Chem S-20ab Purple Book Solutions Week 5 - Page 1 of 37 Carbohydrates 1. When glucose is treated with methanol and acid, the methyl glucoside is formed. Provide a complete curved-arrow mechanism for this transformation: OH OH
HO CH OH, H+ HO O 3 O
H+ HO OH HO OMe
OH OH
OH OH OH
HO HO O HO O O
HO OH HO OMe 2 HO OH OH H OH HOCH3 Sol
2. The correct stereochemistry of only one stereocenter is indicated on the above structure. The actual stereochemistry of methyl glucoside is such that, when it is in a chair form, all of the substituents fall in the equatorial positions, except for the –OMe, which is found in the axial position. Given this information, draw a clear, correct representation of methyl glucoside in a chair conformation, with all substituents properly indicated. In addition, explain why the –OMe group preferably adopts an axial conformation; a diagram will be essential.
OH This is an example of the anomeric effect. HO OH The –OMe goes axial in order that the lone pair O * OH C–O O fromtheringOcandonateinto *C–O of the –OMe group. OMe OMe
3. When glucose is in its open-chain form, its stereochemistry can H O best be represented using a Fischer projection. On the skeleton Fischer projection shown at right, fill in the H and OH groups of the glucose "backbone" with the correct stereochemistry as H OH determined from your structure shown above. HO H
H OH
H OH
CH2OH Chem S-20ab Purple Book Solutions Week 5 - Page 2 of 37 Carbohydrates: Mechanisms
1. Provide a complete curved-arrow mechanism for all steps in the following transformation.
OH
HO HO O O OH OH– catalyst HO HO O H OH OH OH OH glucose HO O H—OH O HO OH OH OH
H—OH HO O Number the carbons carefully to see where they are going! This O HO O is a tricky mechanism! HO OH OH
HO OH OH O
HO OH OH O H HO OH HO OH O HO :O H OH OH HO O HO OH H OH O
:OH
OH OH
HO HO OH OH H—OH HO O HO OH
OH HOH O Chem S-20ab Purple Book Solutions Week 5 - Page 3 of 37
More Carbohydrates Mechanisms
1. Sorbitol is a sugar alcohol that is used as a sweetener in many low-calorie foods. It can be prepared from glucose.
Provide a complete curved-arrow mechanism for the following reaction. You may ignore stereochemistry for the purpose of this mechanism.
CH2OH OH OH HO NaBH O 4 HO OH NaOH , H2O HO OH OH OH
OH glucose OH sorbitol
CH2OH OH OH HO HO O O
OH OH HO O
OH HO H
NaBH4
CH OH CH2OH 2 O H HO H HO O OH "H-"
HO O HO O OH OH Chem S-20ab Purple Book Solutions Week 5 - Page 4 of 37 The Anomeric Effect
1. Draw the most stable chair conformations of the following molecules. Explain your choices in terms of either sterics or molecular orbital interactions.Whichwouldyouexpecttohaveashorter C-Cl bond?
Cl Cl
S
Cl Cl prefers to be axial Cl so that S lone pair can * S C–Cl * donate into C–Cl antibonding orbital Cl prefers to be (the anomeric effect). equatorial (sterics) This overrides the steric preference. Slp
This one would have a longer C–Cl bond because the donation into C–Cl * antibonding orbital will weaken the C–Cl bond.
2. Explain why the molecule H3C—S—S—CH3, when viewed along the S—S bond, prefers to adopt the gauche conformation. (A Newman projection will be helpful). Describe the orbital interaction(s) involved.
CH3 CH3
CH3 CH3 S S
There are two interactions: and S lone pair (in the front) Slonepair(intherear) donating into S–C *(intherear) donating into S–C *(inthefront)
(Note that only one lobe of the S–C * antibonding orbital is shown.)
