CHEM 2750 Test 4 Study Guide (with assignment)
CHEM 2750 Test 4 Study Guide (with assignment) Chem 2750
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This 10 page Study Guide was uploaded by Tyler Ebeling on Saturday March 19, 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 149 views. For similar materials see Organic Chemistry I in Chemistry at East Carolina University.
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Date Created: 03/19/16
CHEM 2750 TEST 4 STUDY GUIDE Hückel’s Rule: An aromatic compound has 4n+2 π electrons. (This means that an aromatic compound ONLY has either 2, 6, 10, 14, 18….etc. π electrons.) Rules of Aromaticity: 1) The molecule must be cyclic 2) The molecule is planar 3) The molecule is fully conjugated 4) The molecule has 4n+2 π electrons (where n=0 or any positive whole number) If a molecule is ANTI-aromatic, then it will match all of the above criteria, but will have 4n electrons instead of 4n+2 electrons (the two examples above that aren’t aromatic are actually anti-aromatic). Here are some examples of aromatic compounds (you should memorize these): Be familiar with 1,2 and 1,4 addition products Know the difference between kinetic and thermodynamic reactions. Kinetic reactions have a lower activation energy and higher-energy end product, while thermodynamic reactions have a higher activation energy, but a lower energy ﬁnal product. This explains why thermodynamic reactions occur when heat is added, because the heat provides enough energy to get over the activation energy “hump” of the thermodynamic reaction. Diels Alder Reaction: You need to know the reactants and products of a diels alder reaction. It’s not as complicated as it seems at ﬁrst. Below is the most basic Diels-Alder reaction, followed by some more complicated examples. The reason this reaction occurs is because it forms new sigma bonds, which are more stable than pi bonds. Nucleophilicity vs Basicity: It’s important to understand the differences between these two, and to be able to rank the strength of certain nucleophiles. Also, it’s important to note that nucleophilicity can depend on what type of solvent a molecule is in. Protic polar solvent: This type of solvent has a hydrogen bonded to an oxygen or nitrogen. (The weakest bases are the best nucleophiles in a protic solvent) Aprotic polar solvent: This type of solvent does not have a hydrogen bonded to an oxygen or nitrogen. In an aprotic polar solvent, the best/strongest base is also the best/ strongest nucleophile. (Basicity and nucleophilicity are positively related) SN2 Reactions: An SN2 reaction is a very important reaction to know for this test. Below is the most basic SN2 reaction, where R represents any group. (Note how the product is INVERTED, relative to the reactant) ALSO, it’s important to understand that SN2 reactions are CONCERTED, meaning that the whole reaction occurs in one single step, the transition state shown below is just to help understand exactly what’s happening in the reaction. SN2 reactions are heavily inﬂuenced by what solvent they are in. They occur the fastest in Polar aprotic solvents (this doesn’t mean that they won’t occur in polar protic solvents, it’s just much slower.) Some examples of commonly used polar aprotic solvents are acetone, DMSO, and DMF. Also, please make note of the fact that SN2 reactions can only occur on primary (1º) or secondary (2º) alkyl halides; NOT tertiary (3º). This means that if you are asked a question about the rate of an SN2 reaction, primary will be the fastest, followed by secondary, and tertiary would be the slowest (so slow that the reaction likely wouldn’t happen). SN1 Reactions: SN1 reactions are similar to SN2 reactions in that they involve substituting one group for another, but different in the way that they occur (two separate steps), and which solvent they work best in. In an SN1 reaction, the leaving group (generally Br or Cl), leaves the molecule ﬁrst. This results in the formation of a carbocation. Then, in the second step, the nucleophile attacks the carbocation, which completes the reaction. (Note how the Br leaving is the slow/rate limiting step) If you’re wondering how to tell whether an SN2 and an SN1 reaction will occur, it’s really quite simple: SN1 reactions can only occur on tertiary alkyl halides. So if you see a primary or secondary alkyl halide, you know that it’s gonna be either SN2 or E2, which we’ll cover shortly. Also, SN1 reactions occur best in Polar Protic Solvents. E1 Reactions: E1 reactions are stereoselective, and are different from SN1/SN2 reactions because they involve elimination rather than substitution. However, E1 reactions are similar to SN1 reactions in that they ONLY occur on tertiary alkyl halides, and that they involve two steps. The ﬁrst step is basically exactly the same as an SN1 reaction: an alkyl halide (leaving group) dissociates, forming a carbocation. But the second step is where E1 and SN1 reactions differ; rather than a nucleophile attacking the carbocation, a base removes a hydrogen/proton from the ß (beta) carbon (the carbon adjacent to the carbocation). Once the proton is removed, the extra electrons are used to form a double bond between the ß carbon and the carbocation. Once again, it’s important to notice that E1 reactions can only occur on tertiary carbons, just like SN1 reactions. (Also note that the Br leaving is the slow/rate limiting step) The chart below would be good to memorize, as it helps you determine which reaction will occur Zaitsev’s Rule: This rule applies to elimination reactions, and says that “an elimination will occur so that the hydrogen being removed will be removed from the carbon with the LEAST amount of hydrogens. You can observe this below by looking at the E2 reaction. E2 Reactions: An E2 reaction is regioselective. It can occur on 3º, 2º, or 1º carbons, although it generally prefers 3º or 2º (1º generally only occurs when steric hindrance is involved. It’s a concerted reaction (1-step), just like SN2 reactions. However, unlike SN2’s it involves elimination of a proton. The products will look basically identical to E1 reaction products, the only difference being how it gets there. As you can see below, instead of the leaving group dissociating, a base (generally a strong one) “steals” a hydrogen, which causes those electrons to form a double bond, which “kicks” Br off of the molecule to exist as an anion. Substitution of alcohols with HX: All alcohols have the OH functional group, which doesn’t make a good leaving group. However, H2O is an excellent leaving group, so when an HX (X being any halide) acid reacts with an alcohol, it forms H2O as a leaving group and a substitution reaction takes place. For alcohols, both SN1 and SN2 reactions can occur, but it still depends on 1º, 2º, or 3º. Substitution of alcohols with PX3 and SOCl2: This is another substitution reaction. In this reaction, the products maintain the same conﬁguration as the reactants. (Pyridine is often used as a solvent because it’s a poor nucleophile but a good base.) Getting Sulfonate Esters from Alcohols: The basic reaction is shown below. Mechanism Assignment: I’m not 100% sure about #4, because no one at the tutoring center knew how to do it so I just guessed based off of the book. Everything else should be pretty close though.
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