Midterm 3 Review Flashcards
Why does the electrophile add to the alpha carbon of the keto instead of the oxygen? (The counter ion is Na+)
The counter ion Na+ is more tightly associated with the oxygen than the alpha carbon thus block the approach of the electrophile
Acetaldehyde ->
Reagent: NaOH
H-carbonyl-deprotonated CH2 + H2O
Aldol Condensation Reaction
Which side predominates in Aldol Condensation Reaction?
Reactants
A) H-carbonyl-deprotonated CH2
B) Acetaldehyde
Which is nucleophilic and which is electrophilic?
A is nucleophilic, B is electrophilic
H-carbonyl-deprotonated CH2 + Acetaldehyde ->
Attack of the Nu: arrow from carbon lone pair to electrophile’s carbonyl carbon, whose double bond goes up to oxygen
Product: H-carbonyl-CH2-C bound to OH, CH3, and H (B-hydroxyaldehyde)
Racemic about OH
In B-hydroxyaldehyde which carbon is alpha and which is beta?
CH2 = alpha, C bound to OH = beta
B-hydroxyaldehyde ->
Reagent: H3O+
OH is a leaving group, double bond forms between alpha and beta carbons
Product: H-carbonyl-CH=CH-CH3 (a-B-unsaturated carbonyl compound)
Both E & Z alkene can be formed, stable
Dehydration
A) Acetaldehyde
B) H-carbonyl-Ph
Which one is more electrophilic and why?
B because Ph EWG makes B’s carbonyl carbon more reactive than A’s
Acetaldehyde + LDA <->
What is the product? Which side predominates?
H-carbonyl-deprotonated CH2
Almost 100% conversion
Products predominate
What molecule will undergo Intramolecular Aldol Condensation? What is the product?
Reagents: 1) NaOH, 2) H3O+
Number the deprotonated carbon as 1, if the last carbon (5 or 6) is an aldehyde you can cyclize! Double bond O from aldehyde on 6th carbon becomes OH. Carbons 1 and 6 are racemic.
Dehydration: OH goes away, double bond forms between 1 and 6 (a-B unsaturated aldehyde)
How many eq LDA are needed to cyclize?
1
How many eq LDA are needed for Aldol Condensation Reaction?
0.5
Retrosynthesize a 6-membered ring with carbonyl and double bond between a and B carbons
It must have come from an Intramolecular Aldol Condensation: break the bonds between a and B and put an aldehyde on B
How many eq LDA are needed for Cross-Aldol Condensation?
1
Retrosynthesize a ketone with Ph attached to one side and terminal alkene attached to other side
Ph attached to ketone + formaldehyde
Ethyl ester ->
Reagents: 1) NaOEt, EtOH
2) H3O+
Double bond O side x2 with alpha carbon on starting material bound to carbonyl carbon on substituent (B-ketoester)
Claisen Condensation
Why must you match the tail of the ester to the tail of the base in Claisen Condensation?
To prevent transesterification
How many eq LDA are needed for Claisen Condensation?
0.5
A) ethyl ester with alpha hydrogens
B) ethyl ester with no alpha hydrogens ->
Reagents: 1) NaOEt, EtOH
2) H3O+
Entirety of A with its alpha carbon bound to B’s carbonyl carbon (B-ketoester), ester O + tail go away
Cross Claisen Condensation
How many eq LDA are needed for Cross Claisen Condensation?
1
What molecule will undergo Dieckmann Condensation (Intramolecular Claisen Condensation)? What is the product?
Reagents: 1) NaOEt, EtOH
2) H3O+
Ester on both sides. Number the alpha carbon as 1, if the last carbon (5 or 6) is a carbonyl you can cyclize! 1 and 6 form a bond. 6’s O + tail is LG. Carbons 1 is racemic. (B-ketoester)
How many eq LDA are needed for Dieckmann Condensation?
