UM CHM 201 - Alkanes and cycloalkanes

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Introductory Organic and BiochemistryQuiz: Alkanes and cycloalkanesAnswer key.Introductory Organic and Biochemistry Quiz: Alkanes and cycloalkanesWhen a question asks for an explanation, you will normally receive zero credit if you give an answer without an explanation. The point is to emphasize reasons. Answers result from reasons; answerrs alone are not interesting. 1. Draw the structure of 3,4-diethylheptane. 2. Draw the structure of 2-bromo-2,3,3-trichloro-4-ethyl-1-iodohexane. 3. Is cyclopropane an isomer of propane? Explain. (There is only one relevant consideration. Be clear, and specific.) 4. Draw the structure of cis-1,3-dimethylcyclopentane. (For practice, try to draw this structure two ways: one using "stereo bonds" (e.g., wedges), and one not using them. Be sure that key features are very clear.) 5. Which has the higher boiling point, pentane or hexane? Why? 6. Which C4H10 isomer (n-butane or methylpropane) has the higher boiling point? Why? (Are there any other compounds with the given molecular formula?) 7. a. Draw the structure of propane. Show all the atoms. b. There are 8 hydrogen atoms in your structure in part a. You could replace each one of them (individually) by a chlorine atom. In each case, you get a chloropropane. However, you don't get8 distinct compounds. * How many distinct (mono) chloropropanes are there? * Draw them, and name them. The following parts reexamine the issues from part b, in a somewhat different way. Before doingthis part, try to see that you have the basics from part b. c. Consider replacing each one of those 8 H from the original compound, propane, with a Cl atom. In how many cases do you get 1-chloropropane? d. Now take the structure of propane that you drew in part a. Mark all of the H that you referred to in part c. For example, you might mark these as Ha. (That is, use Ha to mark all the H which, when replaced with a Cl, give 1-chloropropane.) e. Now, use Hb to mark all the H which, when replaced with Cl, give 2-chloropropane.Parts c-e introduce the idea of "equivalent hydrogens". It is hard to explain in words, without reference to a specific example. The purpose here is for you to discover "equivalent hydrogens" by working through an example. Hydrogens are equivalent if they behave exactly the same way. In this case, you see that substituting equivalent H's gives the same compound. Ouellette develops the idea of equivalent atoms in a structure in the Explorations with MolecularModels.. 8. Which would you expect to be more soluble in water: propane or any (mono)chloropropane? Explain. 9. What is the simplest compound that contains a quaternary C atom? 10. Draw the structure of diiodobutadiyne. Show all the atoms (including all the hydrogens). Give the molecular formula.Answer key. 1. Strategy: Draw heptane, the "root" or "parent" compound as the name indicates. Number the C atoms, 1 to 7. (The circled numbers show the chain numbering.) Then attach the indicated groups. Finally, add enough H so that each C has four bonds. 2. 3. No. Isomers have the same molecular formula, but different structures. These two compounds do not have the same molecular formula. Be sure you have the correct chemicals before trying to answer the question. Don't be misled by the names. 4. Both methyl groups are shown on the same side (up, but down would be fine); the given compound is "cis". It is very important that your structure show clearly that the two methyl groups are on the same side of the ring (both up or both down). Look at some of the structures in the book for ways to show this. I will not give you the benefit of the doubt on a critical issue such as this. (For those who really have trouble drawing it clearly, I suggest you add a little note to say what you mean.) See my web page Rings: Showing cis/trans and axial/equatorial relationships for guidance on making such drawings. 5. Hexane. The two chemicals are of the same type (i.e., both are alkanes), and hexane is bigger heavier. Therefore, there are greater London forces between the molecules. 6. n-butane has the higher boiling point (BP). The two chemicals are of the same type (alkanes) and they have the same mass. So what gives them different BP? Their shapes. The more tightly two molecules can stick together, the harder it is to break them apart, thus the higher the BP. Linear molecules can stick together better than branched ones, which tend to look like balls.Simply quoting the boiling points of these compounds misses the purpose of the question, and would receive no credit on a test. The question discusses an issue, and asks for a reason. Of course, if you do happen to know which has the higher BP, that can help guide you, but per se makes no intellectual contribution. The two compounds stated are the only two compounds with the molecular formula C4H10. 7. a. b. There are only two: 1-chloropropane (with a Cl on a terminal C), and 2-chloropropane (with a Cl on the middle C). If you think you have more than 2... I suggest that you make a model of each. See if you can rotate one model so that it looks identical to another. If so, the two structures you thought were different are actually the same. c. 6. Replacing any one of the H on the -CH3 group at either end gives the same result. d & e. 8. The chloropropane is somewhat more polar than propane, so should be somewhat more soluble. (In this case, neither is very soluble. But it is true that the chloro compound is more soluble.) 9. 2,2-dimethylpropane (common name: neopentane). A quaternary C has 4 other C attached to it. 10. EEEE It is C4I2. No hydrogens. So how did we get that? Why are there no numbers in the name to tell us where the triple bonds or I atoms are? The name is diiodobutadiyne. Let's break that down: di iodo buta di yne. Four C; two triple bonds; two I. So lay out 4 C atoms: CECECEC. Now, two triple bonds. Where? Triple bonds cannot be next to each other; that is, one C cannot have two triple bonds. Therefore the only possible place for the triple bonds is at the ends (i.e., 1,3-butadiyne). Now, look for placesto put I -- or H. Each C forms 4 bonds; the only places left are at the ends, so put the two I atoms there. That's it. There are no more bonds available on the C atoms, so there are no H. This compound is an example where numbers ("locants") are not needed to specify the location of groups, because there is only one possibility. If you were asked to name this compound, you might well include numbers, without


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UM CHM 201 - Alkanes and cycloalkanes

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