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CH 320N > Midterm 2 Review > Flashcards

Midterm 2 Review Flashcards

1
Q

What product predominates under low temperature conditions or when a very sterically-hindered amine is used?

A

Non-zaitsev kinetic

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2
Q

What product predominates in high temperature conditions?

A

Zaitsev alkene thermodynamic

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3
Q

How do 1° amines react under acidic conditions?

A

1° amines can react with aldehydes and ketones to give imines (aka Schiff bases)
• C=N bond

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4
Q

How do 2° amines react with enolizable aldehydes and ketones under acidic conditions?

A

2° amines can react with enolizable aldehydes and ketones (has alpha hydrogens) to give enamines
• C-N bond & C=C bonds

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5
Q

How do 2° amines react with non-enolizable aldehydes and ketones under acidic conditions?

A

2° amines can react with non-enolizable aldehydes and ketones (does not have alpha hydrogens) to give iminium ions
• C=N bond
• (+) charge

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6
Q

Alpha hydrogens

A

Hydrogens on the carbon next door to the carbonyl carbon

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7
Q

Enolizable

A

Alpha hydrogens present

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8
Q

Carbonyl ->
Reagents: NH3 or 1° amine, pH = 4

A

N substitutes carbonyl O, N loses 2 hydrogens, N has lone pair, N has its substituent (if applicable), H2O byproduct

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9
Q

Carbonyl with alpha hydrogens ->
Reagents: 2° amine, pH = 4

A

N substitutes carbonyl O and their bond becomes a single bond, N loses its hydrogen, N has its substituent(s), N has a lone pair, C=C double bond forms between reactant’s most substituted carbons
For intramolecular reactions, number carbons starting with nitrogen as 1 and carbonyl C as last #, form a ring with the number of carbons

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10
Q

Carbonyl with no alpha hydrogens ->
Reagents: 2° amine, pH = 4

A

N substitutes carbonyl O, N loses its hydrogen, N has its substituent(s), N has a + charge (loses its lone pair)

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11
Q

What would reagents 1) LAH, 2) H3O+ work on?

A

Strong reducing agent
Aldehyde -> 1° alcohol
Ketone -> 2° alcohol (racemic)
Imine -> amine (double bond becomes H)
Carboxylic acid -> 1° alcohol
Nitrile -> 1° amine (triple bond N becomes single bond NH2 (RNH2))
Ester -> 1° alcohol (double bond O on ester removed, ether O becomes OH and its non-carbonyl R group leaves bound to OH)
Amide -> amine (double bond O chopped off)

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12
Q

What would reagents 1) NaBH4, 2) H3O+ work on?

A

Mild reducing agent
Aldehyde -> 1° alcohol
Ketone -> 2° alcohol (racemic)
Imine -> amine (double bond becomes H)

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13
Q

What would reagents H2/Pd work on?

A

Alkene -> alkane
Alkyne -> alkane
Aldehyde -> 1° alcohol
Ketone -> 2° alcohol (racemic)
Imine -> amine (double bond becomes H)
Nitrile -> 1° amine (triple bond N becomes single bond NH2 (RNH2))

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14
Q

Why don’t hydride reducing agents react with carbon-carbon double bonds or triple bonds?

A

Alkenes & alkynes do not have an electrophile carbon unlike a carbonyl carbon

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15
Q

What would reagent NaBH3CN work on?

A

Imine -> amine (double bond becomes H)

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16
Q

Whenever you reduce imine -> amine (double bond becomes H), what do you always have to check for?

A

Racemic

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17
Q

Carbonyl ->
Reagents: NH2NH2, KOH

A

Double bond O goes away (works on aldehydes and ketones)
Wolf-Kishner

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18
Q

Carbonyl ->
Reagents: Zn(Hg), HCl

A

Double bond O goes away (works on aldehydes and ketones)
A 3° alcohol will go away and form the most stable C=C double bond from its original point of attachment
Clemmenson Reduction

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19
Q

Enol

A

C=C-OH

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20
Q

Enol to keto tautomerization

A

Reagent: cat acid or base
Non-alcoholic C gains a H, double bond moves to in between C and OH which loses its H so it’s just C=O

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21
Q

In keto-enol tautomerization, which product usually predominates and why?

