Chem S-20ab Purple Book Solutions Week 4 - Page 1 of 38 Aromaticity

1. Given the following structures, circle any that are aromatic. Explain why the other species are not aromatic.

4n electrons not a continuous antiaromatic cyclic system.

NH

not a continuous 4n electrons cyclic system. antiaromatic

2. In the following molecule, which nitrogen atom is more basic (i.e. more willing to accept a proton)? Why?

The other one has its lone pair as part of CH3 the 6 electron aromatic system. Protonating that lone pair would destroy the aromaticity, N which would be bad: H CH3 this one is more basic N N not a continuous cyclic system. N 3. When the following molecule is oxidized, the expected ketone is not formed, but phenol is formed instead. Describe what is going on here, and explain this unusual behavior.

OH O OH

CrO3

oxidation

phenol

Tautomerizes in order to achieve aromaticity. Chem S-20ab Purple Book Solutions Week 4 - Page 2 of 38 Electrophilic Aromatic Substitution: Mechanisms

1. Provide a curved-arrow mechanism for the following transformation. Show resonance structures for the cationic intermediate which make it clear why the substitution goes where it does.

O O

Cl HN CH3 HN CH3

AlCl3 catalyst

Cl AlCl3 Cl

AlCl3 O O O O

HN CH3 HN CH3 HN CH3 HN CH3

H H H

want to put + charge next to :Solvent electron-donating nitrogen

2. Provide a curved-arrow mechanism for the following transformation.

H+ + O: H2SO4 catalyst

H OH :Sol

:Solvent OH2

H

OH H+ OH OH Chem S-20ab Purple Book Solutions Week 4 - Page 3 of 38 Electrophilic Aromatic Substitution: Direction

1. For each of the following species, show the most likely site(s) for electrophilic aromatic substitution, and predict whether the molecule is more reactive than benzene or less reactive than benzene.

NO2

more reactive less reactive

Strong deactivator Cl O

Note: Imagine that these two groups are separate. The fact that they are connected is irrelevant. Weak activator less reactive less reactive

CH3 O OCH3 O

CH3 NO2

more reactive

less reactive Chem S-20ab Purple Book Solutions Week 4 - Page 4 of 38 Electrophilic Aromatic Substitution: Reagents 1. Each of the following transformations can be carried out in one or two steps. Fill in the reagents required for each step. If a second step is not needed, please put an "X" through the second box.

,AlCl 1. Cl 3

Note: Don't try to get fancy with F-C alkylation. You can use it 2. ONLY for an isopropyl or tert-butyl group!

O O

,AlCl 1. Cl 3

Cl Cl2 ,AlCl3 2.

O OH

,AlCl 1. Cl 3

NaBH4,EtOH 2.

CH3 O CH3 SO H ,AlCl 3 1. Cl 3

SO3 ,H2SO4 2. O

O O Cl ,AlCl 1. Cl O 3

O 2. Chem S-20ab Purple Book Solutions Week 4 - Page 5 of 38 Electrophilic Aromatic Substitution: Mechanism

1. Provide a complete curved-arrow mechanism for the following transformation.

CH3 CH3

H2SO4

H3C OH CH3

H+

(aromatic resonance)

CH CH3 3

H3C OH2 CH3

:Solvent

H H3C H3C

CH3 CH3 Chem S-20ab Purple Book Solutions Week 4 - Page 6 of 38 More Mechanisms

1. The following reaction produces a mixture of the two products A and B: Br

Br Br Br

Br2 +

OCH3 OCH3 OCH3

AB a) Provide a complete curved-arrow mechanism that leads to the product A.

Br – Br: Br :Br—Br Br

OCH OCH3 3 OCH3 b) Provide a complete curved-arrow mechanism that leads to the product B. (Hint: The lone pairs of the – OCH3 play an important role in the mechanism. Also, consider the mechanism for electrophilic aromatic substitution.)

