MCAT Chemistry - Organic Chem

Sn2 Rate and Transition State The SN2 reaction is one of the most common in organic chemistry. CH3O- + CH3Br→CH3OCH3 + Br-    The reaction of methoxide ion with methyl bromide proceeds via an SN2mechanism. It has a second order rate law: rate = k [CH3O-] [CH3Br]    The reaction is first order with regards to methoxide ion and methyl bromide, and second order overall. This means that the transition state must involve one molecule of methoxide and one molecule of methyl bromide. The sim plest possible transition state is one where the methoxide nucleophile substitutes for the bromide ion: This is a trigonal bipyramidal transition state. The dashed lines represent forming and decomposing bonds. The δ - symbolizes a partial negative charge. The transition state is formed by the flow of electrons from the nucleophilic methoxide ion into the C-O bond. At the same time, the electrons from the C-Br bond flow onto the bromine leaving group, weaken the bond and place a partial negative charge onto the bromine atom. All of this happens simultaneously, and the reaction is said to be concerted. All SN2 mechanisms are concerted.

SN2 Stereochemistry The rate law indicates which molecules are in the transition state, but it does not specify how they come together. This can be accomplished by attacking a stereocenter: R1 , R2 and R3 represent three different groups. LG is a good Leaving Group.The nucleophile can attack the stereocenter in two ways. In frontside attack, it attacks from the same side as the leaving group. In backside attack, it attacks from the opposite side of the leaving group. These two modes of attack give retenti on and inversion of stereochemical configuration, respectively. Retention and inversion will yield two different stereoisomers. Purely SN2 reactions give 100% inversion of configuration. Thus SN2reactions must occur through backside attack. The phrase "inversion of configuration" may lead you to believe that the absoluteconfiguration must switch after SN2 attack. This is not always true. Recall that absolute stereochemical configuration is determined via the Cahn-Ingold-Prelog (CIP) system, and categorized with the labels "R" and "S." The absolute configuration inverts (from "R" to "S" or vice-versa) only when the nucleophile and leaving group have the same CIP priority relative to the other substituents. Since good nucleophiles and leaving gro ups tend towards equally high CIP priorities, most SN2 reactions do result in a switch of absolute configuration. If the nucleophile and leaving group have different relative CIP priorities, however, the absolute configuration does not necessarily change even though inversion occurs. Keep on your toes.A favorite test question presents a SN2 reaction that results in 100% retention of both absolute and general configuration. In this case, it's likely that two SN2reactions take place. Two inversions yield a net retention of configuration.

Molecular Orbital Explanation of SN2 The stereospecificity of theSN2 immediately begs the question of why it must proceed through backside attack. There are two common explanations: The steric argument states that there's simply more room for the nucleophile to attack from the back. If the nucleophile attacked from the front, it would collide with the leaving group. This steric clash makes front side attack impossible. The molecular orbital hinges on the σ * carbon-leaving group antibond. Recall from molecular orbital bonding theory that electrons donated into an antibondg orbital weaken the corresponding bond. So as the nucleophile donates electron density into the σ * C-LG antibond, the σ C-LG bond will weaken. The σ C-LG bond must break for the leaving group to leave. Since the σ * C-LG antibond is very accessible from the back, MO theory gives a ti dy explanation for backside attack.

The Sn2 Reaction

The E2 Reaction

E2 Rate and Transition State The rate law of the above E2 reaction follows: In the E2 reaction, a base removes a β -hydrogen, forms a double bond and kicks out the leaving group. The reaction occurs through a concerted mechanism and requires a β -hydrogen. This mechanism is also called β -elimination. The E2 rate law is first order for both reactants. Here's the simplest possible transition state:

E2 Rate and Transition State The rate law of the above E2 reaction follows: In the E2 reaction, a base removes a β -hydrogen, forms a double bond and kicks out the leaving group. The reaction occurs through a concerted mechanism and requires a β -hydrogen. This mechanism is also called β -elimination. The E2 rate law is first order for both reactants. Here's the simplest possible transition state:

E2 Stereochemistry As in the SN2 reaction, the mechanism of transition state formation can be deduced using a stereocenter. There are two possible modes of reactivity that yield two different stereoisomers.In syn elimination, the base attacks the β -hydrogen on the same side as the leaving group. In anti elimination, the base attacks the β -hydrogen on the opposite side of the leaving group. It has been experimentally determined that E2 elimination occurs through an antimechanism.

