Test 4
Dr. Bachrach
CHEM 2319
4/23/2008

1) (12 points) Provide the IUPAC name for the following compounds.

2) (15 points) Predict the product(s) of the following reactions. If there is more than one product, indicate which is the major product.

a)

The diene is locked into the s-trans conformation and cannot participate in the Diels-Alder reaction.

b)

c)

3) (12 points) How would you distinguish the following two isomer using ¹H and ¹³C NMR?

¹³C NMR – Both molecules have 5 unique carbon atoms and will therefore have five peaks in the spectrum. One might be able to distinguish the methyl carbons in A being farther downfield than those in B, but that’s pretty iffy. Basically, the ¹³C NMR will not really distinguish these two compounds.

¹H NMR – on the other hand the proton NMR will easily distinguish these two. Both compounds have 3 unique protons, numbered in each structure above. So let’s examine what the signals from each proton will be:

A -

protons 1, singlet δ ~ 4.2ppm

protons 2, singlet, δ ~ 1.4ppm

protons 3, singlet, δ ~ 2.0ppm

B -

protons 1, singlet, δ ~ 1.7ppm

protons 2, triplet, δ ~ 4.0ppm

protons 3, triplet, δ ~ 2.0ppm

So the real simple distinguishing feature will be that the spectrum of A will have three singlets, while the spectrum of B will have a singlet and two triplets!

4) (12 points) Consider (3E)-1,3-pentadiene and (3Z)-1,3-pentadiene. Which isomer will have the greater percentage of s-trans conformer at room temperature? Explain your answer.

The repulsion between the hydrogen and methyl group in the s-cis conformer of the 3Z isomer is much more than the repulsion between the two hydrogen atoms in the s-cis conformation of the 3E isomer. Therefore, the 3Z isomer really does not want to be s-cis, so pretty much all of it will be in the c-trans conformation. Since the repulsion isn’t so bad in the s-cis conformer of the 3E isomer, both conformations will occur, though again most will be c-trans. It’s the relative % of c-trans that matters here – and the 3Z will have a higher % of s-trans than with the 3E. In other words, give the equilibrium constants for both rotations:

K(E) > K(Z).

5) (15 points) The ¹H and ¹³C NMR spectra below is for a compound with molecular formula C₅H₉OCl. Determine the structure of the molecule. Matching each proton with its corresponding peak and indicate what you learned from the ¹³C NMR spectrum.

From the ¹³C NMR we can learn that all five carbon atoms are unique and the signal at 210 ppm is from a carbonyl carbon (C=O).

The proton NMR tells us the following: The triplet at 3.6 means this proton(s) is connected to a chlorine and also a CH₂ group. The singlet at 2.1 is a methyl group connected to the carbonyl (CH₃-C=O). The triplet at 2.6 is from hydrogen(s) on a carbon adjacent to the carbonyl and a CH₂ group (-CH₂-CH-C=O). We still have the hydrogens at 2.1 which looks like a quartet or even messier –probably a CH₂ group with to CH₂ neighbors. Putting this all together gives us ClCH₂CH₂CH₂C=OCH₃.

6) (10 points) Identify all important IR signals. What functional groups are present?

Signal at about 3100 cm^-1 – due to C=C-H stretch if the C-H bond<

Signal at 1720 cm^-1 – due to the C=O stretch

Signal at 1660 cm^-1 – due to the C=C stretch

Also present: 3000 cm^-1- alkyl C-H stretch, 1300-1400 cm^-1 alkyl C-C stretches

4) (10 points) Provide the appropriate reagent to affect the following reactions.

a)

b)

8) (14 points) The reaction of 1,3,5-hexadiene with HCl leads to three different stereoisomers (not including E/Z isomers). What are the three products and provide a mechanism to account for the formation of all three products.