:Br—Br Br Br:– Br Br Br

OCH OCH 3 3 OCH3 OCH3 Chem S-20ab Purple Book Solutions Week 4 - Page 7 of 38

Chemistry S-20ab Nucleophilic Aromatic Substitution: Mechanisms Week 4 Provide a complete curved-arrow mechanism for the following transformation. You must draw the most favorable resonance structure for all intermediates.

:OH O N Cl O N OH 2 KOH 2

O2N O2N

OH OH O2N O2N Cl Cl O O N N O O

Provide a complete curved-arrow mechanism for the following transformation.

tBuO: H OK I O

KOtBu

I

tBuO H

O O Chem S-20ab Purple Book Solutions Week 4 - Page 8 of 38 Nucleophilic Aromatic Substitution: Direction

1. For each pair, choose the species which will react faster with NaOCH3, and explain your reasoning. (Resonance structures will be helpful!)

– CH3O: Cl Cl

NO2 NO2 vs.

H3C faster CH3

H3CO Cl In this resonance structure, the NO2 negative charge is next to the electron-donating CH3 group. This is bad. So the other one is faster; it never has a negative charge next to

CH3 the methyl.

– CH3O: Cl Cl NO 2 NO2 vs.

NC faster CN

H3CO Cl In this resonance structure, the NO negative charge is next to the 2 electron-withdrawing CN group. This is good. So this one will react faster.

CN 2. Explain the unusual product mixture observed in the following reaction.

CH3 CH3 CH3

Cl NaOH, heat OH + This is the "benzyne" mechanism! H OH

–:OH

Cl CH3

–:OH can attack either carbon. Chem S-20ab Purple Book Solutions Week 4 - Page 9 of 38 Nucleophilic Aromatic Substitution: Reactivity

1. Explain the following observations. Writing chemical equations and resonance structures will be more useful than long sentences!

a) Although most hydrocarbons are fairly nonpolar, the following hydrocarbon has a significant dipole moment. (As part of your explanation, please indicate the direction of the dipole moment: which end is positive and which end is negative.)

resonance

This resonance structure is particularly stable. The left half is an aromatic ring with 2 pi-electrons. The right half is an aromatic ring with 6 pi-electrons.

This resonance contributor gives an overall dipole moment to the molecule, with the left side as the positive end and the right side as the negative end.

b) Sodium ethoxide will substitute for chloride in one of the substrates below but shows no reaction with the other, similar substrate. Cl OEt Cl

NaOEt NaOEt no but reaction

This substrate has no EtO:– special mechanism for stabilizing the Cl EtO Cl anionic intermediate.

Notice the aromatic ring with 6 pi-electrons. This stabilizes the anionic intermediate in this nucleophilic aromatic substitution reaction, even without the presence of an electron withdrawing group. Chem S-20ab Purple Book Solutions Week 4 - Page 10 of 38 Carbonyl Reactivity

1. A site on a molecule will be electrophilic if it has a partial positive charge and a low-lying LUMO. Explain, using resonance and molecular orbital theory, why the carbonyl carbon is electrophilic.

From MO point of view, O O we know that a * antibonding orbital will already be a good LUMO, and the electronegative oxygen makesitevenlower,soitisan Resonance shows partial excellent LUMO. electrophilic carbon positive charge on carbon (even though this is not a particularly good res. structure). . 2. Using resonance and molecular orbital theory, rationalize the reactivity of the following carbonyl analogs towards nucleophilic attack: More reactive Less reactive

+ + OH NR2 O NR

OH NR2 O NR

These resonance structures become increasingy worse from left to right, so there's less + charge on carbon. From an MO perspective, the two factors changing here are electronegativity and charge. We know that a more electronegative atom (like O) makes the energy of the LUMO lower, in turn making the molecule more electrophilic. Positive charge has the same effect. It makes electrons more strongly attracted, lowering the energy of the LUMO.