1
Retrosynthesize B-ketoester
Using ester carbonyl to determine a and B carbon, break bond between alpha and beta, add OEt to beta carbon
Carboxylic acids ->
Reagents: excess EtOH, cat H2SO4
EtO substitutes OH on carboxylic acids
Enolate anion ->
Reagent: propyl-Br
Double bond goes away, propyl added to terminal carbon
Alpha-alkylation
Enamine ->
Reagents: 1) propyl-Br, acid chloride, or CH3Br
2) H3O+
Carbonyl that the enamine came from (carbonyl, 2° amine, pH = 4) adds substituent and it’s racemic if tertiary
a-alkylation
In an a-B-unsaturated aldehyde, which carbon is electrophilic? When it resonates to an enolate anion, which carbon is electrophilic?
First the carbonyl carbon, then the terminal/least substituted carbon (not involved in the double bond)
a-B-unsaturated aldehyde ->
Reagents: 1) (CH3)2CuLi, 2) H3O+
Double bond goes away, CH3 added to beta carbon which is racemic if tertiary
Michael Addition
H3C-carbonyl-ethyl ester (2 ketones) (Acetoacetic Ester/B-ketoester) ->
Reagents: 1) NaOEt, EtOH
2) SN2 CH3-I
3) NaOH, H2O
4) H3O+
5) Heat
1) deprotonate the alpha carbon
2) CH3 adds to deprotonated site and is racemic
3-5) ethyl ester goes away
Ketone-ethyl ester ->
Reagents: 1) NaOH, H2O
2) H3O+
Ester tail(s) substituted by H (forming carboxylic acid)
Malonic Ester ->
Reagents: 1) NaOEt, 2) Epoxide, 3) H3O+, 4) H3O+, 5) Heat, 6) H3O+
1) Deprotonate alpha carbon
2) 2 carbons and O- added at deprotonated site
3) O- becomes OH
4) Hydrolyze ester: ester tail(s) substituted by H (forming carboxylic acid)
5) One carboxylic acid goes away
6) Cyclize forming an ester (H goes away on both OHs)
Malonic Ester ->
Reagents: 1) NaOEt, EtOH
2) Ketone on cyclohexane with a-B alkene
3) NaOH, H2O
4) H3O+
5) Heat
1) Deprotonate
2) Form a bond between B carbon and deprotonated site, double bond on substituent goes away
3-4) Ester tail(s) substituted by H
5) One carboxylic acid goes away, B carbon is racemic
1) Ketone on cyclohexane, another ketone at B position
2) Ketone on cyclohexane with a-B alkene
Which is the Michael donor? Michael Acceptor? Electrophile? Nucleophile?
1, 2, 2, 1
1) Ketone on cyclohexane, another ketone at B position
2) Ketone on cyclohexane with a-B alkene
Reagents: NaOH, H2O
1 deprotonates alpha carbon, forms a bond with 2’s B carbon whose double bond moves one toward its O & double bond O becomes O-, double bond within cyclohexane goes away & B carbon is racemic & O- becomes double bond O again
Retrosynthesize an enamine
NH bound to its substituents + ketone without C=C double bond
Ketone with a-B alkene ->
Reagents: 1) Enamine (double bond between “carbonyl” carbon and alpha carbon and a gamma carbon), 2) H3O+
Retrosynthesize the enamine to the carbonyl compound (Donor) and bond the Donor’s alpha carbon to the Michael Acceptor’s beta carbon, acceptor’s C=C double bond goes away, donor’s alpha carbon is racemic (because tertiary)
Michael Acceptor
Always has a carbonyl and a-B alkene
A) Donor with deprotonated alpha carbon (between two carbonyls)
B) acceptor
Reagents: 1) NaOEt, EtOH
2) H3O+ (workup)
Donor’s alpha carbon bonds with acceptor’s beta carbon, acceptor’s C=C double bond goes away, donor’s alpha and acceptor’s beta carbons are racemic (tertiary)
A) acceptor
B) H2N-Et (donor)
What is the product?
