Part 5: STRUCTURE DETERMINATION

a.
Molecular weight = 6 x 12.01 + 10 x 1.008 + 1 x 16.00 = 98.14 g/mol (1 mark)

b. The formula for index of hydrogen deficiency, IHD = 0.5 ( 2c+2-h), therefore, the IHD = 0.5 (2 x 6 + 2 - 10) = 2 (1.5 marks)

c. Functional groups with 1 x O atom and upto 1 unit of unsaturation : aldehyde, ketone, alcohol, ether, epoxide (1.5 marks)

d.i. An IR absorption at 1715 cm-1 suggests a ketone. Cyclohexanone is the easiest compound to match the MF with the right number of C and H types. (2 marks)

cyclohexanone

d.ii. To have the most acidic H being a pKa =25 requires a terminal alkyne. For the most acidic H to be 25, there can't be an alcohol present. (2 marks)

part dii

e. For resonance stabilisation, you need a conjugated system (extended pi system). To be Z,Z neither double bond can be at the end of a chain. A few of the possible answers are shown below: (3 marks)

conjugated c6h10o

f. A chiral center in this structure will require an sp3 C with 4 different groups attached. A few of the possible answers are show below: (2 marks)

cyclic chiral c6h10o

Common errors:

b. Incorrect IHD calculation (but still drew isomers with the right IHD !) : they must be consistent!
c. Naming functional groups with 2 O atoms (the molecular formula had only 1 O).
di. Drawing an aldehyde (incorrect IR frequency).
dii.Drawing internal alkynes instead of terminal alkynes and drawing alkynes with incorrect geometry. They are liner at the alkyne!
e. Drawing pi bonds with incorrect stereochemistry (carbonyls aren't E or Z).
f. Not showing the stereochemistry at the chirality center (read the question).