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Q: On slide 269 (23.5 Using Aldol Reactions in Synthesis), why is it that you get the b-hydroxy carbonyl or the alpha/beta unsaturated carbonyl from the aldol reaction. Is this the mixture of products that can result (it appears that the unsat. carbonyl is the dehydrated product of the b-hydroxy carbonyl??)?

A: We are not talking about getting a mixture of products although that is certainly a possibility. The aldol reaction can give two possible products. It can be stop at either the beta-hydroxy carbonyl stage. The dehydration can be catalyzed by either acid or base. The aldol reaction is often run under catalytic base conditions. If the aldol is done at lower temperatures, say 0° C or below, then the b-hydroxy carbonyl is the product. If the aldol reaction is run at room temperature or above, then the dehydration occurs top give the a,b- unsaturated carbonyl

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Q: When we talk about enolates, are we only talking about enolate ions?

A: Yes, the word enolate refers to the enolate anion.

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Q: LDA in THF at -78 degrees helps to cause E2 elimination as a strong base; but it also is used in the alkylation of ketones/esters/nitriles?? I am trying to understand slide 274 where is shows discrete (in situ) generation of an ester enolate with LDA; what purpose is the enolate serving in this purpose? Is it E2 elimination again or alkylation as well?

A: LDA can be used to affect an E2 elimination reaction. However, the way we have used LDA is to perform a-deprotonation of carbonyl compounds. On slide 274, once the enolate is formed it is undergoing a Claisen condensation with various esters. There is no E2 elimination on this slide.

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Q: So, does the aldol reaction require that one of the reactants has alpha hydrogens whereas the Claisen condensation can do a reaction between a ketone enolate with an ester to form a 1,3 diketone?

A: The aldol and Claisen reactions are very closely related. Any of the enolates that we have discussed can be used for both of these reactions. The only difference is that the electrophile. For the aldol reaction, the enolate reacts with an aldehyde or ketone to give a b-hydroxy carbonyl (which can dehydrate to the a,b-unsaturated carbonyl). For the Claisen reaction, the enolate reacts with a ester to give a 1,3-dicarbonyl (rather that a 3-hydroxyl carbonyl which is the same as a b-hydroxy carbonyl). There is a parallel to the reactivity. The electrophile for the aldol is a ketone or aldehyde and the product is a b-hydroxy carbonyl. The electrophile for the Claisen is a ester (which is in the next higher oxidation than a ketone or aldehyde , the electrophile for the aldol) and the product after reaction with the enolate is a b-keto carbonyl, which is in the next higher oxidation state than a b-hydroxy carbonyl from the aldol reaction.

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Q: On slide 274, how do we know that the Claisen condensation always breaks the bond between the C and the OEt?

A: The Claisen condensation is a nucleophilic acyl substitution reaction between an enolate and an ester. When the enolate adds to the ester, a tetrahedral intermediate is formed. The carbonyl is the re-formed when the EtO(-) is ejected as a leaving group.

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Q: On slide 277, what is the purpose of having EtONa and EtOH in the second step when the bromoalkene is added? (what is that reaction?)

A: This is a a-substitution reaction. The substitute reaction of the hydrogen for an alkyl group must go through the enolate. The EtONa is to generate an enolate.

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Q: If decarboxylation is a loss of CO2 why does the whole of the CO2Et get removed in the addition of H30+ and heat in the reaction on slide 277?

A: This is a two stage reaction. The first is the acid catalyzed hydrolysis of the ester to the carboxylic acid. The second is the loss of CO2; overall, the CO2Et is lost as CO2 and EtOH

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Q: When you were drawing the reaction for the Robinson Annulation you mentioned that of the four alpha carbons/hydrogens one is far more acidic and it was the one between the two carbonyl groups – why was this not the one extracted by EtO-?

A: You are talking about the intermediate after the Michael addition that undergoes intramolecular aldol to give the cyclohexenone product. There are at least four different enolates that are possible from that intermediate. The major enolate from deprotonation between the ketone and ester can react with the other ketone, but this would give a disfavored four member ring aldol product. Deprotonation at either of the methyl groups next to the ketones, gives an enolate that can react with the other ketone to give a much more favored six-membered ring aldol product (see next question). Even though these enolate are present in much lower concentrations, they react to give more stable products, so the aldol to give six-membered ring is more favored than to give the four-membered ring.

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Q: Would two products not have been possible from the Robinson Annulation you showed us (for ex. two products with alkene double bond in two different places?)

A: For the Robinson annulation product on slide 283, it is possible for the double bond to be in conjugation with the ester. However, the base catalyzed dehydration of water that yields this double bond results from a-deprotonation and elimination of hydroxide. One would expect deprotonation alpha to the ketone rather than the ester because it is the more acidic position.

If you look at this intermediate on slide 282, the carbon fragments are color coded. What is shown is the blue methyl ketone undergoing aldol reaction with the green ketone and dehydration to give the product. It is possible that the green methyl ketone can undergo aldol reaction with the blue ketone followed by dehydration to give a cyclohexenone was not shown on the slide in the notees (see below). This is certainly a feasible reaction, which I did not consider. Thamks for pointing this out.

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03-31-2006

Q: On page 863 (middle of the page), the reaction is a mixed aldol reaction; why does it produced an unsaturated product using the same reagents (NaOEt and ethanol) while the reaction at the end of the chapter on p. 881 (1c) shows the
product not having been dehydrated? (yet, the reactants are still the same)

A: As I noted above, the conditions of the aldol reaction determine if the product is a b-hydroxy carbonyl or the a,b-unsaturated carbonyl. You need to be able to recognize that two products are possible for the aldol reaction.

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Q: Is the effect of acetoacetic ester synthesis the addition of three carbons to the carbon chain?

