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CH 336 > Midterm #1 > Flashcards

Midterm #1 Flashcards

1
Q

describe a 1,2- addition

A

ch. 22 slide 6:
a nucleophile attacks the carbonyl carbon and then the double bond goes to the oxygen.

products formed:
FS: enolate and a nucleophile attached to former carbonyl carbon (treat with acid to form…)
SS: hydroxide and a nucleophile (alcohol formed)

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2
Q

describe a 1,4- addition

A

ch. 22 slide 6:
a nucleophile attacks one of the carbons of the double bond. The double bond pushes up to form a double bond with the carbonyl carbon. the double bond of the carbonyl goes up to the oxygen.

products formed:
FS: an enolate is formed and the Nu is attached to where it attacked (treated with acid to form…)
SS: carbonyl regained and Nu attached. (no alcohol formed)

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3
Q

what addition is kinetically controlled

A

1,2- addition and it is generally irreversible

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4
Q

what addition is thermodynamically controlled

A

1,4- addition and it is typically reversible. it is also more stable because carbonyls are more stable

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5
Q

hard vs. soft nucleophiles and examples of each

A

hard sites: have their lone pair and charge very localized and are not polarizable. ex. grignard / organolithium. hard acids prefer hard bases.

soft sites: have their lone pair either quite delocalized, or in a large orbital. ex. organocuperates. soft acids prefer soft bases.

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6
Q

describe orbital reactivity

A
  • orbitals closer in energy react forwardly
  • orbitals further apart in energy don’t
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7
Q

how can the alpha carbon be deprotonated

A

by a reasonably strong base

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8
Q

describe alpha carbon pka values

A

Ch. 22 slide 12

  • due to resonance stability of the anion (charge and space), the pka values are lower
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9
Q

would a carbonyl with an alkene/ketone have a higher/lower pka, compared to a carbonyl with an ester

A

Ch. 22 slide 12

a carbonyl with an alkene/ketone would have a pka around 16-20. This has a lower pka because groups can share electron density.

a carbonyl with an ester would have a pka around 25 because the oxygen is an EDG which destabilizes

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10
Q

how would having two carbonyls on a compound affect pH?

A

ch.22 slide 13

  • the pka would be even lower
    there would be three alpha carbons that have hydrogens. see resonance forms
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11
Q

why would a compound with just two carbonyls have a lower pka than a compound with two carbonyls and an extra oxygen attached?

A

ch.22 slide 13

  • the oxygen is an EDG via resonance, which is destabilizing because it is donating into a negative charge.
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12
Q

keto enol tautomerization in basic conditions

A

ch.22 slide 14

  1. the base attacks a hydrogen on the alpha carbon.
  2. forms an enolate which shifts to form a double bond between carbons
  3. acid workup forms enol form
  • since there are two alpha carbons, the compound could be protonated from either side. follow regioselectivity rules and draw the most stable form as the product. This mechanism is similar to E2, which is direct and with no rearrangements.
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13
Q

keto enol tautomerization in acidic conditions

A

ch.22 slide 14

  1. carbonyl oxygen attacks the acid
  2. unstable carbonyl formed which shifts to form a carbocation via resonance
  3. an enol form was produced
  • under acidic conditions, you don’t get an enolate, you go right to an enol. enolates are helpful because they are easier to control. this is similar to an E1 mechanism because it can rearrange (less certainty).
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14
Q

what reactants push keto / enol tautomerization under basic conditions

A

LDA because it is a strong and big base. small bases don’t work because SN2 would be able to happen.

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15
Q

what reactants push keto / enol tautomerization under acidic conditions

A

an acid. this process is reversible

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16
Q

does the equilibrium favor keto or enol form

A

ch. 22 slide 16

The preference for the keto or enol tautomeric form depends on…
- Resonance stabilization –> keto form is more stable when there is resonance stabilization of the carbonyl group. Ex. conjugated systems or when the carbonyl group is adjacent to an EWG.

