CHEMISTRY 133

Name___________________                  Quiz 6 – Solutions

5.  A scientist is trying to produce 2-chloroacetophenone (C₆H₅C(O)CH₂Cl) but is worried that chlorination occurred on the benzene ring instead (producing ClC₆H₄C(O)CH₃ instead).  Given that ketones tend to fragment between the a carbon and the carbonyl carbon,

a) Indicate two fragment peaks (give the mass to charge ratio and structure) that would be indicative of 2-chloroacetophenone.  What related fragment would be expected from ClC₆H₄C(O)CH₃ instead?  How can the isotope effect (Cl has a prominent M+2 peak) be used to aid in this?

Atomic weights: ¹H – 1.008 amu, ¹²C – 12.000 amu, ¹⁶O – 15.995 amu, and ³⁵Cl – 34.969 amu.

If we break at each a carbon and each fragment can become charged, we can get 77 (splitting off at the benzene ring – both sides), 49 (CH₂Cl fragment at other a carbon), or 105 (C₆H₅CO fragment) + 3 pts

If the Cl is on the benezene ring, we would expect different fragments: 111 and 113 for ClC₆H₄, 139 and 141 for ClC₆H₄CO, 15 for CH₃ (probably not observed), and 43 for CH₃CO.

+ 3 pts

b) Two fragments from 2-chloroacetophenone have the same mass to charge ratio.  What is this ratio?  Describe how you could tell how much came from each fragment.

77 This is from both the benzene ring (78 – 1) or from COCH₂Cl.  One could the fraction coming from each fragment by looking at the isotopic peaks.  The COCH₂Cl fragment produces a significant 79 peak from a ³⁵Cl), while almost none of the C₆H₅ will produce such a peak. Thus by the observed ratio of 79 to 77 could be used to determine the % from COCH₂Cl.

observed M+2/M = (32)(%COCH₂Cl)

+4 pts