Let's draw a 1,3,4-thiadiazole ring (five-membered aromatic; atoms clockwise: N1-N2-C3-S4-C5).

At C3 (exocyclic chain):
  - C3 is bonded to a nitrogen that is double-bonded to a carbon (C-alpha); no E/Z geometry is specified for this C=N bond.
  - C-alpha bears a hydroxyl group (-OH) and a vinylene bridge (-CH=CH-) to a benzene ring. The -CH=CH- double bond is in the E configuration.
  - On the benzene ring, a chlorine atom is placed ortho to the vinylene attachment point.

At C5, 
  - Attach a thiophene ring through a carbon adjacent to the sulfur (alpha/2-position). The thiophene is unsubstituted.