CH3
S In the anti conformation there are no such interactions:
CH3 Chem S-20ab Purple Book Solutions Week 5 - Page 5 of 37 More Anomeric Effect
1. Each of the following observations can be explained by a frontier molecular orbital (FMO) argument. Provide such an explanation for each observation. Be sure to label the relevant donor-acceptor interactions and provide a clear diagram showing the interacting orbitals. (An energy-level diagram is not expected.) a) The following ester prefers conformation A rather than conformation B.(Hint: Consider all thelonepairs in the molecule. A "steric" argument is not acceptable.) O O preferred to CH3 H3C O H3C O CH3 AB
O O
CH3 H3C O H3C O
CH3
In conformation B,thesp2-hybrid In conformation A,thesp2-hybridized oxygen lone pair on the ester oxygen is lone pair is positioned to overlap with *of notabletooverlapwiththe *of the carbonyl (not *).Thisisafavorable the neighboring carbonyl, so there interaction which lowers the overall energy of is no such favorable interaction. this conformation.
b) The two conformers of the nitrite ester shown below have slightly different bond lengths (as indicated). (Note that both conformations are entirely planar. How are the bond lengths different between A and B?) O O R 1.17 Å 1.19 Å 1.39 Å NO NO 1.43 Å R A B
There is an important orbital interaction in conformation B which is not present Olp O R which makes in conformation A: donating the N=O bond longer and the into NO central N–O bond *N–O shorter
In conformation A the orbitals are not aligned properly for this lone pair -> * donation. Chem S-20ab Purple Book Solutions Week 5 - Page 6 of 37 Synthesis and Properties of Amino Acids
Here are three of the naturally-occuring amino acids:
O O O
H2N H2N H2N OH OH OH
OH
serine NH2 OH
lysine O
glutamic acid 1. Draw the structures of these three amino acids as they would appear at pH = 7. Classify each one as "neutral," "basic," or "acidic." O
+ H3N O O O– + + H3N H3N O– O–
O– OH acidic + neutral basic NH3 O
2. Match these amino acids with the following isoelectric points: 3.2 5.8 10.1 Explain your reasoning. glutamic serine lysine acid
Isoelectric point is the pH at which the species will be neutral. Glutamic acid will only be neutral if its sidechain acid is protonated at low pH. Lysine will only be neutral if its sidechain amino group is unprotonated at high pH. Serine will have an isoelectric point near neutrality.
3. Propose a synthesis of serine (as a racemic mixture) using the Strecker synthesis.
+ – H O H2N CN + H3N COO NaCN H3O H NH4Cl OH OH OH Chem S-20ab Purple Book Solutions Week 5 - Page 7 of 37 Peptide Synthesis
1. Provide reagents for protecting and deprotecting an amino acid as shown:
O O – O MeOH, H+ 1. OH H2N H2N H2N OH OMe 2. H+ workup OH R R R
O O
O O O tBuO O OtBu H TFA t H2N BuO N H2N OH OH OH
R (this is known O R R as "Boc-O-Boc")
2. Using these protecting groups, synthesize the following dipeptide using traditional solution-phase chemistry. You may start with amino acids that are already protected in any way you choose.
O Me H tBuO N + OMe OH H2N
O O
DCC
O Me H tBuO N OMe N H O O
1. TFA 2. OH- 3. neutralize O Me
H2N OH N H O glycylalanine Chem S-20ab Purple Book Solutions Week 5 - Page 8 of 37 Complete Peptide Synthesis
1. Provide a multistep synthesis for the desired product. You may use any inorganic reagents, plus any organic reagents with 8 carbons or less, the protected amino acid shown below, and the organic reagent DCC. The best answer will require six or fewer steps. (Hint: Think carefully about protecting-group strategy; it is possible to hydrolyze a t-butyl ester using dry acid without disturbing other esters.) For the purpose of this question you may ignore the stereochemistry of the product.
Starting material: the following partially-protected amino acid, plus any compounds with 8 carbons or less.
O O H O
OH H2N O
+ – 1. NH4 ,CN
O O + 2. H3O
O O O (Strecker)
(Boc-anhydride protecting grp.)
