1. Let’s talk septets for awhile. You’ve already seen one, in the proton spectru
ID: 1009301 • Letter: 1
Question
1. Let’s talk septets for awhile. You’ve already seen one, in the proton spectrum of something as simple as 2-propanol.
a. Sketch the molecule and briefly explain where the septet comes from here.
b. In theory, what are the peak ratios for the peaks in this septet?
We’ve seen another septet in the carbon spectra taken in deuterated acetonitrile, CD3CN.
c. Briefly explain where this septet comes from. (A calculation, like those seen on a video, would be really helpful here).
d. In theory, what are the peak ratios for the peaks in this septet? (Warning: they are NOT the same as in 1b.)
Finally, the PF6— anion phosphorus spectrum featured a lovely septet. e. Briefly explain where this septet comes from.
f. What are the peak ratios in this septet? You might find it helpful to get out a ruler and actually measure the heights of these things … it will give you a rough idea. Which kind of septet does this one resemble the most: the one from the proton spectra of 2-propanol, or the one from the solvent peak of deuterated acetonitrile?
From this you should be able to make a conclusion on when you can, and when you can’t, use Pascal’s Triangle for estimating the peak ratios.
Explanation / Answer
1. Septet is a peak split into seven peaks.
For 2-propanol the septet comes from -CH- which is connected to two CH3 on each end and an -OH on one end.
The peak splitting in proton is based on number of neigbouring H's. So in this case we have 6 neighbouring H's.
The peak splits into = (n + 1) = (6 + 1) = 7 peaks for the -CH- proton coupling with the neighbouring protons.
The ratio of peaks in a septet being = (1 : 6 : 15 : 20 : 15 : 6 : 1)
2. Similarly for PF6, the septet comes for P with 6 neighbouring F atoms.
The peak splitting again is on the number of neighbours = (n + 1) = (6 + 1) = 7
The ratio again for the septet would be thus = (1 : 6 : 15 : 20 : 15 : 6 : 1)
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