Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 4

The following organic compounds cannot exist. Why? a. 2 -chloro-2-butyne b. 2 -methyl-2-propanone c. 1,1 -dimethylbenzene d. 2 -pentanal e. 3 -hexanoic acid f. 5,5 -dibromo-1-cyclobutanol

Short Answer

Expert verified
The following organic compounds cannot exist due to the reasons mentioned: a. 2-chloro-2-butyne: Carbon can't form 5 bonds (a triple bond and chlorine at the same position). b. 2-methyl-2-propanone: No room for a methyl group at C2, which is already bonded to a carbonyl group. c. 1,1-dimethylbenzene: Carbon in benzene ring can't form two bonds with methyl groups. d. 2-pentanal: Aldehyde functional group must always be at the end of a carbon chain. e. 3-hexanoic acid: Carboxylic acid functional group must always be at the end of a carbon chain. f. 5,5-dibromo-1-cyclobutanol: Position 5 does not exist in a cyclobutane ring.

Step by step solution

01

Compound a: 2-chloro-2-butyne

2-chloro-2-butyne indicates that there is a chlorine atom at position 2 on a 4-carbon (butyne) chain with a triple bond between carbons 2 and 3. However, this is not possible, as having both a chlorine atom and a triple bond at the same carbon is not allowed. Carbon can only form a maximum of 4 bonds, and in this case, it would require 5 bonds (1 with the chlorine, 3 with the triple bond, and 1 with the adjacent carbon).
02

Compound b: 2-methyl-2-propanone

2-methyl-2-propanone suggests that there is a methyl group at position 2 on a 3-carbon (propanone) chain with a carbonyl group (C=O) at the second carbon. However, this is not possible since adding a methyl group at the position 2 would mean the carbon already has 4 other attached atoms, one being the carbonyl group and two being hydrogens, leaving no room for another bond with a methyl group.
03

Compound c: 1,1-dimethylbenzene

1,1-dimethylbenzene implies that there are two methyl groups attached to the same carbon in the benzene ring. However, this is not possible as the carbon in the benzene ring can only have one free bond to make a connection with a methyl group since it already has two bonds with other carbons and one with a hydrogen atom.
04

Compound d: 2-pentanal

2-pentanal suggests that there is an aldehyde functional group (-CHO) at position 2 on a 5-carbon (pentane) chain. However, an aldehyde functional group must always be at the end of a carbon chain. The correct name for this compound would be 3-pentanone, a ketone with the carbonyl group at carbon 3.
05

Compound e: 3-hexanoic acid

3-hexanoic acid implies an acid functional group (-COOH) at position 3 on a 6-carbon (hexane) chain. However, similar to the aldehyde in the previous example, the carboxylic acid functional group (-COOH) must always be at the end of a carbon chain. The given name does not represent a valid organic compound.
06

Compound f: 5,5-dibromo-1-cyclobutanol

5,5-dibromo-1-cyclobutanol indicates that there are two bromine atoms attached to the same carbon in the cyclobutane ring at position 5, and an alcohol functional group (-OH) at position 1. However, this is not possible because in a cyclobutane ring, there are only four carbons, so position 5 does not exist. The structure is impossible to form.

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

Draw the structures for two examples of unsaturated hydrocarbons. What structural feature makes a hydrocarbon unsaturated?

In Section \(22.6\), three important classes of biologically important natural polymers are discussed. What are the three classes, what are the monomers used to form the polymers, and why are they biologically important?

Considering your answers to Exercises 130 and 131, how can you justify the existence of proteins and nucleic acids in light of the second law of thermodynamics?

The codons (words) in DNA (that specify which amino acid should be at a particular point in a protein) are three bases long. How many such three- letter words can be made from the four bases adenine, cytosine, guanine, and thymine?

Poly(lauryl methacrylate) is used as an additive in motor oils to counter the loss of viscosity at high temperature. The structure is The long hydrocarbon chain of poly(lauryl methacrylate) makes the polymer soluble in oil (a mixture of hydrocarbons with mostly 12 or more carbon atoms). At low temperatures the polymer is coiled into balls. At higher temperatures the balls uncoil and the polymer exists as long chains. Explain how this helps control the viscosity of oil.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

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