No it does not, because synchronized already has a memory barrier inserted after it, so all threads will see the update that the current thread performs, taking into account that the other threads will synchronize on the same lock.
Volatile, just like synchronized, has memory barriers that are attached to it - depending on the CPU, it is store/load/full barrier that ensures that an update from one thread is visible to the other(s). I assume this is performed with CPU cache invalidation.
EDIT From what I've just read, the store buffers are flushed to the CPU cache, and this is how the visibility is achieved.