CHEM 2750 Test 3 Study Guide (WITH Mechanism worksheet solutions!)
CHEM 2750 Test 3 Study Guide (WITH Mechanism worksheet solutions!) Chem 2750
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This 11 page Study Guide was uploaded by Tyler Ebeling on Friday March 18, 2016. The Study Guide belongs to Chem 2750 at East Carolina University taught by Shouquan Huo in Fall 2016. Since its upload, it has received 307 views. For similar materials see Organic Chemistry I in Chemistry at East Carolina University.
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Date Created: 03/18/16
CHEM 2750 TEST 3 STUDY GUIDE Nucleophile: A nucleophile is a molecule in a reaction that has electrons to give away. It could be negatively charged, or it could be neutral and have a lone electron pair that it’s willing to give up. For this reason, when you are drawing a reaction, you always draw the arrow pointing AWAY FROM the nucleophile. This makes sense if you think about it; the arrow indicates where electrons are going, and nucleophiles are giving away electrons, so naturally the arrow would point away from them. NEVER DRAW AN ARROW POINT TOWARDS A NEGATIVE MOLECULE. Electrophile: An electrophile is the opposite of a nucleophile. An electrophile (electrophile meaning “electron lover”) wants electrons. It doesn’t have as many electrons as it wants, so it happily accepts extras from a nucleophile. Arrows drawn should always point TOWARDS the electrophile, because the electrons are going to it. **An easy way to remember this is that “electrons go to the electrophile”.** Understand the addition of HBr to an alkene In all of these reactions, notice that the Bromine always bonds to the side of the double bond with the MOST carbons and LEAST hydrogens. This is called Markovnikov’s Rule, which says that in the addition of a halide, (HCl, HBr, or HI), the hydrogen from HCl, HBr, or HI bonds to the carbon of the double bond with the most hydrogens, and the halide bonds to the carbon with the most alkyl groups attached to it. Enthalpy (∆Hº) = Heat (easy to remember - ∆H = Heat). Entropy (∆Sº) = Disorder/randomness of a system. ∆Gº = Gibbs’ free energy. This is a measure of whether or not a reaction will happen. If ∆G is negative, then the reaction will happen. If ∆G is positive, then it won’t happen. The only equation that will likely be on the test is ∆Gº = ∆Hº – T∆Sº. (T = temperature in this equation. Just know how a large ∆H/∆S affects the outcome of ∆G. As you can see in the graph above, the peaks represent transition states, while the lower parts, or, “valleys” represent intermediates. The highest point on the graph (relative to the reactant/intermediate) is known as the rate limiting step or the rate determining step. This step controls the rate of the entire reaction, so it’s very important to know what it is. As you can see by the far right of the graph, it is below the initial dotted line that the reactant started at. Because it’s below this line, ∆Gº would be negative, meaning that the reaction is likely to happen. (Remember, ∆Gº works opposite to what you would think: negative means that the reaction will happen, whereas a positive ∆Gº means the reaction won’t happen.) Hammond’s postulate: This rule states that the transition state of a reaction most closely resembles whatever it is closest in energy to. This may be somewhat hard to understand, so to put it more simply, the “hump” of a graph (transition state) goes to the product/reactant highest in energy. First order rate reaction: rate = k[A] Second order rate reaction: rate = k[A][B] k = rate constant (determined experimentally) Catalyst: something that increases the rate of a reaction. (Only affects the rate, not the amount of reactants/products/etc.) CHAPTER 6: REACTIONS OF ALKENES This is by far the hardest chapter we’ve covered, but with a bit of practice and memorization, it’s not as bad as it initially seems. Reaction mechanism of HX (“X” mostly represents Cl, Br, or sometimes I): You need to know the entire reaction mechanism of HX. Below is a picture of a basic reaction of HBr with 2-butene. This is one of the simplest reactions involving HX and an alkene, but it’s important to understand what’s happening, and be able to draw it for other alkenes. Carbocation: A carbocation is a positively charged Carbon atom. You need to understand what affects the stability of a carbocation. The easiest thing to remember is that a tertiary carbocation is more stable than a secondary, is more stable than a primary. In other words, the more carbons the carbocation is connected to, the more stable it is. Hydration and Alcohol reactions: you need to be familiar with hydration and alcohol reactions and how they add themselves to alkenes. Notice, from the ﬁgure below, how the hydrogen from the H2O/alcohol adds itself to the carbon that has the most hydrogens. This follows Markovnikov’s rule. The remaining OH/R–O group adds itself to the carbon of the double bond with the least amount of hydrogens (or the most carbons, depending on how you choose to look at it), which also follows Markovnikov’s rule. Hydroboration-oxidation: This is a very complicated mechanism, and to make matters worse, it actually goes against Markovnikov’s Rule. The BH3 adds itself to the carbon with the most hydrogens, rather than the least hydrogens, and the reason for this is steric repulsion. More simply put, it’s because there is more room at a primary carbon than there is at a secondary carbon. If you are unable to understand exactly what’s happening in this reaction, I think that’s ok; as long as you know and memorize the steps of the reaction below. Epoxidation of alkene: This is a very complicated one-step mechanism, but it’s relatively simple to predict the product. All you really need to know is that the oxygen is bonded to either side of the double bond. Think of the double bond as one side of a triangle. In an epoxidation reaction, the oxygen bonds to form the other two sides of the triangle. Ozonolysis: This is one of the longest, most complicated mechanisms we’ve learned so far. However, lucky for us, it’s incredibly easy to draw the products if given the reactant. While the reaction below may look extremely complicated, look closer. If you didn’t notice on your own, all that’s happening is the two double bonds are being cut in half, and an oxygen is being added to both new ends. So if you were to take the two smaller groups on the top and lower them both straight down, you can see that they line up with the double bonds from the original reactant. I’ve also attached a simpler reaction to hopefully help you better understand the reaction products. Hydrogenation: First things ﬁrst, DO NOT confuse this with hydration. They ARE NOT the same thing. Hydrogenation only involves H2 (hydrogen), and although many of these reactions also include “Pd/C”, this is just a catalyst, and doesn’t appear in the reaction. Hydrogenation, in my opinion, is one of the easier reactions to understand and draw out because basically all that happens is the double bond goes away, as seen below. Oxidation/reduction: There’s an easy way and a hard way to understand this concept, so I’ll tell you the easy way. Oxidation is simply when there is a decrease in the number of C–H bonds, or an increase in the number of C–O bonds, or an equivalent such as C–Cl, C–Br, etc. Reduction is the opposite. There’s either an increase in the number of C–H bonds, or a decrease in the number of C–O/Cl/Br/ect. bonds. Chapter 7: Reaction of alkynes Nomenclature of alkynes: You need to know how to name alkynes and/or draw a structure if given the name. The nomenclature is basically the same as alkenes, with the only difference being ending the name with “yne” instead of “ene”. Priority of functional groups: Dr. Huo barely mentioned priority of functional groups, but he said we need to know it for the test so here’s the ofﬁcial priority order.
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