A: Acetoacetic ester is initally a four carbon unit (not including the OR of the ester). After decarboxylation, a three carbon unit is added.

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Q: Can cycloalkanes be made with acetoacetic ester synthesis like that can intramolecularly with malonic ester synthesis?

A: Yes

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Q: Is the enol-keto tautomerism essentially equivalent to the relationship between resonance structures?

A: No! Enol-keto tautomers there is a movement of a proton. You can only move electron when draeing resonances structure; atoms can not move

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Q: In the carbonyl condensation reaction, why do we show the carbonyl as an enolate ion and not an enol?

A: This depends on the on the reaction conditions. If it acid catalyzed, then the enol reacts, but this is less common and less reactive. For the base catalyzed reaction with ketones or aldehydes, you can draw either the enol or enolate- the enolate is more reactive than the enol, so I usually show the enolate reacting. The pKa pf beta-dicarbonyl compounds is low enough that the enolate is the reactive species when treated with base. If you use a strong base such as LDA, the the carbonyl compound is completely deprotonated and the enolate is generated.

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Q: Is it true that the starting carbonyl in a carbonyl condensation reaction is an electrophile that than becomes a nucleophile upon attacking the second carbonyl compound?

A: It becomes a nucleophile (and stops becoming a electrophile) upon enolate or enol formation.

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Q: Why is the aldol reaction between two ketones the first reaction listed in Chapter 23 if it is so unfavorable? Is it still a reliable reaction?

A: I don’t thing that the way it was presented specified a ketone per se. It could also be an aldehyde. The self-condensation of either a ketone or aldehyde is the basic reaction

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Q: In the acid catalyzed dehydration of aldol, it is possible to go directly from the first resonance structure enol to the final product with E2 elimination and bypass two or three steps, right?

A: If you mean to bypass the resonance structure with the the negative charge on the alpha-carbon, yes. Remember, that neither resoancne structure is truly correct- the the actual structure is a hybrid of the resonance structures.

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Q: On the last reaction of slide 261, should there be two equivalents of the reactant to do a carbonyl condensation reaction followed by an aldol reaction?

A: The slides numbering has gotten messed up so I am not exactly sure what reaction you are talking about. If it is the aldol reaction of cyclohexanone, then yes, there should be a two in front of the reactant.

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Q: Is the carbonyl condensation reaction/mechanism discussed first in Chapter 23 an overall reaction/mechanism for carbonyl compounds or is it its own separate reaction? It was not listed in the summary of reactions at the end of Chapter 23, so I was just wondering what its status was.

A: The reaction shown on page 855, Fig 23.1 is the base-catalyzed aldol reaction. It is the general reaction that is listed at the aldol reaction in the Summary on page 881. It happens to be listed as three specific types.

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Q: What were the conditions for that you gave in slides 262 and 263 (for ex. if one partner has no alpha H or if one of the carbonyl species is way more acid than the other)? Were they conditions for producing one product in the mixed aldol reactions as opposed to 4?

A: The mixed aldol reaction (slide 264 in the latest version) gives four possible products. This uses the base-catalyzed conditions shown. The way to get just one is to discretely generate the enolate from one of the aldehydes (lets say pentanal) with LDA in THF at -78° C; this completely convert pentanal into the enolate. After the enolate is formed, the other aldehyde is added (in this case propanal). This will give just one of the aldol products: 2-(1-hydroxyproyl)penanal (top right) because the pentanl was convereted to the enolate which is going to act as the nucleophile and propanal is the electrophile in the reaction.

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Chapter 22

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02-27-2006

Q: I just have a quick question regarding mechanisms for tomorrow’s exam. Should we know in detail all mechanisms or should we mainly focus on the ones you actually drew and discussed on the board during class? are those the more emphasized mechanisms in the chapter – should we study those in particular?

A: Many of the mechanism are very similar and there were some that I did in class because they might help you understand (and remember) the reactions. As for the last exam, the most likely mechanisms are those that exemplify a major theme of the material.

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Chapter 21

02-27-2006

Q: Do we need to know reactions or mechanisms for acid anhydrides since we did not talk about them?

A: They are exactly the same as for acid chloride.  just substitute a O2CR group for
the Cl.

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Q: I was told of a web site that had unique/fun ways of remembering some of the mechanisms. I thought it was too great to not send to you, so if you get a chance check out the link below:

http://www.oir.ucf.edu/chemistry/s02/frameset41.html

A: Thank you . Unique is a good way to describe it. However, can you sing the song backwards as well?

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Q: Does NaBH4 not reduce any of the carboxylic acid derivatives discussed in Chapter 21? And thus it can be used to reduce other ketones and such in the presence of these carboxylic acid derivatives?

A: It does not react with esters, amides, nitriles or carboxylic acids.  Acid chlorides are incompatible with the NaBH4 reaction conditions which is done in ethanol and will give the ethyl ester of the acid chloride.  So that pretty much rules out carboxylic acid derivatives.

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Q: On page 780 of the text book, in the mechanism of converting a carboxylic acid to an acid chloride, why does the SO2 group leave instead of Cl?  I thought Cl is one of the best leaving groups because it is the conjugate base of a very strong acid.

A: I believe that you are referring to the middle intermediate (in brackets) on the second line of the mechanism.  Losing chloride gets you back to the intermediate on the left. This almost certainly happens, but is an unproductive step. That step should probably be denoted as a equilibrium. Losing SO2 would be an irreversible step since it is a gas and this drives the reaction forward to give the acid chloride.  Losing a neutral gas is almost always a very favorable driving force for a reaction.