3. The electrophilicity of a carbonyl compound is affected by both inductive and resonance effects of neighboring groups. Rationalze the reactivity of the following carbonyl derivatives: More reactive Less reactive

O O O O O O

– Cl H R OR NR2 O electron C-H bonds withdrawing No resonance * by induction or induction donate to lone pairs are electron donating (hyperconjugation)

O O O

OR NHR2 O

increasingly important alternate res. structures

The energy of the * LUMO is affected by the electronegativity of the neighboring group (which is why the Cl lowers the energy of the LUMO), but also by electron donation from neighboring lone pairs (which is why the ester, amide, and carboxylate raise the energy of the LUMO and make it a poorer acceptor.) Chem S-20ab Purple Book Solutions Week 4 - Page 11 of 38 Carbonyl Addition: Mechanisms

1. Provide curved-arrow mechanisms for the following transformations:

O

NaBH4 H C H H C OH 3 EtOH 3

H H OEt BH3 O H

H3C H

O OH MgBr + H /H2O

H3C H H3C H+

CH2 O H3C H

H3C

H+

O + H2O, H HO OH

H3C H H3C H

HO H O OH H H3C H

H3C H H2O

:OH2

+ O HO OH H2O, H

H CH3 H CH3

H2O H O OH H 2 O

H H CH3 Chem S-20ab Purple Book Solutions Week 4 - Page 12 of 38 Reactivity of Carbonyl Addition Reactions

1. Explain the observed ranking of equilibrium constants for carbonyl hydration in the following substrates:

O H2O HO OH [hydrate] K= [carbonyl] R R' R R'

Carbonyl Compound K

O 1x106 electron-withdrawing groups (F's) destabilize the highly electron- F C CF 3 3 deficient C=O. (The hydrate is more O electron rich than the carbonyl.) 3x104

F3C H

O 2x103 H H

O alkyl groups (CH3)aremore 1 electron-donating than H, and H3C H thus stabilize the electron- deficient C=O O

1x10–3 H3C CH3

O

9x10–6 resonance with aromatic CH3 ring stabilizes pi-bonding in C=O

2. The following compound has 2 carbonyl groups. Which one will become hydrated under aqueous conditons? Show the structure of the hydrate. Explain why the hydration equilibrium ratio K is equal to 0.8, that is, significantly larger than the ratio for most unsubstituted ketones such as acetone (above). Ketone is more reactive O HO OH than ester, since ester is H O O 2 O stabilized by donation of H C CH H C CH 3 3 3 3 O lone pairs from –OCH3. O O Ester group is electron- withdrawing group.

3. In the reduction of ketones with NaBH4, cyclobutanone reacts 17 times faster than acetone. Explain this observation.

O H OH O H OH NaBH4 17 times NaBH4 faster than: isopropanol, 0°C isopropanol, 0°C

The4-memberedringismorestrainedwiththesp2 C=O (wants to be 120°) than with the sp3 alcohol which wants to be 109.5°. The alcohol has less ring strain, making the NaBH4 reduction faster. Chem S-20ab Purple Book Solutions Week 4 - Page 13 of 38 Oxidation of Alcohols: Mechanisms

TherearetwomainstepsinthemechanismofCrO3 oxidations. Let's figure out the mechanisms by making analogies to other, similar mechanisms. Draw the "analogy" mechanism first, then try to figure out the chromate mechanism:

Analogy Oxidation Mechanism

1. Formation of Sulfonate Ester 1. Formation of Chromate Ester

RCH2–OH + ClSO2R' RCH2–OSO2R' RCH2–OH + CrO3 RCH2–OCrO3H O O O S Cr Cl O O R' –H+ H+ transfer H H RCH O RCH O 2 O 2 O S Cr O O R' O

2. Elimination (E2)toformC=C 2. "Elimination" to form C=O

H H H H O O H H Base Base R Br R R CrO3H R

Base: Base:

H H O H H R Br R CrO3H

3. Under aqueous conditions, primary alcohols (such as the one above) are further oxidized to carboxylic acids because the aldehydes become hydrated. Provide a complete curved-arrow mechanism for the acid- catalyzed hydration of an aldehyde, and show how it can be oxidized to a carboxylic acid. (You can make an analogy to the acid-catalyzed hydration of an alkene.) O H + H H ,H2O OH CrO3 ox. O R OH OH R H+ R Oxidation is same mech. Sol as above: H and OH H :OH2 become C=O. H H OH

O OH R R H Chem S-20ab Purple Book Solutions Week 4 - Page 14 of 38 Carbonyl Additions: Reactions and Synthesis

1. Provide a synthesis of the following compound using starting materials with 6 or fewer carbons.

O MgBr Several other pathways are + H possible.

