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CH 320N > Midterm 2 Review > Flashcards

Midterm 2 Review Flashcards

(156 cards)

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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
In a ketone, what is the nucleophile and what is the electrophile? When an enolate anion resonates into a keto, what is the nucleophile?
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
26
What makes a carbonyl carbon electrophilic?
The pull of the oxygen
27
Enolate anion
Enol but without the H that makes it an alcohol, so O would have an extra lone pair and - charge
28
Aldehyde or ketone Reagents: Br2, CH3COOH
Alpha carbon loses one H and forms a bond with Br
29
IUPAC priority rankings
Carboxylic acid > aldehyde > ketone > hydroxy > alkene > alkyne > R, NO2, X, OR, Ph
30
Which is more acidic and why? A) FCH2COOH B) CH3COOH
A due to inductive effects (pull of electron density through sigma bonds)
31
Which is more acidic? A) CH3COOH B) PhCOOH
B
32
Which is more acidic and why? A) PhCOOH B) ClPhCOOH
B due to inductive effects
33
Which is more acidic and why? A) PhCOOH B) H3CPhCOOH
A because the H3C in B is electron donating
34
Which is more acidic? A) PhCOOH B) O2NPhCOOH
B
35
In YPhCOOH, where the substituent Y could be Cl, H, CH3, NO2, or COH (aldehyde), rank the substituents from most acidic to least acidic.
NO2, COH (aldehyde), Cl, H, CH3
36
1° alcohol -> Reagent: H2CrO4
Carboxylic acid
37
Aldehyde -> Reagent: H2CrO4
Carboxylic acid
38
PhMgBr -> Reagents: 1) CO2, 2) H3O+
PhCOOH
39
PhCOOH + 1° alcohol <-> Reagent: cat H2SO4
PhCOO bound to alcohol’s substituent (H leaves PhCOOH, OH leaves 1° alcohol) H2O byproduct Fischer Esterification
40
HO bound to 5C chain, last C is an aldehyde <-> Reagent: cat H2SO4
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
41
What is the difference between Reagents: 1) NaBH4, 2) H2O Reagents: 1) NaBH4, 3) H3O+
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.
42
When double bond O is at the beta position, and reagent: warm, what happens to carboxylic acid?
It goes away (CO2)
43
1° alcohol -> Reagents: SOCl2, pyridine Et3N
Cl substitutes OH
44
Carboxylic acid -> Reagent: SOCl2
Cl substitutes OH (acid chloride)
45
List the carboxylic acid derivatives from most to least reactive. What is the trend for stability?
1) Acid Chloride 2) Acid Anhydride 3) Ester 4) Amide 5) Nitrile Stability increases as you go down the pyramid (opposite of reactivity)
46
Lower the pKa, ____acidic the acid, ____ stable the anion/conjugate base
More acidic More stable
47
Nucleophilic Acyl Substitution Reaction C bound to R, double bond O, and LG
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
48
Nucleophilic Acyl Addition Reaction C bound to R, double bond O, and H
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
49
Acid Chloride + H2O ->
Cl on Acid Chloride substituted by OH (forming RCOOH, a carboxylic acid) + HCl
50
Acid Anhydride + H2O -> Reagent: H3O+ cat acid or cat H2SO4
Ether splits so that side with O adds H and side without O adds OH Product: 2 RCOOH (carboxylic acid)
51
Ester + H2O <-> Reagent: cat H2SO4
OR substituted by OH (forming RCOOH, a carboxylic acid) + HOR
52
Ester + H2O -> Reagent: NaOH base promoted
R leaves so single-bonded O gains a lone pair and - charge, + ROH Saponification
53
Amide + H2O -> Reagent: acid promoted
Products: NH2 substituted by OH, NH4+
54
Amide + H2O -> Reagent: base promoted
Products: NH2 substituted by O- (3 lone pairs), neutral NH3
55
Nitrile -> Reagents: H2O, acid & base cat
Imidic acid intermediate: HO added to C, triple bond becomes double bond, H added to N Tautomerization: amide
56
N in a ring bound to one H and two C, one of the C has a double bond O -> Reagent: H3O+
Double bond O side gains OH (forms carboxylic acid), N drops down and becomes NH3+
57
Ester in a ring bound to two C, one of the C has a double bond O -> Reagent: NaOH
Double bond O side gains O- (3 lone pairs), other side drops down and becomes OH
58
N in a ring bound to one H and two C, one of the C has a double bond O -> Reagent: NaOH
Double bond O side gains O- (3 lone pairs), N drops down and becomes neutral NH2
59
CN on end of carbon chain Reagents: cat H2SO4, excess water