1° amine loses one H, and N forms a bond with acceptor’s beta carbon whose C=C double bond goes away
A) carbonyl compound with C [triple bond] N (nitrile) at the beta position
B) acceptor
Reagents: 1) NaOEt, 2) H3O+ (workup)
The donor A’s alpha carbon forms a bond with acceptor’s beta carbon, acceptor’s C=C double bond goes away, donor’s alpha carbon is racemic because it’s tertiary
Ketone on cyclohexane ->
Reagents: 1) 1 Eq LDA, 2) ketone with a-B alkene, 3) NaOH
1) Deprotonate alpha carbon (forming Enolate Anion)
2-3) Michael Addition, starting material’s alpha carbon is racemic
Cyclize: number the carbon chain so that 6 is the starting material’s carbonyl carbon, the O double-bonded to 6 goes away so double bond with 1 forms (not racemic)
Enamine ->
1) Ketone with a-B alkene
2) H3O+
1) acceptor’s C=C double bond goes away and it bonds with donor’s alpha carbon, donor’s C=C double bond goes away (forming iminium ion)
2) N + its substituents become O, and if there is a tertiary carbon it is racemic
What gives better yields with Gilman Reagent?
1° Haloalkane
Aryl Haloalkane
Steps of Robinson Annulation
Michael Addition -> Aldol Condensation -> dehydrate of aldol product
A) B ketoester
B) Ketone with a-B alkene ->
Reagents: NaOEt, EtOH
Michael Addition
Cyclize: number the carbon chain so that 6 is the starting material’s carbonyl carbon (from ketone), the O double-bonded to 6 goes away so double bond with 1 forms (not racemic), quaternary carbon racemic
1) Ketone on cyclohexane with methyl at alpha position and another ketone at beta position
2) Ketone with a-B alkene
Reagents: NaOH, H2O
Michael Addition
Cyclize: number the carbon chain so that 6 is the starting material’s carbonyl carbon (from ketone), the O double-bonded to 6 goes away so double bond with 1 forms (not racemic), quaternary carbon racemic
In a retrosynthesis problem, how would you recognize an Aldol bond? Michael bond? How would you retrosynthesize?
Aldol bond = a-B alkene
Michael bond = from quaternary carbon, it is the bond that is uniquely within the same ring as the Aldol bond
Retrosynthesize by turning the Aldol bond into a double bond O (separated from Aldol bond ring) + take out Michael bond and make its adjacent bond a double (2 reactants ?)
Malonic Ester ->
Reagents: 1) NaOEt, EtOH
2) Br-CH2-CH2-CH2-CH2-Br
3) NaOEt, EtOH
1) one H goes away by deprotonation
2) deprotonated site grabs substituent by one Br
3) other H goes away by deprotonation
Cyclize: Br goes to deprotonated site and forms a ring (pentagon)
Ester ->
Reagents: 1) DIBALH, -78°C
2) H3O+
O bound to non-carbonyl tail breaks off bound to H (alcohol), 2nd product is gains an H where ester O once was
Reactant: cyclohexane with ketone and alpha racemic methyl
What type of product is formed when the reagent is >1 Eq LDA?
Kinetic Product
Double bond O becomes single bond O-, double bond forms between “carbonyl” carbon and least substituted carbon
Reactant: cyclohexane with ketone and alpha racemic methyl
What type of product is formed when the reagent is >1 Eq LDA?
Thermodynamic Product
Double bond O becomes single bond O-, double bond forms between “carbonyl” carbon and most substituted carbon
What happens to the Kinetic Product in the presence of an excess of ketone?
The reaction becomes reversible to generate a Thermodynamic Enolate Anion
of MO =
of atomic orbital
Conjugated diene
Butadiene (2 terminal alkenes)
Conjugated molecule absorbs light in the ___ region (___-___ nm)
UV-visible (200-700 nm)
Which has a smaller delta E (HOMO-LUMO gap): butadiene or hexatriene?
Hexatriene
Whenever you add HBr to a conjugated diene, what are the 2 possible products?
