16 = A, 12 = B, 8 = C. (Ignores +/-)

Mean = 16.6, median = 17. Top scores: 20.5, 20. All stats are for basic part (excluding bonus; 4 on the bonus was very good).

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 (7/14), due next class (7/21). 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 Ch 3-4, 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. The non-Word structures shown below were drawn either with that program or with ChemSketch. See my web page for an introduction to these chem drawing programs.

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. 2,2-dichloro-5-ethyloctane
b. 6-bromo-1-hexene

2. Ethene. One can do a variety of controlled reactions with alkenes, not with alkanes.

Always beware of generalizations. In fact, in the specific case of ethane, a simple symmetrical molecule, the poorly controlled chlorination reaction (Sect 3.9, which we skipped) is useful, because only one (mono)chlorination product is possible (for same reason as discussed below for B2).

3. n-pentane. The more extended chain has a larger surface area, thus allowing more contact ‑‑ and therefore greater London forces -- between adjacent molecules. (Fig 2.3)

Important. This question is about physical properties -- and intermolecular forces. Chemical reactivity and strength of ordinary covalent bonds are irrelevant.

4. a.
		Note the use of up and down wedges to show the trans nature of this compound.

An alternative is to show an edge view of the ring, with clear up and down. (Sect 3.5 for both ways.)	

b. Many answers possible. Key issue is that it must have one double bond. The original compound is short 2 H because it has one ring. The other way to be short 2 H is to have a double bond. Since the original compound is C12, one simple answer would be 1-dodecene.

c. The terminal -CH3 groups are primary. The ring -CH2- groups are secondary (and there is also one in the ethyl group).

d. 120 angles are around the double bond in part b. There are no 90 angles.

There are no 90 angles in organic chemistry, period. That is not entirely true, but almost. The only “natural” angles (nominal, no strain) around C are 109, 120, 180; Section 1.7. (Cyclobutane has nominal 90 angles, but actually it is distorted a bit, to avoid eclipsing all the H’s; p 82 bottom.) 

There was some difficulty here with unclear answers. You were asked to mark angles. But it was sometimes hard to tell what you were marking. In some cases you seemed to mark bonds, or just generally had an arrow pointing vaguely towards some C. An angle is a very precise idea. A bond angle requires designating, somehow, the three atoms that form the angle.

e. C12H24 (l) + 18 O2 (g)  12 CO2 (g) + 12 H2O (g)

f. 

The key point is that you cannot put both alkyl groups equatorial in this structure. So you put the bigger one equatorial.

In drawing chair structures, try to use the corner positions as much as possible; things are clearer there.

In both parts a and f, it helps to show both groups at any substituted positions. That is why I asked you to show the H here. It may be clear enough without them in professionally drawn structures, but with handwritten structures it can be hard to tell which way things are pointing if you show only one.

The question noted the assumption of a chair. As you can see, there is a conflict here. Because of that, the actual conformation will be something more twisted, with both of the bigger groups close to equatorial. The question as stated helps you to see the problem; the actual solution it suggests is not entirely correct.

5.
	
	The main difficulty here was with the side chain. 1-butene is C‑C‑C=C. Thus, 1-butenyl is C‑C‑C=C‑. 

(The possible group C‑C‑C‑C= is called butylidene, and has not come up.)	

6.
        H3C     Cl *hi
           \   /
            C=C
    *hi    /   \
C-C-C-C-CH2     CH3
8           3 2 1	At the right hand end of the double bond, Cl>C by atomic number. At the left hand end, the first atom is C for both groups. The upper C is attached to HHH, the lower C is attached to CHH; the lower C is priority. Thus the priority groups (*hi) are down at the left, up at the right; they are on opposite sides = entgegen = E.	

7. a. pentane

You did not need to draw the reactant in order to answer this, but if you did, you should draw it correctly. Note that the cis-ness of the reactant has no particular relevance to what happens, and to put cis in the name of the product is just wrong.

b. 1,2-dibromo-3-propylcyclohexane (Add Br-Br across the double bond.)
c. (Reminder: Question is about an alkane; “hexene” in the question should read “hexane”; the name should simply be propylcyclohexane.) no reaction (alkanes are inert)

8. 3,4-dimethyl-3-hexene + HCl. Simple addition to give desired product. 

(Use of the analogous 2-hexene compound leads to a mixture of products. Markovnikov’s rule, which was not supposed to be on this test, would help you some. But using the symmetrical 3‑hexene is best.)

B1. Key is to draw the Lewis structures -- and count the electrons properly. A carbocation has three bonds and no extra electrons around the central C. Thus it is “trigonal planar” (120 angles around the C). A carbanion has three bonds plus a lone pair. Thus the electron geometry is tetrahedral, and the ion itself is a “trigonal pyramid” (like ammonia, Fig 1.3) (nominal 109 angles).

B2. a. C8H8 (one H at each corner)
b. One; the compound is entirely symmetrical, and all 8 H are equivalent. Put the Cl at any corner; doesn’t matter.

c.
   C      step 1: a
           quaternary C
CCC      (“C5 core” see below)
   
   C	   CC        final answer, 
               C skeleton
CCCC C#1
      
   CC C#5	

The H count tells you that you are short 5 pairs of H, so you will need 5 “extra” bonds (beyond single). Start by drawing a quaternary C (“step 1” above). That uses 5 of your 8 C, and allows for no extra bonds. So you now have 3 C left and need 5 extra bonds. The only way to do that is to attach those 3 C with two triple bonds and one double bond to your “C5 core”. Check that the “final answer” shown above indeed has 8 H.

d. 3-ethynyl-3-methylpent-1-ene-4-yne

That the double bond belongs in the parent chain is a subtlety of the priority rules. I did accept some reasonable variations of this, but the name I gave is the proper name. (Try it with ChemSketch or ISIS/Draw.)

e. It really is called cubane. 

cq\bq1a	Summer 2003
6/29/03