  • Steric effects –> Steric hindrance affects the stability of the enol form. If there are bulky groups near the potential enol site, the enol form may be disfavored due to steric repulsion.
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17
Q

describe the halogenation of the alpha carbon (acidic)

A

ch. 22 slide 18
halogens are electrophilic - the enol / enolates are nucleophilic

  • extra acidic hydrogens can continue to react to form halogenated products, however, these new EWGs make it difficult to keep halogenating. its especially bad to have EWGs and a carbocation right next to each other
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18
Q

explain how alpha halogenation to alpha beta unsaturated ketones occurs

A

ch.22 slide 19

  • a monosubstituted halogenation can easily produced an alpha-beta unsaturated ketone
  • the halogen is a good LG for E2
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19
Q

explain the halogenation of the alpha carbon under basic conditions

A

Ch.22 slide 21

reagents: X2, basic

FS: the enolate is in equilibrium with enol
- the X is an EWG, which is good because you’re distributing electron density. this makes it more reactive, which is why it can proceed further than in acidic conditions.
Produces: a haloform

SS: the haloform (C-X3) is a great LG, so OH comes in to form a tetrahedral intermediate. this leaves a carboxylic acid derivative.

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20
Q

describe LDA use in reactions via an enolate

A

ch.22 slide 22

  • 1 equiv of LDA and aprotic THF are used to produce products
  • with LDA, you have a lot of control. you can stop it.
  • with NaOH, its more disorganized. LDA is preferred.
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21
Q

describe the reaction via an enolate

A
  • enolates are formed through a base mediated reaction
  • one of the best bases for this is LDA - strong and non-nucleophilic
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22
Q

what happens to the products if the pka of the products are the same

A

you get a mixture of products

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23
Q

how does the equivalents of LDA used affect the halogenation

A

1 equiv gives monohalogenated
2 equiv gives dihalogenated

this is because using LDA prevents the cleavage of the haloform

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24
Q

explain how enolates are great nucleophiles and what do you have to worry about for the reactant when using them

A
  • enolates can be used to form carbon-carbon bonds (i.e., an alkylation)
  • if the reactant is symmetrical, you dont have to worry about kinetic / thermodynamic control
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25
Q

where is the enolate nucleophilic

A

at the alpha carbon

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26
Q

what does dess martin (DMP) do

A

oxidizes –> turns -OH into double bond O

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27
Q

what does NaH do to an -OH

A

becomes ONa

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28
Q

What does CH3I do when added to an enolate

A

adds the CH3 to the O-

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29
Q

kinetic vs. thermodynamic

A

see Ch.22 slide 28

Kinetic control: A reaction in which the product ratio is determined by the rate at which the products are formed. “sterics” (irreversible)

Thermodynamic product: The more stable product formed in a chemical reaction.
Thermodynamic control: A reaction in which the product ratio is determined by the relative stability of the products. “electronics” (reversible)

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30
Q

what temp must LDA be at and why

A

Ch. 22 slide 29

  • LDA must be at -78 degrees celsius. this is because kinetic control must be irreversible. even though LDA is very bulky and it should go to the less steric (kinetic position), due to the pka it goes to the thermodynamic position. so it must be cooled.
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31
Q

what product is favored when the base pka is close to the alpha C-H pka

A

the thermodynamic product is favored

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32
Q

explain how carbonyls can react with themselves

A

see Ch.23 slide 5

  • carbonyls can act as both nucleophiles and electrophiles, so they can react with themselves (one attacks the equivalent of another)
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33
Q

what groups do keto and enol forms of molecules react as

A

keto –> reacts like a carbonyl group
enol –> reacts like an alkene

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34
Q

what is the keto form

A

H3C-carbonyl- H

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35
Q

what is the enol form

A

H2C=C-OH
|
H

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36
Q

draw the mechanism for the aldol reaction

A

draw

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37
Q

what is the best group to use for an aldol reaction

A

if E+ is an aldehyde, it works better because it is less sterically hindered. a ketone still works, though

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38
Q

describe an aldol reaction

A

it is a reaction between a nucleophile and an electrophile. a new bond is formed between the alpha carbon and the carbn that was formerly the carbonyl carbon. it forms conjugated double bonds between two carbons and a water is produced

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39
Q

where does the carbon double bond for an aldol condensation go?

