You could use a wildcard Find where:
Find = <BRAND®[!\(]{1,5}\([!\)]@\)

__________________
Cheers,
Paul Edstein
[Fmr MS MVP - Word]

Excellent. That seems to work perfectly, thanks!

I did make one small change, since I need to use that optional part in a replace function (I added parentheses around the optional 1-5 range), and I have that working. My requirement is to remove the '®' and the parentheses text so that 'BRAND® XXL (sub brand text)' becomes 'BRAND XXL'

The only issue is that it ends up with an extra space in the replaced value (since that is part of the captured value), but since this will be run in a scripted find/replace, I should be able to remove that extra space using another replace on the range.

(e.g. 'BRAND® XXL (sub brand text) more text...' becomes 'BRAND XXL more text...' with 2 spaces)

I don't see any way around that, and I don't foresee any problems using another find/replace on the found range, so hopefully I should be able to accomplish this.

Thanks again!