Organic 2 Study Guide PDF

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"Master Organic Chemistry"

Introduction to Alcohols and Ethers Ethers from alcohols and alkyl halides The Williamson Ether synthesis: alcohol, base, alkyl halide (or tosylate) base (e.g. NaOH) OH

O Br

+ H 2O

+ NaBr

This is an SN 2 reaction; it works best for primary alkyl halides (and alkyl tosylates)

Conversion of alcohols to good leaving groups

Due to ring strain, epoxides are highly reactive towards nucleophiles. They will react with nucleophiles under both acidic and basic conditions. However the patterns are different.

The hydroxide group (HO –) of alcohols is a strong base and a poor leaving group. Converting to a halogen or "sulfonate" (e.g. tosylate or mesylate) greatly facilitates substitution reactions. Alcohols to alkyl chlorides

Under basic conditions Under basic conditions, nucleophiles will attack epoxides at the least sterically hindered position (primary [fastest] > secondary > tertiary [slowest]) The reaction is essentially an SN 2 reaction! "Nu"

R

O

ROH

O

OH Nu

The reaction is similar to the hydration of alkenes with aqueous acid. Acid leads to the formation of a carbocation, which is then trapped by the alcohol as solvent. Carbocation rearrangements (hydride and alkyl shifts) can occur in certain cases.

OH

1) CH3MgBr

O

2) NaBH 4

+ HOAc + Hg (s) + NaOAc + BH 3

H 2SO4 OH

OH

O

+ H 2O

Aqueous acid (e.g. H 2O

OH

O H

H δ+ O preferred site of nucleophilic attack: the "most more substituted substituted" carbonpositive charge of the epoxide more stabilized

m-CPBA (meta-chloroperoxybenzoic acid, a peroxyacid) converts alkenes to epoxides, a cyclic ether. Other peroxyacids can be used (e.g. CH3CO 3H)

PCC

O

Alternative reagents for oxidation of primary alcohols to aldehydes and secondary alcohols to ketones (not seen in all courses): Dess-Martin Periodinane (DMP) AcO

OH

Epoxides from halohydrins

Swern oxidation

O

Formation of the halohydrin from the alkene is stereospecific for the trans product. Deprotonation of the alcohol by base results in SN 2 (with inversion at carbon bearing the leaving

Oxidation of primary alcohols to carboxylic acids OH

KMnO 4 or H 2CrO4

•oxidizes secondary alcohols to ketones

OH

PBr 3

Br

+ HOPBr 2

PBr 3

Br inversion!

+ HOPBr 2

O

O OH

retention!

Alcohols through reduction

O

2) H 2O

OH

Reduction of carboxylic acids by lithium aluminum hydride (LiAlH 4) 1) LiAlH 4 OH

2) acid

OH

Reduction of aldehydes by sodium borohydride (NaBH 4) LiAlH 4 can also do this reaction. O NaBH 4 H

NEt 3 (often just written, "Swern") (COCl) 2 is oxalyl chloride DMSO is dimethyl sulfoxide and NEt 3 is triethylamine (base)

+ HCl

Reduction of esters by lithium aluminum hydride (LiAlH 4) O 1) LiAlH 4

O •oxidizes primary alcohols to aldehydes •oxidizes secondary alcohols to ketones

(COCl) 2, DMSO

O

Br

Br

One thing to note - these reactions do not change the stereochemistry of the alcohol. OH OTs TsCl + HCl

•oxidizes primary alcohols to aldehydes

OH

Can also use KMnO 4 or H 2CrO 4 (or DMP or Swern, see right) base

OAc I OAc O O

Oxidation of secondary alcohols to ketones

OH

SOBr 2

p-toluenesulfonyl chloride (Tosyl chloride, TsCl) O Cl S O OH OTs

PCC

Br 2, H 2O

+ HBr + SO2

OH

Methanesulfonyl chloride (mesyl chloride, MsCl) O Cl S CH 3 O OH OMs + HCl

less substituted positive charge less stabilized

O H Cl Cr O N H O PCC = pyridinium chlorochromate

O

+ HOPCl 2

Alcohols to tosylates and mesylates ("sulfonate esters")

