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CH 320N > Midterm 1 Review > Flashcards

Midterm 1 Review Flashcards

1
Q

Penultimate carbon

A

Second to last carbon furthest away from aldehyde

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2
Q

D-Glucose

A

When OH on penultimate carbon is on the right

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3
Q

L-Glucose

A

When OH on penultimate carbon is on the left

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4
Q

Anomeric carbon

A

Carbon bonded to oxygen and OH

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5
Q

Alpha-anomer

A

When OH on anomeric carbon is axial

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6
Q

Beta-anomer

A

When OH on anomeric carbon is equatorial

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7
Q

Simple ylides give _ alkenes governed by ___

A

Z, kinetics

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8
Q

Resonance-stabilized ylides give _ alkenes governed by ___

A

E, thermodynamics

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9
Q

3 lines substituent

A

Propyl

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10
Q

4 lines substituent

A

Butyl

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11
Q

Substituent with 4 lines attached to one carbon

A

Tert-butyl

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12
Q

IUPAC priorities

A

Aldehyde > ketone > hydroxy > alkene > alkyne > X, R, OR

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13
Q

Simplest ketone

A

Acetone

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14
Q

Simplest carbonyl

A

Formaldehyde

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15
Q

Simplest aldehyde

A

Acetaldehyde

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16
Q

Benzene ring attached to simple aldehyde

A

Benzaldehyde

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17
Q

Benzene ring attached to simple ketone

A

Acetophenone

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18
Q

Gilman reagent reactivity compared to grignard and organolithium reagents

A

Gilman reagent less reactive, more selective

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19
Q

What reagent(s) would react with epoxides in the presence of acidic functional groups (-OH)?

A

Gilman reagents, not grignard or organolithium reagents

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20
Q

___ pka predominates

A

Higher

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21
Q

Carbocation: electrophile or nucleophile?

A

Electrophile

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22
Q

Carbanion: electrophile or nucleophile?

A

Nucleophile

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23
Q

Grignard reagents: strong or weak bases?

A

Strong

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24
Q

Organolithium reagents: strong or weak bases?

A

Strong

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25
Q

Gilman reagents: strong or weak bases?

A

Weak

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26
Q

Trans alkene + reagents: CH2I2 Zn(Cu)

A

Triangle forms where double bond was with anti substituents (one wedge one dash), racemic

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27
Q

Cis alkene + reagents: CH2I2 Zn(Cu)

A

Triangle forms where double bond was, triangle arms are syn racemic (both wedge or both dash)

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28
Q

Grignard reagent reaction with an epoxide mechanism

A

C-MgBr bond attacks least substituted carbon, that carbon’s bond to O breaks off O
Arrow label: attack of the nu
Product 2: O shifts over with additional lone pair (now 3) and - charge, substituent excluding MgBr added, MgBr with + charge byproduct, a lone pair on O attacks a proton from given acid whose bond breaks off of its O
Arrow label: add a proton
Product 3: H replaces one of three lone pairs (now neutrally charged), H2O byproduct

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29
Q

What is special about grignard reagent reaction with an achiral meso epoxide?

A

Achiral meso epoxides are symmetrical, substituent-MgBr can attack from either side, 2 racemic products

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30
Q

Grignard (or organolithium) reagent with an aldehyde or a ketone mechanism

A

C-MgBr bond attacks carbonyl carbon, that carbon’s bond to O breaks off O
Arrow label: attack of the nu
Product 2: O gains a lone pair (now 3) and - charge and now on a single bond, substituent excluding MgBr now on other side of ex-carbonyl carbon, MgBr with + charge byproduct, a lone pair on O attacks a proton from given acid whose bond breaks off of its O
Arrow label: add a proton
Product 3: H replaces one of three lone pairs (now neutrally charged), H2O byproduct, OH can be on a wedge or dash (racemic)

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31
Q

Key recognition element: -OH group and the new carbon-nucleophile are on the same carbon

A

Grignard (or organolithium) reagent with an aldehyde or a ketone

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32
Q

How can you make a carboxylic acid with a Grignard reagent?

A

CO2

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33
Q

Alkyne anion reacting with an aldehyde or ketone mechanism

A

Lone pair on deprotonated alkyne attacks carbonyl carbon, that carbon’s bond to O breaks off O
Arrow label: attack of the nu
Product 2: O gains a lone pair (now 3) and - charge and now on a single bond, substituent (- charge and lone pair went away) now on other side of electrophilic carbon bound by a wedge or dash (racemic), Na+ byproduct, a lone pair on O attacks a proton from given acid whose bond breaks off of its O
Arrow label: add a proton
Product 3: H replaces one of three lone pairs (now neutrally charged), H2O byproduct, product remains racemic

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34
Q

HCN reacting with an aldehyde or ketone mechanism

A

Lone pair on deprotonated carbon attacks carbonyl carbon, that carbon’s bond to O breaks off O
Arrow label: attack of the nu
Product 2: O gains a lone pair (now 3) and - charge and now on a single bond, C (- charge and lone pair went away) triple-bonded to N now on other side of electrophilic carbon, a lone pair on O attacks a proton from given acid whose bond breaks off of its bonded atom
Arrow label: add a proton
Product 3: H replaces one of three lone pairs (now neutrally charged), C (with lone pair and - charge) triple-bonded to N byproduct, OH could be on a wedge and H on electrophilic carbon would be on a dash or vice versa (racemic)

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35
Q

After HCN reacting with an aldehyde or ketone mechanism, what happens when you react H2/Ni OR LiAlH4 and H2O?

