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.

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▪ SN2 – Second Order Nucleophilic Substitution

o Nucleophile just knocks off the halogen and takes its place

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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.

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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

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• DMSO (dimethylsulfoxide)

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• Acetonitrile

CH3CN

• DMF (dimethylformamide)

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▪ 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.

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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)

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▪ 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

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▪ 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

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o Deprotonation finishes it off

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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

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o Same first step as SN1, so they’ll both happen at the same time

▪ Rearrangement still possible

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o A hydrogen next-door to the + falls off as a proton, leaving it’s electrons behind to form a double bond

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o Get all possible stereoisomers

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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.

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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.

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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.

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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.

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o Now rearrange the carbocation.

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▪ 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.

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▪ 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

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▪ 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

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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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