Organic Chemistry

546 CHAPTER 14 NMR Spectroscopy

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C OCHCH 3

O

CH 3 CH 2 CH 2

CH 3

ad b

b

c e

δ (ppm)

frequency

Figure 14.16N
NMR spectrum of
isopropyl butanoate.

1 H

There are two signals in the NMR spectrum of 1,3-dibromopropane (Figure 14.15).

The signal for the protons is split into a triplet by the two hydrogens on the adjacent

carbon. The protons have two adjacent carbons that are bonded to protons. The

protons on one adjacent carbon are equivalent to the protons on the other adjacent car-

bon. Because the two sets of protons are equivalent, the rule is applied to both

sets at the same time. In other words,Nis equal to the sum of the equivalent protons on

both carbons. So the signal for the protons is split into a quintet Inte-

gration confirms that two methylene groups contribute to the signal because twice as

many protons give rise to the Hbsignal as to the Hasignal.

Hb

Ha 14 + 1 = 52.

N+ 1

Ha

Hb

1 H

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Br CH 2 CH 2 CH 2 Br

b abab

δ (ppm)

frequency

Figure 14.15N
NMR spectrum of
1,3-dibromopropane.

1 H

The NMR spectrum of isopropyl butanoate shows five signals (Figure 14.16).

The signal for the protons is split into a triplet by the protons. The signal for the

protons is split into a doublet by the proton. The signal for the protons is

split into a triplet by the protons, and the signal for the proton is split into a

septet by the protons. The signal for the protons is split by both the and

protons. Because the and protons are not equivalent, the rule has to be

applied separately to each set. Thus, the signal for the Hcprotons will be split into a

Ha Hd N+ 1

Hb Hc Ha Hd

Hc He

Hb He Hd

Ha Hc

1 H