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O Chem 2 > Exam Unit 4 > Flashcards

Exam Unit 4 Flashcards

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0
Q

enamines form from

A

secondary amines

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1
Q

aldol addition

A

aldehyde + enol = beta hydroxyaldehyde

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2
Q

imines form from

A

ammonia or primary amines

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3
Q

Make primary amines (4 ways)

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

Making Secondary amines

A

Reductive Amination Only!

NaCNBH3 + aldehyde (R group attaches!)

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5
Q

ketones in aldol rxns

A

Only in acid or intramolecular.

Must be made irreversible.

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6
Q

intramolecular aldol addition

A

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

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7
Q

selectivity in aldol rxns

A

NaOEt is not selective.

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

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8
Q

claisen rxn

A

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

Match bases to avoid transesterification, or do in acid.

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9
Q

mixed claisen

A

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

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

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10
Q

michael rxn

A

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

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11
Q

good michael acceptors

A

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

michael acceptors get attacked beta.

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12
Q

bad michael acceptors

A

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

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13
Q

robinson annulation

A

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.
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14
Q

stork enamine synthesis

A

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)
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15
Q

mannich rxn

A

makes beta-ketoamines

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

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16
Q

keys to mannich

A
  • **nucleophilicity of amine competes with aldol addition
  • *can be intramolecular
  • iminium formation fastest at pH 4.5
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17
Q

reductive amination

A

make secondary amine by treating primary with NaCNBH3 and formaldehyde.

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18
Q

Gabriel amine sythesis

A

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-)

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19
Q

Azide Reduction

A
  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)
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20
Q

Hoffman Rearrangement

A

Amide to primary amine, loss of one carbon.

NaOH, Br2, H2O

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21
Q

Curtius Rearrangement

A

Acid Chloride to primary amine.

NaN3, H2O

22
Q

Hoffman elimination

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

generates least substituted product.

23
Q

Making tertiary amines

A

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

24
Q

Nucleophilic Aromatic Substitution: Halogen to Amine

A

Use NaN3, then reduce

25
Q

When to dehydrate?

A

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)

26
Q

1,4 carbon nucleophiles

A

must be gilman reagents (Me2CuLi)

27
Q

Grignard reagents attack beta?

A

No! 1, 2 only

28
Q

Why are reaarrangements favored?

A

thermodynamics: get rid of CO2

29
Q

most to least basic amines.

A
  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
30
Q

Do inductive EWGs change selectivity of an aldol addition?

A

Nope. Still four products.

31
Q

Reopening a ring: Four steps

A
  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.
32
Q

Replace ester with amine

A

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

33
Q

Who needs a workup?

A

Gilman
Wolff Kishner
All additions/condensations

34
Q

Ethers hang off after addition?

A

No, and no conjugation. Just another carbonyl.

35
Q

Wolff-Kishner

A

Turns carbonyl into double bond.

  1. H2NNH2, H+ [-H2O]
  2. KOH/H2O, workup.
36
Q

POCl3

A

with pyridine, turns alcohols into alkenes.

37
Q

Trade an OH for an I

A

PI3, pyr.

38
Q

Carbonyl to Cyano group

A

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

39
Q

Why does the Hofmann product form?

A

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

40
Q

greater s character?

A

less basicity

41
Q

Adding NH2 at alpha position

A
  1. LDA, Br2
  2. NaN3
  3. reduction: PPh3 and heat, LAH, or H2 Pd/C
42
Q

Replace carbonyl O with amine

A

add desired amine, H+, NaCNBH3, workup.

43
Q

Hoffman v. Curtius

A

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

44
Q

Ousted amine in Hoffman elimination

A

has three constituent alkyl groups.

45
Q

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

A

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

46
Q

More basic: NEt2H or NEtH2

A

NEt2H, but no one knows why.

47
Q

Stork alklyation

A

NH in 5 ring, +H [-H2O]

then desired alkyl halide, workup

48
Q

Why is NaCNBH3 different than NaBH4?

A

EWD CN makes the H- ion less reactive

49
Q

cross aldol selectively?

A

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

50
Q

stork acylation v. alkylation

A

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

51
Q

Giving functionality from nothing but a double bond

A

O3, Me2S!

52
Q

Adding an amine to beta position (alpha beta unsaturated)

A

Desired amine, pyridine and heat.

O Chem 2 (5 decks)

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