A

N becomes NH2, product remains racemic

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36
Q

The Wittig reaction with simple ylide mechanism

A

Ph3P’s lone pair on P attacks carbon on haloalkane, carbon’s bond breaks off LG atom
Arrow label: simultaneous attack of the nu and departure of the LG
Product 2: Ph3P+ replaces LG and one H on C is suggestively drawn, LG now with four lone pairs and - charge byproduct, lone pair on deprotonated base attacks suggestively drawn H whose bond with C breaks
Arrow label: remove a proton
Product 3: suggestively drawn H goes away and C gains a lone pair and - charge, Li+ byproduct, resonates to Ph3P=CH2

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37
Q

The Wittig reaction with resonance-stabilized ylide mechanism

A

Lone pair on C attached to Ph3P attacks carbon on resonance-stabilized ylide, carbon’s bond breaks off O
Arrow label: attack of the nu
Product 2: O gains a lone pair and - charge and now single-bonded to C
Arrow label: n/a
Product 3: resonates, bond between C and O points to O, O loses a lone pair and forms a bond with Ph3P (both become neutrally charged), bond between Ph3P and CH2 points to bond between that CH2 and the electrophilic carbon
Arrow label: n/a
Product 4: O- bonded to Ph3P+ leave, double bond forms between electrophilic carbon and CH2

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38
Q

Acid catalyzed hemiacetal and acetal formation from an aldehyde and ketone mechanism

A

Lone pair on carbonyl oxygen attacks an H, which breaks off O
Arrow label: add a proton
Product 2: byproduct and major product form. Carbonyl double bond points to carbonyl O and resonates to give O a lone pair. Lone pair on nucleophile attacks carbocation.
Arrow label: attack of the nu
Product 3: major product forms, lone pair on given base removes an H from ex-nucleophile
Arrow label: remove a proton
Product 4 (hemiacetal intermediate): byproduct and major product form, lone pair on O from carbonyl steals a proton
Arrow label: add a proton
Product 5: byproduct and major product form, arrow drawn for H2O LG to leave
Arrow label: departure of the LG
Product 6: H2O leaves, + charge where LG broke off, O lone pair points to its bond with carbocation, resonates to double bond and + charge moves to O, lone pair on new given nucleophile attacks ex-carbocation which breaks off the O it is double-bonded to
Arrow label: attack of the nu
Product 7: major product forms, given base steals proton from newly-added nucleophile
Arrow label: remove a proton
Product 8: byproduct and major product form

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39
Q

Why no SN2 (attack of nu rather than remove a proton) in the last step of acid catalyzed hemiacetal and acetal formation from an aldehyde and ketone?

A

Because of sterics

40
Q

Acid catalyzed formation of cyclic hemiacetal mechanism

A

Electrophilic oxygen steals proton
Arrow label: add a proton
Product 2: byproduct and major product form, lone pair on OH (not double-bonded) attacks more electrophilic carbon, one of the double bonds goes back to oxygen
Arrow label: attack of the nu
Product 3: ring forms, given base removes a proton
Arrow label: remove a proton
Product 4: byproduct and major product form, racemic about OH (where the double bond O was)

41
Q

Base-catalyzed formation of a hemiacetal mechanism

A

Lone pair on base removes a proton attached to O
Arrow label: remove a proton
Product 2: byproduct and major product form, deprotonated oxygen attacks given electrophilic carbon, double bond gives oxygen a lone pair
Arrow label: attack of the nu
Product 3: O- attacks given proton
Arrow label: add a proton
Product 4: byproduct and major product form, asterisk on chiral center

42
Q

Carbon radical: electrophile or nucleophile?

A

Electrophile

43
Q

Carbene: electrophile or nucleophile?

A

Both

44
Q

Primary OH + PCC

A

Aldehyde

45
Q

Secondary OH + PCC or H2CrO4

A

Ketone

46
Q

Alkene + 1) O3
2) (CH3)2S

A

Ozonolysis

47
Q

Terminal alkyne + 1) (sia)2BH
2) H2O2, NaOH

A

Aldehyde on terminal carbon, carbon chain of single bonds

48
Q

Terminal alkyne + HgSO4
H2SO4, H2O

A

Ketone on internal (most substituted) carbon from alkyne, carbon chain of single bonds

49
Q

Alkene + 1) BH3
2) H2O2, NaOH

A

Non-Markovnikov OH

50
Q

Alkene + H2O (or ROH)
Cat H2SO4

A

Markovnikov OH (or -OR)

