16 = A, 12 = B, 8 = C; same as Test 1. (Ignores +/-)

Mean = 16.9, median = 17.5. Top scores: 2 at 19.5.

To estimate your current course grade, average your two test letter grades so far. If relevant, round toward Test #2. I will apply the same basic strategy throughout the course. (I am happy to discuss individual situations with you.)

Test 3 has a reputation of being considerably harder, presumably because of the amount of material behind us -- and the complexity of certain things such as acetal formation. You are forewarned.
 
A brief, partial answer key follows. Alternative answers are possible in some cases. I have tried to minimize drawing structures here. (You can ask me about any of these, in class or privately.)

1. a. 2-decanol, due to the polarity (specifically, ability to hydrogen bond) of the alcohol (‑OH).
b. 2-propanol. Each molecule has the same polar group. The smaller propanol has a smaller nonpolar part; the nonpolar part tends to reduce the solubility.
c. 2-decanol. It is more polar than the decane, and heavier than either of the others. Thus it can stick together by both London forces and hydrogen bonding, more than either of the others.

Actual data for BP (in C): 2-decanol, 211; decane, 174; 2-propanol, 82. For solubility, both of the C12 compounds are listed as insoluble, but we would expect a (very) slight solubility of the alcohol. 2-propanol is miscible with water.

2. 
C-C-C-C-CH2OH
   
  O C-C

Some of you had the wrong starting compound here. If your starting compound did not include both a 1 and 2 -OH, then you were missing the point of the question.

Based on M W Pelter et al, A microscale oxidation puzzle. J Chem Educ 77:1481, 11/00. The oxidizing agent is sodium hypochlorite in acetic acid.

3. The main chain includes the double bond and both I; the double bond is priority. So, ignoring the chirality, the compound is 3,6-diiodo-1-hexene.

The ranking of the four groups at the stereocenter are, from #1: -I, CH=CH2, CH2CH2CH2I, ‑H. Thus the stereocenter is R, and the final complete name is (R)-3,6-diiodo-1-hexene.

For the rankings, I>C>H by atomic number. But this does not resolve #2-3; both of those have a C attached to the stereocenter, so we must look at the next atoms. The double-bonded C is attached to CCH; the single-bonded C is attached to CHH. CCH>CHH.

4. a. The main chain must include the priority -OH group (which will be at C#1), and can easily include both ring groups. The compound is therefore 4-benyzl-2-butyl-3-phenylheptan-1-ol.

Someone asked CH3O- is methoxy; HOCH2- is hydroxymethyl. In both cases the information is in order in the name. However, neither is relevant in this case; the -OH must be in the main chain.

b. C #2, 3, 4 of the main chain.
c. For three stereocenters, there are 23 = 8 stereoisomers.

5. There are only three isomers possible for C3H8O, but there are 72 isomers possible for C8H18O. Of course, you don’t know those exact numbers in advance, but you should realize that they are “few” and “many”. Draw a C3 skeleton; there is only one way. Then, there are three (non-equivalent) places one can put an O:
•	on a primary carbon, giving a primary alcohol;
•	on a secondary C, giving a secondary alcohol;
•	between two C’s, giving an ether.

6. (R)-2-pentanol and (S)-2-pentanol are enantiomers. They must have opposite signs of optical rotation. (You have no way to know which is which, only that they cannot both be +. That is why the R/S, or D/L, and +/- are sometimes given, as additional information.) (This is similar to the problem with questions 6.21-22, which I alerted you to in the handout.)

3-pentanol is not chiral, so the R is incorrect. (I objected to some of you saying that the R is “unnecessary”. That would imply that it is “ok”, i.e., optional. No, it is just wrong.)

7. Infinite number of possible answers.
a &b. One simple logical way to approach this is to a write a C, then put four different CH groups on it: -H, methyl, ethyl, propyl. That gives 3-methylhexane.
c. A simple approach is to take all the C you used in part a and put them in one unbranched chain (heptane, in my example).

8. benzene + nitric acid  nitrobenzene + water

9. a.
H     Ph (phenyl group)
 \   /
  C=C
 /   \
H     H

b. Hydrophobic (repels water); it is entirely nonpolar.

There are two kinds of petri dishes for use with bacteria. The original type was glass, which is hydrophilic; these have largely been replaced by polystyrene dishes, which are hydrophobic. Waters “wets” (sticks to) glass, but not to the polystyrene. For bacteria, both kinds work fine. But when people started growing animal cells in petri dishes, they found that glass dishes worked and plastic (polystyrene) ones did not. Turns out that (many) animal cells require a hydrophilic surface. Plastic dishes for animal cell work have been developed, by special treatment of the surface to provide a hydrophilic layer.

Chemical reactivity is not relevant to the question. And there is no reactivity. Polystyrene contains no (non-aromatic) double bonds, and in general, no reactive sites.

B1. a. 
H       H
 \     /
  C=C=C
 /     \
H       H

b. sp. The middle C thus has two sp orbitals and two p orbitals. That central C forms a double bond with each terminal C, using one sp orbital ( bond) and one p orbital ( bond). Using sp orbitals gives the linear geometry that you would expect here from VSEPR.

c. No, as with any double bonds, there is a major restriction of rotation.

d. 90. (Fig 1.7.)

e. 2. This is a chiral molecule, despite not having the usual type of stereocenter. The molecule is rigid and tetrahedral. The four groups attached to C1 + C3 are “just as tetrahedral” as the four groups attached to a single C. Thus the given molecule does not have a symmetry plane.

B2. a. First, one needs to recognize that three products are possible, with the Br at 2, 5, or 4-or-6 (4 & 6 are equivalent). Bromination at 5 is definitely least likely, because this is meta to both methyl groups, which are o,p-directing. The other three positions are all favored by the o,p-directing -CH3 groups. Of those, the 2 position is least likely, because of steric hindrance (it is crowded there). Thus the best answer would be with the Br ortho to one -CH3 and para to the other.

b. 1,4-dimethylbenzene (p-xylene). It is symmetric, and the only possible product is 2-bromo-1,4-dimethylbenzene.

cq\bq2a	Summer 2003
7/26/03