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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 six-carbon enediyne fragment is drawn as follows: \[ H_{2}C = CH - C \equiv C - CH_{2} - CH_{3} \] This structure contains an alternating double bond (ene) and triple bond (diyne).

Step by step solution

01

Draw the carbon chain

Start by drawing a straight chain of six carbon atoms in a row: \[ \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} \]
02

Add the ene (double bond)

Now add a double bond between the first two carbon atoms: \[ \underset{C}{\bullet} = \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} \]
03

Add the diyne (triple bond)

Add a triple bond between the third and fourth carbon atoms, creating the characteristic enediyne structure: \[ \underset{C}{\bullet} = \underset{C}{\bullet} - \underset{C}{\bullet} \equiv \underset{C}{\bullet} - \underset{C}{\bullet} - \underset{C}{\bullet} \]
04

Complete the structure with hydrogen atoms

Finally, add hydrogen atoms to the carbon atoms in order to achieve the appropriate number of bonds. Each carbon atom should have four bonds. \[ H_{2}C = CH - C \equiv C - CH_{2} - CH_{3} \] That's it! We have drawn a six-carbon enediyne fragment containing an alternating double bond and triple bond.

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Most popular questions from this chapter

Give the molecular formula of a hydrocarbon containing five carbon atoms that is (a) an alkane, \((\mathbf{b})\) a cycloalkane, (c) an alkene,(d) an alkyne.

(a) Draw the condensed structural formulas for the cis and trans isomers of 3-methyl-3-hexene. (b) Can cyclohexene exhibit cis-trans isomerism? Explain. (c) Does 2 -hexyne have geometric isomers? Explain.

Name or write the condensed structural formula for the following compounds: (a) cis-2-pentene (b) 1,6-dichloro-3-hexyne (c) \(\mathrm{BrC} \equiv \mathrm{CBr}\) (d) 1,3 -Benzenediol (e) $\mathrm{CH}_{2}=\mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}=\mathrm{CH}_{2}$

(a) Are \(\alpha\) -glucose and \(\beta\) -glucose enantiomers? (b) Show the condensation of two glucose molecules to form a disaccharide with an \(\alpha\) linkage. (c) Repeat part (b) but with a \(\beta\) linkage.

Give the name or condensed structural formula, as appropriate: (a) 3-phenylpentane (b) 2,3-dimethylhexane (c) 3,3-dimethyloctane (d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ e.
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