In order to match the given range using a wildcard mask of 0.0.0.15, the network address specified on the ACL has to be a multiple of 16. If you enter the ACL as you have posted above into a router, it will change it to:
access-list 100 deny ip 10.246.32.224 0.0.0.15 any
The important thing here is to use the right bit boundary when configuring ACLs. With a mask of 0.0.0.15, the last 4 bits of your network address have to be zero. When converted to binary, 230 is 11100110 whereas 224 in binary is 11100000 (the last 4 bits are zero).
Therfore, the answer to your question is that the ACl you have specified will not work.
We are pleased to announce availability of Beta software for 16.6.3. 16.6.3 will be the second rebuild on the 16.6 release train targeted towards Catalyst 9500/9400/9300/3850/3650 switching platforms. We are looking for early feedback from custome...