So, @PetSerAl is absolutely right that the correct answer to this (if we are looking for a lambda-calculus-esque solution) is:
(lambda (x y) (b y x))))
Thus: ((not true) 1 2) gives 2 and ((not false) 1 2) gives 1.
((not true) 1 2)
((not false) 1 2)
But since you didn't specify lambda calculus in your question, but instead tagged both Racket and Scheme, I will give you an answer that works there.
Consider if you have true and false defined as above. Then writing not is as simple as:
(cond [(equal? x true) false]
[(equal? x false) true]
[else (error "Value is not true or false")])))
This has the same semantics listed above: ((not true) 1 2) gives 2 and ((not false) 1 2) gives 1. But it has the added benefit that you can actually directly examine it to see the value, rather than just applying it to another procedure. For example:
> (not true)
> (not false)
You can even now use equal? (or really eq? on them), whereas before you would have always gotten #f if you tried that:
> (equal? (not true) false)
> (equal? (not true) true)
> (equal? (not false) true)
> (equal? (not false) false)
This leads to a whole discussion of PL theory about what can and cannot be decided with the lambda calculus. That I will elide here. But if you are interested, I recommend you check out this book, or that book.