Part 5: STRUCTURE DETERMINATION

a. Use the molecular weight of X = 120.19 g/mol and the % of C and H to calculate the formula: (2 marks)

C: 120.19 x 89.94% g/mol C = 108.098 g/mol C. Now divide by atomic weight of C, 108.098 / 12.011 = 9

H: 120.19 x 10.06% g/mol H = 12.09 g/mol H. Now divide by atomic weight of H, 12.09 / 1.008 = 12

Hence the molecular formula is C₉H₁₂ where the simplest ratio of C to H, the empirical formula is C₃H₄.

b. C₉H₁₂ (see above) (2 marks)

c. For a hydrocarbon, the formula for the index of hydrogen deficiency, IHD = 0.5 ( 2c+2-h), therefore, the IHD = 0.5 (2 x 9 + 2 - 12) = 4 (2 marks)

d. What information do we have ? (structure 5 marks, name 2 marks)

the molecular formula, C9H12 IHD = 4 IR data : 3100 cm-1 (sp2 C-H), 3000 cm-1 (sp3 C-H) and 1600 & 1500 cm-1 (Ar C=C) 5 H and 6 C types 2 monochlorination products implies only 2 types of sp3 C-H

Common errors:

Did not know that an empirical formula was the simplest ratio of the elements. (High school / 1st yr knowledge)
Not checking molecular formula against the molecular weight provided.
Incorrectly counting or not checking the number of H and or C types in X
Drawing structures for X that did not fit the molecular formula of C₉H₁₂, the IHD and the IR data.
Note that IHD defines the number of rings + pi bonds present for a given molecular formula.
Could not name X correctly : incorrect numbering, incorrect alphabetisation, wrong root name (e.g. wrong root, mixing up phenyl with benzyl).