Exam Unit 4

Question 0

enamines form from

Answer 0

secondary amines

Question 1

aldol addition

Answer 1

aldehyde + enol = beta hydroxyaldehyde

Question 2

imines form from

Answer 2

ammonia or primary amines

Question 3

Make primary amines (4 ways)

Answer 3

1. CN + LAH
2. amide + LAH
3. Acid Chloride + NaN3 (Curtius)
4. amide + Br3 KOH (Hoffman)

Question 4

Making Secondary amines

Answer 4

Reductive Amination Only!

NaCNBH3 + aldehyde (R group attaches!)

Question 5

ketones in aldol rxns

Answer 5

Only in acid or intramolecular.

Must be made irreversible.

Question 6

intramolecular aldol addition

Answer 6

5 or 6 membered rings only
less hindered C is attacked.
more hindered C is attacked with ringed N bases.

Question 7

selectivity in aldol rxns

Answer 7

NaOEt is not selective.

Use LDA to select enolate formation. H30+ will quench LDA and dehydrate alcohol

Question 8

claisen rxn

Answer 8

ethers and ketones! Final enolization drives reaction, so 2 alpha protons on attacking group required.

Match bases to avoid transesterification, or do in acid.

Question 9

mixed claisen

Answer 9

enolate forms from more acidic partner. (ketone over ester).

can’t do claisen with an aldehyde, unless ester is first selectively enolated before mixing.

Question 10

michael rxn

Answer 10

makes 1,4 dicarbonyls. enolate attacks beta position on unsaturated michael acceptor

Question 11

good michael acceptors

Answer 11

conjugated aldehydes, cyano, nitro, tertiary amides, esters, ketones

michael acceptors get attacked beta.

Question 12

bad michael acceptors

Answer 12

primary, secondary amides, carboxylic acids (protons will quench rxn), acid chlorides (funnels to (1,2) product)

Question 13

robinson annulation

Answer 13

michael then enol condensation (first protonate then deprotonate to move enolate to other side). Makes 6 membered ring.

• *limited to cyclic ketones: second enolation selectivity
• **don’t use a strong base like LDA, or you’ll get stuck w/out a ring. Use NaOEt.

Question 14

stork enamine synthesis

Answer 14

acid catalyzed condensation of secondary amine with aldehyde or ketone

• can be added to, alpha, by alkyl halide
• can be added to enone via michael-like rxn
• acyl chlorides can add alpha, but need 2 eqs (to clear N-acylated product)

Question 15

mannich rxn

Answer 15

makes beta-ketoamines

three parts: ketone, aldehyde and amine. amine and aldehyde form imminium ion, ketone enolates and attacks.

Question 16

keys to mannich

Answer 16

• **nucleophilicity of amine competes with aldol addition
• *can be intramolecular
• iminium formation fastest at pH 4.5

Question 17

reductive amination

Answer 17

make secondary amine by treating primary with NaCNBH3 and formaldehyde.

Question 18

Gabriel amine sythesis

Answer 18

use alkyl halides to add R groups to “masked” amines to avoid multiple acylations.

cleave with H2N-NH2, heat (forms attached amines in ring)
or NaOH, H2O (leaves open ring with O-)

Question 19

Azide Reduction

Answer 19

1. LAH, H3O+ (will reduce carbonyls, too)
2. H2Pd/C (will reduce double bonds, and benzylic ketones, too)
3. PPH3, H2O (will substitute alkyl halides, too)

Question 20

Hoffman Rearrangement

Answer 20

Amide to primary amine, loss of one carbon.

NaOH, Br2, H2O

Question 21

Curtius Rearrangement

Answer 21

Acid Chloride to primary amine.

NaN3, H2O

Question 22

Hoffman elimination

Answer 22

1. MeI excess, pyr.
2. Ag2O, H2O

generates least substituted product.

Question 23

Making tertiary amines

Answer 23

make secondary (via reductive amination), then add another through R.A., or RX, NaH

Question 24

Nucleophilic Aromatic Substitution: Halogen to Amine

Answer 24

Use NaN3, then reduce

Question 25

When to dehydrate?

Answer 25

ketone and ester additions aren't favored until dehydrated. in acid? always dehydrate. in base? secondary alcohol + heat = condensation tertiary alcohol will not condense! (NR in base)

Question 26

1,4 carbon nucleophiles

Answer 26

must be gilman reagents (Me2CuLi)

Question 27

Grignard reagents attack beta?

Answer 27

No! 1, 2 only

Question 28

Why are reaarrangements favored?

Answer 28

thermodynamics: get rid of CO2

Question 29

most to least basic amines.

Answer 29

1. on ring (primary most, then secondary) 2. near inductive EWGs 3. aromatic with EDG O/P 4. aromatic with EDG M 5. aromatic with t-butyl O/P 6. aromatic with halogen O/P 7. aromatic 8. aromatic with EWG M 9. aromatic with EWG O/P

Question 30

Do inductive EWGs change selectivity of an aldol addition?

Answer 30

Nope. Still four products.

Question 31

Reopening a ring: Four steps

Answer 31

1. OH- attacks DB, sends - up to carbonyl 2. O- sends down charge, enolate DB grabs proton from solvent 3. base deprotonates hydroxyl 4. O- sends down charge to break bond.

Question 32

Replace ester with amine

Answer 32

add amine you want with an extra H (HNMe2, for example)

Question 33

Who needs a workup?

Answer 33

Gilman Wolff Kishner All additions/condensations

Question 34

Ethers hang off after addition?

Answer 34

No, and no conjugation. Just another carbonyl.

Question 35

Wolff-Kishner

Answer 35

Turns carbonyl into double bond. 1. H2NNH2, H+ [-H2O] 2. KOH/H2O, workup.

Question 36

POCl3

Answer 36

with pyridine, turns alcohols into alkenes.

Question 37

Trade an OH for an I

Answer 37

PI3, pyr.

Question 38

Carbonyl to Cyano group

Answer 38

Mannich, Br to KCN 18-C-6 NH3 and heat

Question 39

Why does the Hofmann product form?

Answer 39

a beta H must be removed in elimination. Since the N species is so bulky, it goes for the least hindered proton.

Question 40

greater s character?

Answer 40

less basicity

Question 41

Adding NH2 at alpha position

Answer 41

1. LDA, Br2 2. NaN3 3. reduction: PPh3 and heat, LAH, or H2 Pd/C

Question 42

Replace carbonyl O with amine

Answer 42

add desired amine, H+, NaCNBH3, workup.

Question 43

Hoffman v. Curtius

Answer 43

Hoffman, with Br2 and whatnot, is less selective than Curtius.

Question 44

Ousted amine in Hoffman elimination

Answer 44

has three constituent alkyl groups.

Question 45

More basic: aromatic with N or heteroaromatic with N and S

Answer 45

N and S is more basic. No one knows why.

Question 46

More basic: NEt2H or NEtH2

Answer 46

NEt2H, but no one knows why.

Question 47

Stork alklyation

Answer 47

NH in 5 ring, +H [-H2O] | then desired alkyl halide, workup

Question 48

Why is NaCNBH3 different than NaBH4?

Answer 48

EWD CN makes the H- ion less reactive

Question 49

cross aldol selectively?

Answer 49

One partner does not have 2 alpha Hs, and the second is dripped in slowly, to prevent rxn with itself.

Question 50

stork acylation v. alkylation

Answer 50

acylation can self correct, so it has higher laboratory yields.

Question 51

Giving functionality from nothing but a double bond

Answer 51

O3, Me2S!

Question 52

Adding an amine to beta position (alpha beta unsaturated)

Answer 52

Desired amine, pyridine and heat.