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Q: For problem 21.38, is the reason that there is no reactions for addition of cyclohexanol to methyl propanoate because cyclohexanol is a better leaving group than the -OCH3 of methyl propanoate

A: This is not necessarily a straightforward answer. I think the point of the problem is to stress the reactivity order of acyl derivatives; less reactive acyl derivatives can be prepared from more reactive acyl derivatives. For 21.37e, you made cyclohexyl propanoate from the acid chloride and cyclohexanol, which clearly can be done according to the reactivity order; the starting acid chloride is more reactive than the product ester, so the reaction is favored. In 21.38e you are asked if the same reaction can be done starting from methyl propanoate (an ester). Since the starting compound, methyl propanoate, is not a more reactive acyl derivative than the product, cyclohexyl propanoate (they are both esters) you were to conclude that this is not a favorable reaction- so there is no reaction. Your initial thought of looking at the relative leaving group abilities is a very good one. The pKa of methanol is ~16, but cyclohexanol (a secondary alcohol) would probably not be much different (~17), so the leaving group abilities would be roughly the same.

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Q: Slide 133 shows the dipole strength of each carbonyl compound. Do these dipole strengths correspond to reactivity?

A: To a degree but no one parameter can explain everything. The prediction that ketones are more reactive than aldehydes which are more reactive than esters holds pretty well.  But amides, which have one of the larger dipole are the least reactive.

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Q: Are carboxylic acids are the only carbonyl compound reduced by Borane?  I know that it is more selective.

A: No, aldehydes and ketones are also reduced and I suspect that an acid chloride would be reduced as well.  Amides and nitriles are reduced to the amines.  Ester do not react.  S you note. the point is that BH3 can do selective chemistry.   It reacts very quickly with carboxylic acids, so a carboxylic acid can be selectively reduced in the presence of virtually any other functional groups with BH3 without the need to use a protecting group.

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02-26-2006

Q: Regarding problem 21.35 e, is it possible to get 1-butene from butanoic acid by first using LiAlH4 to form an alcahol and then form the double bond with POCl3?

A: That isn’t a great way to do it.  LiAlH4 (or BH3) is correct for the reduction of butanoic acid to 1-butanol, but primary alcohols are very difficult to dehydrate to alkenes directly.  So the way to go would be to convert it to a halide or tosylate followed by E2 elimination.  POCl3 works for secondary and tertiary alcohols.

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Q: You mention on slide 185 that LiAlH4 reduces alkyl halides and epoxides. Are alkyl halides (RCH2X) reduced to RCH3 ? For epoxides, is one carbon reduced to a CH while the other becomes a C-OH? Is this reduction selective (i.e., does a certain carbon of the epoxide generally become the alcohol like we saw for acidic/basic hydrolysis of epoxides)?

A: Yes, alkyl halides are reduced to the hydrocarbon and epoxides are opened to alcohols. These reactions are likely to be SN2. For epoxides, the hydride will attack the less hindered carbon to give the more substituted alcohol.

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Q: On slide 186, you mention that the carbonyl oxygen of an amide is nucleophilic and can react with electrophiles. Is this not true of all carbonyl compounds (due to the substantial dipole moment of the carbonyl functional group)?

A: Yes, but if you treat a ketone with SOCl2 nothing happens. The dipolar structure explains the reactivity of carbonyl towards most nucleophiles, which add at the C. It also explains that protons (and other strong Lewis acids) will react with a carbonyl at O, however the reactivity of the carbonyl O toward electrophiles is very modest compared to amides.

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Q: What would happen if one attempted to dehydrate a secondary/tertiary amide with POCl3 or SOCl2 ? (no reaction?) Is there any way to convert these compounds to carboxylic acids?

Because there is a group or groups on the nitrogen, the dehydration can not proceed unless you break the bond between the N and the group. You end up getting a species know as a Vilsmeier salt (see attached) which is a very powerful electrophile and can be used in F-C acylation reactions. If there is nothing for this salt to react with, it will be hydrolyzed back to the original amide when you add water at the end of the reaction; the ocerall result is no reaction

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Q: Why is the hydride ion the only nucleophile capable of adding twice to a nitrile? (do all other nucleophiles give the imine? is the imine stable unless hydrolyzed by aqueous acid/base to give the ketone?) I also have in my notes that only hydride ions from LiAlH4 are capable of adding twice to a nitrile. What about BH3?

A: Borane reduces nitriles to amines. I don’t know about all other nucleophiles, but since the intermediate (imine anion) from the first addition already has a negative charge, it is much harder to add a second nucleophile to give the di-anion. LiAlH4 is the only reagent reactive enough to add again. Borane is a slightly different case. Because the boron is such a powerful Lewis acid, it is strongly associated with the nitrogen- this bond has a lot of covalent character it it so the initial negative charge is largely neutralized and this allows the second hydride addition to occur.

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Q: When a nitrile is reduced with DIBAL to give an imine, will the imine hydrolyze spontaneously to the aldehyde (if in aqueous solution), much like the product of Grignard addition to a nitrile spontaneously hydrolyzes to give a ketone? (does this mean that carbonyls are more stable than imines?)

A: Yes. When aqueous acid is added at the end of the DIBAH reduction or Grignard reaction, a imine that is formally derived from an aldehyde and ammonia, is obtained. This spontaneously hydrolyzes in aqueous solution to give the aldehyde and ammonia. I am not sure about stability (i.e., bond energies, etc.) but there are other considerations. The nitrogen of an imine is basic and is easily be protonated, even at neutral pH. The gives a positively charges iminium ion to which water will added very rapidily. In aqueous solution, the equilibrium is pushed to aldehyde and ammonia.

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Q: If a set of atoms are conjugated, must each one of these atoms have the same hybridization?