A

the more stable (conjugated) location is favored (ex. C-CH3 is more favored than C-H)

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40
Q

what reactants are used for the aldol condensation

A
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41
Q

describe the E1cB process of the aldol condensation

A

see ch. 23 slide 9

  • due to the conjugation (i.e., increased stability) condensation can occur under basic conditions, as well. However, the mechanism is neither E1 nor E2

draw mechanism

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42
Q

when can the aldol condensation occur without heat

A

ch. 23 slide 10

if the compound produced has increased conjugation, the condensation occurs spontaneously. this is due to resonance

draw resonance examples

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43
Q

where will the aldol condensation reaction attached to if reacted with LDA at -78 degrees C

A

the kinetic side because LDA

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44
Q

what does [H] symbolize

A

reduction (gaining atoms)

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45
Q

what does [O-]

A

oxidation (losing atoms)

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46
Q

how can you control a crossed aldol condensation reaction without LDA

A

two main routes:
1. use one carbonyl that lacks alpha-hydrogens
2. use kinetic control wit LDA
3. make your enolate first (slowly)
4. use knowledge of EWGs / EDGs to facilitate reactivity

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47
Q

how do you control a crossed aldol condensation by adding the enolate first

A

add one equivalent of LDA on only the reactant you want it to (to form the enolate), then add the other reagent

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48
Q

how can you control a crossed aldol condensation by using a carbonyl that lacks alpha-hydrogens

A

add a coupling partner that cannot form an enolate

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49
Q

how do you control a crossed aldol condensation using pka

A

ch.23 slide 17

ex. a compound with lower pka would deprotonate most easily and would be the donor, the other compound would be the acceptor

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50
Q

describe the claisen condensation

A

its similar to an aldol reaction, though with esters –> so a tetrahedral intermediate will be present.
- crossed claisens are controlled in the same manner as an aldol

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51
Q

draw the claisen condensation reaction

A

draw it

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52
Q

what is the difference between the claisen condensation and the aldol addition rxn

A

ch. 23 slide 22

claisen condensation:
- forms a pi bond by elimination of RO-
- produces a beta-keto ester
- forms a tetrahedral intermediate

aldol addition:
- the protonation of O-
- produces a beta-hydroxyaldehyde

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53
Q

how do unstable vs. stable intermediates impact claisen rxns

A

The stability of tetrahedral intermediate depends on the ability of the groups attached to the new tetrahedral carbon atom to leave with the negative charge.

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54
Q

how do enolates change if they are conjugated

A

see slide 24 ch. 23

  • conjugate addition is usually soft/soft interaction, so a soft base can facilitate conjugate addition
  • ” harder” enolate is one with just the enolate ion
  • “softer” enolate is one with the enolate and a carbonyl right next to it

*both are considered soft Nus, but one is a bit harder
**both are considered 1,4- additions

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55
Q

describe a michael addition

A
  • a 1,4- addition to a conjugated enone/enal –> one needs a weak Nu
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56
Q

what is an enone

A

C=C-carbonyl-C

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57
Q

what is an enal

A

C=C-carbonyl-H

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58
Q

draw the michael addition mechanism

A

draw it

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59
Q

what is a quick way to identify if a michael addition has occurred

A

see ch.23 slide 27

if a 1,4- addition has a occurred then its michael

ex. a double bond oxygen is a good indication that 1,4- occurred vs. just the -OH (which is 1,2-)

60
Q

tert butyl pka

A

18

61
Q

when does an amine act as an acid

A

-N-
H

pka = 36-38

62
Q

when do amines act as bases or nucleophiles

A

-N(+)-
H2

pka = 11

63
Q

why cant amines exist biologically in their acidic form

A

their pka is too high, so leveling effect can occur –> the hydrogen would deprotonate water

64
Q

what bonding types do primary amines engage in

A
  • H-bonds
  • dipole-dipole
  • LDF
65
Q

what bonding types do secondary amines engage in

A
  • H-bonds
  • dipole-dipole
  • LDF
66
Q

what bonding types do tertiary amines engage in

A
  • no-h-bond
  • dipole-dipole
  • LDF
67
Q

draw primary vs. secondary vs. tertiary amines

A

ch. 24 slide 4

68
Q

identify the pkas of the amines in slide 6 of ch.24

A
69
Q

what pka tell you about this reaction: HA + H2O <–> H3O+ + A-

A

pka tells you the acidity to form these products (as stability increases, the pka decreases)

70
Q

how do primary and secondary amines differ in terms of resonance and induction stabilization?

