17 = A, 13 = B, 8 = C, 4 = D. My experience with classes such as this suggests that 80% = A, 60% = B, etc, are good cutoffs for the kind of tests I give. (Scores very near those cutoffs are given a + or -. For example, the range 13-16.5 includes all B, from B- to B+.)

I am not showing the class statistics; that seems appropriate for a small class, but I am happy to discuss the issue.

For each question, the circled number is your total points for that question. (Points for parts may be shown, and are then added together to get the circled total for the question.) Check the arithmetic!

Be sure to read the comments on the test, regardless of scoring. It is often appropriate to comment on certain features of your answer even if I do not take off points. Learn from the feedback!

I encourage you to go back and read the “Test follow-up” section of the “Sample test” handout, especially if you are not happy with how you have done so far. Happy to discuss. Remember, you are the judge of how you are doing relative to your background, needs, and time commitment.

Test 2. Take home test, handed out class 7 (3/20), due next class (3/27). Test 2 should cover Ch 5 through 9; we will decide for sure at the end of class 6. (But note that it will also review the earlier Ch, simply because the material is inherently cumulative. It’s quite worthwhile to work on any major issues that bothered you from those earlier chapters.) Test date is firm, even if chapter coverage slips. Test 2 will be similar in style and length to Test 1. However, former students would point out that reactions play an increasingly important role as we go on (and are probably more complex than the few so far).

Think about the note sheet. Did it help you during the test? How might it have helped more? Did it help you during preparation for the test? For many people, that turns out to be more important than its use during the test. 

Chapter coverage will be shown on the cover page of the test (and posted on the Updates page at the web site). Remember that sample tests beyond the one handed out are at the web site, with answer keys. The cover sheet for the take home test is there, too.

The chair template provided on the test is from ISIS/Draw. My non-Word structures shown in class materials are drawn either with that program or with ChemSketch. See my web page for an introduction to these chem drawing programs. (I have them installed on the classroom computer, and can give brief demos.)

A brief, partial answer key follows. I have skimped on drawing structures. (You can ask me about any of these, in class or privately.) Alternative answers are possible in some cases.

1. a.
	
	b. I is larger than Cl, so gets the more spacious equatorial position. Cl is forced to be axial.

In drawing chair structures, remember that axial positions alternate up/down as you go around. Also, it is generally easier to show clear ax/eq at the four corner positions of the chairs. You will usually make things easier for yourself if you use those corners as much as possible.

2. There are only two: 1-bromopropane and 2-bromopropane.

3. a. CH2=C=CH2 (propadiene; common name allene). (Also: carbon dioxide.)
b. No. C forms 4 bonds. Two double bonds is fine (leaving no room for anything else). But two triple bonds would exceed the bonding capacity of C.

4. The upper line (diamond symbols) is for unbranched alkanes. As a good rule of thumb, the unbranched isomer has more surface area, and hence greater London forces between molecules, thus leading to higher BP. In other words, branching makes the molecules “more compact”, thus harder to pack together and make good contacts with each other.

Once again, note the fairly small magnitude of the branching effect. The difference between the two lines is small compared to the difference between consecutive C numbers on the same line.

5. Combustion is the only reaction typical of (almost) all organic compounds. It is the only reaction of alkanes that we consider. I will use propene, CH3CH=CH2, C3H6, for my example: 
C3H6 (g) + 9/2 O2 (g)  3 CO2 (g) + 3 H2O (g).

6.
            CH3
            
C-C=C-C-C-C-C-C + H-Cl  
          
  H CH3     CH3
1 2 3       7 8	  H Cl      CH3
          
C-C-C-C-C-C-C-C
          
  H CH3     CH3
8   6       2 1	

I showed the relevant H in the reactant, and underlined the adding groups in both reactant and product. Addition follows Markovnikov rule, with the adding H going to C#2, which has more H (1 vs 0).

b. C# 3, 4, 5, 7 are all 2. C#2 is 4. Note numbering; see part c.

c. The reactant is numbered from the left, since the double bond takes precedence. But the product is numbered from the right, with the dimethyl determining the numbering. So it is
6-chloro-2,2,6-trimethyloctane.