O O O H2N OH O H2N OMe OH + O NH MeOH, H+ O (Protect) O
DCC coupling O
HO O O H N H2N OCH3 O O + H H (dry) O N deprotect O NH OMe aspartame tert-butyls ® O (NutraSweet ) O Chem S-20ab Purple Book Solutions Week 5 - Page 9 of 37 Solid-Phase Peptide Synthesis
1. Now show how you could use solid-phase peptide synthesis to synthesize the same dipeptide, below. Again, you may start with amino acids which are protected in any way you choose. O Me
OH Cl tBuO N Merrifield H Resin O
Boc-protected Cs2CO3 Alanine O Me O– Cs+ tBuO N H O
O Me
O tBuO N H O
TFA (remove tBoc)
Me
O H2N O
O H t Boc-protected BuO N glycine OH DCC O
O Me H tBuO N O N H O O
O Me HF H2N OH N (remove H Boc O and detach from glycylalanine resin) Chem S-20ab Purple Book Solutions Week 5 - Page 10 of 37
More Complete Peptide Synthesis
1. Provide a complete synthesis of the dipeptide, lysylglycine, shown below. You may start with unprotected amino acids only. All of the carbon atoms from these amino acids must end up in the product. You may use any organic or inorganic reagents; you may use a Merrifield resin (but it is not required). The best answer will require seven or fewer steps.
O
H2N OH OH H2N
(CH2)4NH2 O
Boc-O-Boc MeOH, H+ (protect) (protect)
O H t OMe BuO N H N OH 2 O O (CH2)4NH
O tBuO
DCC
O O H t BuO N OMe H2N OH N N H H O O 1. TFA O
2. OH- 3. neutralize
HN O NH2 lysylglycine OtBu (as a racemic mixture) Chem S-20ab Purple Book Solutions Week 5 - Page 11 of 37
Putting It Together: Synthesis
1. Provide a multistep synthesis for the desired product (an unnatural amino acid) from the indicated starting material. You may use any inorganic or organic reagents. The best answer will require eight or fewer steps.
Starting material:
O O 1. DIBAL, -78°C H H EtO 2. H+ workup H OO OO
H2C PPh3
H
OO
+ H3O
O
H 1. NaCN, NH4Cl O
+ O 2. H3O NH3
Desired product (an unnatural amino acid) Chem S-20ab Purple Book Solutions Week 5 - Page 12 of 37 Acidity of Carbonyl Compounds
1. All of the following compounds have acidic C-H bonds. The pKa value for each compound is indicated:
Compound pKa Conjugate Base
O O O O O O 11 etc. R OR R OR R OR H H
O O O O O O etc. 13 RO OR RO OR RO OR H H
Less resonance O O O stabilization than R the examples above, R 20 R R so less stable R R conj. base means H H weaker acid.
O Ester oxygen is an O O EDG by resonance, and destabilizes the R 25 R R RO RO RO conj. base. H H Nitrogen is less N N electronegative than C R C R N oxygen, so this "enolate" 31 C R resonance f orm is a less significant contributor H H than in the case of For each of the above compounds: oxygen. draw each conjugate base. draw resonance structures showing the stabilization of each conjugate base. rationalize the order of acidity indicated.
2. Provide a complete curved-arrow mechanism for the keto-enol tautomerizations below.
O H H OH OH O :OH2 H H
HOH O O :OH OH H Chem S-20ab Purple Book Solutions Week 5 - Page 13 of 37 More Practice with Acidity of Carbonyl Compounds
1. The following molecules are both ketones, and are superficially similar in structure. However, they display strikingly different acidity. Explain this observation using frontier molecular orbital (FMO) theory. O O O * No overlap C-O possible!
pKa ~20
pKa ~40 C-H * C-O O Not only would deprotonation and delocalization of this species result in an C-H enolate that violates Bredt's rule, the * ketone's C-O orbitals are positioned such that the overlap required for this Deprotonation in this case allows f or delocalization is impossible in the first resonance stabilization of the conj. place! Thus the alpha protons are not at all acidic. base because the C-H electrons are positioned such that they can overlap * O with (and thus donate into) C-O, Ack! Not possible allowing enolate f ormation. to form a bridge- head alkene!
2. Provide a complete curved-arrow mechanism and FMO theory explanation that accounts for the observation that only the indicated proton exchanges with deuterium in the compound shown below.