1. Mix + 2. H3O workup

OH

CrO3

O H3C OH

1. CH3MgBr

+ 2. H3O workup

2. Explain the selectivity observed in the following hydride reduction. (Ph = phenyl = C6H5)

Ph Ph Ph Ph NaBH4

O OH O N O N

Me Me

The hydride nucleophile must attack the p* antibonding orbital of the C=O. This trajectory is blocked by the phenyl groups in the case of attacking the leftmost C=O.

trajectory trajectory is blocked is OK Ph

Ph O N O Me Chem S-20ab Purple Book Solutions Week 4 - Page 15 of 38 Arrow-PushingwithOandN

Oxygen and Nitrogen are uniquely powerful arrow-pushers, because they have 2 useful properties:

1. Lone pairs which can act as a base or a nucleophile.

2. High electronegativity which makes them good leaving groups (electrophiles). Let's look at these types of reactivity. Note: In all the structures shown below, the O or N is shown without any protons, since the details of protonation may vary. The group "X" is any potential leaving group, the group "Nu:" is any nucleophile, and the group "E+" is any electrophile (such as CH3Br).

NOTE: These illustrations are only schematic, they are not complete. The actual molecules would be protonated on O or N, and could have multiple bonds (C=O or C=N).

LONE PAIR reactions: the atom prefers to be unprotonated and/or negatively charged. Protonation Reaction with Electrophile "Lone Pair Push" H+ E+

O O O

H H H X X X

H+ E+

N N N

H H H X X X

LEAVING GROUP reactions: the atom prefers to be protonated and/or positively charged. Leaving Reaction with Nucleophile "Neigboring Proton Pull"

("SN1") ("SN2") ("E1 /E2")

O O O B: H H H X X X

Nu:

N N N B: H H H X X X

Nu: 1. Describe each of the steps in the following mechanism in terms of the above ideas. Look at ALL the atoms! H O+ H + 2 H + + O: OH react OH OH prot.OH2 LP push deprot.

prot. with deprot. Nu: + + H2N HN NH HN N CH3 CH3 CH3 CH3 CH3 H2NCH3 H2O: H2O: We want O to be prot. in order to react with Nu:. Then, before the LP push, we have to adjust the protons again Chem S-20ab Purple Book Solutions Week 4 - Page 16 of 38 Addition-Elimination: Acetals and Imines

1. Provide curved-arrow mechanisms for the following transformations. Why are these called "addition-elimination" mechanisms?

H+

O HO OH OO

H C H H+ 3 H3C H

OH Sol OOH H C H 3 HO OH

H3C H

HO H O H 2 H O proton H3C O H3C O OH H transfer H3C H OH OH

H+ CH2CH3 N O CH3CH2NH2

H+ (pH = 5) H C H H3C H 3

OH

H3C H Sol H2N H CH2CH3 N

H3C H

HO H H2O H proton H3C N H3C NH H 2 transfer Chem S-20ab Purple Book Solutions Week 4 - Page 17 of 38 Synthesis with Acetal Protecting Groups

1. Show the reagents required to add an acetal protecting group to an aldehyde or ketone and then remove it.

O MeO OMe O + + MeOH, H H2O, H

2. Provide a complete synthetic pathway from the starting material to the product. Your synthesis will require an acetal protecting group.

O O

O

HO OH + H2O, H H+

OO

OO O

1) NaH 2) EtBr 1) BH3,THF - 2) H2O2,OH

OO

OH

OO

OH r gB M d Et ry 1) p C ku rO + w 3 H O 2) OO

H Chem S-20ab Purple Book Solutions Week 4 - Page 18 of 38 Aldehydes and Ketones: Reagents 1. Each of the following transformations can be carried out in one or two steps. Fill in the reagents required for each step. If a second step is not needed, please put an "X" through the second box.

