\)][-OH] Tertiary carbocation + OH^- 2. Nucleophilic attack After the tertiary carbocation is formed, it can act as a great electrophile. The diol (1,2-diol or 1,3-diol) has two hydroxyl groups which can act as nucleophiles. The nucleophilic hydroxyl will attack the carbocation, forming a bond with the electrophilic carbon center. This bond formation will result in a protected diol molecule with an acetal group. Tertiary carbocation + Hydroxyl group of diol -> Acetal-protected diol
02

Part (b): Removing the protecting group

To remove the protecting group and regenerate the unprotected diol, we need to carry out an experimental procedure that reverses the steps mentioned in part (a). This can be done using hydrolysis under acidic conditions. 1. Add an aqueous acid The first step is to add an aqueous acid, like hydrochloric acid (HCl), to the reaction mixture containing the protected diol. The acidic conditions will help promote hydrolysis and cleaving of the acetal protecting group. 2. Heat the mixture Heat the reaction mixture to facilitate hydrolysis and promote the acid-catalyzed cleavage of the protecting group. This will generate the carbocation and the regenerated diol with two free hydroxyl groups. 3. Isolate the diol The last step is to isolate the unprotected diol from the reaction mixture. This can be achieved by using different techniques like extraction or distillation, depending on the specific properties of the diol. The overall process can be summarized as: Acetal-protected diol->[HCl][Heat] Unprotected diol + 2-methoxypropene

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

In Section \(11.5\) we saw that ethers, such as diethyl ether and tetrahydrofuran, are quite resistant to the action of dilute acids and require hot concentrated HI or HBr for cleavage. However, acetals in which two ether groups are linked to the same carbon undergo hydrolysis readily, even in dilute aqueous acid. How do you account for this marked difference in chemical reactivity toward dilute aqueous acid between ethers and acetals?

Treatment of \(\beta\)-D-glucose with methanol in the presence of an acid catalyst converts it into a mixture of two compounds called methyl glucosides (Section 25.3A). In these representations, the six-membered rings are drawn as planar hexagons. (a) Propose a mechanism for this conversion, and account for the fact that only the - \(\mathrm{OH}\) on carbon 1 is transformed into an \(-\mathrm{OCH}_{3}\) group. (b) Draw the more stable chair conformation for each product. (c) Which of the two products has the chair conformation of greater stability? Explain.

When cis-2-decalone is dissolved in ether containing a trace of \(\mathrm{HCl}\), an equilibrium is established with trans-2-decalone. The latter ketone predominates in the equilibrium mixture.

Using your roadmaps as a guide, show how to convert (2-bromoethyl)benzene into 2-chloro-1-phenylethanone. Show all reagents and all molecules synthesized along the way.

Starting with acetylene and 1-bromobutane as the only sources of carbon atoms, show how to synthesize the following. (a) meso-5,6-Decanediol (b) racemic 5,6 -Decanediol (c) 5-Decanone (d) 5,6 -Epoxydecane (e) 5-Decanol (f) Decane (g) 6-Methyl-5-decanol (h) 6-Methyl-5-decanone

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