A: Not necessarily; a C-C triple bond can be conjugated with a carbonyl, a nitrile can be in conjugation with an aromatic ring. The hydridization of the atoms must be such, that allows for an overlapping array of p-orbitals. So for an amide, the hybridization of the nitrogen is sp2 because this allows the lone pair of the nitrogen to be in a p-orbital that can conjugate with the pi-bond on the carbonyl (which are also in p-orbitals). This also true of the nitrogen of pyrrole and of an enamine (see below).

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Q: You mentioned that NaBH4 is more basic than LiAlH4 and will undergo acid-base chemistry with carboxylic acids. What accounts for the basicity of NaBH4 (as opposed to lithium alumnium hydride)?

A: Any anionic species has the potential to be a base. Both are NaBH4 and LiAlH4 are basic reagents and the first mol of hydride from either will deprotonate the acid to give a carboxylate. NaBH4 is not a strong enough hydride source to react with the carboxylate further. As noted above with regard to the nitrile reduction, LiAlH4 (which is in an aprotic solvent) is reactive enough to reduce the carboxylate ion of the acid.

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Q: Why are aldehydes and ketones more downfield in 13C NMR than carboxylic acids and their derivatives?

A: As we have noted the C=O bond has a large dipole moment and the carbon is deshielded as a result. For aldehydes and ketones, the alkyl groups or -H do not compensate much in terms of electron donation (which will shield the nuclei) for the large pull of the carbonyl oxygen. Carboxylic acid derivatives where the Y group can be OH, OR, NR2 or -Cl have lone pairs of electrons and can donate electron density though resonance; thus, the carbonyl carbon is not as electron deficient and consequently not as deshielded. Also, each of these group are electronegative in their own right. The individual bond dipole of the -Y group to a degree opposes the C=O bond dipole. Thus, the carbonyl carbon bond dipole is reduced and the carbon is less deshielded.

Nitrile are much more shielded for a different reason. If you remember the cryptic explanation of ring currents for aromatic rings that cause aromatic protons to be unusually deshielded, but a proton in the middle of the ring would be unusually shielded. There is a similar explaination for triples bonds. There is a ring rurrent for triple bonds that shield the carbons of nitriles and acetylenes which is why 13C and 1H resonances of these compounds are upfield compaied to alkenes.

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Chapter 19

02-28-2006

Q: For the conjugate addition of alkyl groups via Gilman reagents, I was wondering if the Gilman’s reagent is limited to alkyl groups. I know that we said alkynyl groups don’t work in Gilman’s Reagents, but what about something like (HOCH2CH2)2CuLi ? Can functional groups such as an alcohol be added through this reaction or is it strictly alkyl groups?

A: Gilman reagents can be alkyl, vinyl or aryl.

Any basic reagent such as Grignards, Gillmans and Wittigs are going to have compatibility issues with acidic functional groups either as part of the reagent or the substrate it is reacting with. Acid base chemistry will be much faster. The key to all of these reagents is that they are negatively charged (and hence, their basic nature) nucleophiles. If they do acid base chemistry, their charge and nucleophilicity are gone. So the Gillman reagent you wrote above wouldn’t fly because of the hydroxyl group- you would not be able to generate the organilithium species. If a Wittig reagent has a ketone or aldehyde as part of the ylide, it could react with itself rather than with the substrate.

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Q: I had written in class that the Wittig reaction is not compatible with acidic groups such as alcohols. Just to be sure, this means that if an alcohol were on the same molecule as the ketone reactant, the Wittig reaction wouldn’t take place?

A: Yes, the negative charge of the ylide would be re-protonated by the acidic functionality. The ylide and ketone/aldehyde substrate should be free of acidic protons. The ylide should probably not have aldehyde or ketones either since the Witting reagent would react with itself, unless this is by design.

02-27-2006

Q: The equilibrium for hydration of aldehydes and ketones favors the carbonyl compound (except with the least hindered carbonyls such as formaldehyde and acetaldehyde). Because of this, is acid or base catalyzed hydration an acceptable reaction for the synthesis of gem-diols?  If not, what other methods do we know to synthesize a gem diol?

A: This is something of a model or a prototype reaction for the nucleophilic addition to carbonyls.  Hydrates of ketones and aldehyde are intermediates for some reactions, i.e., Jones oxidation.  Gem-diols really can not be synthesized. You put and aldehyde or ketone in water and you get an equilibrium with the hydrate.  You remove the water and you get the carbonyl back.

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Q: On page 689 the book shows 2 different variations that a nucleophilic addition reaction can follow.  What determines which variation the reaction will take?  Is it the presence of a hydrogen on the nucleophile?

A: The nature of the nucleophile.  The top reactivity is for hydrides and most carbon anion like Grignard reagents.  The chemistry is pretty much limited to the addition step and an alcohols is usually the product.  The bottom reactivity is for nucleophiles that have additions lone pairs of the electron or a proton that can be lost after the initial addition (the loss of the protons unveils a lone pair of electrons).  In these cases, the lone pair can puch out the hydroxyl group (usually after protonation as H2O) to give a C=Nu.  These nucelophiles have either O or N as the nucleophilic atom.  For N, the C=N- is an imine and is stable or can rearrange to an enamine.  Of O, the C=O-(+) is not a stable end product but is an intermediate for another nucleophilic addition (i.e. acetal formation)

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02-26-2006

Q: Do you know why aldehyde/ketone hydration is more favored for less substituted molecules (e.g., formaldehyde)?

A: This probably has to so mostly with sterics.

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Q: Can one control the direction of the hydration equilibrium?

A: If you remove the water, the carbonyl is produced.  I don’t think that you can push the equilibrium toward hydrate.

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Q: Are there any other nucleophilic addition reactions of H-Y that are favorable besides cyanohydrin formation?Do you know why does cyanohydrin formation favor the cyanohydrin?