DO: CH3 CH3 H NaOD D
D2O CH3 O CH3 O
Deuteration takes place in CH3 an axial orientation to avoid DOD pushing the methyl group up into the axial position, thus minimizing unf avorable 1,3- diaxial interactions. CH3 O
Deprotonation occurs at the axial alpha-proton because the C-H orbital is f avorably positioned tooverlapwiththe *C-O of the carbonyl. The equatorial proton lacks the appropriate orientation f or overlap to f orm an enolate.
So: (Makeamodelifyou Me Me still have trouble seeing C-H * C-H these!) t C-O t Bu BUT Bu * Me O Me O C-O Chem S-20ab Purple Book Solutions Week 5 - Page 14 of 37 Alpha Bromination Mechanisms
1. Provide a curved-arrow mechanism for each of the following transformations.
H+ OO Br2 OO Br trace H+
Sol
OO H O OH Br
OH OH OH O O O
H Br Br Br :Solvent
O O Br OH H Br Br2,NaOH
OH Br Br O Br Br O O H Br Br
More acidic than starting material --> can't stop! Chem S-20ab Purple Book Solutions Week 5 - Page 15 of 37 Basic Aldol Mechanisms
1. Predict the product, and provide a curved-arrow mechanism for each of the following transformations.
O O O
H H H
H OEt
O O
H H
OH O OH O O O HOEt H H H H OEt
O O O NaOMe H + H H
H
:OMe O Aldol Condensation H
H—OMe OH O
O O OH O H
H H
H :OMe Chem S-20ab Purple Book Solutions Week 5 - Page 16 of 37 Basic Claisen Mechanisms
1. Predict the product, and provide a curved-arrow mechanism for each of the following transformations.
O O O 1) NaOEt
OEt 2) H+ wkup OEt H OEt
H+ O O O O
OEt OEt OEt
O O
O O OEt OEt
OEt H OEt
MeO O O
OMe 1) NaOMe MeO 2) H+ wkup O H O
–:OMe
O + OMe H MeO MeO O O
O
:OMe MeO O MeO O O H OMe O Chem S-20ab Purple Book Solutions Week 5 - Page 17 of 37 More Aldol and Claisen Mechanisms
1. Provide curved-arrow mechanisms for the following transformations: O
O O NaOH
H
H :OH Why deprotonate at that carbon? Any other possibilities would give less stable 4-membered ring. 5- and 6-membered rings are nearly always preferred. O
O O
H OH
O O H :OH
O– H—OH OH
OH O H+ O O H O+ 3 H H H Acid-catalyzed Aldol addition!
:OH2 H OH O OH O
H H H H
:OH2
H+ OH O OH OH
H H H H Chem S-20ab Purple Book Solutions Week 5 - Page 18 of 37 Still More Aldol and Claisen Mechanisms
1. Provide curved-arrow mechanisms for the following transformations: O O
NaOH
OH O :OH O :OH condensation O H
OH
O – O :OH O O O O H reverse Aldol!
HO H OH O O O
forward Aldol H—OH O O O O OMe NaOMe OMe OMe
O MeO O O O
OMe OMe reverse Claisen! H O O OMe H—OMe MeO: OMe MeO O O
O OMe O (turn the molecule OMe around...) OMe H O O
MeO OMe
H MeO: Chem S-20ab Purple Book Solutions Week 5 - Page 19 of 37 Synthesis using Aldol and Claisen Reactions
1. Propose synthetic pathways for the following multistep transformations. An aldol or claisen reaction will form a crucial step in each synthesis.
O O– O Couldalsopossibly be done under RO- LDA + RO and ROH conditions.
O O O O
H+ workup
(any acyclic starting material)
O O O O
NaOMe O
OH OH OH 1. LiAlH4 CrO3 (dry) OH OH H 2. H+ workup O O
O
Cl
O
O O O KOtBu O tBuOH
H coumarin (first synthetic perfume) (It would be possible to do the esterification and aldol condensation in 1 step.) Chem S-20ab Purple Book Solutions Week 5 - Page 20 of 37 Decarboxylation: Mechanisms
1. Provide a curved-arrow mechanism for the following transformation.
O O – O NaOCH3 1. OH + Br Br (excess) + OCH3 2. H3O , H Sol :OCH3 H O
O– O
H+ OCH 3 OH
Br
Br
H O O
O O O
OCH3 H redraw in new :OCH3 conformation
Br O O
H O
O– O
OCH3 H+ O O
Br O
OH OH O O O O O O OH OCH3 H O OCH3 Chem S-20ab Purple Book Solutions Week 5 - Page 21 of 37 Decarboxylation: Synthesis
2. Using starting materials containing no more than 3 carbons, synthesize the following target using a route that involves decarboxylation as the final step.