A

Keto predominates because C=O bond stronger than C=C bond

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22
Q

Keto

A

Ketone functional group - aldehydes and ketones

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23
Q

In keto-enol tautomerization, when does enol predominate?

A

In B-diketones, enol predominates due to delocalization of pi bonds
Conjugated hydrogen bonding interaction

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24
Q

Keto to enol tautomerization

A

Reagent: cat acid or base
Double bond on one keto moves to in between that carbonyl C and most stable position, ex-carbonyl O gains a H

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25
Q

In a ketone, what is the nucleophile and what is the electrophile? When an enolate anion resonates into a keto, what is the nucleophile?

A

In a ketone, O nucleophile, C electrophile from carbonyl. When an enolate anion resonates into a keto, alpha carbon is nucleophilic (with lone pair and - charge)
Can make carbon carbon bonds

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26
Q

What makes a carbonyl carbon electrophilic?

A

The pull of the oxygen

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27
Q

Enolate anion

A

Enol but without the H that makes it an alcohol, so O would have an extra lone pair and - charge

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28
Q

Aldehyde or ketone
Reagents: Br2, CH3COOH

A

Alpha carbon loses one H and forms a bond with Br

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29
Q

IUPAC priority rankings

A

Carboxylic acid > aldehyde > ketone > hydroxy > alkene > alkyne > R, NO2, X, OR, Ph

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30
Q

Which is more acidic and why?
A) FCH2COOH
B) CH3COOH

A

A due to inductive effects (pull of electron density through sigma bonds)

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31
Q

Which is more acidic?
A) CH3COOH
B) PhCOOH

A

B

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32
Q

Which is more acidic and why?
A) PhCOOH
B) ClPhCOOH

A

B due to inductive effects

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33
Q

Which is more acidic and why?
A) PhCOOH
B) H3CPhCOOH

A

A because the H3C in B is electron donating

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34
Q

Which is more acidic?
A) PhCOOH
B) O2NPhCOOH

A

B

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35
Q

In YPhCOOH, where the substituent Y could be Cl, H, CH3, NO2, or COH (aldehyde), rank the substituents from most acidic to least acidic.

A

NO2, COH (aldehyde), Cl, H, CH3

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36
Q

1° alcohol ->
Reagent: H2CrO4

A

Carboxylic acid

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37
Q

Aldehyde ->
Reagent: H2CrO4

A

Carboxylic acid

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38
Q

PhMgBr ->
Reagents: 1) CO2, 2) H3O+

A

PhCOOH

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39
Q

PhCOOH + 1° alcohol <->
Reagent: cat H2SO4

A

PhCOO bound to alcohol’s substituent (H leaves PhCOOH, OH leaves 1° alcohol)
H2O byproduct
Fischer Esterification

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40
Q

HO bound to 5C chain, last C is an aldehyde <->
Reagent: cat H2SO4

A

You don’t draw H on OH, that O is the point of cyclization
All of the carbons form the ring, double bond O from aldehyde becomes single bond OH, and it’s racemic
Hemiacetal

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41
Q

What is the difference between
Reagents: 1) NaBH4, 2) H2O
Reagents: 1) NaBH4, 3) H3O+

A

In the first instance, double bond O (not carboxylic acid) would become racemic OH, and cat H2SO4 must be used to cyclize. The second instance does all of this in one synthetic step.

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42
Q

When double bond O is at the beta position, and reagent: warm, what happens to carboxylic acid?

A

It goes away (CO2)

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43
Q

1° alcohol ->
Reagents: SOCl2, pyridine Et3N

A

Cl substitutes OH

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44
Q

Carboxylic acid ->
Reagent: SOCl2

A

Cl substitutes OH (acid chloride)

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45
Q

List the carboxylic acid derivatives from most to least reactive. What is the trend for stability?

