1. Give the formula and structure of the compound with this IR and a molecular ion peak at 116.

2. How many peaks in the 13 C NMR spectra? 234 3 6 4 2

3. Which compound matches this 13 C NMR spectrum?

4. A B C A B C B CA Rank the labeled C’s of each compound in order of increasing chemical shift? B ~ 33ppm C ~ 35ppm A ~ 115ppm A ~ 8ppm B ~ 37ppm C ~ 210ppm C ~ 9ppm B ~ 30ppm A ~ 177ppm

7. Information gained from a 1 H NMR Spectrum Number of Signals Position of Signals Intensity of Signals Spin-Spin Splitting of Signals

8. How many different types of protons are in each compound? HAHA HAHA 1 type HAHA HBHB 2 types HAHA HBHB HCHC 3 types HAHAHAHA HBHB HCHC HDHD HEHE 5 types

9. Determining the number of different protons in compounds with  bonds. cis to Cl 1 type cis to Cl trans to Cl 2 types cis to Cl cis to H 3 types

10. Determining the number of different protons in compounds with rings.. All H’s are equivalent, 1 type cis to Cl cis to H 3 types Protons on methyls are equivalent Each of these is cis to a methyl 2 types

17. Spin-Spin Coupling

18. When determining the spin-spin coupling, look at the number of protons on the adjacent carbon. For the methyl group, look at the methylene group. There are 2 protons, so using the N+1 rule tells us that the peak should be a triplet in a 1:2:1 ratio.

19. For the methylene group, look at the methyl group. There are 3 protons, so using the N+1 rule tells us that the peak should be a quartet in a 1:3:3:1 ratio.

20. Protons attached via a double bond show a unique splitting pattern., a doublet of doublets.

23.  Hc  Ha  Hb

26. An unknown molecule A has 4 signals in the 1 H NMR spectrum. Which of the following corresponds to molecule A

27. How many nonequivalent protons does the following structure have? 4

28. Reading from left to right, what multiplicity would be found for the three nonequivalent sets of protons in the 1 H NMR spectrum of the following compound? d, d, s

29. Introduction Homolytic bond cleavage leads to the formation of radicals(also called free radicals) Radicals are highly reactive,, short lived species Single headed arrows are used to show the movement fo single electrons. Production of Radicals Homolysis of relatively weak bonds such as O-O and X-X bonds can occur with the addition of energy in the form of heat or light.

30. Carbon radicals are categorized as primary (1°), secondary (2°) and tertiary (3°) based on the number of attached R groups. 1°1° 2°2°3°3° A carbon radical is sp 2 hybridized with a trigonal planar geometry with the unpaired electron in the unhybridized p orbital. Bond dissociation energy is used as a measure of radical stability.

31. Two different radicals can be formed with the cleavage of a C-H bond. Basically, the more alkyl groups attached to the radical carbon the more stable it is. Also the more stable the radical, the less energy it takes to break the C-H bond.

33. What type of radical are each of the following? 1°1° 2°2° 3°3° Of these three radicals, which is the most stable?

34. Radical Reactions of Alkanes Abstraction of a H from a C-H  bond in which one electron is sued to form H-X while the other is left on the new alkyl radical. A radical can also add to a alkene by adding onto a double bond and leaving the other carbon that was part fo the double bond as a radical.

35. Radicals are highly reactive and unstable and usually react quickly with a sigma or pi bond. However sometimes they can react with another radical. When oxygen, a diradical, is present it acts as a radical inhibitor or scavenger. Meaning it prevents the radical from attacking any alkanes or alkenes.

36. In the presence of heat and light, alkanes and halogens will react to form alkyl halides.

37. Predict the products from the monobromination of the foloowing compound?

38. Step 1 - Initiation Step 2 – Propagation Step 3 - Termination

39. In each step of the propagation a bond is broken and formed. And because the overall step has a -  H it is exothermic. Step 1 is called the rate determini g step because it is higher in energy.

40. Transition States Cl----H----CH 2 CH 3 Cl---Cl---- CH 2 CH 3

41. There are 6 Methyl H’s and 2 Methylene H s. Based on this, the ratio fo the two products should be 3:1(primary to secondary). However, the ratio is 1:1. The more stable the radical being formed is, the easier it is to cleave the C-H bond.

42. Which C-H bond in each compound is most reactive?

43. Chlorination Vs. Bromination 1:1

44. 99%

45. Chlorination is faster and nonselective. This is due to it’s rate determining step being exothermic. Bromination is slower and chooses the most stablew radical. This due to it’s rate determining step being endothermic.

46. Halogenation is useful in the formation of alkenes. An elimination in the presence of a strong base is responsible for the formation of the alkene.

47. Conversion of the alkene to an alcohol via nucleophillic substitution is an extension of the utility of radical halogenation.

48. Oxymercuration-demercuration of an alkene results in the formation of an ether.

49. Radical halogenations give a racemic mixture of pRodcuts when possible. This halogenation of an achiral compound results in 3 products. A primary and secondary alkyl halide. The secondary halide exist as a pair of enantiomers due to the creation of a stereogenic center upon halogenation.

50. enantiomers diastereomers

51. Only achieve enantiomers or diastereomers if the halogenation takes place at a stereogenic center.