O H + δ

Oxidation of primary alcohols to aldehydes OH

m-CPBA

Cl

OH

Oxidation of alcohols

Epoxides from alkenes

PCl 3

HCl, HBr, or HI R Cl R Br R I R OH These reactions can proceed through an SN1 or SN 2 pathway depending on the structure of the alcohol

heat Strong acid (and heat) leads to protonation of the alcohol, followed by nucleophilic attack of a second molecule of alcohol to give the ether. Only practical for the synthesis of symmetrical ethers.

OH

Alcohols to alkyl halides by using acids

H 2SO4) The nucleophile will attack the carbon best able to stabilize positive charge - which is the more substituted carbon Just like Markovnikov's rule!

Ethers from alcohols through dehydration

+ HCl + SO 2

Under acidic conditions

O

The reaction is similar to the hydration of alkenes with aqueous acid. The key difference is that it does not proceed through a carbocation, so no rearrangements can occur.

Cl

OH

Under acidic conditions, the epoxide oxygen is protonated:

R

SOCl2

One thing to note - these reactions occur with inversion of configuration. For example:

2) acid workup (e.g. H +, H 3O+, H 2O)

Ethers from alkenes through oxymercuration 1) Hg(OAc) 2, ROH

OH

Alcohols to alkyl bromides

Example: reaction of epoxides with Grignard reagents

H 2SO4

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

Opening of epoxides

O Ethers from alkenes

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OH

Reduction of ketones by sodium borohydride (NaBH 4) LiAlH 4 can also do this reaction. O

NaBH 4

OH

group) to give the epoxide.

Common source of confusion: Another way of writing H 2CrO 4 is K 2Cr2O7 / H 2SO4 or Na 2Cr2O7/H2SO 4

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"Master Organic Chemistry"

Introduction to the Diels Alder Reaction

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

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1. The Basics The Diels-Alder reaction is a concerted reaction between a compound with two adjacent double bonds (a "diene") and an alkene (the "dienophile"), usually attached to an electron withdrawing group. The transition state is "pericyclic", meaning it has cyclic geometry and is concerted The Diels-Alder reaction always forms a six-membered ring with an alkene. A Very Simple Diels Alder 1

1

2

2

6

6

3

5

3

5

4

4

Form

Break

C1 –C6

C1 –C2 ( π)

C 2–C3 ( π)

C 3–C4 ( π)

C 4–C5

C5–C6 ( π)

Rule #3: The two "outside" groups ( red )and the two "inside" groups (blue) always end up on the same side of the new cyclohexene

Rule #1

1 2

the C2-C3 bond can rotate freely

When the diene is "locked" in an s-trans configuration, no Diels-Alder reaction is possible (the two ends are too far apart)

1 2

3

3 4

4

note how the two alkenes are on the same side of the C 2–C3 sigma bond

note how the two alkenes are on opposite sides of the C 2–C3 sigma bond

The end carbons (C1 and C 4 ) are close together

The end carbons (C1 and C 4 ) are too far apart no Diels-Alder reaction possible

6 3

5

3

top view

5

side view 7

1

1 7

3

2

6

1

2 3

5

4

5

4

4

6

7

3

5

8

6 8

8

H 3C H

top view

enantiomers

A O

CH 3 O

side view

C1 –C2 ( π)

C 2–C3 ( π)

C 3–C4 ( π)

C 4–C5

C5–C6 ( π)

note how the pattern of bonds forming and bonds breaking is exactly the same as for the simple case, above.

