Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 26: Problem 2

Draw a structural formula for each of the following compounds. (a) 3 -isopropyloctane; (b) 2 -chloro- 3 -methylpentane; (c) 2 -pentene; (d) dipropyl ether.

Short Answer

Expert verified
The drawn structures of the compounds are: (a) 3 -isopropyloctane: An octane chain with an isopropyl group attached to the third carbon. (b) 2 -chloro- 3 -methylpentane: A pentane chain with a chlorine atom attached to the second carbon and a methyl group attached to the third carbon. (c) 2 -pentene: A pentene chain with a double bond between the second and third carbons. (d) Dipropyl ether: Two propyl groups attached to an oxygen atom.

Step by step solution

01

Understand the nomenclature

In organic chemistry nomenclature, the first part of the name generally speaks about the parent hydrocarbon chain. The prefixes denote the position of the groups attached to the parent chain. For example, '3-isopropyloctane' means that on the third carbon of the octane chain an isopropyl group is attached
02

Draw the parent chain

Draw the main or parent hydrocarbon chain. For instance, octane is an eight-carbon chain. This will form the 'backbone' of our structure.
03

Attach the groups to the parent chain

Now it is time to attach the groups to our main chain as indicated by the nomenclature. In our example, attach an isopropyl group to the third carbon of the octane chain.
04

Repeat steps 2 and 3 for other compounds

Repeat the previous steps for other compounds: '2 -chloro- 3 -methylpentane,' '2 -pentene,' 'dipropyl ether.' Understand the nomenclature before starting to draw.
05

Revise the drawn structures

After drawing the structures, revise them to ensure that they match with the given names and there are no loose bonds.

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

With appropriate sketches, represent chemical bonding in terms of the overlap of hybridized and unhybridized atomic orbitals in the following molecules. (a) \(\mathrm{C}_{4} \mathrm{H}_{10} ;\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCl} ;\) (c) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\).

Give the isomers of the carboxylic acid with molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}\).

Give the isomers of the carboxylic acid with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\).

Which hydrocarbon has the greater number of isomers, \(\mathrm{C}_{4} \mathrm{H}_{8}\) or \(\mathrm{C}_{4} \mathrm{H}_{10} ?\) Explain your choice.

For each of the following pairs, indicate which substance has (a) the higher boiling point, \(\mathrm{C}_{6} \mathrm{H}_{12}\) or \(\mathrm{C}_{6} \mathrm{H}_{6}\) (b) the greater solubility in water, \(\bar{C}_{3} H_{7} O H\) or \(\mathrm{C}_{7} \mathrm{H}_{15} \mathrm{OH}\) (c) the greater acidity in aqueous solution, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}\) or \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\).
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

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