Table of Contents Table of Contents
Previous Page  486 / 912 Next Page
Basic version Information
Show Menu
Previous Page 486 / 912 Next Page
Page Background

478 Chapter 14

Copyright © 2017 Pearson Education, Inc.

d.

The group attached to the oxygen in the fourth spectrum has only one kind of hydrogen. The carbon

attached to the oxygen is not bonded to any hydrogens, because there is no signal at

4.0.

C

O

O

CCH

3

CH

3

CH

3

CH

3

75.

CH

3

CHCHCH

3

OH

CH

3

HBr

CH

3

CH

2

CCH

3

CH

3

Br

compound

A

compound

B

76.

The DEPT

13

C

NMR spectrum indicates that the compound has six carbons. The molecular formula shows

that the compound has six carbons and an oxygen. The signal at 220 ppm suggests a ketone. The following

ketones have the single

CH

3

group, the single CH group, and the three

CH

2

groups indicated by the DEPT

spectrum. The highest frequency signals are a

CH

2

group and a CH group, indicating that these are the

groups attached to the carbonyl carbon. Thus, the compound on the right is responsible for the spectrum.

O

O

or

O

or

77.

a.

The IR spectrum indicates that the compound is a ketone. The molecular formula shows that there are

12 hydrogens in the compound. Therefore, the signals in the NMR spectrum are due to 2, 4, and 6 protons.

The doublet in the NMR spectrum at

0.9

ppm that integrates to 6 protons suggests an isopropyl group.

There is a singlet at

2.1

ppm on top of a multiplet, which is due to the methine proton of an isopropyl

group plus a three-hydrogen singlet for a methyl group adjacent to the carbonyl carbon. The doublet at

2.2

ppm that integrates to 2 protons suggests a

CH

2

group adjacent to the carbonyl group that couples to

an adjacent methine proton. Knowing that the compound contains six carbons helps in the identification.

O

b.

The IR spectrum indicates that this oxygen-containing compound is not a carbonyl compound or an

alcohol; the absorption band at

1000 cm

-

1

suggests that it is an ether. The doublet at

1.1

ppm in

the

1

H

NMR spectrum that integrates to 6 protons and the septet that integrates to 1 proton indicate an

isopropyl group. Because there are no other signals in the NMR spectrum, the compound must be a

symmetrical ether. The compound is

diisopropyl ether

.

O