Lose some FDs in 2nf normalization?
Yes, the breaking is not good. 

The reason is that you FDs (B-&gt;C and C-&gt;D) also imply B-&gt;D. So B-&gt;CD (both B-&gt;C and B-&gt;D) break 2NF and should be used for the table breaking. 

So, the relations, in order to be 2NF should be: (A,B) and (B,C,D). 

(and then the C-&gt;D can be used to break the second further for 3NF.)