/*! 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 26 Write structural formulas for ea... [FREE SOLUTION] | 91影视

91影视

Write structural formulas for each of the following: (a) 3,4 -Octadiene (b) \((E, E)-3,5\) -Octadiene (c) \((Z, Z)-1,3\) -Cyclooctadiene (d) \((Z, Z)-1,4\) -Cyclooctadiene (e) \((E, E)-1,5\) -Cyclooctadiene (f) \((2 E, 4 Z, 6 E)-2,4,6\) -Octatriene (g) 5-Allyl-1,3-cyclopentadiene (h) trans-1,2-Divinylcyclopropane (1) 2,4 -Dimethyl \(-1,3\) -pentadiene

Short Answer

Expert verified
Structural formulas vary for each compound based on carbon placement, geometry, and special groups. Always place double bonds correctly according to number and geometry rules.

Step by step solution

01

Understanding Structural Formulas

Structural formulas provide a graphical representation of molecules, showing how atoms are arranged and bonded together. For unsaturated hydrocarbons, such as dienes and trienes, pay special attention to the position and geometry (E/Z or trans/cis) of double bonds.
02

Drawing 3,4-Octadiene

The molecular formula for octadiene is C鈧圚鈧佲倓. In 3,4-octadiene, double bonds start at the third and fourth carbon in the chain. The structural formula is CH鈧-CH鈧-CH=CH-CH=CH-CH鈧-CH鈧. Ensure that these bonds are depicted between the correct carbon atoms without specifying stereochemistry.
03

Drawing (E,E)-3,5-Octadiene

The (E,E) notation refers to the trans configuration of double bonds. For (E,E)-3,5-octadiene, draw the molecule with trans double bonds between carbon 3 and 4, as well as carbon 5 and 6 to ensure the largest groups attached are on opposite sides: CH鈧-CH鈧-CH=CH-CH=CH-CH鈧-CH鈧 with E-geometry for each.
04

Drawing (Z,Z)-1,3-Cyclooctadiene

This is an 8-membered ring with double bonds at positions 1 and 3. The (Z,Z) configuration means both double bonds are in cis (same side) form: Draw a cyclo-octane structure and insert cis double bonds converting relevant single bonds into double bonds at C1 and C3.
05

Drawing (Z,Z)-1,4-Cyclooctadiene

Similar to the previous step, but with double bonds at positions 1 and 4 with z-configuration. Keep both as cis: Draw a cyclo-octane and add cis double bonds between C1 and C2, and C4 and C5.
06

Drawing (E,E)-1,5-Cyclooctadiene

In contrast, this has double bonds at positions 1 and 5. This creates a trans geometry making it planar between those double bonds: Structure it to have anti orientation ensuring those carbons in double bonds 1 and 5 have the biggest groups on opposite sides.
07

Drawing (2E,4Z,6E)-2,4,6-Octatriene

It involves both diene and triene characteristics with mixed geometries - E at positions 2 and 6, and Z at position 4. CH鈧=CH-CH=CH-CH=CH-CH鈧-CH鈧 with those geometries considered.
08

Drawing 5-Allyl-1,3-Cyclopentadiene

Here, an allyl group (CH鈧=CH-CH鈧-) is attached to a cyclopentadiene, which has double bonds at C1 and C3. Position the allyl group such that its methylene links it to the fifth carbon.
09

Drawing trans-1,2-Divinylcyclopropane

Trans-1,2-divinylcyclopropane features a cyclopropane ring with vinyl (CH=CH鈧) groups at C1 and C2, oriented trans to each other. Depict the small ring and show the vinyl groups extending outwards in opposite directions.
10

Drawing 2,4-Dimethyl-1,3-pentadiene

This diene consists of a five-carbon chain with methyl groups at C2 and C4, and double bonds at C1 and C3. Depict it as CH鈧-C(CH鈧)=CH-CH(CH鈧)=CH鈧 keeping in mind the diene framework once substituents are added on designated carbons.

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!