N goes away and carboxylic acid added, cyclizes if possible (need OH)
60
How can you move up the pyramid?
Fischer Esterification: carboxylic acid to acid anhydride SOCl2: carboxylic acid to acid chloride
61
Acid chloride + ROH ->
Cl on Acid Chloride substituted by OR (forming RCOOR, an ester) + HCl
62
Acid anhydride + ROH ->
Ether splits so that side with O (leaving group) adds H and side without O adds OR Product: RCOOR (ester) + RCOOH (carboxylic acid)
63
Ester + ROH <-> Reagent: cat H2SO4
OR on ester substituted by OR from ROH (forming a new ester) + HOR Transesterification
64
Amide + ROH ->
No reaction
65
Acid chloride + 2 eq NH3 ->
Products: amide, NH4+, Cl-
66
Acid anhydride + 2 eq NH3 ->
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
Ester + NH3 ->
Amide + ROH
68
Reactants: acid chloride, RCOO-, Na+ ->
Acid anhydride
69
EtCl -> Reagent: SN2 NH3 excess
NH2 substitutes Cl
70
Exhaustive alkylation
Adding EtCl to EtNH2 a bunch of times (up to 3 times)
71
Reductive amination
EtNH2 -> aldehyde + NH3
72
Ester -> Reagents: 1) 2 eq or excess CH3-MgBr, 2) H3O+
• 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
Acid chloride -> Reagents: 1) 2 eq CH3MgBr, 2) H3O+
Cl goes away, double bond O becomes OH with two methyls attached (3° alcohol)
74
Acid chloride -> Reagents: 1) (CH3)2CuLi, 2) H3O+
CH3 substitutes Cl (ketone) Gilman reagent special
75
Ester -> Reagents: 1) (CH3)2CuLi, 2) H3O+
No reaction
76
Epoxide -> Reagents: 1) (CH3)2CuLi, 2) H3O+
HOCH2CH2CH3 Gilman reagent special
77
CH3-CH=CH-I -> Reagents: 1) (CH3)2CuLi, 2) H3O+
CH3 substitutes I Gilman reagent special
78
H-COOH
Formic acid
79
H3C-COOH
Acetic acid
80
Ph-COOH
Benzoic acid
81
HO-carbonyl-carbonyl-OH
Oxalic acid
82
HO-carbonyl-CH2-carbonyl-OH
Malonic acid
83
HO-carbonyl-CH2-CH2-carbonyl-OH
Succinic acid
84
HO-carbonyl-COH-COH-carbonyl-OH COHs are racemic
Tartaric acid
85
Simplest ketone
Acetone
86
Simplest carbonyl
Formaldehyde
87
Simplest aldehyde
Acetaldehyde
88
Benzene ring attached to simple aldehyde
Benzaldehyde
89
Benzene ring attached to simple ketone
Acetophenone
90
Ester -> Reagents: 1) DiBAlH -78°C, 2) H3O+
H substitutes ester O, non-carbonyl side of ester breaks off attached to HO
91
2° amine with next door carbonyl -> Reagents: 1) LiAlH4, 2) H3O+
Chop off double bond O
92
Nitrile -> Reagents: excess H2O, NaOH
Triple bond N goes away and C becomes bound to O- and double bond O, NH3 byproduct
93
3 lines substituent
Propyl
94
4 lines substituent
Butyl
95
Substituent with 4 lines attached to one carbon
Tert-butyl
96
Simplest ketone
Acetone
97
Simplest carbonyl
Formaldehyde
98
Simplest aldehyde
Acetaldehyde
99
Benzene ring attached to simple aldehyde
Benzaldehyde
100
Benzene ring attached to simple ketone
Acetophenone
101
Trans alkene -> Reagents: CH2I2, Zn(Cu)
Triangle forms where double bond was with anti substituents (one wedge one dash), racemic
102
Cis alkene -> Reagents: CH2I2, Zn(Cu)
Triangle forms where double bond was, triangle arms are syn racemic (both wedge or both dash)
103
Primary OH -> Reagent: PCC
Aldehyde
104
Secondary OH -> Reagent: PCC or H2CrO4
Ketone
105
Alkene -> Reagents: 1) O3, 2) (CH3)2S
Ozonolysis
106
Terminal alkyne -> Reagents: 1) (sia)2BH, 2) H2O2, NaOH
Aldehyde on terminal carbon, carbon chain of single bonds
107
Terminal alkyne -> Reagents: HgSO4, H2SO4, H2O
Ketone on internal (most substituted) carbon from alkyne, carbon chain of single bonds
108
Alkene -> Reagents: 1) BH3, 2) H2O2, NaOH
Non-Markovnikov OH
109
Alkene -> Reagents: H2O (or ROH), Cat H2SO4
Markovnikov OH (or -OR)
110
Alkene -> Reagents: 1) Hg(OAc)2, H2O, 2) NaBH4
Racemic Markovnikov OH
111
Alkene -> Reagent: HBr
Markovnikov Br
112
Alkene -> Reagent: Br2
Br on both carbons, racemic anti product
113
Br on both carbons racemic anti product -> Reagents: 1) 3NaNH2, 2) H3O+
Brs go away and in between those, an alkyne forms
114
Alkene -> Reagents: 1) OsO4, 2) NaHSO3
OH on both carbons, racemic syn product
115
Alkene -> Reagents: H2/Pd
Alkane syn product
116
Alkene -> Reagents: Br2, H2O
OH on the most substituted carbon, Br on the least substituted carbon, racemic anti product
117
OH on the most substituted carbon Br on the least substituted carbon racemic anti product + NaOH
Racemic syn epoxide