OH CrO3 OCH 1. 3

OCH3

+ HOCH3 ,H 2.

O 1. OsO4,H2O

O O ,H+ cat. 2.

H2N O 1. ,pH5 N

2.

HO + O O H3O 1.

NaBH4,EtOH HO 2. OH

OCH 3 H O+ OCH 3 3 1.

H2NNH2 ,KOH,heat 2. Chem S-20ab Purple Book Solutions Week 4 - Page 19 of 38 Introduction to Multistep Synthesis

Now that you know more organic reactions than ever before, it is possible for you to attempt much more complicated multistep syntheses. For all of these syntheses, it is crucial that you learn to work retrosynthetically (backwards) from the desired product. This requires you to examine the functional groups and carbon-carbon bonds in the target molecule and ask: what reactions do I know that can synthesize this functional group or make this carbon-carbon bond (or, in a best-case scenario, both simultaneously)?

Consider the following synthetic target. Working retrosynthetically from this target, there are at least three possible immediate precursors to this target. Find three possibilities, and then evaluate each one to see whether it makes good synthetic sense!

O Seeing that this is simply a functional group interconversion (FGI), this is a correct retrosynthesis, but would not be particularly usef ul if we can't make the rest of the carbon structure! Br

OH

Br

O

O Cl

H Br Br This Friedel-Crafts disconnection is probably our best bet, as it allows us This Grignard disconnection is to access the most convergent theoretically excellent, but there synthetic pathway with reactions that is def initely going to be a problem will work well. with the bromide already present on the ring...

There may be other ways, too! Ask your TF, Chem S-20ab Purple Book Solutions Week 4 - Page 20 of 38 Aldehydes and Ketones: Multistep Synthesis

1. Propose a synthetic pathway for the following multistep transformation.

NaBH4 1. NaH

H 2. EtBr

O HO O starting material

1. O3

2. Me2S

H O OH O MgBr 1. CrO3

+ 2. H3O workup O O O desired product

2. Propose a synthetic pathway for the following multistep transformation.

OH O + H3O CrO3

starting material

MgBr 1.

+ 2. H3O workup

OH OH 1. BH3 OH

– 2. H2O2 ,OH

desired product Chem S-20ab Purple Book Solutions Week 4 - Page 21 of 38 The Wittig Alkene Synthesis

1. Write a complete arrow-pushing mechanism for each step of the following reaction sequence. What is the final product? O

Br 1. Ph3P 2. BuLi 3.

Ph3P:

PPh PPh3 3 PPh3 PPh3

H H O O

Bu: O

2. Propose synthetic pathways for the following multistep transformation.

Br + OH O Ph PCH Br H ,H2O CrO3 3 2 2 Br

1. MeMgBr Ph mCPBA CrO3 Ph Ph Ph + 2. H wkup O O Epoxide goes into page OH because of steric interaction between mcPBA and phenyl substituent.

Ph3PCH2

Ph Chem S-20ab Purple Book Solutions Week 4 - Page 22 of 38

More Wittig Mechnisms

1.

a) Provide a complete curved-arrow mechanism for all the steps in the following multi-step transformation.

O

Ph3P n-BuLi H I

O O H P Ph3

PPh3 PPh3 O H PPh3

n-Bu

b)Thefollowingsimilarattemptedmulti-steptransformationdidnot work. It turned out that the ylide that was formed was not reactive. Provide a curved-arrow mechanism for the formation of this ylide and explain briefly why it did not react. O

Ph3P n-BuLi H I

was not produced

PPh3 PPh3 H This ylide is a stable, aromatic molecule with 6 electron (4n + 2). It is not a good nucleophile and will not react with the aldehyde. n-Bu Chem S-20ab Purple Book Solutions Week 4 - Page 23 of 38 Acidity of Carboxylic Acids