One double bond goes away, carbocation added to carbon 2
1,2 addition: Br added to carbocation
1,4 addition: double bond shifts over one to the middle/more stable, carbocation added to carbon 4, Br added to carbocation
What is the difference between the Kinetic Product and the Thermodynamic Product formed from the conjugated diene?
The kinetic product has the less substituted alkene, and the thermodynamic product has the more substituted alkene ALWAYS! Regardless of 1,2 or 1,4 addition
Which product is more likely to form at -78°C? 40°C?
Kinetic, thermodynamic
Whenever you add Br2 at a high temperature to a conjugated diene, what are the 2 possible products?
One double bond goes away
1,2 addition: Brs added to carbons 1 and 2
1,4 addition: double bond shifts over one to the middle/more stable, Brs added to carbons 1 and 4
Pericyclic Reaction
Occurs in a single step with a close loop of molecular orbitals. Reaction takes place in single step where ions or radicals are not made. Thus have significant stereochemical control.
Diene + dienophile
Reagent: High temp/warm
One of dienophile’s double bonds comes off and bonds with diene’s top carbon, one of diene’s top double bonds shifts down one to the middle/more stable, one of diene’s bottom double bonds comes off and bonds with dienophile’s bottom carbon (tertiary carbon will be racemic if applicable)
5 EWGs
-X, -NO2, -CN, -carbonyl-, Ph
What lowers the energy of the LUMO?
EWG
What raises the energy of the LUMO?
More electron rich
What does it mean for Diels-Alder Reaction to be a stereo specific reaction?
Retain the stereochemistry of the dienophile. If the dienophile is trans, its substituents will be racemic (one wedge one dash). Switch the wedge and dash for the second product but the wedge bridge (if applicable) will stay wedge. If the dienophile is cis, its substituents will be wedge.
What is a symmetrical product called?
Meso
Cyclopentadiene + alkene bound to CO2Et ->
Bicyclic product of the Diels-Alder Reaction with wedge bridge, dash CO2Et, double bond and CO2Et on opposite sides of the bridge, and tertiary carbon(s) will be racemic if applicable
Between Exo and Endo, which is the major product when dienophile has a substituent? Why? Where does substituent point?
Endo with substituent pointing down/axial (substituent points up/equatorial in Exo) because secondary orbital interaction is an additional interaction between the HOMO & LUMO with the [T.S.] therefore stabilizing
Stereochemistry of Diels-Alder Reaction product with two rings sharing a bond
Wedge bridge (from diene) both of those carbons are racemic, dashes coming off of first ring (diene) that may make up a second ring (dienophile), the carbons of the shared bond are also racemic, any other substituents coming off of the shared bond that do not make up any rings are wedge
Stereochemistry of the diene for a substituent coming off of the 1 or 4 carbon forming an E alkene (R group sticking out)? Z alkene (R group pointed in)?
E alkene: dash
Z alkene: wedge
If the diene has one of each it’s racemic
Stereochemistry of Diels-Alder Reaction product with two rings sharing a bond when ring from diene breaks
The arms coming off of the shared bond that formed the diene ring are dashes
What are arenes and 4 characteristics?
Benzene as a functional group
Most stable
Conjugate system
Each carbon is sp2
Close loop p-orbital
What is the structure in which each double bond in the benzene ring moves over one position?
Kekul structure
Aromatic Molecules
Most Stable
Have electron in the bonding molecular orbital which are lower in energy as compared to the atomic orbitals
Non-Aromatic molecules
As stable as alkene or conjugated allenes
Anti-aromatic molecules
Unstable
Electron in the non-bonding molecular orbitals
Is cyclobutadiene aromatic, non-aromatic, or antiaromatic?
Antiaromatic (4 pi)
Jahn Teller Distortion
Cyclobutadiene is so unstable that it tries to stabilize itself by distorting its shape
Hückel Rule of Aromaticity
1) Cyclic
2) All atoms in the ring have to be sp2 hybridized-close loop of p-orbital
3) Flat planar
How many pi electrons for aromatic? Antiaromatic?