118
Alkene -> Reagents: RCO3H or mCPBA
Racemic syn epoxide
119
Alkene -> Reagents: NBS, hv
Double bond moves away one to be more stable, Br adds to least substituted carbon (allylic bromoalkene)
120
Alkene -> Reagents: HBr, H2O2
Non-Markovnikov Br
121
Alkene -> Reagents: 1) OsO4, 2) NaHSO3
OH on both carbons, racemic syn product
122
Br on both carbons racemic anti product -> Reagent: 3NaNH2
Brs go away and in between those, a deprotonated alkyne forms
123
Internal alkyne -> Reagents: H2, Lindlar
Cis alkene
124
Internal alkyne -> Reagents: Na°, NH3
Trans alkene
125
Carbon chain of single bonds with primary Br -> Reagents: E2 tBuOK
Br goes away, double bond forms between the carbon Br was connected to and the carbon next to it
126
Secondary or tertiary Br -> Reagents: tBuOK E2
Br goes away, double bond forms between the carbon Br was connected to and the least substituted carbon (non-Zaitsev)
127
Secondary Br -> Reagents: E2 NaOH
Br goes away, double bond forms between the carbon Br was connected to and the most substituted carbon
128
Primary Br -> Reagents: NH3 SN2
NH2 substitutes Br
129
Primary OH -> Reagents: PBr3 SN2
Br substitutes OH
130
Secondary OH -> Reagents: SOCl2, pyridine
Cl substitutes OH and stereochemistry inverted
131
Tertiary OH -> Reagents: HBr SN2
Br substitutes OH
132
Primary OH -> Reagents: conc H2SO4 E2
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
Primary OH -> Reagent: H2CrO4
Double bond O added to carbon next door to OH
134
Syn epoxide -> Reagent: NH3
NH2 added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
135
Syn epoxide -> Reagent: NaOH
OH added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
136
Syn epoxide -> Reagent: NaOCH3
OCH3 added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
137
Syn epoxide -> Reagent: NaCN
CN added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
138
Syn epoxide -> Reagents: 1) LiAlH4, 2) H3O+
H added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
139
Syn epoxide -> Reagent: NaN3
N3 added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
140
Syn epoxide -> Reagents: 1) RC [triple bond] C : -, 2) H3O+
Substituent added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
141
Syn epoxide -> Reagents: 1) C-C-MgBr, 2) H3O+
Ethyl added to least substituted carbon in an anti fashion, O from epoxide becomes OH and retains stereochemistry
142
Syn epoxide -> Reagents: CH3OH, Cat H2SO4
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
Syn epoxide -> Reagents: H2O, Cat H2SO4
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
Br on both carbons racemic anti product -> Reagent: 2NaNH2
Internal alkyne
145
Alkane -> Reagents: Br2, hv or heat
Br added to most substituted carbon
146
Alkene -> Reagents: HBr, ROOR hv or heat
Br added to least substituted carbon of double bond, and double bond goes away
147
Secondary OH -> Reagent: H2SO4
Double bond forms between carbon OH was attached to and the carbon next to it, OH goes away
148
Primary Br -> Reagents: 1) Ph3P, 2) butylLi
+PPh3 substitutes Br, lone pair and - charge on the carbon next to it
149
The product is ___ reactive as compared to the starting material in acid monohalogenation because
Less, halogen is an EWG and decreases the basicity of the carbonyl oxygen
150
In base halogenation the product is ___ reactive as compared to the starting material because
More, the halogen increases the acidity of the alpha hydrogen thus making the product more reactive which leads to polyhalogenation
151
What is the driving force for base promoted hydrolysis of esters (saponification)?
Acid base reaction
152
Do reactants or products predominate in base promoted hydrolysis of esters (saponification)?
Products
153
What amount of base is required for base promoted hydrolysis of esters (saponification)?
Stoichiometric
154
Is base promoted hydrolysis of esters (saponification) reversible and why?
Irreversible as carboxylic acid anion is a weak electrophile and is not attacked by R-OH which is a weak nucleophile
155
In aqueous base promoted hydrolysis of amides the reaction is driven to completion by
The acid base reaction in the end
156
Higher or lower pKa predominates?
Higher

CH 320N (5 decks)

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