A

see slide 7 of Ch. 24 and draw resonance structures

71
Q

how does induction impact the pka of amines

A

see slide 8 of Ch. 24

EWG stabilizes amine (since they would have a pos charge if the proton was removed) - pka of groups with EWG (more induction) are lower

72
Q

how does hybrid character affect the pka of amines

A

see slide 9 of Ch. 24

more sp^2 character means more s character. more s character means greater EN on carbon which makes it a greater EWG, stabilizing the compound (lower pka)

73
Q

answer practice question on slide 10 of ch. 24

A
74
Q

how to determine which amine would be the strongest base

A

ch. 23 slide11

look for:
- conjugation (where positive charge would be based on resonance) and how to destabilizes that charge
OR
- where the lone pair is most reactive

75
Q

how can you change the solubility of amines in aqueous systems, and at what pH would water solubility be greatest

A

ch. 24 slide 13

deprotonating –> increases pka
protonating –> decreases pka

lower pkas are more water soluble

76
Q

draw mechanisms on slide 14 of ch. 24 (-NH2 + CH3Br)

A

CH3-N(+)H2
|
R

77
Q

how do you form an imine and an enamine from an aldehyde / ketone

A

see slide 15 of ch.24

78
Q

draw the gabriel synthesis

A

see ch. 24 slide18

79
Q

what makes an NH hydrogene between two carbonyls more acidic

A

resonance stabilization

80
Q

how can you reduce an imine to an amine

A

slide 19 ch. 24

reduce it [H]

81
Q

what does 1) R2NH 2) NaBH3CN do to a five membered ring attached to double bond O

A

slide 19 ch. 24

creates a tertiary amine

82
Q

how to turn NO2 to NH2

A

H2, Pd/C

83
Q

how to turn CN to primary NH2

A

[H]

84
Q

what does Br2, hv do to a benzene with a methyl

A

ch.24 slide 21

add a Br to the end of the methyl

85
Q

what does NaCN do to a benzene with CH2-Br attached

A

ch.24 slide 21

adds CN in place of Br

86
Q

what does NaBH4 and water do to a benzene with CH2-CN attached

A

ch.24 slide 21

becomes CH2-CH2-NH2

87
Q

what happens when you add Mg, Et2O to a benzene with Br attached

A

ch. 24 slide 22

becomes benzene with MgBr

88
Q

whats an imine

A

R-N=C-R2

89
Q

whats an enamine

A

R2N-C=CR-R2

90
Q

what does an aldehyde and a primary amine react to produce

A

an imine

91
Q

what does a ketone and a secondary amine react to produce

A

an enamine

92
Q

how do you produce an imine

A

ch. 24 slide 15/16

aldehyde and a primary amine with trace acid

93
Q

how do you produce an enamine

A

ch. 24 slide 15/16

ketone and a secondary amine with trace acid

94
Q

explain the conjugate addition of enones/enals with an amine

A

ch.24 slide 17

1,4- addition

95
Q

how can amines be synthesized via SN2

A

the gabriel synthesis for primary amines
direct amine SN2 for tertiary amines

96
Q

explain the gabriel synthesis

A

ch.24 slide 18

  • prepares primary amines
  • H is acidic due to resonance stabilization
97
Q

what are the reactants for a gabriel synthesis

A

ch.24 slide 18

Compound and then

  1. MOH
  2. R-LG
98
Q

what are the reactants for a gabriel synthesis

A

ch.24 slide 18

R2NH + R-LG

99
Q

if given a cyclopentanone, how do you make an imine

A

ch.24 slide 19

NH3

100
Q

if given cylopentanimine, how do you make a primary amine

A

ch.24 slide 19

NaBH3CN (H –> reduce it)

101
Q

if given cyclopentanone, how do you make a secondary amine

A

ch.24 slide 19

  1. RNH2
  2. NaBH3CN
102
Q

-NO2 to -NH2

A

H2, Pd/C

102
Q

if given cyclopentanone, how do you make a tertiary amine

A
  1. R2NH
  2. NaBH3CN
103
Q

-CN to -NH2

A

[H]

OR

  1. NaBH4
  2. H2O
104
Q

a benzene is reacted with
1. Br2, FeBr3
2. Mg, Et2O

what is produced?