There is a potential pitfall when I ask you to name a product you have drawn. What if you have drawn the wrong compound? My general approach is to grade your name based on the compound you drew, so long as it is “reasonable” -- that is, a valid compound and a reasonable answer to the original question, and of comparable difficulty to name. For example, if you violated the M rule here, I took off in part a for having the wrong product. But it was then fine to name the product you drew. There is some judgment of what to accept as “reasonable”, though most cases seem fairly clear to me.

7.
Reaction is addition of H-OH across the double bond, acid-catalyzed. There would be a mixture of products, since the H-OH could add in either orientation. Either one, alone, was fine.

Products are named 3-ethyl-4-heptylcyclohexanol and 2-ethyl-3-heptylcyclohexanol. The priority group in cyclo compounds, double bond or alcohol, is understood to be at the 1-position. That is, the priority group defines C#1. Naming of alcohols is in Ch 8.

8. a. 

C-C=C-C-C + Cl-Cl  
          
  H H
1 2 3 4 5  	 Cl Cl
   
C-C-C-C-C
   
  H H
1 2 3 4 5	

I have drawn this using the same basic conventions as in #6, above. But note that the geometry at the double bond is now important, since the question specified cis.

b. 2,3-dichloropentane

c. 120 (typical at C=C)

d. Many possible; perhaps simplest is cyclopentane. 

Key logic is that C=C and rings have same effect on H-count: each “costs” 2H per extra C-C bond formed. If you don’t see that right away, figure out the molecular formula of the original compound, and then make sure your proposed isomer has the same molecular formula -- but different structure.

e. No reaction. Alkanes are inert under standard conditions for simple addition reactions.

All of the reactions on this test are simple examples illustrating the major reaction characteristics of the major classes of organic compounds that we have discussed so far. If you got significantly less than full credit on these, this should be of concern. 

Bonus. They are intended as “more challenging”. You benefit most from them by trying them (struggling with them) before just reading the answers. If you didn’t have time to give them full attention yet, you might do that before reading further.

B1.
	   Cl
   
Cl-Pt-NH3
   
   NH3	The name illustrates the idea of cis, in a different context. It would be possible to have an isomer with the two Cl opposite each other -- trans. In this isomer, they are next to each other, on same side -- cis.	

B2. Octane is more volatile than decane (same chemical family, only difference is size, and the octane is smaller, thus has fewer electrons and lower London forces). (We say it has a higher “vapor pressure”.) Thus the vapor phase in the container -- the bottle or, in the real case, the fuel tanks of the plane -- would have >30% octane.

Analysis of the accident involved the question of how flammable the vapor in the fuel tank was. That required estimating the composition of the vapor. This question explores the logic of that estimation. Of course, the actual composition was more complex.

J Shepherd, Learning from a tragedy: Explosions and flight 800. Engineering & Science Vol 61 #2, p 19. 1998. A very readable article about the National Transportation Safety Board’s investigation. Free online at: http://pr.caltech.edu/periodicals/EandS/archives/LXI_2.html. Scroll down to the article. I have posted a live link to this on the X402 Updates page, with the Class 5 handouts.

B3. There are many possible answers. Here is one example. The top part of the figure shows two isoprene units. For the right hand unit, I have shown the adding units, the “A-B” for this addition, by explicitly showing the C and H of one bond. Those atoms add across the right hand double bond of the left hand isoprene, giving the result shown. (I asked you to show the addition with arrows. That is hard to do here, but would be clearer than my attempt to describe it in words!)	

The question asked for the molecular formula of the product. C10H16. You can answer that without doing anything else, simply because the question is about an addition of one C5H8 to another C5H8.

B4. cyclooctane. Why? Go look at the sign.

cq\bq1a	Spring 2006
3/4/06