C

D

CH 3 O

CH 3 O

CH 3

CH 3

+

+

Break

C1 –C6

B CH 3 O

CH 3

CH 3

CH 3 enantiomers

Let's just pick one set of diastereomers. A and C. C A CH 3 O

enantiomers Exo/endo examples: O

CH 3 O CH 3 +

H H

3. Stereochemistry: The Dienophile How the stereochemistry of the dienophile works: Rule #2: Stereochemistry in the dienophile is always preserved Example:

cis relationship is preserved in product O O

O

CH 3 Endo

CH 3 Exo

Note how in A, the two methyl groups are on the same side of the ring as the electron withdrawing group (ketone), whereas in C, they are on the opposite side of the ring

Rule #4: Under normal conditions, endo products dominate O

O

these are the same molecule (it's meso)

O

trans relationship is preserved in product O O +

How to tell if a product is exo or endo One way to do it: if the outside groups (red) end up on the same side of the ring as the electron withdrawing group, the product is endo. If the outside groups end up on the opposite side of the ring as the electron withdrawing group, the product is exo.

CH 3

O

CH 3

O

CH 3

O

CH 3

O

O

CH 3

A is referred to as the endo product. C is referred to as the exo product (note: B is also endo, and D is also exo)

or O cis dienophile

H 3C H

Note how the two "outside" groups (CH 3 and H) end up on the same side of the ring, as do the two "inside" groups (CH 3 and H)

Note how the two "outside" groups (CH 3 and CH 3 ) end up on the same side of the ring, as do the two "inside" groups (H and H)

6

Form

4

4

4

2

H CH 3 and

CH 3

H 3C H

CH 3

1 2

7 5

H 3C H

Up to 4 products are possible!

7

1 2

7 3

CH 3

H CH 3 CH 3 H

=

Example:

Diels-Alder of cyclic dienes 6

H

H 3C H

What if there's a substituted diene and a substituted dienophile?

These look a little weird, but they're no different than a normal Diels-Alder

1

H 3C H

H H

5. The Endo Rule

2. Special Cases Diels-Alder Reactions of Cyclic Dienes:

2

CH 3

these are the same molecule These dienes cannot participate in the Diels-Alder

"s-trans"

"s-cis"

Example 2:

Example 1:

The diene must always be in the "s-cis" conformation

What this means:

4. Stereochemistry: The Diene How the stereochemistry of the diene works:

endo O

H 2

6

1

3

4

H OCH 3

2

7

7

O

1

1

6

5

CH 2

exo

OCH 3

7

3

5

3

H

2

endo (side view) 7

CO 2Me

1

6

2

OCH 3

7 5

3

5

H 3CO

endo (top view) H O 1

O

4

4

H

6

2

3

6 4

5

4

H exo (top view)

EWG

exo (side view)

EWG Omissions, Mistakes, Suggestions?

O trans dienophile

O these are enantiomers

O

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"Master Organic Chemistry"

Introduction to Aromaticity 1. An introduction to resonance energy and aromaticity

3. How to tell if a molecule is aromatic? Rule 4. It must be flat

Hydrogenation of alkenes liberates 119 kJ/mol of energy ΔH = –119 kJ/mol (28.4 kcal/mol)

Pd/C H2

Rule 1. It must be a ring. No acyclic molecule is aromatic. Ever. Most molecules that obey the first 3 rules are also flat. One exception is [10]-annulene, which is bent due to repulsion of the hydrogens.

We would expect hydrogenation of benzene to liberate 3 × 119 = 357 kJ/mol. Pd/C H2

ΔH = –207 kJ/mol (49 kcal/mol)

1,3,5 hexatriene not aromatic

N H pyrrole aromatic

butadiene not aromatic

benzene aromatic

H H not aromatic

Instead, 207 kJ/mol is liberated (150 kJ mol less than we expect!) So it is 150 kJ/mol (36 kcal/mol) more stable.

