Start with a pyridine ring (six-membered aromatic containing one nitrogen). Number the ring conventionally with the nitrogen as position 1, and proceed clockwise to positions 2-6.

There are two substituents on this pyridine:
  
1. Position 2 bears a pyrazole (1,2-diazole; five-membered aromatic ring with three carbons and two adjacent nitrogens), attached through the nitrogen that originally carries the hydrogen.

2. Position 5 bears an amide chain constructed as follows (moving outward from the ring carbon):
   - The ring carbon is bonded to an NH group.  
   - That nitrogen is part of an amide: -NH-C(=O)-.  
   - The carbonyl carbon of this amide is single-bonded to another carbonyl carbon, giving -NH-C(=O)-C(=O)-.  
   - The second carbonyl carbon is attached to a tertiary nitrogen.  
   - This tertiary nitrogen bears two alkyl groups:
       a) one methyl group (-CH3)  
       b) one 2-ethoxy-ethyl group, written -CH2-CH2-O-CH2-CH3.

No stereocenters or defined double-bond geometries are present besides the carbonyl (C=O) groups described.