A to Z Directory – Virginia Commonwealth University
Chapter 6 – Reactions of Alkyl Halides
▪ Nucleophiles
o Nucleophiles are Lewis Bases, so they must have a lone pair.
o Negatively charged nucleophiles are stronger than their neutral counterparts.
o Nucleophile strength decreases from left to right on periodic table.
o Nucleophile strength increases going down the periodic table.
o Bulky groups decrease nucleophile strength.
▪ Ex. Hydroxide is a better leaving group than t-butoxide because the t-butoxide is so large that it has difficulty getting to the site of reactivity.
▪ Leaving groups – halide ions in this chapter
o Must be electron-withdrawing to create a partial positive on the carbon
o Take their electrons with them
▪ They’re that kid on the playground who takes the ball when he doesn’t get his way.
o The weaker the base, the better the leaving group
▪ Substitutions vs. Elimination
o These are words in the English language and they don’t have new definitions for chemistry.
▪ With substitution, your nucleophile comes in and takes the place of the leaving group.
• If you had a substitute teacher in high school, a new teacher came in and replaced your regular teacher.
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▪ With elimination you are getting rid of the leaving group and the nothing takes its place.
• If the teacher had been eliminated, you would have come into class and not had a teacher at all.
[pic]
▪ SN2 – Second Order Nucleophilic Substitution
o Nucleophile just knocks off the halogen and takes its place
[pic]
o One-step
▪ No intermediate, so no rearrangement
▪ rate=k[RX][nuc]
• Both the alkyl halide and nucleophile are involved in the rate-limiting step (the only step in this reaction) so they both affect the rate.
o Backside attack leads to inversion of configuration
o Most of the time your question won’t look like the mechanism drawn above.
▪ When you are given a substrate in line-angle form and asked to draw the product, draw the product with the nucleophile where the halogen was, but with a dash where a wedge once was or vice versa.
[pic]
o Needs strong nucleophile
▪ At this point in the course the strong nucleophiles given will all be negatively-charged, though ammonia and amines are also strong nucleophiles which can undergo SN2 reactions.
o Polar, aprotic solvents can help this along
▪ Aprotic means that the solvent is not a good source of protons.
• No hydrogen bonding.
▪ Mr. Baker requires you to know four of these polar, aprotic solvents.
• Acetone
[pic]
• DMSO (dimethylsulfoxide)
[pic]
• Acetonitrile
CH3CN
• DMF (dimethylformamide)
[pic]
▪ A polar, aprotic solvent is not required for this reaction.
• So if he gives you a reaction and doesn’t give a solvent, still do the reaction.
[pic]
o Suitability of alkyl halides as substrates
▪ Methyl>1°>2°>>>>>>>3° (wont’ happen)
▪ This should make sense as adding carbon groups will increase the steric hindrance.
▪ The rate will be faster with more suitable substrates (so an SN2 on a methyl will go faster than on a 2°).
▪ SN1
o First, the halogen falls off (slow step)
[pic]
▪ Sometimes you’ll see AgNO3 here.
• The silver complexes with the halide and precipitates out of solution.
• It’s just a hint that you have SN1/E1 conditions.
• Do not write Ag or NO3 on your product (unless for some reason you’re asked for the inorganic byproducts.
o Now you have a carbocation, which can rearrange
[pic]
▪ Rearrangements will only occur when the resulting carbocation is more stable than the initial carbocation.
• Ex. If a secondary carbocation becomes tertiary after shift, tertiary carbocation becomes tertiary allylic, etc.
▪ There are two types of rearrangements:
• Hydride shifts
o In a hydride shift, a hydrogen next door moves over taking its electrons with it.
• Alkyl shifts
o In an alkyl shift, an alkyl group next door moves over taking its electrons with it.
o You only do alkyl shifts when there’s a quaternary position next door to the initial carbocation.
o The nucleophile (often a polar, protic solvent) attaches to the positively-charged carbon
[pic]
o Deprotonation finishes it off
[pic]
o Major substitution product always comes from the rearranged carbocation
o Happens with weak nucleophiles
▪ For this test the weak nucleophile will be your polar, protic solvent
• Water and alcohols
o Order of reactivity
▪ 3°>2°>>1°>>>>methyl (won’t happen)
• This should make sense, as the slow step is forming the carbocation.