Benzene has a particularly large resonance energy, which leads us to classify it as "aromatic". Resonance energy of some compounds:

Meaning: there must be a continuous line of p orbitals around the ring. p orbitals can come from 1) π-bonds 2) lone pairs 3) carbocations

O

Conjugated

2. Two major ways in which reactions of aromatic compounds differ from alkenes

H

easy

H

Reaction with electrophiles (such as bromine): aromatics give substitution, not addition Br

Br

Br 2

Br 2

AlBr3 Br Addition reaction

benzene

Substitution reaction difficult

easy

2 aromatic

not aromatic (antiaromatic)

4 not aromatic (antiaromatic)

6

4

O 6 [NOT 8] aromatic

N 6 [NOT 8] aromatic

aromatic

N H 6 [NOT 8] 6 aromatic aromatic

H 6 aromatic

10

H N N 8 not aromatic

6 [NOT 8]

O 6 [NOT 8]

5. Some physical evidence for aromaticity: H

1. Shows a different reactivity profile than for alkenes (see section on reactivity at left)

Rule 3. There must be 4n + 2 π electrons

6

2

All bonds in benzene are the same length (1.4 Angstrom) i.e. 2, 6, 10, 14.... π electrons

π electrons can come from double bonds or lone pairs. Note: A carbocation indicates the absence of π electrons. One tricky part: for a given atom, you can only count electrons from a lone pair if the atom is not part of a π bond. And in that case you can only count a maximum of one lone pair. This is due to the fact that each atom can only share one p orbital with the π system of the molecule. the electrons from the lone pair on N are at 90° to the π system, and do not contribute to aromaticity.

N pyridine 6 π electrons (3 π bonds, 2 electrons per π bond) ignore lone pair for this calculation

Similar example: lone pair on this anion does not contribute towards aromaticity.

furan 6 π electrons (2 π bonds, 2 electrons per π bond) + 2 electrons from lone pair only count one lone pair for this calculation

Compare this to cyclobutadiene, which has short double bonds and long single bonds - like a rectangle. 3. Ring currents in NMR Resonances for aromatic protons in NMR typically show up in the region 6.8–8.0 ppm, whereas those for "normal" alkenes show up in the region from 5.0–6.5 ppm.

6. "Frost circles" - a trick for obtaining the molecular orbital structures of aromatic rings. General idea: Inscribe a polygon of n sides in a circle. Make sure one of the apices is pointing down. Then, each apex will represent a level in the molecular orbital energy diagram. Frost, J. Chem. Phys. 1953, 21, 572 Examples:

only one of the lone pairs can align itself with the pi system at any given time, so only one lone pair is counted towards aromaticity.

O

aromatic

extremely unstable

2. All π bonds are of the same length & do not alternate

no reaction without a catalyst such as AlCl3

π electrons - some examples

•cyclic •conjugated •4π electrons

cyclooctatetraene has 8π electrons, but can adopt a "tub" shape, "escaping" antiaromaticity.

H

Not Conjugated

no reaction at normal temperature and pressure

Alkene

Conjugated

difficult

benzene

Alkene

H

O

antiaromatic

N

Pd-C H2

H 3C CH3

H2

Conjugated

Conjugated H

Hydrogenation: more difficult with aromatic compounds Pd-C

N H

H

H Conjugated

H 2C CH2

Molecules that obey rules 1,2 and 4 but have (4n) π electrons instead of (4n +2) π electrons have special instabiity. This special instability is called "antiaromaticity".

A good test. Can you push electrons all the way around the ring through resonance? If not, it's not conjugated.

N O N H 121 kJ/mol 67 kJ/mol 92 kJ/mol 12.2 kJ/mol (29 kcal/mol) (16 kcal/mol) (22 kcal/mol) (3 kcal/mol) aromatic aromatic aromatic not aromatic

aromatic

must adopt a twisted structure to avoid steric repulsion!

4. Antiaromaticity Rule 2. The molecule must be conjugated

The extra stability of benzene is called the "resonance energy".

150 kJ/mol (36 kcal/mol)

Note - this sheet is not meant to be comprehensive. Your course may provide additional material, or may not cover some of the reactions shown here. Your course instructor is the final authority.

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Similar examples: only count one lone pair

S

N

3

4

5

6

7

8

this shows the arrangement of molecular orbitals for benzene

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aromatic

aromatic