and you will get an error.
It's not a valid Unicode character (which consist of 4 digits from 0-F each).
and it compiles !
No it doesn't, because the Unicode codepoint at 1 can not be the first character in an identifier. Something like "\u00612345" would compile, and would correspond to an identifier named "a2345" (which you can check by putting "System.out.println(a2345);" behind it.
The compiler performs a substitution of Unicode characters before the compilation. In the former case, you'd end with 2 characters, which can't be assigned to a char. In the latter case you end up with a valid identifier (a2345) if you use the sequence I suggested.