/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Enediynes are a class of compoun... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 25

Enediynes are a class of compounds that include some antibiotic drugs. Draw the structure of an "enediyne" fragment that contains six carbons in a row.

Short Answer

Expert verified
The enediyene fragment containing six carbons in a row can be represented by the following structure: H H | | H-C=C-C鈮-C鈮 | | H H

Step by step solution

01

Identify the Alkene and Diyne Positions

To create a six-carbon enediyene fragment, we need to first consider the positions of the alkene and diyne bonds in the structure. Since there are only six carbons, we can start by identifying pi-bond positions to create the enediyene fragment. The alkene can be positioned between carbons 1 and 2, and the diyne can be positioned between carbons 3 and 4, and carbons 5 and 6.
02

Illustrate the Bonds

Now that the positions of the alkene and diyne have been identified, the next step is to draw the structure. Start by drawing 6 carbon atoms in a row, and then add the appropriate bonds between these carbons: a single bond between carbons 2 and 3 and a single bond between carbons 4 and 5. Add a double bond between carbons 1 and 2, create two triple bonds between carbons 3 and 4, and between carbons 5 and 6.
03

Add Hydrogen Atoms

After completing the carbon framework, it's essential to add the necessary hydrogen atoms bonded to each carbon, fulfilling the tetravalency of carbon atoms. Add hydrogen atoms to each of the carbon atoms so that each carbon atom has four bonds. There should be two hydrogen atoms on carbon 1, one hydrogen atom on carbon 2, only one hydrogen atom on both carbon 3 and 6, and no hydrogen atoms on carbon 4 and 5. The resulting structure for the enediyene fragment containing six carbons in a row is as follows: H H | | H-C=C-C鈮-C鈮 | | H H

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 91影视!

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

Indicate whether each statement is true or false: (a) Disaccharides are a type of carbohydrate. (b) Sucrose is a monosaccharide. (c) All carbohydrates have the formula \(\mathrm{C}_{\mathrm{a}} \mathrm{H}_{2 \mathrm{~m}} \mathrm{O}_{\mathrm{m}}\).

Locate the chiral carbon atoms, if any, in each molecule: (a) \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CCH}_{2} \mathrm{OH}\) (b) O=C(O)CO (c) CCC(C)C(N)=O

The structural formula for the linear form of galactose is O=CC(O)C(O)C(O)C(O)CO (a) Is this molecule a sugar? (b) How many chiral carbons are present in the molecule? (c) Draw the structure of the sixmember-ring form of this molecule.

Write the condensed structural formula for each of the following compounds: (a) 2-pentanol, (b) 1,2-propanediol, (c) ethyl acetate, (d) diphenyl ketone, (e) methyl ethyl ether.

Name or write the condensed structural formula for the following compounds: (a) trans-2-pentene (b) 2,5-dimethyl-4-octene (c) CCC=CCC(C)CC (d) Brc1ccc(Br)cc1 (c) C#CCC(C)(C)CC
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

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.