1. Match the carboxylic acids below with the list of pKa's at right. Explain your ranking. O O Cl 0.70 OH OH Cl Cl More electron- O withdrawing Cl O Cl substituents stabilize the 1.48 OH carboxylate anion and Cl favor acid dissociation OH Cl (stronger acid) O O 2.85 Cl Cl OH OH O Cl O 2.86 OH Cl OH Cl Cl Closer electron- Cl Cl O withdrawing Cl O 4.05 substituents stabilize the OH carboxylate anion and favor acid dissociation OH O (stronger acid) Cl Cl O 4.52 Cl OH OH O O 4.76 OH compared with Cl acetic acid OH

2. Explain the observed trend in pKa's for formic acid, acetic acid, and carbonic acid:

O O O

H OH H3C OH HO OH

pKa: 3.75 4.76 6.37

Methyl is a moderate The oxygen is a good electron donor: electron electron donor (by resonance) donation will destabilize which destabilizes the the carboxylate anion carboxylate anion even more (compared with formate) (compared with acetate) and will make acetic and makes carbonic acid acid less acidic. even less acidic. Chem S-20ab Purple Book Solutions Week 4 - Page 24 of 38 Fischer Esterification and Ester Hydrolysis Mechanisms

1. Provide a complete curved-arrow mechanism for this Fischer esterification reaction:

O O + H ,CH3OH

OH OCH3

OH H Sol O OH

OCH3 CH3OH

HO OH HO OH2 H+ transfer OH O

CH3 CH3

2. Provide a complete curved arrow mechanism for this hydrolysis reaction. Make sure that your mechanism shows why the acid workup step is required.

O O 1. NaOH +EtOH OEt OH 2. H+ wkup

OH

O OH

OEt

H+ O

O O OEt

H O

Note: This step has an enormous equilibrium constant (1010 or more), driving the reaction to completion. Chem S-20ab Purple Book Solutions Week 4 - Page 25 of 38 More Mechanisms

1. Provide a complete curved-arrow mechanism for the following transformation:

H+

O + H ,CH3OH O O O HO O OCH3

Sol

H O H HO O O O O OCH3

OHCH3 O H HO O OH O OCH3 O

H+ transfer

CH3OH

OCH3 HO O OH H HO O OCH3 MeO OMe O

O H H3C

+ H transfer HO OCH OMe O 3 H O H + HO +H O OCH3 O HO OH OCH3 H O

OCH3 Chem S-20ab Purple Book Solutions Week 4 - Page 26 of 38 Carboxylic Acid Derivatives: Mechanisms 1. Propose curved-arrow mechanisms for the following transformations. + O H + O H2O, H +

N OH H3N H

OH sol H O N H H2N OH H2O: H+

OH OH proton

H2O N transfer HO N H H2

O "H:–" 1. LiAlH4 O HO OH 2. H2O

H—OH

H O

O O H H OH

O

H O

"H:–" O H H O H—OH Chem S-20ab Purple Book Solutions Week 4 - Page 27 of 38 Carboxylic Acid Derivatives: Reagents 1. Each of the following transformations can be carried out in one or two steps. Fill in the reagents required for each step. If a second step is not needed, please put an "X" through the second box.

O SOCl2 O 1. OH N HN (xs) 2.

+ O H3O O 1.

NH2 O HO H+ (This can also be done in just 2. one step with ethanol and acid, but it is very slow.)

O 1. LiAlH4 O OH

H+ workup 2.

O O CrO3 ,H2O 1. H Cl

SOCl2 2.

OH O MgBr O 1.

+ H3O workup 2. Chem S-20ab Purple Book Solutions Week 4 - Page 28 of 38 Removing the Carbonyl Group

1. The carbonyl group can be completely removed (down to CH2)foraldehydes and ketones using one of three different methods.

a) What reaction can be used to accomplish this under strongly acidic conditions?

Clemmensen Reduction (Zn, HCl, H2O)

b) What reaction can be used to accomplish this under strongly basic conditions?

Wolff-Kishner Reduction (H2NNH2,KOH,heat)

c) What reaction can be used to accomplish this under (almost) neutral conditions?