51
Q

Alkene + 1) Hg(OAc)2, H2O
2) NaBH4

A

Racemic Markovnikov OH

52
Q

Alkene + HBr

A

Markovnikov Br

53
Q

Alkene + Br2

A

Br on both carbons, racemic anti product

54
Q

Br on both carbons racemic anti product + 1) 3NaNH2
2) H3O+

A

Brs go away and in between those, an alkyne forms

55
Q

Alkene + 1) OsO4
2) NaHSO3

A

OH on both carbons, racemic syn product

56
Q

Alkene + H2/Pd

A

Alkane syn product

57
Q

Alkene + Br2
H2O

A

OH on the most substituted carbon, Br on the least substituted carbon, racemic anti product

58
Q

OH on the most substituted carbon Br on the least substituted carbon racemic anti product + NaOH

A

Racemic syn epoxide

59
Q

Alkene + RCO3H
mCPBA

A

Racemic syn epoxide

60
Q

Alkene + NBS, hv

A

Double bond moves away one to be more stable, Br adds to least substituted carbon (allylic bromoalkene)

61
Q

Alkene + HBr
H2O2

A

Non-Markovnikov Br

62
Q

Alkene + 1) OsO4
2) NaHSO3

A

OH on both carbons, racemic syn product

63
Q

Br on both carbons racemic anti product + 3NaNH2

A

Brs go away and in between those, a deprotonated alkyne forms

64
Q

Internal alkyne + H2
Lindlar

A

Cis alkene

65
Q

Internal alkyne + Na°
NH3

A

Trans alkene

66
Q

Carbon chain of single bonds with primary Br + E2
tBuOK

A

Br goes away, double bond forms between the carbon Br was connected to and the carbon next to it

67
Q

Secondary or tertiary Br + tBuOK
E2

A

Br goes away, double bond forms between the carbon Br was connected to and the least substituted carbon (non-Zaitsev)

68
Q

Secondary Br + E2
NaOH

A

Br goes away, double bond forms between the carbon Br was connected to and the most substituted carbon

69
Q

Primary Br + NH3
SN2

A

NH2 substitutes Br

70
Q

Primary OH + PBr3
SN2

A

Br substitutes OH

71
Q

Secondary OH + SOCl2
pyridine

A

Cl substitutes OH and stereochemistry inverted

72
Q

Tertiary OH + HBr
SN2

A

Br substitutes OH

73
Q

Primary OH + conc H2SO4
E2

A

OH goes away, double bond forms between the carbon OH was connected to and the carbon next to it *double bond could resonate to be more stable

74
Q

Primary OH + H2CrO4

A

Double bond O added to carbon next door to OH

75
Q

Syn epoxide + NH3

A

NH2 added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

76
Q

Syn epoxide + NaOH

A

OH added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

77
Q

Syn epoxide + NaOCH3

A

OCH3 added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

78
Q

Syn epoxide + NaCN

A

CN added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

79
Q

Syn epoxide + 1) LiAlH4
2) H3O+

A

H added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

80
Q

Syn epoxide + NaN3

A

N3 added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

81
Q

Syn epoxide + 1) RC [triple bond] C : -
2) H3O+

A

Substituent added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

82
Q

Syn epoxide + 1) C-C-MgBr
2) H3O+

A

Ethyl added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry

83
Q

Syn epoxide + CH3OH
Cat H2SO4

A

O from epoxide becomes OH on least substituted carbon and retains stereochemistry, OCH3 added to most substituted carbon in an anti fashion, methyl on most substituted carbon gets its stereochemistry inverted

84
Q

Syn epoxide + H2O
Cat H2SO4

A

O from epoxide becomes OH on least substituted carbon and retains stereochemistry, OH added to most substituted carbon in an anti fashion, methyl on most substituted carbon gets its stereochemistry inverted

85
Q

Br on both carbons racemic anti product + 2NaNH2

A

Internal alkyne

86
Q

Alkane + Br2
hv or heat

A

Br added to most substituted carbon

87
Q

Alkene + HBr
ROOR hv or heat

A

Br added to least substituted carbon of double bond, and double bond goes away

88
Q

Secondary OH + H2SO4

A

Double bond forms between carbon OH was attached to and the carbon next to it, OH goes away

89
Q

Key recognition element: what does the formation of a carbon-carbon double bond mean?

A

Wittig Reaction

90
Q

Primary Br + 1) Ph3P
2) butylLi

A

+PPh3 substitutes Br, lone pair and - charge on the carbon next to it

91
Q

Carbonyl reactant + H2O <-> hydrate product
Which side predominates?

A

Side with less sterics predominates

92
Q

When Fs are present in carbonyl reactant + H2O <-> hydrate product, which side predominates?

A

Product side predominates because carbonyl carbon is more electrophilic and inductive effects

93
Q

What reagents do Grignard reagents react with?

A

Epoxides, aldehydes/ketones

94
Q

What reagents do organolithium reagents react with?

A

Epoxides, aldehydes/ketones

95
Q

What reagents do Gilman reagents react with?

A

Epoxides, 1° and 2° alkyl halide

CH 320N (5 decks)

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