4n + 2 pi electrons = aromatic
n is any integer
4n pi electrons = antiaromatic
Is benzene aromatic, non-aromatic, or antiaromatic?
Aromatic (6 pi)
Is cyclooctatetraene aromatic, non-aromatic, or antiaromatic?
Anti-aromatic
Cyclic✅
All sp2✅
Flat✅
8 pi
Is cyclooctatetraene in a tub conformation aromatic, non-aromatic, or antiaromatic?
Non-aromatic
When a cycloheptatriene has a lone pair and - charge on one carbon, is it aromatic, non-aromatic, or antiaromatic? What is the hybridization of that carbon?
Anti-aromatic (8 pi electrons) sp2
When a cycloheptatriene has nothing on one carbon, is it aromatic, non-aromatic, or antiaromatic? What is the hybridization of that carbon?
Non-aromatic
sp3
When a cycloheptatriene has a cation on one carbon, is it aromatic, non-aromatic, or antiaromatic?
Aromatic (6 pi)
A) cyclopentane
B) cyclopentadiene
Which is the stronger acid?
B
Is furane (O in a ring with 4 C and 2 double bonds) aromatic, non-aromatic, or antiaromatic?
Aromatic; although O is sp3 at first, when you resonate, every point can be sp2✅
Cyclic✅
Frat✅
6 pi electrons
Heterocyclic Molecules
Cyclic compounds that have atoms of at least two different elements as members of their rings (ex. Furane)
Is pyrole (N in a ring with 4 C and 2 double bonds) aromatic, non-aromatic, or antiaromatic?
Aromatic; although N is sp3 at first, when you resonate, every point can be sp2
6 pi electrons
Is B in a ring with 4 C and 2 double bonds aromatic, non-aromatic, or antiaromatic?
Antiaromatic
Cyclic✅
sp2✅
Flat✅
4 pi
Is Napthalene (a Polyaromatic Compound where a benzene shares a double bond with a cyclohexadiene) aromatic, non-aromatic, or antiaromatic?
Aromatic (10 pi)
Dif between [10]-Annule and Bridged [10]-Annulene
1) Non-Aromatic
2) Aromatic
Benzene bound to COOH
Benzoic Acid
Benzene bound to CHO
Benzaldehyde
Benzene bound to OH
Phenol
Benzene bound to NH2
Aniline
Benzene bound to CH3
Toluene
Benzene bound to OCH3
Anisole
Benzene bound to C=C
Styrene
Benzene bound to NO2 on one carbon, CH3 on the next carbon, NO2 on the next carbon, skip a carbon, and NO2 on the next carbon
Trinitrotoluene (TNT)
Benzene with CH3 coming off of each carbon of a double bond
o-xylene
Benzene bound to CH3 on one carbon, skip a carbon, and CH3 on the next carbon
m-xylene
Benzene bound to CH3 on one carbon, skip 2 carbons, and CH3 on the next carbon
p-xylene
S in a ring with 4 C and 2 double bonds
Thiophene
N-C=N-C=C- in a ring
Imidazole
Is pyridine (N-C=C-C=C-C= in a ring) aromatic, non-aromatic, or antiaromatic?
Aromatic (6 pi)
Monosubstituted
Benzene bound to one substituent
Disubstituted
A substituent bound to each carbon in a double bond in a benzene
Polysubstituted
Benzene with a substituent on each carbon
Benzene bound to Br
Bromobenzene
Benzene bound to NO2
Nitrobenzene
Benzene bound to ethyl
Ethylbenzene
When is a benzene named as a phenyl group?
1) If the alkyl chain attached to the benzene ring >= 6 carbons (ex. (S)-2-phenylheptane)
2) If the alkyl chain had the higher priority gp (ex. 2-phenylethan-1-ol)
IUPAC priority rankings
1) COOH
2) -CHO
3) -C=O -
4) -OH
5) NH2
6) Alkene
7) Alkyne
8) R, X, phenyl, NO2, OR
Is each ring in Azulene (cycloheptatriene sharing a bond with cyclopentadiene) aromatic, non-aromatic, or antiaromatic?