1. Thioacetal formation (R–SH) 2. Raney Ni reduction

2. The following transformation looks like it might be accomplished by one of the three reaction sequences described above. Explain why that is not possible, and propose a sequence of reactions that can accomplish the following transformation. Include curved-arrow mechanisms in your answer.

O 1. LiAlH4 + 2. H workup O O 3. NaH "–:H" Br 4. Br O– O O H

H O H H H+ H H O O

"–:H"

The three methods above only work on ketones or aldehydes. The starting material here is an ester. Going from ester to ether is completely different from going from ketone to alkane! Chem S-20ab Purple Book Solutions Week 4 - Page 29 of 38 Alternate Mechanisms for Carboxylic Acid Derivatives 1. Provide a complete curved-arrow mechanism for each of the following transformations. O O

H CH2N2 O OMe

H2CNN

O

O: H3CNN

H—OCOCF3 O O CF3COOH + O OH

OH

O H Sol

I

:I O I O O 2 I OH

O O O H I I - OH OCO2

You could also 2- deprotonate with CO3 before the nucleophilic attack. Chem S-20ab Purple Book Solutions Week 4 - Page 30 of 38 Synthesis with Acetal Protecting Groups

1. Provide synthetic pathways for the following multi-step transformations. Some of the syntheses may require acetal protecting groups.

O OH NaBH4

EtOH OEt OEt

O O

O EtO OEt EtO OEt

HOEt LiAlH4

+ OEt H OEt O–

O O + H3O (workup and hydrolyze acetal)

O

OH

O EtO OEt EtO OEt HOEt 1. LiAlH4 + + 2. H workup OEt H OEt OH

O O

dry CrO3, ** Note: there's an easier way using some special tricks from the next 2 pages; figure it out!

EtO OEt O O 1. MgBr CrO3

+ 2. H3O O OH O (workup and hydrolyze acetal) H Chem S-20ab Purple Book Solutions Week 4 - Page 31 of 38 Special Hydride Reductions: Synthesizing Aldehydes

1. Provide the internediates and products of the following reaction sequences.

O 1. LiAlH4 + CH3 OH OH 2. H wkup

O + O CH3OH, H 1. LiAlH4 CH3 OH + OH CH3 OCH3 2. H wkup

O O 1. SOCl2 CH 1. LiAlH4 3 CH3 H3C N H3C N + H OH 2. CH3NH2(xs) H 2. H wkup

2. Provide the internediates and products of the following special reaction sequences. What makes these reactions different?

O (iBu)2 Al O + CH3OH, H DIBAL O OCH3 H3C OCH3 OH H C 3 H (Don't worry about the mechanism for the collapse of the tetrahedral intermediate) The DIBAL will react with the ester to form a stable tetrahedral intermediate which, upon workup, yields the aldehyde. O + H3O

H3C H

O O OCH O AlH2 1. SOCl 3 LiAlH H 2 H3C N 4 OCH3 HN OCH CH3 OH 2. 3 H3C N (Weinreb amide) CH 3 CH3 (Don't worry about the mechanism for the collapse of the tetrahedral intermediate) The tetrahedral intermediate is stabilized by the Weinreb amide, so that, after workup, an aldehyde is produced.

O + H3O

H3C H Chem S-20ab Purple Book Solutions Week 4 - Page 32 of 38 Special Organometallic Additions: Synthesizing Ketones 1. Provide the intermediates and products of the following reaction sequences.

O O CH3MgBr +CH4 – OH O

O O + HO CH3 CH3OH, H CH3MgBr

OH OCH3 CH3 (after H+ workup)

+CH4 O 1. SOCl O O 2 CH3MgBr CH3 CH3 OH 2. CH3NH2(xs) N N H

2. Provide the internediates and products of the following special reaction sequences. What makes these reactions different?

O O O 1. SOCl2 CH3ZnBr

OH Cl CH3

The zinc reagent will react with the very reactive acid chloride to form a ketone, but the resulting ketone will not react with the zinc reagent, so you can stop at the ketone.