O O O O 1. NaOCH3 + + Br + OCH3 OCH3 OCH3 2. H wkup
O O NaOCH Br + 3 OCH3
O O O NaOCH3 1. OH– OCH3 + 2. H3O ,
2. Show how the two starting materials shown below could be used to make the indicated product. What is the importance of the extra ester group in this synthtic route?
O 1) NaOMe O OMe O O + Ph + 2) H wkup Ph O MeO Ph
Ph O The ester group is required in order to make the enolate nucleophile do the conjugate (Michael) addition.
If we used an enolate made with LDA, it would add directly to the carbonyl. 1) NaOH O O O LDA + + 2) H3O ,heat Ph Ph Ph
O
O Ph O O Ph Ph Ph Chem S-20ab Purple Book Solutions Week 5 - Page 22 of 37 Controlling Enolate Formation
Each of the following enolates can be formed from a specific carbonyl precursor. Draw the structure of the carbonyl precursor that could be used, and give specific conditions necessary to form that enolate.
O 0.9 eq. LDA OLi
OLi O 1.1 eq. LDA
O O 1.1 eq. LDA OLi O OMe OMe *ketones are more acidic than esters*
LDA CN C NLi Chem S-20ab Purple Book Solutions Week 5 - Page 23 of 37 Using Lithium Enolates
1. Provide a synthesis and draw the curved-arrow mechanism.
O OH O
H NR2 H2CO H2O H work-up
– H2CO O – H—OH Aldol O O H Addition H O O
H
Make kinetic enolate first, then thermodynamic over time.
2. Explain why only one of the three possible products below is observed:
1. LDA (1.1 eq.) 2. MeI Me no H O O
LDA Attack of MeI CH3 H3C from top is blocked
H so yes O: MeImustattack LDA from bottom Me O
giving observed product
This enolate is impossible. That H can't be removed because its bonding no orbital does not overlap with * O: C-O Me O Chem S-20ab Purple Book Solutions Week 5 - Page 24 of 37 Simple Enolate Reactions 1. Each of the following transformations can be carried out in one to three steps. Fill in the reagents required for each step. Be specific! If a second step is not needed, please put an "X" through the second box.
0.9 eq. LDA O (thermo. enolate) O 1.
Br 2.
O 1.1 eq. LDA O 1.
2. Br
O O + CH3 Br2 ,H cat. CH3 1. Br
(Note: base catalysis would add more Br.) 2.
O O 1. Zn OCH3 OCH3 2. H Br HO 3. H+ wkup O
N C O LDA H 1. C N + H workup OH 2. Chem S-20ab Purple Book Solutions Week 5 - Page 25 of 37 Conjugate Additions: Mechanisms
1. Provide a complete curved-arrow mechanism for all steps in the following multi-step transformation.
O O
1. Me2CuLi
- H 2. H2O, OH
O "Me-"
O O OH
H
O
O O O OH HOH
OH O OH H Chem S-20ab Purple Book Solutions Week 5 - Page 26 of 37 More Synthesis
1. Provide a multistep synthesis for the desired products from each indicated starting material. You may use any inorganic reagents.
Starting material: OH
1. BH3
– O 2. H2O2,OH O
On this synthesis it's crucial to work backwards: look at the product, and observe the aldol condensation ( -unsaturated carbonyl). Then work back from there! O Base O (NaOCH3, CrO3 NaOH, etc.)
(Aldol Condensation) O Desired product
Starting material:
O O O 1. CuLi 2 1. O3
2. H+ workup 2. Me2S
H
O
There are other ways to make that key aldehyde intermediate, like starting with an acetal...