A: Not that I can think of.  Cyanohydrin formation is pretty unique.

It defies conventional wisdom because H-CN is a stronger acid that either water or an alcohol, so you would predict that the equilibrium would not favor cyanohydrin formation.  However, I think the driving force must be the added energy of the new C-C bond that is being formed.  Interpreting equilibrium can be tenuous because very small energy differences can significantly tip the balance of an equilibrium.  An energy difference of zero means a 1:1 mixture of products and reactants.  A change of just ~14 KJ/mole (which is a very small amount of energy) change the equilibrium to > 99.9 : 0.1.

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Q: Are carbinolamines stable in the absence of catalyst?

A: I think you mis-spoke here.  As you know, the catalyst only affects the rates of formation, not the stability of the products.

Some carbinolamines are stable and observable.  Usually, there is some structural effect that holds it together, like being constrained in a ring.

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Q: What is implied by the isoelectronicity of an imine with a carbonyl? That the two should exhibit very similar chemistry? Enamines are isoelectronic with enolate anions?  Does this imply that the two also have similar chemistry?

A: Yes.  The imine is in the same oxidation state as a carbonyl and the bonding arrangement of a C=O and C=NR are similar. So there is overlapping chemistry.  A Grignard reagent will add to an imine to get an amine; they can be reduced to amines, the alpha-protons are acidic, etc. However, there is hydrolytic chemistry of imines which often competes with these reactions, which is not similar to a ketone or aldehyde

We will in the next Chapter 23 that enamines are used a enolate equivalents.

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Q: Is the “alpha nitrogen” of a hydrazone (such as those generated in the Wolff-Kishner reaction) acidic (due to the neighboring imine group)? Would the acid catalyst from the previous step (hydrazone formation) have to be neutralized first?  Would one just add excess base?

A: The hydrazone N-H is much more acidic than a normal N-H.  There is a pH range where hydrazones can be formed.  I believe that the optimal pH is just on the acidic side of neutral.  But just like imines, hydrazone can be formed at basic pH, although the rate is slower (the book mentions this, but I thought that the discussion was somewhat opaque and tmight cause more questions than clarity).  So the W-K reduction is most often done without a discrete acid-catalyzed formation of the hydrazone step. McMurry presents the reaction as simply hydrazine and KOH and my notes did as well.  This is somewhat inconsistent with the previous acid catalyzed formation of imines, however.

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Q: Are there no other functional groups that are reduced (or interfere with) the Wolff-Kishner reaction?

A: Groups that can not tolerate strongly basic conditions would not be compatible with the W-K reduction.  For instance, a ester would be hydrolyzed, alkyl halides would eliminate, etc.  However, these are reduction conditions that unique for ketones and aldehydes.  Almost any other reducable group will not be affected by the W-K reduction, unless they were hydroxide sensitive (so an acid chloride is going to be hydrolyzed).

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Q: On slide 157, a ketone is selectively reduces in the presence of an ester with NaBH4.  Am I correct in saying that this would still work if the ester were a carboxylic acid instead?  While the ketone would again be converted to a ketal, the acidic conditions would cause the carboxylic acid to be converted to an ester by a Fischer esterification reaction. This ester could then be reduced with LiAlH4 and the protecting group could be removed with aqueous acid.  Would this be a viable way to selectively reduce a acid?

A: NaBH4 does not reduce esters, carboxylic acids, amides or nitriles.  It depends on the reaction conditions, however, if the substrate on that slide were a keto-acid, I believe that you would get the acid out at the end of the reaction rather that the ester of ethylene glycol.  This reaction is done in a Dean-Stark trap with a stoicheometric amount of ethylene glycol and benzene as the solvent.  The Fischer esterification generally requires huge excesses of the alcohol to drive it forward.  So the ketalization reaction would be faster because the ketone is more reactive.  Ethylene glycol esters would be minor products of the reaction.

We noted in the Chapter 20 that borane is a method for selective reduction of a carboxylic acid.  So al alternative to that shown in Slide 157 is to hydrolyzed the keto exter to a keto acid and treat with borane.  This would give the keto alcohols.

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02-25-2006

Q: Do imines and enamines have synthetic usefulness?  I recall you saying that imines are of crucial importance in biochemistry.  Can imines or enamines be converted to nitriles or amides?  Can imines be reduced to amines with lithium alumnium hydride?

A: Imines are generally used as precursors for amines.  We will cover a reaction called a reductive amination in which a ketone or aldehyde is reacted with an amine under reducing conditions (such as H2, Pd/C).  The amine and ketone form an iminium ion (protonated imine), which is very easily reduced to give an amine.  This is one of the best ways for making substituted amines. We will see in the next chapter that enamines are the equavalent of enolates. Oximes, which are related to imines can be dehydrated to nitriles.

In biochemistry, an imine is referred to as a Schiff base. The reactions of an entire class of enzymes that are dependent on the coenzyme pyridoxal phosphate, go through a critical imine linkage between the coenzyme and the substrate.  As also in the class of retinal binding to opsin, imines are often transient linkages between a receptor and a protein or enzyme and substrate.

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Q: How does the chemistry of organolithiums compare to that of Grignard and Gilman reagents?  The text frequently discusses the latter two but rarely mentions organolithiums.

A: I probably should just stop mentioning organolithium reagents (I guess I don’t always remember what is in the first half of the text).  In general, the reactivity of a Grignard’s parallels that of an organolithium (organolithiums are actually more reactive). Gillman reagents are somewhat less reactive, but they possess unique reactivities that other organometallic reagents don’t have.  The coupling reaction with an acid halide being one of them.

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Q: DIBAL allows partial reduction of esters to aldehydes.  Does it partially reduce any other functional groups?