Aromatic; it can resonate so that each ring is 6 pi
Benzene with a substituent: what is the position one carbon away called? Two away? Three?
Ortho, meta, para
Benzene with Br on each carbon from a double bond
o-dibromobenzene
Benzene with NO2 on one carbon, skip a carbon, and Br on the next carbon
m-bromonitrobenzene
Benzene with ethyl on one carbon, skip two carbons, and Cl on the next carbon
p-chloroethylbenzene
Benzene with COOH on one carbon, skip a carbon, and NO2 on the next carbon
m-nitrobenzoic acid
Benzene with CHO on one carbon, skip two carbons, and CH3 on the next carbon
p-methylbenzaldehyde
Benzene with OH on one carbon and Br on the adjacent carbon
o-bromophenol
Benzene with NH2 on one carbon, skip a carbon, and COOH on the next carbon
m-aminobenzoic acid
Benzene with CHO on one carbon, skip two carbons, and OH on the next carbon
p-hydroxybenzaldehyde
Benzene with COOH on one carbon, skip a carbon, Br on the next carbon, skip a carbon, and OH on the next carbon
3-bromo-5-hydroxybenzoic acid
Which is more acidic: phenol bound to OH or cyclohexane bound to OH?
Phenol bound to OH
Which is more acidic: Phenoxide Anion (phenyl bound to O-) or cyclohexane bound to O-? Why?
Phenoxide Anion because resonance stabilize = more stable anion = more acidic
If benzene is bound to OH, order the following from least acidic to more acidic:
1) Cl at the para position
2) Cl at the ortho position
3) Cl at the meta position
4) Cl at both ortho positions and the para position
What is your reasoning for this order?
1, 3, 2, 4
Inductive Effects
If benzene is bound to OH, which is more acidic: Cl at the para position or CH3 at the para position? Why?
Cl at the para position because Cl is an EWG, while CH3 is an EDG
If benzene is bound to OH, which is more acidic: Cl at the para position or NO2 at the para position? Why?
NO2 because extra resonance
If benzene is bound to OH, which is more acidic: NO2 at the para position or NO2 at the meta position?
NO2 at the para position (p-nitrophenol) is more acidic
If benzene is bound to OH, which is more acidic: CN at the para position or Cl at the para position? Why?
CN because extra resonance
Ethyl-OH ->
Reagents: 1) Na°, 2) ethyl-Br (1° Haloalkane)
1) Ethyl-(O-)
2) Ether
Benzene-OH ->
Reagents: 1) NaOH, 2) ethyl-Br
Which side predominates?
Product 1: benzene-(O-) and H2O
Product 2: benzene-O-ethyl
Products predominate
In general, which side does equilibrium favor in terms or acid strength and pKa?
Side of weaker acid/higher pKa
Benzene-OH ->
Reagent: H2CrO4
-OH becomes =O, another =O added at para position, double bonds within ring reduced to 2: one on right and one on left
p-Qunione
Benzene-OH with another -OH at the ortho position ->
Reagent: H2CrO4
Both -OH become =O, double bonds within ring reduced to 2 so that no double bond is touching a carbonyl
o-Qunione
Benzene-OH with another -OH at the para position ->
Reagent: H2CrO4
Both -OH become =O, double bonds within ring reduced to 2: one on right and one on left
p-Qunione
Benzene-CH3 ->
Reagent: H2CrO4
COOH substitutes CH3 (benzoic acid)
Benzene-ethyl ->
Reagent: H2CrO4
Benzene-COOH
Benzene-isopropyl ->
Reagent: H2CrO4
Benzene-COOH
Benzene—tert-butyl ->
Reagent: H2CrO4
No Reaction
Radical stability trend
Methyl < 1° < 2° < 3° < allylic radical ~= Benzylic radical