Key Concepts

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

Unsaturated Hydrocarbons
Unsaturated hydrocarbons are organic compounds characterized by the presence of double or triple bonds between carbon atoms. These bonds introduce kinks in the molecular structure affecting the compound's properties. The simplest unsaturated hydrocarbons are alkenes, with one or more double bonds (dienes, trienes, etc.), and alkynes with one or more triple bonds. The presence of multiple bonds influences the reactivity of these compounds. For example:
  • Dienes contain two double bonds.
  • Trienes have three double bonds.
  • Due to these bonds, they do not have as many hydrogen atoms as their saturated counterparts.
The double and triple bonds in these molecules can also provide sites for reactions such as polymerization and hydrogenation. Unsaturated hydrocarbons can be identified by their distinctive physical and chemical behaviors, such as their ability to react with halogens.
E/Z Notation
E/Z notation is used to describe the stereochemistry of double-bonded atoms within unsaturated hydrocarbons. This notation helps clarify the spatial arrangement of substituent groups around each double bond based on the Cahn-Ingold-Prelog priority rules. The letters stand for the German words 'Entgegen' meaning opposite and 'Zusammen' meaning together.
  • If the highest priority groups on each carbon atom of the double bond are on opposite sides, the bond has an E (Entgegen) configuration.
  • If they are on the same side, the molecule has a Z (Zusammen) configuration.
The E/Z notation is particularly useful in differentiating isomers, which can significantly differ in their properties. For example, (E,E)-3,5-octadiene means both double bonds are in the E configuration.
Organic Chemistry
Organic chemistry is the branch of science that studies the structure, properties, compositions, reactions, and synthesis of carbon-containing compounds, which include both hydrocarbons and their derivatives. Its principles and reactions are vital for developing a vast array of materials used in daily life, such as pharmaceuticals, plastics, and fuels.
This field is fundamentally dependent on understanding how carbon atoms bond together to form chains and rings, leading to different structures and functions of molecules. Here are a couple of critical points:
  • Carbon's ability to form four bonds allows for the construction of complex molecules with various functional groups.
  • The presence of unsaturated bonds (like double and triple bonds) adds layers of complexity to these molecules, as they influence molecule shape and reactivity.
  • Understanding structural formulas, as seen in unsaturated hydrocarbons, is an essential skill for organic chemistry students.
Diene and Triene Structures
Dienes and trienes are unsaturated hydrocarbons containing two and three double bonds, respectively. Knowing how to represent these structures with structural formulas accurately is crucial for understanding their chemistry.
Here are some essential aspects to consider for diene and triene structures:
  • Position of double bonds: The locations of these multiple bonds influence the molecule's shape and stability.
  • Steric effects: The spatial arrangement of atoms can affect how these compounds interact with other molecules, impacting their reactivity and properties.
  • Stereochemistry: As seen in the exercises, locating the appropriate E/Z or cis/trans configurations helps depict accurate molecular geometry.
Correctly interpreting these structural formulas opens possibilities for predicting molecule reactions, understanding physical properties, and synthesizing new compounds in organic chemistry.

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

Another way in which energies of isomers may be compared is by their heats of combustion. Match the heat of combustion with the appropriate diene. Dienes: \(\quad 1,2-\) Pentadiene, \((E)-1,3\) -pentadiene, 1,4 -pentadiene Heats of combustion: \(3186 \mathrm{~kJ} / \mathrm{mol}(761 \mathrm{kcal} / \mathrm{mol}), 3217 \mathrm{~kJ} / \mathrm{mol}(769 \mathrm{kcal} / \mathrm{mol}), 3251 \mathrm{~kJ} / \mathrm{mol}(776 \mathrm{kcal} / \mathrm{mol})\)

What is the 1,3 -dipolar cycloaddition product formed in the reaction between diazomethane and maleic anhydride? (0)

(a) Describe the molecular geometry expected for \(1,2,3\) -butatriene \(\left(\mathrm{H}_{2} \mathrm{C} \square \mathrm{C} \square \mathrm{C} \square \mathrm{CH}_{2}\right)\). (b) Two stereoisomers are expected for \(2,3,4\) -hexatriene \(\left(\mathrm{CH}_{3} \mathrm{CH} \square \mathrm{C} \square \mathrm{C} \square \mathrm{CHCH}_{3}\right)\). What should be the relationship between these two stereoisomers?

Bromination of 1,5 -cyclooctadiene with \(N\) -bromosuccinimide (NBS) gives a mixture of two constitutional isomers of \(\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{Br}\). Suggest reasonable structures for these two isomers

In the second step of the mechanism of ozonolysis of alkenes, the ozonide described earlier undergoes a fragmentationrecombination process to give a \(1,2.4\) -trioxolane. The fragmentation step is the reverse of a 1,3 -dipolar cycloaddition and yields a carbonyl compound and a 1,3 -dipolar compound \(\mathrm{X}\). The carbonyl compound and \(\mathrm{X}\) then react to give the \(1,2,4\) -trioxolane by 1,3 -dipolar cycloaddition. What is the structure of \(\mathrm{X}\) ?

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.