A: Lactone which are cyclic esters are reduced to aldehydes.  In this case the product has an appended alcohol group, which comes from the other oxygen of the lactone.  Nitriles are reduced to imine by DIBAH, which hydrolyze to aldehydes. DIBAL also reduced aldehydes and ketones and is used for many types of reductions.Why are acidic groups not tolerated in the Wittig reaction?

Since the reactive form of the Wittig reagent is the ylide, the negative charge
can potentially do acid-base chemistry with acidic groups.  If the ylide is re-
protonated, then the phosphonium salt results.

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Q: Aryl ketones are reduced to methylene groups via hydrogen over palladium.  Does this reaction also work with benzaldehyde?

A: Yes

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Q: What is the mechanism for heat-induced expulsion of water from carboxylic acids to form an anhydride?

A: The mechanism I would write is one eqivalent of acid protonates a second equivalent to give a RC(=OH+)-OH and a carboxylate ion. Now draw the acyl substitution reaction as we did today for other reactions, to give the anhydride and water as the product.  The reaction must be done at a temperature were the water is driven out of the reaction. If water is present, the water will simply add to the anhydride and the reverse reaction will proceed.

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Q: Do anhydrides react with Grignard reagents? (it is not listed as such on page 788 of the text in Figure 21.7)

A: Yes they do.  Grignard addition will give the tertiray alcohol just as for acid chlorides and esters.  For the most part, anhydrides are used interchangably with acid chlorides (except for the reaction with Gillman reagents).  The book notes, however, that an anhydride is actually an inefficient reagent in that two mols of the starting acid are used to make the anhydride, but if you react an anhydride with, say an amine to give an amine, only half of the anhydride reacts.  The other half is lost as the carboxylate.  That is OK for simple anhydrides such as acetyl anydride.  However, if you wanted to make an amide from a carboxylic acid, but the carboxylic acid had to be synthesized from other starting materials, you would not want to make an anhydride out of it before reacting it with an amine.  The acid chloride would be the way to go.

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Q: Why is the NH2 group more nucleophilic than the OH group?  Also, in the synthesis of aspirin, why does the alcohol -OH of salicylic acid rather than the carboxyl -OH react with acetic anhydride?

A: I am not sure I have a good answer for that.  In general, amino groups are much more reactive than alcohols toward electrophiles.  When an neutral amino group reacts with a electrophile the initial product has a positive charge on nitrogen. Nitrogen handles having a positive change very well as there are many stable ammonium salts.  While oxygen can handle positive change transiently, there are really not many stable species that have a positive change on oxygen.  Also, if you consider what makes for a good nucleophile (Chapter 11), good nuclephiles are the conjugate bases of weak acids (i.e., strong bases). The pka of the ammonium ion (NH4+) is about 10, while the pKa of H3O+ (a surrogate for ROH2+) is -1.7.

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Q: Is the basicity of an amine comparable to the acidity of an alcohol?

A: The pKa of ammonium ions are in the 9-10 range.  Most alcohols are in the 16-18 range.  So there is not much acid base chemistry between an amine and a normal alcohol.

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Q: Why does the book say that the conversion of acid chlorides to esters is “more general” than the conversion of acids to esters? The text claims that the Fischer esterification reaction is limited to the synthesis of methyl, ethyl, propyl, and butyl esters because of the need to use excess liquid alcohol as solvent.  Why is this limiting?

The Fischer esterification is practically limited to making esters with simple alcohols because they are used a solvent.  Also, as the alcohols get larger, the boiling point gets higher, so there is a practical consideration of how to get the product out of the reaction.  If you wanted to make an acetate ester of an elaborate alcohol that required many steps to synthesize, you couldn’t practically use it as solvent for the Fischer esterification with acetic acid (since acetic acid would be the limiting reagent in that case).  So the more efficient way to go would be to use your fancy alcohol as the limiting reagent and react it with an excess of acetyl chloide (or anhydride) in pyridine solvent.

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Q: The acid chloride reduction by lithium alumnium hydride proceeds through an aldehyde intermediate before the primary alcohol is produced.  Can the aldehyde be isolated with DIBAL at -78 degrees celsius?  (out of curiosity, why is -78 degrees celsius often used as a reaction temperature in the book?  Is this because the freezing point of CO2 (dry ice) is around this temperature and therefore is a particularly easy temperature to attain?)

A: No.  With DIBAH and an ester, the intial addition of a hydride give an tetrahedral intermediate with a negative charge on oxygen.  The Al of DIBAH is coordinated to this oxygen.  This Al is also coodinated to the othe oxygen of the ester linkage. The bi-dentate coordination is critical at stabilizng this intermediate.  if it breaks down, OR is expelled and the aldehyde is generated which will react fruther with more DIBAL to give the primary alcohol.  The Al-coordinated intermediate is still in an aldehyde oxidation state and in fact can be viewed as a acetal of sorts-  it does not react with nucleophiles.  For an acid chloride, this intermediate does not have sufficient stability partly because chloride is such a good leaving group and partly because -Cl is not as good a ligand for Al as oxygen.  So the aldehyde is generated as an intermediate in the reaction and is further reduced.  When LiAlH4 is used, the nature of the Al is much different than in DIBAH, so the coordination phenomena I described above does not occur.

When you powder dry ice and add isopropyl alcohol, the temeperature of the resulting slush bath is -78° C.

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Q: Can the Grignard addition to esters be paused at the ketone stage?

A: In general no.  The ketone intermediate is actually more reactive toward nuleophilic addition than the starting ester.  The the second equv. adds faster than the first.

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Q:  For the reduction of amides to amines, why does the oxygen (O-) leave instead of the amine group (as occurs in reductions of other carboxylic acid derivatives)?  Is the O- a better leaving group than an amine?