O O O MgBr H C 1. SOCl2 CH3MgBr 3 OCH3 OCH3 OH 2. HN OCH3 N N CH CH 3 CH3 (Don't worry about the 3 mechanism for the + (Weinreb amide) H workup collapse of the tetrahedral intermediate) O

CH3 The tetrahedral intermediate is stabilized by the Weinreb amide, so that, after workup, a ketone is produced. Chem S-20ab Purple Book Solutions Week 4 - Page 33 of 38 Putting It Together: Mechanisms

1. Provide curved-arrow mechanisms for the following transformations.

H+ N MeO OMe H+

H NH2

:Solvent HOMe H MeO N

H NH2

OMe H2 H MeO N MeOH N proton H NH2 transfer

MeO OMe + O H2O, H

OMe OH H+ Sol H O MeO OMe This is a good exercise in OH HOMe protonation and deprotonation. Remember: protonate the group that you want to have leave in each step. HO HOMe OMe OH OMe :OH2 H2O: proton transfer

MeO OMe MeO HOMe OMe proton H2O OMe OH2 transfer OH OH OH Chem S-20ab Purple Book Solutions Week 4 - Page 34 of 38 More Mechanisms

1. Provide a curved-arrow mechanism for all steps in the indicated transformation.

O

H CO H CO + O 3 3 1. LiAlH4 2. H (excess) O O

"H:– "

H O:

H3CO H3CO O

:Solvent H O O O

H H3CO O

"H:– "

H+ H O:

OH H H3CO O

O

H OH H+ OH H H3CO H O H3CO O Chem S-20ab Purple Book Solutions Week 4 - Page 35 of 38 Carbonyl Synthesis: Starting Materials

1. Fill in each box with a starting material that would provide the product indicated by the transformation.

1. CrO3 OH 2.

PPh3

+ O 1. CrO3 ,H3O

OH 2. SOCl2 N(CH3)2 3. HN(CH3)2, excess (aldehyde is also OK)

OCH3

1. CrO3 (dry) OCH OH 3 + 2. CH3OH, H cat.

OH O O + 1. CH3OH, H cat. CH3 OH OEt CH3 2. CH3MgBr (excess) or 3. H+

O O 1. K2CO3

OCH3 OH 2. CH3I Chem S-20ab Purple Book Solutions Week 4 - Page 36 of 38 Carbonyl Synthesis: Products

Fill in each box with the major organic product(s) of the indicated transformation.

O O

OH NaOCH3 O

O 1. mCPBA HO Ph

2. PhMgBr HO Ph 3. H+ wkup

O

CH3 Cl 1. AlCl3 CH3

2. H2NNH2, KOH,

O O 1. LiAlH4 CH3 OCH3 CH3 N + H 2. H3O CH3

O OH O 1. LiAlH4 + OH O 2. H3O O Chem S-20ab Purple Book Solutions Week 4 - Page 37 of 38 Carboxylic Acid Derivatives in Synthesis

Provide a synthetic pathway for the following multistep transformation. You may ignore stereochemistry.

O O O + MeOH MeO O H3O O NH2 OH MeO OMe H+ cat.

1. DIBAL, 78 C 2. H+ workup

O PPh MeO 3 MeO MeO MeO H

+ H3O

O

OsO4,H2O

OH O OH

NaCN, HCN

OH HO OH NC

Desired Product Chem S-20ab Purple Book Solutions Week 4 - Page 38 of 38 Multistep Synthesis Practice

1. Propose a synthetic pathway for the following multistep transformation.

O O O O o mCPBA 1. DIBAL, -78 C OH H 2. H+ wkup

1. NaH 2. MeI

O OH

OMe 1. PhMgBr OMe H Ph 2. H+ wkup

desired product

2. Propose a synthetic pathway for the following multistep transformation. Br MgBr

Br2 Mg AlCl3

FeBr3 Et2O Cl

O

1. p ku w + H 2. O There are other ways to make this OH alcohol (e.g. Friedel-Crafts acylation O followed by reduction followed by O elimination followed by hydro- H+ cat. boration-oxidation), but this is the OH most efficient.) O or Cl