• The more stable the resulting carbocation, the more likely it is that the first step will happen.
o Partial racemization
▪ Means you’ll get both R and S where you added the nucleophile.
• You’ll also get both R and S where the methyl moved if you have a methyl shift.
▪ You will see throughout the course that if a reaction goes through a trigonal planar intermediate, then there is no way to select for R or S.
• This time the trigonal planar intermediate is a carbocation.
o rate=k[RX]
▪ The first step is the slow step and only the alkyl halide is involved with that.
▪ E1
[pic]
o Same first step as SN1, so they’ll both happen at the same time
▪ Rearrangement still possible
[pic]
o A hydrogen next-door to the + falls off as a proton, leaving it’s electrons behind to form a double bond
[pic]
o Get all possible stereoisomers
[pic]
o Choosing the major elimination product
▪ Look at the products that come from the rearranged carbocation
▪ Zaitsev’s rule
• The more carbons there are coming off a double bond, the more stable the double bond is.
• More stable products are more likely to form.
▪ If there’s a tie, look to see if one of the more substituted products was reached from two different paths
• If so, then that’s the major product.
o rate=k[RX]
▪ Drawing all the products when you have SN1/E1 conditions
o You should expect to see a problem like this at some point.
[pic]
o I think this is easiest when you make a template for yourself.
▪ Here’s my template:
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o So now you go through and fill out the templateTo get the initial carbocation, take off the halogen and put a positive charge there.
[pic]
o Then to get the SN1 product(s) just let your solvent plop on there and if the position is chiral remember you get both R and S.
[pic]
o To get the E1 product(s) make a double bond in any possible direction from the carbocation and then consider whether you have E/Z double bonds. If you do, you’ll get both E and Z.
[pic]
o Now rearrange the carbocation.
[pic]
▪ Just move the positive charge to the more substituted spot next door.
▪ Notice that you lose the wedge to that methyl and now it’s flat.
• That is because carbocations are trigonal planar.
o And to get the new products just do what you did with the unrearranged products.
[pic]
▪ Notice that you get one of the E1 products from both the rearranged carbocation and the unrearranged carbocation.
o Picking the major products.
▪ Substitution
• This one’s easy.
• It’s always the rearranged product.
▪ Elimination
• There are two factors which can compete.
o The most substituted double bond is the most stable.
▪ More stable products are more likely to form.
o There is always an E1 product that you get from both the unrearranged and the rearranged carbocation.
▪ If you get to it through multiple paths then it is more likely to form.
• If the factors compete, he will not ask you for the major E1 product.
o In this situation he would ask you for the most stable E1 product; go with Zaitsev!
▪ E2
o Happens with secondary and tertiary alkyl halides combined with strong bases.
▪ At this point in the course, the only strong bases you are responsible for knowing are hydroxide and alkoxides.
• -OH and -OR
o One step
[pic]
▪ Strong base abstracts proton next-door to halogen
▪ The electrons that were attached to the hydrogen kick off the halogen
o The hydrogen and halogen must be anti-coplanar to each other
[pic]
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o rate=k[RX][base]
▪ Zaitsev’s Rule
o Applies to elimination reactions
o Whenever possible, the more substituted double bond will form
o When your base is particularly bulky (think t-butoxide or similar), you will get the less substituted product.
▪ Which reaction should I do?
|Alkyl Halide |Strong Nucleophile |Weak Nucleophile |
| |Weak Base Strong Base |(H2O, ROH) |
| |(RCOO-, -CN, -SH (-OR, -OH) | |
| |-SR, N3-, X-, NH3 | |
| |or amines) | |
|Methyl |SN2 SN2 |NR |
|1° |SN2 SN2 |NR |
|2° |SN2 E2 |SN1/E1 |
|3° |NR E2 |SN1/E1 |
When you have things in solution that tell you to do different reactions, the strong thing wins!
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• The strong nucleophile wins and we do an SN2.
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