A: Yes, but with LiAlH4 it is not a simple alkoxy group, it is the oxygen coordinated to the aluminium of LiAlH4.  It leaves to form some aluminium oxide or something.  For the reduction with borane, the oxygen leaves are a borate.

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02-23-2006

Q: Could you go over problem #27 from chapter 19?

A: The geometry of the nitrogen atom is trigonal planar and it is sp2 hybridized.

The lone pair of electrons on the nitrogen are in an unhybridized p-orbital.

In the geometry shown, all carbons and the nitrogen are (or nearly are) co-planar. This geometry allows for the lone-pair of electrons on Nitrogen to beconjugated with the pi-electrons of the C=C. This is a stabilizing interaction.An important resonance structure of enamines is where there is a negative chargeon the carbon and a N=C double bond. We will see this is Chapter 22.

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02-22-2006

Q: At the bottom of page 689 in the book, for the second variation > of nucleophilic addition to a ketone, why does the -OH leave. I thought -OH was a bad leaving group. Does the proton transfer occur from the nucleophile to protonate the -OH?

A: This is acyually an abbreviated mechanism. Note that the -OH actually ends up as H2O. There is a proton transfer for the -Nu-H to the -OH to give -OH2 as the leaving group. This is better shown in Fig. 19.8 (page 687) for the mechanism of imine formation

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Q: In the mechanism for the Grignard reaction (p695), why does water remove the MgX?

A: The -O(-) MgX+ is more or less an ionic bond. When water or H3O+ is added at the end of the Grignard reaction the -O(-) is protonated to give the -OH, which is a covalent bond. The simple answer is that the overall bonding is energetically more favorable.

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02-17-2006

Q: On slide 159, is the Wittig reagent supposed to be the phosphonium salt or the ylide because CH2 is added, not CH3?

A: That scheme shows the reagents needed to generate the ylide.  So showing the phsophonium salt plus a strong base, in this case methyllithium, indicates that the ylide is being formed.  Methyl lithium deprotonates the phosphonium salt to give the ylide.

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Q: Between the ylide and phosphorane shown on slide 158, does one resonance form predominate?

A: I don’t know if one resonance form contributes more than the other to the resonance hybrid.  Keep in mind that that neither is the actual structure.  I would say that it is a personal preference to denote the ylide or phosphorane. They are equivalent, although the ylide probably more acturatly depicts the reactive species.

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Q: Regarding the cyclic acetal shown on slide 157, what is the hybridization of the C attached to both oxygens?

A: The carbon of any acetal or ketal is Sp3 hydridized.  However, it is still in the same oxidation state as a ketone or aldehyde.  On slide 88 (Ch. 17) there is a diagram that shows increasing oxidation state.  The geminal dichloride is the same as a ketone/aldehyde and imine.  The oxidation state of the acetal/ketal or hydrate is the same as the geminal dichloride.  However, acetals and ketals do NOT have the same reactivity as the carbonyl

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Q: On slide 155, can any strong base (NaOH or KOH) be used in the Wolff- Kishner reaction?

A: KOH is interchangeable with NaOH for the W-K reduction

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Q: On slide 154, is it a base that adds to the iminium intermediate to convert it to the enamine form?

A: Yes.  Water can act as a base.  In this examples the acid that was used was H3O+, so H2O is the conjugate base.

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Q: What does the phrase on slide 152 “imines are isoelectronic with a carbonyl” mean?

A: The C=N double bond of an imine has similar bonding properties with a C=O of a carbonyl and consequently they have similar reactivity.  We will see some reactions of C=N later.  Imines can be reduced (with hydride) and reacts with nucleophiles such as Grignard reagents to give amines in an analogous fashion as addition reactions to carbonyls.

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Q: On slide 147, the reaction shows a ketone, but you can also hydrate an aldehyde? Are base or acid catalyzed reactions occur at the same rate when you have the choice as shown on slid 147 (with one having the hydroxide as a better nucleophile and the other having the protonated carbonyl as a better electrophile)?

A: The carbonyl in the substrate were not meant to be specific, so it can be either a ketone or aldehyde.  This is the way the text denoted a general carbonyl compound.  In fact, aldehydes hydrate much more readily than ketones.

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Q: On slide 143, as the last reaction is written using the Gilman’s reagent, is this an SN2 reaction? If the object to make a ketone, why did our notes show production of a benzene with R-group or an alkene with R group attached?

A: No, this is not an SN2-  SN2 reaction only occur when the whe the C-X has an Sp3 carbon.  I don’t thing the mechanism is straigntforward for this reaction but is may be a nucleophilic acy substitution, which will be covered in Chapter 21.

The “R” denotes any organic groups and is meant to show that the reaction is general.  The top of slide 143 is the Friedel-Crafts acylation which is limited to the preparation of aryl ketones.  The reaction of an acid chloride with a Gilman reagent is a much more general way to make ketones, includinh a,b-unsaturated ketones and aryl ketones.  Just about any acid chloride can be used, and we subsequently said that the Gilman reagent can be any have alkyl groups, vinyl groups and aryl groups (but not alkynyl).  So this reaction can potentially make a ketone with a wide variety of groups on either side of the carbonyl depending on the starting acid chloride and Gilman reagent (which can be made from alkyl, vinyl and aryl halides)

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Exam 1 (02-08-2006)

Q: The book mentions alchols, enols, and phenols. Do ynols exist (i.e., alcohols bonded to sp-hybridized carbon atoms)?

A: Yes, they are tautomers of a class of compounds known as ketenes R2C=C=O. Ketenes have a specialized chemistry. The ynol themselves are not usually observed.

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Q: Is there a special name for protonated phenols? (or are they also called oxonium ions).

