Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 34: Problem 764

Cholesterol can be reduced either to cholestanol or coprostanol. Equilibration of cholesterol with a trace of cholestanone and base gives 90 per cent cholestanol and 10 per cent of a stereoisomer known as epicholestanol. Similar equilibration of coprostanol (in the presence of coprostanone) gives 10 per cent coprostanol and 90 per cent epicoprostanol. Write the configurations of each of these compounds and explain the orders of stablities that are observed.

Short Answer

Expert verified
The configurations of cholestanol, epicholestanol, coprostanol, and epicoprostanol are as follows: 1. Cholestanol: \(\alpha\) at C3 and \(\beta\) at C5 2. Epicholestanol: \(\beta\) at C3 and \(\beta\) at C5 3. Coprostanol: \(\alpha\) at C3 and \(\alpha\) at C5 4. Epicoprostanol: \(\beta\) at C3 and \(\alpha\) at C5 The order of stabilities is Cholestanol > Epicoprostanol > Coprostanol > Epicholestanol. This can be attributed to the differences in orientations of the hydroxyl groups, with trans configurations being more stable due to less steric hindrance than cis configurations.

Step by step solution

01

Understanding the compounds

Cholesterol is a sterol, and its structure includes a hydrocarbon chain and a hydroxyl group. To understand the configurations of each of these compounds, we can look at the orientation of the hydroxyl group on C3 and C5 (carbon number 3 and 5) of the steroid ring system. Cholestanol and epicholestanol are stereoisomers of each other, which means they have the same molecular formula but a different spatial arrangement of their atoms. Similarly, coprostanol and epicoprostanol are also stereoisomers. We can denote the orientation of the hydroxyl group using α (alpha) for the upward direction and β (beta) for the downward direction.
02

Writing the configurations of the compounds

Using the given information and α, β notation, we can write the configurations of the compounds as follows: 1. Cholestanol: \(\alpha\) at C3 and \(\beta\) at C5 2. Epicholestanol: \(\beta\) at C3 and \(\beta\) at C5 3. Coprostanol: \(\alpha\) at C3 and \(\alpha\) at C5 4. Epicoprostanol: \(\beta\) at C3 and \(\alpha\) at C5
03

Analyzing the equilibriums and determining order of stabilities

Now we can use the given equilibriums to determine the order of stabilities: Equilibrium 1: Cholesterol + Cholestanone + Base → 90% Cholestanol + 10% Epicholestanol This tells us that Cholestanol is more stable than Epicholestanol. Equilibrium 2: Coprostanol + Coprostanone → 10% Coprostanol + 90% Epicoprostanol This tells us that Epicoprostanol is more stable than Coprostanol. Now, we can rank them in order of stability: 1. Cholestanol 2. Epicoprostanol 3. Coprostanol 4. Epicholestanol
04

Explaining the order of stabilities

The order of stabilities can be explained by the difference in orientations of the hydroxyl groups. In Cholestanol and Epicoprostanol, the hydroxyl groups are in a trans configuration, leading to less steric hindrance and a more stable compound. On the other hand, Coprostanol and Epicholestanol have hydroxyl groups in a cis configuration, which creates more steric hindrance and results in less stable compounds.

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.

Start your free trial

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

If at this point the student is still unconvinced as to the importance of organic chemistry in daily life, let one consider glycolysis, the process whereby glucose is broken down into \(\mathrm{CO}_{2}, \mathrm{ATP}\), and \(\mathrm{H}_{2} \mathrm{O}\) by the same nebulous mechanisms that the beginning organic student has been trying to master for two semesters. (a) Consider the initial step of the Krebs Cycle; the reaction of oxaloacetate with acetyl CoA, in the presence of base and water to yield citric acid and CoASH. What would be the mechanism of this reaction if it occurs as written below: (b) Vitamin \(\mathrm{B}_{1}\) (Thiamine) also plays an important role in the Krebs cycle, catalyzing several important glycolysis reactions. One reaction it catalyzes in particular is the decarboxylation of \(\alpha\) -ketoglutaric acid to succinic acid. Given the following reactions, propose a mechanism for the decarboxylation of \(\alpha\) -ketoglutaric acid to succinic acid.

What are the products, and their relative ratios, resulting from the ozonolysis of \(\beta\) -carotene? (Assume reductive workup.)

The blood of lobsters contains a blue copper porphyrin complex. As a model for the lobster blood pigment, draw the structure of the \(\mathrm{Cu}^{2+}\) pigment of octamethylporphin. What charge, if any, does this complex have?

Tropinic acid, \(\mathrm{C}_{8} \mathrm{H}_{13} \mathrm{O}_{4} \mathrm{~N}\), is a degradation product of atropine, an alkaloid of the deadly nightshade, Atropa belladonna. It has a neutralization equivalent of \(94 \pm 1 .\) It does not react with benzenesulfonyl chloride, cold dilute \(\mathrm{KMnO}_{4}\), or \(\mathrm{Br}_{2} / \mathrm{CCl}_{4} .\) Exhaustive methylation gives the following results: tropinic acid \(+\mathrm{CH}_{3} \mathrm{I} \rightarrow \mathrm{EEE}\left(\mathrm{C}_{9} \mathrm{H}_{16} \mathrm{O}_{4} \mathrm{NI}\right)\) \(\mathrm{EEE}+\mathrm{Ag}_{2} \mathrm{O}\), then strong heat \(\rightarrow \mathrm{FFF}\left(\mathrm{C}_{9} \mathrm{H}_{15} \mathrm{O}_{4} \mathrm{~N}\right)+\mathrm{H}_{2} \mathrm{O}\) \(+\mathrm{AgIFFF}+\mathrm{CH}_{3} \mathrm{I} \rightarrow \mathrm{GGG}\left(\mathrm{C}_{10} \mathrm{H}_{18} \mathrm{O}_{4} \mathrm{NI}\right)\) \(\mathrm{GGG}+\operatorname{Ag} \mathrm{O}\), then strong heat \(\rightarrow \mathrm{HHH}\left(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}_{4}\right)+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) \(+\mathrm{H}_{2} \mathrm{O}+\mathrm{AgI}\) \(\mathrm{HHH}+\mathrm{H}_{2}, \mathrm{Ni} \rightarrow\) heptanedioic acid (pimelic acid) (a) What structures are likely for tropinic acid? (b) Tropinic acid is formed by oxidation with \(\mathrm{CrO}_{3}\) of tropinone, whose structure has been shown by synthesis to be Now what is the most likely structure for tropinic acid?

Citral, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}\), is a terpene that is the major constituent of lemongrass oil. It reacts with hydroxylamine to yield a compound of formula \(\mathrm{C}_{10} \mathrm{H}_{17} \mathrm{ON}\), and with Tollens' reagent to give a silver mirror and a compound of formula \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}_{2}\). Upon vigorous oxidation citral yields acetone, oxalic acid \((\mathrm{HOOC}-\mathrm{COOH})\), and levulinic acid \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right)\). (a) Propose a structure for citral that is consistent with these facts and with the isoprene rule. (b) Citral has two isomers, geranial and neral, which yield the same oxidation products. What is the most likely structural difference between these two isomers? (c) Geranial is obtained by mild oxidation of geraniol; neral is obtained in a similar way from nerol. On this basis assign structures to geranial and neral.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation
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