CHEM14C_f09_e02_key

CHEM14C_f09_e02_key - Chemistry 14C Fall 2009 Exam 2...

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Chemistry 14C Fall 2009 Exam 2 Solutions Page 1 Statistics : High score, average, and low score will be posted on the course web site after exam grading is complete. The exam is ready to be picked up when these numbers are posted. Some questions have more than one answer, even though only one answer may be listed here. To see the projected course grade cutoffs, consult the grading scale on the Chemistry 14C course web page. 1. Cannot determine. The base peak is the ion with the highest relative abundance (not the highest m/z). Because we do not know the relative abundance of other ions in the spectra (they might be more than 100%), we cannot determine which peak is the base peak . 2. (9 x 12) + (16 x 1) + (1 x 16) = 140 3. (9 x 1.107) + (16 x 0.015) + (1 x 0.037) = 10.24 ~ 10.2 Regardless of rounding, the answer choice remains the same . 4. Zero. No S, Cl, or Br present . 5. Fragmentation. The (E) and (Z) isomers have identical chemical formula, so the m/z and relative abundances for their M, M+1, and M+2 ions are identical. Fragmentation is controlled by structure; (E) and (Z) isomers will have similar, but not identical, fragmentation patterns . 6. O OH The IR suggests the aldehyde is converted into a carboxylic acid . 7. (a) C C C C C C C C C O H (b) C C C C C C C C C O H Most deshielded = highest chemical shift . (c) C C C C C C C C C O H H H Most influenced by magnetic induction = closest to pi bond . 8. (a) ( E )-non-2-enal has 13 hydrogens that are equivalent to at least one other hydrogen in the molecule. The three methyl group hydrogens are equivalent to each other, and each hydrogen of a methylene group is equivalent to the other hydrogen in the same methylene group . (b) The 1 H-NMR spectrum of ( E )-non-2-enal has 9 signals. Of these signals, 2 are triplets. The aldehyde hydrogen and mrthyl group signals are triplets . (c) The 13 C-NMR spectrum of ( E )-non-2-enal has 9 signals. Of these signals, 5 are triplets. None of the carbon atoms are equivalent to any other carbons in the molecule. A carbon directly bonded to two hydrogens (i.e., a CH 2 ) is a triplet in the 13 C-NMR spectrum
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This note was uploaded on 03/02/2010 for the course CHEM CHEM14C taught by Professor Hardinger during the Winter '10 term at UCLA.

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CHEM14C_f09_e02_key - Chemistry 14C Fall 2009 Exam 2...

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