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Reactions of Alkyl Halides

This is probably the most confusing chapter in the first semester of organic chemistry, the

reactions of alkyl halides. Alkyl halides undergo two basic types of reactions in organic

chemistry, including substitutions and eliminations. There are two types of substitution reactions

and two types of elimination reactions. We will look at each individually and then try to compare and contrast so you know what identifying characteristics to look for, to help you

recognize which reaction is most likely to occur. Remember though, that there are exceptions to most rules. As defined and black and white as we might wish life to be, sometimes we have to deal in grey.

Substitutions:

Ex. AB + CD β AC + BD

There are two types of Substitutions reactions.

1. The S

N

2 reaction

Substitution : one species replaces another (self-explanatory) Nucleophilic : the substitution occurs as the result of attack by an electron-rich species Bimolecular : two species are involved in the rate-determining step (the slowest of the mechanism) Bimolecular: Dealing with the rate of the reaction. The rate of any S N

2 reaction is directly linked

to the concentration of two species, the Nucleophile (Nuc) and the alkyl halide (RX) undergoing substitution. Both of these species are involved in the rate-determining (thus slowest) step. The rate equation would be: Rate = k[Nuc][RX], where k is a constant based on what species are

actually involved. It is important to remember that, as a result, if you double the concentration of

either, you double the reaction's rate. If you cut the concentration in half for either, the rate will

also be cut in half. The rate of the reaction is directly linked to the number of times these two species can collide to react together. Change the concentration and you change the number of

possible collisions!

An example of an S

N

2 reaction:

Br H 3 CH 2 C H 3 C H + OH HO CH 2 CH 3 CH 3 H + Br

Mechanism of SN

2: There are actually two potential ways the nucleophile could attack:

Frontside attack:

B r H 3 CH 2 C H 3 C H + OH The attack from the frontside (the side the leaving group is on) is disfavored for two reasons. First, there's sterics - the nucleophile is trying to attack in the same space the leaving group is trying to leave. Second, there's electronics - the nucleophile is trying to attack bearing a full negative charge and the leaving group is trying to leave with a full negative charge, and they will repel each other (a.k.a. electrostatic repulsions)

Backside attack:

Br H 3 CH 2 C H 3 C H OH Attack from the backside, where the nucleophile comes in from the opposite side of the leaving

group, alleviates both the electronic and steric issues seen in the frontside attack. This is a one-

step concerted (i.e. happens all at once) process where all the bond formations and breakages occur simultaneously. There aren't any intermediates formed (no anions, cations, etc) and it is surmised based on the results that the following is an accurate depiction of the transition state of the process: Br H 3 CH 2 C H 3 C H + OH HO CH 2 CH 3 CH 3 H + Br H 3 C H BrHO CH 2 CH 3 Any chiral center that was once perhaps the "S" configuration, as shown in the starting material here, winds up inverting its stereochemistry and becoming the "R" configuration. Centers that are not chiral cannot "show" this inversion but the inversion process is still occurring.

Transition State of S

N

2: The original starting material was sp

3 hybridized but note that the transition state resembles sp 2 hybridization (turned on its side - with the nucleophile, OH, and the group that's leaving, Br, where the perpendicular p orbital might once sit). It is this planar geometry and the need to put the other groups through the inversion into this "planar" state that define what alkyl halides can undergo the S N

2 reaction.

Transition State:

H 3 C H BrHO CH 2 CH 3 You can see how both the Nuc and the Br are involved in this rate-determining step.

Bimolecular!

Factors that affect the S

N 2:

1. Alkyl Halide: Steric congestion around the carbon atom undergoing the inversion process will

slow down the S N

2 reaction. Less congestion = faster reaction!

CH 3 -X > RCH 2 -X > R 2

CH-X >>>>>>>>>>>> R

3 C-X Methyl halides and 1¡ halides are the best at undergoing S N

2 reactions, 2¡ halides are OK but 3¡

halides cannot go through the inversion process and will never do this reaction. The transition state is too crowded.

2. Leaving Group: The leaving group is almost always expelled with a full negative charge. As

a result, the best leaving groups are those that can best stabilize an anion (i.e. a weak base). Best: OMs,

OTos > I

, Br > Cl > F OH, NH 2 ) (Worst) OMs = Mesylate group (methanesulfonate group) [shown below, left] OTos = Tosylate group (para-toluenesulfonate group) [shown below, right] Both are extremely good leaving groups due to resonance stabilization: CH 3 S O Oquotesdbs_dbs2.pdfusesText_3