A

1) Acid Chloride
2) Acid Anhydride
3) Ester
4) Amide
5) Nitrile
Stability increases as you go down the pyramid (opposite of reactivity)

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46
Q

Lower the pKa, ____acidic the acid, ____ stable the anion/conjugate base

A

More acidic
More stable

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47
Q

Nucleophilic Acyl Substitution Reaction
C bound to R, double bond O, and LG

A

Step 1: attack of the nu, double bond on O becomes a lone pair with - charge
Step 2: departure of the LG, double bond on O is restored

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48
Q

Nucleophilic Acyl Addition Reaction
C bound to R, double bond O, and H

A

Step 1: attack of the nu, double bond on O becomes a lone pair with - charge
Step 2: add a proton, O grabs a H from H3O+ to become OH

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49
Q

Acid Chloride + H2O ->

A

Cl on Acid Chloride substituted by OH (forming RCOOH, a carboxylic acid) + HCl

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50
Q

Acid Anhydride + H2O ->
Reagent: H3O+ cat acid or cat H2SO4

A

Ether splits so that side with O adds H and side without O adds OH
Product: 2 RCOOH (carboxylic acid)

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51
Q

Ester + H2O <->
Reagent: cat H2SO4

A

OR substituted by OH (forming RCOOH, a carboxylic acid) + HOR

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52
Q

Ester + H2O ->
Reagent: NaOH base promoted

A

R leaves so single-bonded O gains a lone pair and - charge, + ROH
Saponification

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53
Q

Amide + H2O ->
Reagent: acid promoted

A

Products: NH2 substituted by OH, NH4+

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54
Q

Amide + H2O ->
Reagent: base promoted

A

Products: NH2 substituted by O- (3 lone pairs), neutral NH3

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55
Q

Nitrile ->
Reagents: H2O, acid & base cat

A

Imidic acid intermediate: HO added to C, triple bond becomes double bond, H added to N
Tautomerization: amide

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56
Q

N in a ring bound to one H and two C, one of the C has a double bond O ->
Reagent: H3O+

A

Double bond O side gains OH (forms carboxylic acid), N drops down and becomes NH3+

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57
Q

Ester in a ring bound to two C, one of the C has a double bond O ->
Reagent: NaOH

A

Double bond O side gains O- (3 lone pairs), other side drops down and becomes OH

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58
Q

N in a ring bound to one H and two C, one of the C has a double bond O ->
Reagent: NaOH

A

Double bond O side gains O- (3 lone pairs), N drops down and becomes neutral NH2

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59
Q

CN on end of carbon chain
Reagents: cat H2SO4, excess water

A

N goes away and carboxylic acid added, cyclizes if possible (need OH)

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60
Q

How can you move up the pyramid?

A

Fischer Esterification: carboxylic acid to acid anhydride
SOCl2: carboxylic acid to acid chloride

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61
Q

Acid chloride + ROH ->

A

Cl on Acid Chloride substituted by OR (forming RCOOR, an ester) + HCl

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62
Q

Acid anhydride + ROH ->

A

Ether splits so that side with O (leaving group) adds H and side without O adds OR
Product: RCOOR (ester) + RCOOH (carboxylic acid)

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63
Q

Ester + ROH <->
Reagent: cat H2SO4

A

OR on ester substituted by OR from ROH (forming a new ester) + HOR
Transesterification

64
Q

Amide + ROH ->

A

No reaction

65
Q

Acid chloride + 2 eq NH3 ->

A

Products: amide, NH4+, Cl-

66
Q

Acid anhydride + 2 eq NH3 ->

A

Ether splits so that side with O has 3 lone pairs and - charge, and side without O adds NH2
Product: amide, other half of acid anhydride with O-, NH4+

67
Q

Ester + NH3 ->

A

Amide + ROH

68
Q

Reactants: acid chloride, RCOO-, Na+ ->

A

Acid anhydride

69
Q

EtCl ->
Reagent: SN2 NH3 excess

A

NH2 substitutes Cl

70
Q

Exhaustive alkylation

A

Adding EtCl to EtNH2 a bunch of times (up to 3 times)

71
Q

Reductive amination

A

EtNH2 -> aldehyde + NH3

72
Q

Ester ->
Reagents: 1) 2 eq or excess CH3-MgBr, 2) H3O+

A

• Ester O with substituent leaves and adds H (ROH), double bond O becomes OH with two methyls attached (3° alcohol)
• If reactant is cyclized with ester O next door to carbonyl, bond between ester O and carbonyl carbon breaks, and ester O becomes OH

