If you write a collatz-transformation this way $ \small a_{k+1} = (3 a_k + 1)/2^A $ where **A** is the number of all following even (divide-by-2) steps, then also all $\small a_k$ must be odd.      

With this write for one transformation $\small a_{k+1} = T(a_k;A) $ and for more $\small a_{k+m} = T(a_k;A_1,A_2,\ldots,A_m) $.    
For $\small A=1 $ these steps are increasing and if $\small A>1$ the steps are decreasing.     

A cycle occurs, if $\small a_m=T(a_0;A_0,A_1,\ldots,A_m) = a_0 $ thus $\small a_m = a_0 $.     

Now let's look at two different types of cycles - the "general" one, where the exponents $\small A_k $ have no a priori restriction, and the "primitive" one, where all but the last $ \small A_k = 1 $, thus $\small a_0 = T(a_0;1,1,1,1,\ldots,1,A) $ with say $\small N-1 $ times *1* and only one $\small A_{N} \gt 1$      

For the primitive cycle of the length *N* the smallest element $\small a_0 $ must have the form $\small a_0 = 2^{N-1} \cdot 2w - 1$ where *w* is some positive odd integer and that primitive cycle increases then up to $\small a_{N-1} = 3^{N-1}\cdot 2w-1 $ from where it must decrease by a consecutive set of *A* "even" transformations.    

The [proof of Ray Steiner][1] (and the subsequent proofs of J.Simons and B.de Weger) use that requirement of the "primitive cycle" (in their nomenclature "1-cycle", see [wikipedia][1]) to show that such a cycle *cannot exist* except $\small a_0 = T(a_0;2) \qquad a_0=1 $ (where no exponents of value *1* occur - the degenerate case (also called "circuit").     


For the general cycle this is much more complicated and does definitely not have the properties which you sketch in your op.   
<hr>
[added] Answering to the comment.   A characterization of the general cycle of *m* collatz-steps (*m=N + S* in my notation, where I refer -writing "steps"- to the number *N* of abbreviated steps $\small T(a;A)$      

Assume again one step as on the form $\small a_{k+1} = (3 a_k +1)/2^{ A_k} = T(a_k;A_k) $. Then to have a cycle this means (ignore the index *k* here) $\small a=(3a +1)/2^A $ and also $\small 2^A=(3a +1)/a = (3 + 1/a) $ This can only be solved if *a=1* and *A=2*, so there is only one general-cycle of length *N=1*, *S=2* and *m=N+S=3* and we see a characteristic of the exponent *A*, which is much descriptive: the number of even steps of the original notation of the collatz-transformation is *2*.      
Now we assume a 2-step cycle $\small a =  T(a;A,B) $ and dissolve this in two steps: $\small b=T(a;A) \qquad a=T(b;B) $ thus $\small b = (3 a +1)/2^A \qquad a=(3b+1)/2^B $. I'm used to write *S* for the sum *A+B* meaning the whole number of even steps, and *N* for the number of "odd steps", both wrt the original Collatz-notation, and in my notation *N* is the number of steps (and the power of *3* involved). If we write the (trivial) product of the two involved elements *a* and *b* in their direct notation and in their transformed expression we get $\small a\cdot b = (3b+1)/2^A \cdot (3a+1)/2^B $ and this can be rewritten as $\small 2^{A+B} = (3+1/a)(3+1/b) $ This is an interesting form and easily generalized for the analysis with bigger *N* (and respectively *m*) . Here we see, that a 2-step general cycle can only exist if $\small (3+1/a)(3+1/b) $ is a perfect power of *2*; now considering the whole set of odd positive integers for *a* and *b* we see, that that product can vary only between $\small 9 \ldots 16 $ and thus must be *S=A+B=4* and this requires *a = b = 1* and thus *A=B=2* (which is then only a concatenation of the trivial cycle $\small 1=T(1;2,2) $ . Here *m=N+S=2+4=6* .      

This way you may proceed studying longer assumed cycles. For instance it shows that the whole distance between two numbers $\small a_k - a_j = 2^S$ where *S* is the number of even transformations can never occur because there are always "odd Collatz steps" interspersed; even less can the distance between the minimal and maximal member of a loop be $\small 2^{N+S} = 2^m $, which were the whole length of the cycle in the counting of original Collatz-steps.     

This formula describes pretty well important properties of the exponents of *2* in relation to the length of a "general cycle" , so it might be useful for the answering of your question. However - I can't relate anything in that formula to the asked property of a connection between the distance minimal...maximal member  and *2^m* where *m* is the number of all Collatz steps and actually is $\small m=S+N$ nd so I think there is none.     

Addendum : It might be interesting to look at the *existing* cycles in the domain of *negative* odd numbers; they can also be identified using the procedere exercised above. The 2-step cycle was $\small 2^{A+B} = (3+1/a)(3+1/b) $ and allowing negative *a* and *b* gives the example: $\small 2^{A+B} = (3-1/5)(3-1/7) = {14 \over 5} \cdot {20 \over 7} = {8 \over 1} = 2^3 $ which is a perfect power of *2* as required. Then $\small A+B=3$ and one of them must equal *1*. In fact we have the 2-step-cycle $\small -5=T(-5;1,2) $

Finally: a short treatize using the notation here is in that [article of mine][2] 

<hr> 

  [1] Steiner,R.P.; A theorem on  the syracuse problem, Proceedings of the 7th Manitoba Conference on Numerical Mathematics,pages 553...559, 1977


  [1]: http://en.wikipedia.org/wiki/Collatz_conjecture#m-cycles_cannot_occur
  [2]: http://go.helms-net.de/math/collatz/Collatz061102.pdf