Can you explain the logic behind this chord substitution by Joe Pass
Question: I recently watched a Joe Pass video in which he said that the cycle Bb Eb Ab Db could substitute for a 1 6 2 5 progression in C, namely C Am Dm G7. I can see how Eb Ab and Db are tritone substitutions for Am Dm and G7- but that doesn't explain any logic behind Bb substituting for C. Can you help explain this? It was a Bb13 in the video-- which does contain the notes C and G, but in the color tones... does this really work? -Gary

Answer: When you get past the 7th it's all color, in my opinion. Functional harmony really only counts up to the seventh; the 9th, 11th, and 13th are just frosting when they're not actually making the harmony ambiguous.

But Jazz has no real rules, as near as I can tell. Jazz has the underpinnings of traditional theory and then all the rest is color and gesture. So Joe can do what he wants.

I do think he has some logic working. Were I to attempt finding that logic in his advice it might go like this:

1 6 2 5 (to music students, I, vi, ii, V: one of the most classic pop progressions) is based on a progression of rising fourths from vi. Here it is shifted a tritone and the mode abandoned for a series of "secondary dominants" - moreover, he makes the first chord part of the chain too.

For example, let's convert the original progression to secondary dominants starting with the vi:

I, VI, II, V (in other words: I, "V of ii," " V of V," V)

Now we make the first chord a secondary dominant also:

V of vi, V of ii, V of V, V

Or, in chord names: E, A, D, G

Shift that a diminished fifth (equivalent to a tritone) and you've got your Bb, Eb, Ab, Db.

Makes as much sense as anything.

Bonus followup question:

Thanks Jeff- I did see the circle of fifth progression and yes it makes sense in the same way you describe it, but I think that swapping the I chord out for the V/VI could throw off someone soloing if the chart called for a I? It would make more sense to sub the I for a chord a tritone away- a Gb, or for a simple vi chord - an Am for example. That's what I was getting at- that the odd one out was the sub for the one chord. It would make sense only if no one was soloing at the time, right? Or maybe it's enough that the E chord does share chord tones, so maybe the E chords tritone implies enough?-- Guess I'd have to experiment and just use my ears to see what happens to someone soloing with certain substitution expectations. :)

Bonus followup reply:

You have a good point about throwing off a soloist. It looks like Joe really did just extend the chain of fifths, but perhaps it is close enough for the soloist since the substituted chord shares those chord tones.

This is a good place to confess that I am no jazz expert. Results may vary. To me the concept of tritone substitution is already far out in the wilderness. But rootwise, the Bb makes sense, and given the presence of C and G in the color tones I think we can assume Joe believes one can get away with it.

Even more followup from Gary:

By the way-- 7 chord tritone substitution does make good sense if you think about it this way: the 7 chord a tritone away will contain the same 3rd and 7th notes as the chord opposite it (6 steps away, in the circle of fifths), just swapped. A G7's 3rd and 7th notes equal the 7th and 3rd of the Db7 a tritone away.. and the 3rd and 7th interval in a 7 chord is the tritone interval-- the unstable interval that wants to resolve.

Even more reply:

Indeed that is a good way to approach it, speaking of course of music in which enharmonics like B and Cb are not distinguished. See, I learn things from time to time.

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