Part 5: STRUCTURE DETERMINATION

Note that there is no point showing all possible answers here, so the most likely and potentially simplest answers are shown.

a. C₆H₁₀O has 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 mark)

c. Any 3 of the following : alcohol, ether, epoxide, aldehyde, ketone (0.5 marks each)

d. IR of 1715cm^-1 suggests a simple ketone. 4 types of C and 3 types of H requires some symmetry. (2.5 marks)

ei. Acyclic implies that it does not contain a ring. A chirality center will require 4 different groups to be attached at an sp3 C. (4 marks)

ii. See above (2 marks)

iii. See above (1 mark)

Common errors:

• incorrect IHD calculation (e.g. forget to divide by 2 or didn't read the MF correctly)
• not understanding that isomers are molecules with the same molecular formula
• drawing structures that violated the rules of valence (recall line diagrams require that H are heteratoms e.g. O are shown).
• in part e (i) drawing cyclic structures, (ii) didn't understand a chirality center (e.g. assigning an sp2 C in an alkene or carbonyl as being chiral or not having 4 different groups attached to an sp3 C, and (iii) couldn't apply the Cahn-Ingold-Prelog rules to assign configuration correctly (nomenclature !)