O
NaOH
(or NaOR)
Desired product Chem S-20ab Purple Book Solutions Week 5 - Page 27 of 37 Robinson Annulation: Mechanism
1. The Robinson annulation can be used to produce a six-membered ring. Show the product formed when the following two molecules are reacted with a base:
O
O O O NaOEt + EtOH OEt
CO2Et
Provide a curved-arrow mechanism for the transformation you indicated above: O O O O H OEt + OEt
CO2Et O O O
OEt
HO
CO2Et
O
O HOEt
O
CO2Et EtO H
HO
CO2Et
O
H O O EtO H EtO CO2Et O O
O CO2Et
CO2Et Chem S-20ab Purple Book Solutions Week 5 - Page 28 of 37 Robinson Annulation: Retrosynthesis
1. Each of the following molecules can be synthesized in a single step via the Robinson annulation. Suggest the appropriate starting materials to employ in each case.
a) EtO2C O O CO2Et + O
b)
MeO C 2 O O
MeO2C + OMe O O CO2Me
c) O
CO Et 2 O O O + Ph Ph OEt Ph Ph Chem S-20ab Purple Book Solutions Week 5 - Page 29 of 37 Conjugate Addition: Reactions
Work through the steps of the following reactions, showing all intermediates and the final products. The molecular formula of each of the final products is given for you to check your answer.
O
CuLi 1. 2 C16H20O 2. Ph Br
Cu
O
O
Br
:OMe O O H O CO2Me 1. NaOMe, C10H14O 2. NaOH, H2O 3. H+, O O O
CO2Me tautomerize*
H O O hydrolysis* O
O O O CO Me HO 2 OMe HOMe
CO2Me O
:OMe OMe O O CO2Me H MeO: O H O O H H O HO CO2Me O *Note: You are NOT responsible f or this type CO2M of decarboxylation mechanism. However, you should CO2Me e recognize that H+ and heat is a clue to decarboxylate. Chem S-20ab Purple Book Solutions Week 5 - Page 30 of 37 Conjugate Addition: Mechanisms
Provide a curved-arrow mechanism for each of the following transformations.
O O O2N Ph NaOCH3 Ph
NO2 O O OCH3 CH3OH H O O Ph N N O O Ph
NO2
O O O O H OCH3
Ph Ph Ph Ph H NO2 NO2 N N O O CH3O O O
O O O O NaOCH3 + Cl CH3O OCH3 CH3OH Cl Cl
O O O O OCH3 H Cl O CH3O CH3O Cl Cl Cl CH3O Cl Cl Chem S-20ab Purple Book Solutions Week 5 - Page 31 of 37 Conjugate Addition: More Mechanisms
1. Provide a complete curved-arrow mechanism for the following transformation.
N NH O H2NNH2 H+
H2N :Solvent O NH2
Sol H N NH
H2N O NH
H—Solvent
+ H H2N O NH
HN NH H2O
H2N HO NH NH + H2N H transfer HO Chem S-20ab Purple Book Solutions Week 5 - Page 32 of 37 Mechanisms Practice 2
1. Consider the following multi-step transformation: O O
H3CO CuLi + – Anionic H3O OH + Intermediate X H3CO 2 H CH3 CH3 a) In the box below, draw the structure of the "Anionic Intermediate X."
O–
OCH3
CH3 OCH3
b) Provide a complete curved-arrow mechanism for the transformation of "Anionic Intermediate X"intothe observed final product under the given conditions (aqueous acid, followed by aqueous base).
There are essentially four parts to this mechanism; I'll sketch out the steps but won't draw every arrow.
O– O H+
OCH3 Step 4: Eliminate H O H 2 (aldol CH3 OCH3 CH3 condensation) Step 1: Protonate the enolate O OH O
H OCH3 CH3
CH3 OCH3 O Step 3: Aldol addition (Form an enolate at the ketone, Step 2: then attack the aldehyde at the Acidic hydrolysis O end of the chain to form of the acetal the5-memberedring) (Most people forget this step! Notice the acetal!) H CH3 H Chem S-20ab Purple Book Solutions Week 5 - Page 33 of 37 Putting It Together: Recognizing Acyclic "Aldols"
1. Identify the starting materials which could produce each of the "skeleton" products below, give the name of the reaction, and provide the reagents necessary to carry out both steps of the transformation.
