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Chapter 9: Problem 16

Write mechanisms for the following two reactions. (a) 1,4 -Butanediol \(+\mathrm{H}^{+} \rightarrow\) oxacyclopentane (tetrahydrofuran); (b) 5-methyl-1,5-hexanediol \(+\mathrm{H}^{+} \rightarrow 2,2\) -dimethyloxacyclohexane \((2,2-\) dimethyltetrahydropyran).

Short Answer

Expert verified
1,4-Butanediol forms tetrahydrofuran via intramolecular nucleophilic substitution; 5-methyl-1,5-hexanediol forms 2,2-dimethyltetrahydropyran via similar mechanism.

Step by step solution

01

Protonation of the Hydroxyl Group

In acidic medium, the hydroxyl group of 1,4-Butanediol is protonated by \(\text{H}^+\). This forms a better leaving group (water).
02

Formation of an Oxonium Ion

After protonation, the molecule forms an oxonium ion, \(\text{R-OH}_2^+\).
03

Intramolecular Nucleophilic Substitution

The nucleophile (other alcohol group) attacks the carbocation formed after water leaves leading to the formation of a five-membered ring, oxacyclopentane (tetrahydrofuran).
04

Protonation of the Hydroxyl Group for 5-Methyl-1,5-Hexanediol

Similarly, in acidic medium, the hydroxyl group of 5-methyl-1,5-hexanediol is protonated to form \(\text{R-OH}_2^+\).
05

Carbocation Formation and Nucleophilic Attack

After the water leaves, a carbocation is formed, and the other hydroxyl group attacks it. This forms a new six-membered ring, 2,2-dimethyltetrahydropyran.
06

Proton Loss to Form Stable Ring

Finally, the proton is lost from the oxonium ion to form the stable products: 1,4-butanediol into oxacyclopentane (tetrahydrofuran) and 5-methyl-1,5-hexanediol into 2,2-dimethyltetrahydropyran.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Protonation
In an acidic environment, the first step in these reactions involves protonation. Here, the \( \text{H}^{+} \) protonates the hydroxyl group (OH) of the molecule. This makes the hydroxyl group more reactive and converts it to a better leaving group. For instance, in 1,4-Butanediol, the hydroxyl group is protonated to form water, \( \text{H}_{2}\text{O} \), which is a superior leaving group compared to \( \text{OH}^{-} \). By protonating the alcohols, this step sets the stage for the subsequent transformations. Protonation is crucial because it facilitates later steps in the reaction by increasing the electrophilicity of the molecule.
Oxonium Ion Formation
Following protonation, an oxonium ion forms. The oxonium ion, denoted as \( \text{R-OH}_{2}^{+} \), is highly reactive. It results when the protonated hydroxyl group (OH) now exists with a positively charged oxygen. In our example, 1,4-Butanediol after protonation results in an oxonium ion, which makes the carbon to which the oxonium ion is attached more electron-deficient. This deficiency is necessary for the subsequent steps, where nucleophilic attack and other reactions will take place.
Nucleophilic Substitution
Once the oxonium ion is formed, nucleophilic substitution occurs. In the case of 1,4-Butanediol, one of the alcohol groups on the molecule attacks the carbocation center after the water (a great leaving group now) leaves. This intramolecular reaction results in the formation of a cyclic structure. For example, in 1,4-Butanediol, the nucleophilic OH attacks the carbocation generated from the previous leaving water, forming oxacyclopentane (tetrahydrofuran).
Carbocation Intermediate
Carbocation intermediates are pivotal in our reactions. After the formation of the oxonium ion and the departure of the water molecule, a positively charged carbocation is left. This carbocation is very reactive and can be attacked by nearby nucleophiles. In 5-methyl-1,5-Hexanediol, for example, after water leaves, a carbocation is formed. This positively charged carbon is then attacked by the second hydroxyl group leading to ring formation. The carbocation intermediate is stabilized by the groups attached to it, and its reactivity guides the nucleophilic attack.
Ring Formation
Finally, ring formation in these reactions represents the culmination of the previous steps. In protonated 1,4-Butanediol, after the carbocation is attacked by the remaining nucleophile, a five-membered ring, oxacyclopentane (tetrahydrofuran) forms. Similarly, in 5-methyl-1,5-Hexanediol, after the nucleophilic attack on the carbocation, a six-membered ring, 2,2-dimethyltetrahydropyran, forms. These ring structures are stable and complete our mechanism. Each step from protonation to ring formation is essential to successfully forming these cyclic ethers.

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Most popular questions from this chapter

Write the structure of the products that you expect from the reaction of 1 -methylcyclohexanol with (a) concentrated \(\mathrm{HCl}\) and (b) concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Compare and contrast the mechanisms of the two processes. (Hint: Compare the relative nucleophilicity of \(\mathrm{Cl}^{-}\) and \(\mathrm{HSO}_{4}^{-}\). Caution: When writing mechanisms, use "arrow pushing" to depict electron flow; write out every step separately; formulate complete structures, including charges and relevant electron pairs; and draw explicit reaction arrows to connect starting materials or intermediates with their respective products. Don't use shortcuts, and don't be sloppy!)

Sugars, being polyhydroxylic compounds (Chapter 24), undergo reactions characteristic of alcohols. In one of the later steps in glycolysis (the metabolism of glucose), one of the glucose metabolites with a remaining hydroxy group, 2 -phosphoglyceric acid, is converted into 2-phosphoenolpyruvic acid. This reaction is catalyzed by the enzyme enolase in the presence of a Lewis acid such as \(\mathrm{Mg}^{2+}\). (a) How would you classify this reaction? (b) What is the possible role of the Lewis acidic metal ion?

In an early study of the stereochemistry of nucleophilic displacements, optically pure \((R)-1\) -deuterio-1-pentanol was treated with 4 -methylphenylsulfonyl (tosyl) chloride to make the corresponding tosylate. The tosylate was then treated with excess ammonia to convert it to I-deuterio-1-pentanamine: (a) Describe the stereochemistry that you expect to observe at \(\mathrm{C} 1\) of both the intermediate tosylate and the final amine. (b) When the reaction sequence is actually carried out, the expected results are not obtained. Instead, the final amine is isolated as a \(70: 30\) mixture of \((S)-\) and \((R)-1\) -deuterio- 1 -pentanamine. Suggest a mechanistic explanation. (Hint: Recall that reaction of an alcohol with a sulfonyl chloride displaces chloride ion, which is a nucleophile.)

Give detailed mechanisms and final products for the reaction of 3-methyl-2-pentanol with each of the reagents that follow. (a) \(\mathrm{NaH}\) (b) Concentrated \(\mathrm{HBr}\) (c) \(\mathrm{PBr}_{3}\) (d) \(\mathrm{SOCl}_{2}\) (e) Concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) at \(130^{\circ} \mathrm{C}\) (f) Dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\)

At \(100^{\circ} \mathrm{C}, 3,3\) -dimethyl-2-butanol gives three products of \(\mathrm{E} 1\) reaction: one derived from the carbocation present prior to rearrangement, the other two from that formed after an alkyl shift has taken place. Give the structures of these elimination products.
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