Steric and Molecular Orbital Explanation of Anti Elimination Just as with the SN2 backside attack mechanism, there are steric and molecular orbital explanations for E2 anti elimination.The steric argument notes that the syn transition state is an eclipsed conformation. The anti state is in the more energetically stable staggered conformation. Since the rate of a reaction is inversely proportional to the stability of the transition state, the anti route should be much more favorable.The E2 molecular orbital argument hinges on the σ * C-LG antibond, just like theSN2 MO argument. In this case, the σ * C-LG antibond is only accessible to the electrons of the σ C-H bond if th e two bonds are antiperiplanar. In other words the C-H and C-LG bonds must point opposite directions in parallel planes. If this condition is met, the electrons of the σ C-H bond donate into the σ * C-LG antibond. The C-H σ electrons attack from the "back," of the C-LG bond in a manner similar to SN2 's backside attack.In the syn elimination, the bonds are not antiperiplanar, and the electrons of the σC-H bond cannot attack the σ * C-LG antibond. Thus E2 cannot occur through synelimination.

Saytzeff's Rule Recall the first E2 reaction presented in this section: There's no reason why the methoxide ion can't attack a β hydrogen on the "right" methyl group. The product of both β -eliminations is the same for this reaction, but there are many E2 reactions that can yield enantiomers or diastereomers depending on which β hydrogen is eliminated. The three products that result from the β -elimination of the three underlined β-hydrogens. Product 3 is the major product. But wait! The observed products seem to contradict the antiperiplanar argument. There are three products and three β -hydrogens, but only two of those hydrogens may be antiperiplanar to the C-OTs bond at one time. How can the two hydrogens off carbon B give two products, even though only one can be antiperiplanar?While only one hydrogen off carbon B can be antiperiplanar at one time, both hydrogens spend some time in the antiperiplanar position. Thus the elimination of the hydrogens off carbon B gives the cis and trans products. This is possible because alkanes that are not inordinately hindered rotate about their C-C bonds. For an example of a structure that cannot rotate freely about its C-C bond, see problem #4. A deeper discussion of C-C bond rotation can be found underconformational analysis.Product 3 is the major product because an E2 elimination favors the formation of the most stable alkene. The stability of an alkene is proportional to the degree of substitution at its double-bond carbons. The double-bonds of products 1 and 2 are less branched than the double-bonds of product 1. Thus there will be more of product 3 than products 1 and 2 after the reaction has run to completion. E2 's preference for the more stable (and more highly branched) alkene is called Saytzeff's Rule.

Sn2 VS. E2

SN2 and E2 reactions share a number of similarities. Both require good leaving groups, and both mechanisms are concerted. SN2 reactions require a good nucleophile and E2 reactions require a strong base. However, a good nucleophile is often a strong base. Since the two reactions share many of the same conditions, they often compete with each other. The the outcome of the competition is determined by three factors: the presence of antiperiplanar β -hydrogens, the degree of α and β branching, and the nucleophilicity vs. basicity of the reactant species. In order for an E2 elimination to occur, there must be antiperiplanar β -hydrogens to eliminate. If there are none, theSN2 reaction will dominate. On the same token, the SN2nucleophile needs an free path to the σ * C-LG antibond. α andβ branching block this path and reduce the proportion ofSN2 relative to E2 . E2 occurs even with extensive branching because it relies on the β -hydrogens, which are much more accessible than the σ * C-LG antibond.The identity of the nucleophile or base also determines which mechanism is favored. E2 reactions require strong bases. SN2 reactions require good nucleophiles. Therefore a good nucleophile that is a weak base will favor SN2while a weak nucleophile that is a strong base will favor E2 . Bulky nucleophiles have a hard time getting to the α -carbon, and thus increase the proportion of E2to SN2 . Polar, aprotic solvents increase nucleophilicity, and thus increase the rate of SN2 .SN2 Requires an unhindered path to the back of the α carbon α and β branching block the path and hinder SN2 Requires a good nucleophile Polar, aprotic solvents increase nucleophilicity Bulky groups on the nucleophile decrease nucleophilicity E2 Requires an antiperiplanar β -hydrogen Enhanced by α and β -branching Requires a strong base

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