Objectives

After completing this course, you will be able to:

• Convert a primary alcohol to an alkyl bromide using an SN2

reaction
• Investigate some factors that influence the rate of SN1
reaction
• Involve the transfer of a carbon group from a weak base, the
leaving group, to stronger base, the nucleophile.

What are Alkyl halides?

Alkyl halides are organic molecules containing a halogen atom bonded to an sp 3
hybridized carbon atom. Alkyl halides are classified as primary (1°), secondary (2°), or
tertiary (3°), depending on the number of carbons bonded to the carbon with the halogen
atom. The halogen atom in halides is often denoted by the symbol “X”.
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Nucleophiles
• Nucleophiles are Lewis Bases, so they must have a lone pair.
• Negatively charged nucleophiles are stronger than their neutral counterparts.
• Nucleophile strength decreases from left to right on periodic table.
• Nucleophile strength increases going down the periodic table.
• Bulky groups decrease nucleophile strength.

Example: Hydroxide is a better leaving group than t-butoxide because the butoxide is so
large that it has difficulty getting to the site of reactivity.

Leaving groups - Halide ions in this chapter

• Must be electron-withdrawing to create a partial positive on the carbon.
• Take their electrons with them
• The weaker the base, the better the leaving group.

Substitutions vs. Elimination

• With substitution, your nucleophile comes in and takes the place of the leaving
group.

• With elimination you are getting rid of the leaving group and the nothing takes its
place.

SN2 – Second Order Nucleophilic Substitution

Nucleophile just knocks off the halogen and takes its place

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• 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.
• Backside attack leads to inversion of configuration
• 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.

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.
• 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 – Also called 2-propanone or dimethyl ketone, organic solvent of
industrial and chemical significance, the simplest and most important of the
aliphatic fat derived ketones.

• DMSO (Dimethyl Sulfoxide) – An organosulfur compound with the formula

(CH3)2SO. This colorless liquid is an important polar aprotic solvent that
dissolves both polar and nonpolar compound and is miscible in a wide range
of organic solvents as well as water.

• Acetonitrile – Often abbreviated MeCN, is the chemical compound with the

formula CH3CN. This colorless liquid is the simplest organic nitrile. It is
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produced mainly as a byproduct of acrylonitrile manufacture. It is used as a

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polar aprotic solvent in organic synthesis and in the purification of

butadiene.
• DMF (dimethylformamide) – An organic compound with the formula
(CH3)2NCH. Commonly abbreviated as DMF, this colorless liquid is miscible
with the water and majority of organic liquid. DMF is a common solvent for
chemical reaction.

✓ 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.

• 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

1. First, the halogen falls off (slow step)

• 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.

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2. Now you have a carbocation, which can rearrange.

• 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 allelic, etc.
• There are two types of rearrangements:
✓ Hydride shifts - In a hydride shift, a hydrogen next door moves over
taking its electrons with it.
✓ Alkyl shifts - In an alkyl shift, an alkyl group next door moves over taking
its electrons with it. You only do alkyl shifts when there’s a quaternary
position next door to the initial carbocation.

3. The nucleophile (often a polar, protic solvent) attaches to the positively-charged

carbon.

4. Deprotonation finishes it off

5. Major substitution product always comes from the rearranged carbocation.

6. Happens with weak nucleophiles

• For this test the weak nucleophile will be your polar, protic solvent
o water and alcohols.

7. 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
5 carbocation, the more likely it is that the first step will happen.

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8. 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.

9. rate=k[RX] - The first step is the slow step and only the alkyl halide is involved with
that.

E1

1. Same first step as SN1, so they’ll both happen at the same time. Rearrangement
still possible.

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

3. Get all possible stereoisomers

4. Choosing the major elimination product

✓ Look at the products that come from the rearranged carbocation
✓ Zaitsev’s rule
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❖ 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.
5. rate=k[RX]

Drawing all the products when you have SN1/E1 conditions

1. You should expect to see a problem like this at some point.

2. Make a template.

3. Fill out the template. To get the initial carbocation, take off the halogen and put a
positive charge there.

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4. 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.

5. 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.

6. 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.

7. Get the new products. Do what you did with the un-rearranged products.

• Notice that you get one of the E1 products from both the rearranged carbocation
and the un-rearranged carbocation.

Picking the major products

• Substitution
✓ This one’s easy.
✓ It’s always the rearranged product.

• Elimination
✓ There are two factors which can compete.

❖ The most substituted double bond is the most stable.

o More stable products are more likely to form.

❖ There is always an E1 product that you get from both the un-
rearranged and the rearranged carbocation.
o 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. In this
situation he would ask you for the most stable E1 product; go with Zaitsev!
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E2 - 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.

One step

• Strong base abstracts proton next-door to halogen

• The electrons that were attached to the hydrogen kick off the halogen

The hydrogen and halogen must be anti-coplanar to each other.

Zaitsev’s Rule - applies to elimination reactions. Whenever possible, the more

substituted double bond will form. When your base is particularly bulky think t-butoxide or
similar, you will get the less substituted product.

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Strong Nucleophile
Weak Base Strong Base
Weak Nucleophile
Alkyl Halide (RCOO-, -CN, -SH (-OR, -OH)
-SR, N -, X-, NH (H2O, ROH)
3 3
or amines)
Methyl SN2 SN2 NR
1° SN2 SN2 NR
2° SN2 E2 SN1/E1
3° NR E2 SN1/E1

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Video links:

Nucleophilic Substitution Reaction - SN1 and SN2 Mechanic, Organic Chemistry

• https://youtu.be/yrvV85H737o
Choosing between SN1/SN2/E1/E2 Mechanisms
• https://youtu.be/IOViLLuDMTs
Organic Chemistry - Reaction Mechanisms - Addition, Elimination, Substitution and
Rearrangement
• https://youtu.be/Efh5GkVbhEc
Reference:

• https://crab.rutgers.edu/~alroche/Ch06.pdf
• http://chem.yonsei.ac.kr/chem/upload/CHE2001-02/120692008232858.pdf
• https://www.jtcc.edu/downloads/services/math-center/Chapter06.pdf

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