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1. Here are the names of two of them: 2-chloro-1-butene, 3-chloro-1-butene.
2. The above compounds have 1 double bond, thus have two H less than the maximum possible. What other way can one lose 2 H?
One possible answer is chlorocyclobutane. There are four others; can you draw them -- and name them?
3. What is the difference? One has a C=C double bond, the other has a ring. What is a characteristic reaction of one of these? See Ouellette Sect 4.8. Many reactions would work in principle. In practice, the reaction with Br2 is particularly easy: just add a few drops of bromine solution, which is red. If the color disappears, it means the bromine has reacted. And therefore what do you conclude?
4. Draw the compound you want to make, and think backwards. You need to add one Br atom. Thus, using Br2 would not work. What would you use? What alkene would you add it to? Think about Markovnikov's rule. Is your proposed reaction still valid?
5. Can you have a C with two triple bonds? Explain.
6. An example: 3-methylcyclohexene. Molecular formula is C7H12.
Complete combustion is the reaction with molecular oxygen, to form the major stable oxides (Ouellette Sect 3.8).
C7H12 + 10 O2 --> 7 CO2 + 6 H2O
7. a. The two methyl groups are on the same side, thus are "cis" to each other.
b. The way the parent chain crosses the double bond is "trans". In shorthand, we might say that the parent chain is "trans". (It is important that you recognize this point, regardless of any naming issues.)
c. E. And if you didn't think to use the E,Z system here (despite the hint), go back and try to figure it out before reading the explanation that follows. The point is that the E,Z system has no ambiguity, and thus is clearly called for when there seems to be some ambiguity about applying cis-trans.
How do you figure out E vs Z here? At the left end of the double bond, -CH3 ("upper" group) is higher priority than -H (C is higher atomic number than H). At the right end of the double bond, -COOH ("lower" group) is higher priority than -CH3 (the C are equal, but one has OOO, which is above HHH in priority). Thus the two priority groups are on opposite sides of the double bond (one upper, one lower); opposite = entgegen = E.
Ouellette presents this example in part on p 109 in Example 4.4. He made a remark about cis-trans for the compound, but it struck me as rather incomplete and maybe misleading. That led to this question. For the record, the compound would best be called trans, since it is the parent chain that counts. But I suspect not everyone remembers that. E,Z is better!
d. (E)-2-methyl-2-butenoic acid or (E)-2-methylbut-2-enoic acid
The common name of this compound is tiglic acid. The Z-isomer is known as angelic acid.
8. For the non-ring (upper) double bond... At the top, Br > H; at the bottom, O > C. Thus the priority groups are both on the same side, and this double bond has configuration Z = zusammen = together.
For the ring double bond (lower, left)... At the top, Br > C; at the bottom, C > H. Thus the priority groups are on opposite sides, and this double bond has configuration E = entgegen = opposite.
Usually one would not bother to specify the configuration at the ring double bond here. With a five-membered ring, the ring must be cis.
The compound in this question is a synthetic compound designed to mimic a similar compound isolated from a red alga which interfered with bacterial cell-to-cell communication. W D Bauer et al, Eukaryotes Deal with Bacterial Quorum Sensing. ASM News 71:129, 3/05. The article is archived at http://web.archive.org/web/20070715230515/http://www.asm.org/ASM/files/ccLibraryFiles/FILENAME/000000001442/znw00305000129.pdf. For more about such communication between cells, see my page Unusual microbes, section Bacteria that can count -- and talk.
9. 1, 2, 3, 3, 3, 120.
Note how the first five numbers are all interrelated. The bond angle comes from VSEPR theory, Ouellette Sect 1.7. Ouellette discusses the orbital model for alkenes in Ch 1 Sect 8.
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Last update: October 04, 2011