73
Q

Acid chloride ->
Reagents: 1) 2 eq CH3MgBr, 2) H3O+

A

Cl goes away, double bond O becomes OH with two methyls attached (3° alcohol)

74
Q

Acid chloride ->
Reagents: 1) (CH3)2CuLi, 2) H3O+

A

CH3 substitutes Cl (ketone)
Gilman reagent special

75
Q

Ester ->
Reagents: 1) (CH3)2CuLi, 2) H3O+

A

No reaction

76
Q

Epoxide ->
Reagents: 1) (CH3)2CuLi, 2) H3O+

A

HOCH2CH2CH3
Gilman reagent special

77
Q

CH3-CH=CH-I ->
Reagents: 1) (CH3)2CuLi, 2) H3O+

A

CH3 substitutes I
Gilman reagent special

78
Q

H-COOH

A

Formic acid

79
Q

H3C-COOH

A

Acetic acid

80
Q

Ph-COOH

A

Benzoic acid

81
Q

HO-carbonyl-carbonyl-OH

A

Oxalic acid

82
Q

HO-carbonyl-CH2-carbonyl-OH

A

Malonic acid

83
Q

HO-carbonyl-CH2-CH2-carbonyl-OH

A

Succinic acid

84
Q

HO-carbonyl-COH-COH-carbonyl-OH
COHs are racemic

A

Tartaric acid

85
Q

Simplest ketone

A

Acetone

86
Q

Simplest carbonyl

A

Formaldehyde

87
Q

Simplest aldehyde

A

Acetaldehyde

88
Q

Benzene ring attached to simple aldehyde

A

Benzaldehyde

89
Q

Benzene ring attached to simple ketone

A

Acetophenone

90
Q

Ester ->
Reagents: 1) DiBAlH -78°C, 2) H3O+

A

H substitutes ester O, non-carbonyl side of ester breaks off attached to HO

91
Q

2° amine with next door carbonyl ->
Reagents: 1) LiAlH4, 2) H3O+

A

Chop off double bond O

92
Q

Nitrile ->
Reagents: excess H2O, NaOH

A

Triple bond N goes away and C becomes bound to O- and double bond O, NH3 byproduct

93
Q

3 lines substituent

A

Propyl

94
Q

4 lines substituent

A

Butyl

95
Q

Substituent with 4 lines attached to one carbon

A

Tert-butyl

96
Q

Simplest ketone

A

Acetone

97
Q

Simplest carbonyl

A

Formaldehyde

98
Q

Simplest aldehyde

A

Acetaldehyde

99
Q

Benzene ring attached to simple aldehyde

A

Benzaldehyde

100
Q

Benzene ring attached to simple ketone

A

Acetophenone

101
Q

Trans alkene ->
Reagents: CH2I2, Zn(Cu)

A

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

102
Q

Cis alkene ->
Reagents: CH2I2, Zn(Cu)

A

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

103
Q

Primary OH ->
Reagent: PCC

A

Aldehyde

104
Q

Secondary OH ->
Reagent: PCC or H2CrO4

A

Ketone

105
Q

Alkene ->
Reagents: 1) O3, 2) (CH3)2S

A

Ozonolysis

106
Q

Terminal alkyne ->
Reagents: 1) (sia)2BH, 2) H2O2, NaOH

A

Aldehyde on terminal carbon, carbon chain of single bonds

107
Q

Terminal alkyne ->
Reagents: HgSO4, H2SO4, H2O

A

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

108
Q

Alkene ->
Reagents: 1) BH3, 2) H2O2, NaOH

A

Non-Markovnikov OH

109
Q

Alkene ->
Reagents: H2O (or ROH), Cat H2SO4

A

Markovnikov OH (or -OR)