A

see ch.24 slide 22

105
Q

biggest thing to look for when identifying aromatic conpounds

A

if an sp^3 carbon is present –> NOT AROMATIC

106
Q

what is pyrrole

A

ch.24 slide 26

five mem ring with NH

107
Q

resonance vs. induction in the context of pyrrole vs. pyrrolidine (aromatics)

A

ch.24 slide 27

resonance sustains electron dense core in the ring; induction has electronegativity where the EN atom is

108
Q

pyrrole

A

ch.24 slide 27

5 mem ring with double bonds and NH

109
Q

pyrrolidine

A

ch.24 slide 27

5 mem ring, no double bonds, NH

110
Q

what is delocalization energy

A

the extra stability a compound has as a result of having delocalized electrons

111
Q

what compounds have the most delocalization energy

options:
furan
thiophene
5 mem ring with carboanion
pyrrole
benzene

A

(what compounds are the most aromatic) 1=most aromatic

  1. benzene
  2. 5 mem ring with carboanion
  3. thiophene
  4. pyrrole
  5. furan
112
Q

what compounds are the most reactive?

options:
furan
thiophene
5 mem ring with carboanion
pyrrole
benzene

A

what are the least delocalized compounds 1=most reactive

  1. pyrrole
  2. furan
  3. thiophene
  4. benzene
113
Q

draw the resonance structure for pyrrole

A

ch. 24 slide 29

114
Q

where does bonding occur for pyrrole

A

ch. 24 slide 30

C-2

do not use LP on nitrogen for bonding with an electrophile because it is not nucleophilic (resonance structure)

115
Q

what is the pka of protonated pyrrole

A

-3.8

116
Q

why should you protonate at C-2? can you protonate at C-3?

A

ch. 24 slide 33

if you protonate at the nitrogen you destroy aromaticity. protonating at C-2 is more stable because there are three resonance contributors. C-3 only has two resonance contributors that would move the positive charge, which is why C-2 is more stable than C-3.

117
Q

where would furan protanate

A

C-2

118
Q

what is the major product for furan with a methyl on C-2 reacted with Br2

A

ch. 24 slide 35

methyl on one C-2, Br on the other C-2

119
Q

benzene attached to a five mem. ring with NH and a double bone, where would Br2 attach

A

ch. 24 slide 36

C-3 (C-2 would break benzene aromaticity)
when a benzene is attached, protonating at C-3 leaves the ring intact.

120
Q

pka of hydroiodic acid

A

~10

121
Q

pka of hydrobromic acid

A

~9

122
Q

pka of hydrochloric acid

A

~6

123
Q

pka of sulfuric acid

A

~3

124
Q

pka of a hydronium ion

A

H2O(+)-H

~1.7

125
Q

pka of sulfonic acids

A

~1

126
Q

pka of hydrofluoric acid

A

3.2

127
Q

pka of carboxylic acids

A

4

128
Q

pka of protonated amines

A

9-11

129
Q

pka of thiols

A

13

130
Q

pka of malonates

A

13

131
Q

pka of water

A

14

132
Q

pka of alohol

A

17

133
Q

pka of ketone / aldehyde

A

20-24

134
Q

pka of nitrile

A

25

135
Q

pka of ester

A

25

136
Q

pka of alkyne

A

25

137
Q

pka of sulfoxide

A

31

138
Q

pka of amine

A

~35

139
Q

pka of hydrogen

A

36

140
Q

pka of alkene

A

~43

141
Q

pka of alkane

A

~50

142
Q

periodic table acid trend

A

as EN increases and atomic size increases, relative acidity increases.

however, large, but high EN negatively charged atoms are more basic (ex. F is more acidic then I or Br) because larger atoms can stabilize the negative charge easier.

143
Q

what makes a compound a stronger base

A

more stable, the stronger the base

144
Q

pka and stability rule

A

the more stable a compound is after it loses a hydrogen atom, the lower it’s pKa

CH 336 (6 decks)

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