1. Short Answer: 1) What type of electromagnetic radiation is used in nuclear magnetic resonance? radio 2) What is the most abundant peak in a mass spectrum called? Base peak 3) What would the proton NMR peak look like for the indicated hydrogen? Because the two sets of adjacent protons are equivalent this peak would follow the n+1 rule and be a septet.

2. Nuclear Magnetic Resonance Spectroscopy 1 H NMR—Spin-Spin Splitting When two sets of adjacent protons are different from each other (n protons on one adjacent carbon and m protons on the other), the number of peaks in an NMR signal = (n + 1)(m + 1). Figure 14.8 A splitting diagram for the H b protons in 1-bromopropane

3. 4) To which end of an alkene does the hydrogen add in hydrohalogenation without a radical initiator? The hydrogen adds to the least substituted end of the double bond so that the most stable carbocation is formed. 5) What is another name for the 1,4-addition product? Conjugate addition product

4. Predict the products. 1) This cation is 2° and 3°. Other cation is 2° and 2°. Thermodynamic product is most substituted alkene.

5. 2)

6. 3)

7. 4)

8. 5)

9. 6)

10. Mechanism: Draw out the Mechanism for the following reaction. 1) Step 1 - Initiation

11. Step 2- Propagation Step 3 - Termination

12. 2)

13. Spectroscopy 1) Using the MS and IR spectra attached (1A and 1B) propose the formula and structure of this compound.

14. MS shows a molecular ion peak at 106 and a M+2 peak at 108.. So 106-35=71 so 71/12=5 carbons so 71-60=11 hydrogens so C 5 H 11 Cl 2(5)+2-11-1=0 However, there is a carbonyl peak in the IR

15. So need to add an oxygen. -CH 4 gives C 4 H 7 ClO so 2(4)+2-7-1=2/2=1 So this is taken by the C=O bond. One more thing, there is no peak at 2750 so no aldehyde, our carbonyl is a ketone The first one can be eliminated because of the base peak at 43 in the MS, a loss of 63 accounts for the loss of a –C 2 H 4 Cl group.

16. 2) Using the MS, IR and proton NMR (2A, 2B, 2C) propose a possible formula and structure. So the molecular ion peak is 165. Odd number means a nitrogen. 165-14=151/12=12 carbons so 151-144=7 hydrogens So C 12 H 7 N 2(12)+2-7+1=20/2=10 way to high So lets take off a C and add 12 Hs C 11 H 19 N 2(11)+2-19+1=6/3 Now lets look at the IR. There is a small peak at 3243 which tell sus thee is some sort of amine present. Also there is a peak at 1649 and with a nitrogen present this tells us there is an amide.Also there is a C=C double bond at 1622. So we need to add an oxygen -CH4 and Get C 10 H 15 NO so 2(10)+2-15+1=8/2=4

17. Alright now looking at the proton NMR we see a doublet of doublets in the aromatic region (6 to 7ppm) which tells that we have a para substituted benzene ring. So the last structure we came up with,, had a DOUS of 4. This only covers the benzene ring so we need to ad one more degree for the C=O in the amide. We know adding an oxygen well take care of this and give us C 9 H 11 NO 2 so 2(9)+2-11+1=10/2=5 So there are 5 peaks in the proton NMR, two of those are the doublet of doublets in the aromatic region and another one is the small peak the furthest downfield which is the amine proton.

18. So that leaves 2 carbons and 1 oxygen. And we know that there is only the para position on the ring to add anything to. So lets look at the remaining two peaks in the NMR they are both singlets which tells us there are no adjacent protons that have an effect. This means that they are either attached to an oxygen or nitrogen. So that only leaves one place to add the other carbon.