Part 5: STRUCTURE DETERMINATION

a. %C = 5 x 12.011 / (5 x 12.011 + 8 x 1.008 + 2 x 15.999) = 59.68% (2 mark)

b. The formula for index of hydrogen deficiency, IHD = 0.5 ( 2c+2-h), therefore, the IHD = 0.5 (2 x 5 + 2 - 8) = 2 (1.5 marks)

c. Functional groups with 2 x O atom and upto 2 units of unsaturation could include : aldehyde, ketone, alcohol, ether, epoxide, ester, carboxylic acid, alkene, alkyne (1.5 marks)

d. Three possible example are shown below: (2 marks)

e. Acyclic = not a ring, IR at 1735 cm-1 = C=O, pKa = 25 suggests some sort of an ester with alpha-H, e.g. (2 marks)

f. The base will need to be the conjuate base of an acid with a pKa of at least 27. e.g. NaNH₂, LiN(CH(CH₃)₂)₂, LiCH₃, NaH (2 marks)

g. To be meso, the structure needs 2 chirality centers that have the same set of 4 groups attached such that they are mirror images, e.g. (2 marks)

Common errors:

• incorrect IHD calculation (e.g. forget to divide by 2)
• indentifying FG that did not match MF (e.g. anyhydride)
• incorrect numbers of types in part d
• incorrect acidity (pKa) in part e & part f
• incorrect structures of bases (wrong MF, incorrect charges etc.)
• not drawing a meso structure, not showing chirality using wedge-hash diagram (or alternative)