O O– O O OH O NaOR base + (condens.)
aldol addition aldol condens. Mixing w/base gives condensation directly
O O 1. NaOEt O O 1. OH– O + + 2. H w.u. + OEt OEt OEt 2. H3O ,
Claisen condensation decarboxylation
O O O O O 1. NaOEt 1. OH– + RBr + OEt 2. H w.u. OEt + 2. H3O , R R
enolate alkylation (acetoacetic ester synthesis) decarboxylation
O O NaOEt O O 1. OH– O + R' Br OEt OEt + 2. H3O , R R R' R R'
enolate alkylation decarboxylation (acetoacetic ester synthesis)
O O O O O O NaOEt 1. OH– + + 2. H3O ,
O OEt O OEt decarboxylation
Michael addition Chem S-20ab Purple Book Solutions Week 5 - Page 34 of 37 Putting It Together: Recognizing Cyclic "Aldols"
1. Identify the starting materials which could produce each of the "skeleton" products below, give the name of the reaction, and provide the reagents necessary to carry out both steps of the transformation. (The strange curved loops below are meant to represent a ring of any size; most common are 5- and 6-membered rings.)
O O O OH O NaOR base
(condens.)
aldol addition aldol condens.
O OH NaOR base O O O (condens.)
aldol addition aldol condens.
O O O O O
NaOEt 1. OH– OEt Br Br OEt
+ + 2. H3O ,
enolate alkylation decarboxylation (acetoacetic ester synthesis)
O O O O OEt O 1. NaOEt 1. OH– OEt OEt 2. H+ w.u. + 2. H3O ,
Dieckmann synthesis decarboxylation (intramolecular Claisen)
O O O O R O 1. OH– NaOEt R + RBr + OEt OEt 2. H3O ,
enolate alkylation (acetoacetic ester synthesis) decarboxylation Chem S-20ab Purple Book Solutions Week 5 - Page 35 of 37 Synthesis
Provide a multistep synthesis for the desired product. You may use any inorganic reagents and any acyclic (NO RINGS) organic compounds, but you must also use the indicated starting material. The best answer will require seven or fewer steps. O Starting material: O O O O (A) O H 1. O3 MeO OMe NaOMe 2. Me S MeOH 2
MeO2C CO2Me MeO2C CO2Me
To make (A): O O NaOMe
MeOH H NaOMe, O A MeOH
O
MeO2C CO2Me (+/-) Desired product Chem S-20ab Purple Book Solutions Week 5 - Page 36 of 37 Synthesis
Provide a multistep synthesis for the desired product. You may use any inorganic reagents and any organic compounds with 7 or fewer carbons, but you must also use the indicated starting material. The best answer will require seven or fewer steps.
Starting material: O O O O NaOMe OMe MeO OMe MeOH
O
NaOMe, MeOH
O O
O NaOMe
HOMe CO2Me CO2Me
1. Me2CuLi
2. Ph Br
Ph O Ph OH
NaBH4,EtOH
CO2Me CO2Me
Desired product Chem S-20ab Purple Book Solutions Week 5 - Page 37 of 37 Synthesis
Provide a multistep synthesis for the desired product. You may use any inorganic reagents and any organic compounds with 7 or fewer carbons, but you must also use the indicated starting material. The best answer will require nine or fewer steps.
Starting material: O O
O O OMe Compound A O OMe NaOMe, MeOH Ph 1. NaOH, H2O + 2. H3O ,heat Note: We O O decarboxylate before NaOMe + O completing the aldol condensation because H Ph MeOH forming the enolate Ph we need is not Compound A possible as long as Note: We can't start with the -dicarbonyl is compound A because it has present. too many carbons. O O O
NaOMe H2,Pd/C O MeOH Ph Ph Ph
O O O O MeO O H2C PPh3 O
Ph Ph Ph There's an alternative synthesis along these lines . . .
H HO HO O O OsO4,H2O O H (+/-) Ph H+ Ph Ph
Desired product