110
Q

Alkene ->
Reagents: 1) Hg(OAc)2, H2O, 2) NaBH4

A

Racemic Markovnikov OH

111
Q

Alkene ->
Reagent: HBr

A

Markovnikov Br

112
Q

Alkene ->
Reagent: Br2

A

Br on both carbons, racemic anti product

113
Q

Br on both carbons racemic anti product ->
Reagents: 1) 3NaNH2, 2) H3O+

A

Brs go away and in between those, an alkyne forms

114
Q

Alkene ->
Reagents: 1) OsO4, 2) NaHSO3

A

OH on both carbons, racemic syn product

115
Q

Alkene ->
Reagents: H2/Pd

A

Alkane syn product

116
Q

Alkene ->
Reagents: Br2, H2O

A

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

117
Q

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

A

Racemic syn epoxide

118
Q

Alkene ->
Reagents: RCO3H or mCPBA

A

Racemic syn epoxide

119
Q

Alkene ->
Reagents: NBS, hv

A

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

120
Q

Alkene ->
Reagents: HBr, H2O2

A

Non-Markovnikov Br

121
Q

Alkene ->
Reagents: 1) OsO4, 2) NaHSO3

A

OH on both carbons, racemic syn product

122
Q

Br on both carbons racemic anti product ->
Reagent: 3NaNH2

A

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

123
Q

Internal alkyne ->
Reagents: H2, Lindlar

A

Cis alkene

124
Q

Internal alkyne ->
Reagents: Na°, NH3

A

Trans alkene

125
Q

Carbon chain of single bonds with primary Br ->
Reagents: E2 tBuOK

A

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

126
Q

Secondary or tertiary Br ->
Reagents: tBuOK E2

A

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

127
Q

Secondary Br ->
Reagents: E2 NaOH

A

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

128
Q

Primary Br ->
Reagents: NH3 SN2

A

NH2 substitutes Br

129
Q

Primary OH ->
Reagents: PBr3 SN2

A

Br substitutes OH

130
Q

Secondary OH ->
Reagents: SOCl2, pyridine

A

Cl substitutes OH and stereochemistry inverted

131
Q

Tertiary OH ->
Reagents: HBr SN2

A

Br substitutes OH

132
Q

Primary OH ->
Reagents: 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

133
Q

Primary OH ->
Reagent: H2CrO4

A

Double bond O added to carbon next door to OH

134
Q

Syn epoxide ->
Reagent: NH3

A

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

135
Q

Syn epoxide ->
Reagent: NaOH

A

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

136
Q

Syn epoxide ->
Reagent: NaOCH3

A

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

137
Q

Syn epoxide ->
Reagent: NaCN

A

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

138
Q

Syn epoxide ->
Reagents: 1) LiAlH4, 2) H3O+

A

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

139
Q

Syn epoxide ->
Reagent: NaN3

A

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

140
Q

Syn epoxide ->
Reagents: 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

141
Q

Syn epoxide ->
Reagents: 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

142
Q

Syn epoxide ->
Reagents: 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

143
Q

Syn epoxide ->
Reagents: 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

144
Q

Br on both carbons racemic anti product ->
Reagent: 2NaNH2

A

Internal alkyne

145
Q

Alkane ->
Reagents: Br2, hv or heat

A

Br added to most substituted carbon

146
Q

Alkene ->
Reagents: HBr, ROOR hv or heat

A

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

147
Q

Secondary OH ->
Reagent: H2SO4

A

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

148
Q

Primary Br ->
Reagents: 1) Ph3P, 2) butylLi

A

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

149
Q

The product is ___ reactive as compared to the starting material in acid monohalogenation because

A

Less, halogen is an EWG and decreases the basicity of the carbonyl oxygen

150
Q

In base halogenation the product is ___ reactive as compared to the starting material because

A

More, the halogen increases the acidity of the alpha hydrogen thus making the product more reactive which leads to polyhalogenation

151
Q

What is the driving force for base promoted hydrolysis of esters (saponification)?

A

Acid base reaction

152
Q

Do reactants or products predominate in base promoted hydrolysis of esters (saponification)?

A

Products

153
Q

What amount of base is required for base promoted hydrolysis of esters (saponification)?

A

Stoichiometric

154
Q

Is base promoted hydrolysis of esters (saponification) reversible and why?

A

Irreversible as carboxylic acid anion is a weak electrophile and is not attacked by R-OH which is a weak nucleophile

155
Q

In aqueous base promoted hydrolysis of amides the reaction is driven to completion by

A

The acid base reaction in the end

156
Q

Higher or lower pKa predominates?

A

Higher

CH 320N (5 decks)

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