Organic Chemistry

Finding that the signal for the protons of 1-chloro-3-iodopropane is a quintet indicates

that and have about the same value. The splitting diagram shows that a quintet

results if

We can conclude that when two different sets of protons split a signal, the multiplicity

of the signal should be determined by using the rule separately for each set of

hydrogens when the coupling constants for the two sets are different. When the cou-

pling constants are similar, however, the multiplicity of a signal can be determined by

treating both sets of adjacent hydrogens as if they were equivalent.In other words, the

rule can be applied to both sets of protons simultaneously.

PROBLEM 30 SOLVED

The two hydrogens of a methylene group adjacent to an asymmetric carbon are not

equivalent hydrogens because they are in different environments due to the asymmetric

carbon. (You can verify this statement by examining molecular models.) Applying the

N+ 1 rule to these two diastereotopic hydrogens (Section 5.16) separately in determining

N+ 1

N+ 1

Jab

Ha

Jac

ClCH 2 CH 2 CH 2 I

b ac

Jab=Jac.

Jab Jac

Ha

556 CHAPTER 14 NMR Spectroscopy

We would expect the signal for the protons of 1-chloro-3-iodopropane to be a

triplet of triplets (split into nine peaks) because the signal would be split into a triplet by

the protons and each of the resulting peaks into a triplet by the protons. The signal,

however, is a quintet (Figure 14.26).

Hb Hc

Ha

87654 3210

ClCH 2 CH 2 CH 2 I

cba

δ (ppm)

frequency

Figure 14.26

1 HNMR spectrum of 1-chloro-3-iodopropane.