A: I am not aware of a special name for a protonated phenol. As it turns out phenols are much less basic than aliphatic alcohols and protonation of the phenol oxygen doesn’t lead to any special chemistry. For instance, phenol will not ionize to a benzene cation like other alcohols nor will it dehydrate to benzyne.

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Q: I notice from the top figure on page 599 that LiAlH4 does not reduce alkenes. Is this also true for NaBH4?

A: NaBH4 is fairly specific for aldehydes and ketones (not esters, carboxylic acids, or nitriles). It does not reduce C-C pi-bonds. LiAlH4 (LAH) is much more reactive than NaBH4 and will reduced carboxylic acid derivatives but in general does not reduce C-C pi-bonds. This is not in the book and you do not need to remember this, but LAH will reduce propargyl alcohol, which has an OH on the carbon next to a C-C triple bond, to a trans allylic alcohol.

NaBH4 will reduce C=N bonds as we will see in a later chapter.

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Q: On page 601, the text lists molecules which are capable of being turned into Grignard reagents. They mention primary, secondary, and tertiary alkyl, aryl, and vinylic. Are alkynl able to partake in the reaction?

A: Alkynyl halides are known (a halide on the Sp hybridized carbon of an acetylene). However, because terminal acetylenes are acidic, they can be converted into acetylide anions by treatment with strong base (Chapter 8.8). The lithium acetylide reacts as a carbon anion like Grignard reagents do. If you want MgX to be the counterion, the terminal acetylene is treated with another Grignard reagents, say methyl magneisum bromide. The methyl anion will act as a base and deprotonated the terminal acteylene and give the acetylide with MgBr+ as the counter ion.

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Q: On page 603, the text does not list esters as being incompatible with Grignard formation?

A: Esters will react with Grignard reagents and should be listed. So the ester could be include as the -C(=O)R, where the R = OR’.

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Q: The text does not list addition of methylmagnesiumbromide (CH3MgBr) to a benzylic ester (Ph-CO2R) as a possibility for the solution to Practice Problem 17.3 on page 603-604. I believe this should work because two equivalents of the Grignard reagent will add to an ester.

A: Addition of methylmagniesium bromide to any ester of benzoic acids (i.e., ethyl benzoate) is a perfectly good way to make 2-phenyl-2-propanol.

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Q: Could 17.10 (b) be synthesized with a di-Grignard reagent (if such compounds exist)? For example, one might use a ethanoate (CH3-CO2R) with 1,5-dichloropentane in which both chlorines on the molecule are converted to Grignard reagents. This could then be used to form the cyclohexane in 17.10 (b). Or would such a reaction be non-viable? This is more a question of curiosity.

A: Yes, that bis-Grignard reagent is actually known. Although it was not used as you described, its reaction with methyl acetate would give 1-methylcyclohexanol.

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Q: Is there no way to dehydrate primary alcohols? The text provides methods to dehydrate tertiary and secondary alcohols.

A: POCl3 reacts with alcohols to phosphotriesters (RO)3PO and dehydration to the alkene is only a minor product. The way to dehydrate an alcohol is to convert them to the primary bromide or tosylate and treat with a strong base (E2 reaction) such a sodium alkoxide or lithium amide- potassium tert-butoxide is the best but is not mentioned in the text.

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Chapter 16

03-09-2006

Q: I was reviewing electrophilic aromatic substitution (a great spring break activity…), and looking at figure 16.15 and it has chlorine as the electrophile. How can a negatively charged species be considered an electrophile? The chlorine ion already has it’s eight valence electrons.

A: You are correct, chloride cannot (and does not) act as the electrophile. Chlorine (or more generally X2) is the electrophile. The electrophilic halogenation of benzene has overlap with the halogenation of alkenes (addition of X2 across a double bond). In that case, the pi-bond of the alkene acts as the nucleophile and attacks X2 with X- (halide ion) acting as a leaving group; for the addition to alkenes a halonium ion then results)

For electrophilic substitution of benzene, the pi-bond of benzene is significantly less nucleophilic than an alkene; using a pi-bond of benzene in a reaction will disrupt the aromaticity of the system which accounts for the low reactivity of benzenes. In order to get benzene to act as an nucleophile, the electrophile (in the case X2) has to be made more reactivity and this is done by the addition of a Lewis acid, i.e., FeCl3 for chlorination or FeBr3 for bromination. Please look at the mechanism for electrophilic bromnination on page 530 (or slides 39 and 40). The Lewis acid catalysts has a vacant oribal and can accept a pair of electrons from Br2 (which is now acting as a Lewis base). The Lewis acid-Lewis base complex (Br3Fe-Br-Br) is a much more reactive source of an electrophilic bromine because the Lewis acid complexation leads to a weaker Br-Br bond with the electrons being polarized towrd the Br that is coordinated to the Fe. Benzene then acts as a nucleophile and reacts with the Br with the partial positive charge and Br3Fe assists Br- to act as a leaving group. In the mechanism box on page 531 (Fig 3), imagine FeBr3 coordinated to the green bromine which makes the red Br carry a partial positive charge. This gives a resonance stabilized carbocation as an intermediate (middle of page 530). To complete the substitution reaction, Br- then acts as a base and deprotonated this carbocation intermediate to restore aromaticity.

02-07-2006

Q: Does benzylic bromination occur by Markovnikov addition of the Br?

A: Markovnikov addition is only for the addition across a C=C pi bond.  The benzylic bromination occurs at an Sp3 hybridized carbon by a radical abstratcion of a benzylic hydrogen, then benzylic radical abstracting bromine atom from Br2.  This gives the benzylic bromide and generates bromine radical

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Q: On slide 75, which reaction is it that uses KMnO4 to change an aryl ketone into a carboxylic acid?

A: The KMnO4 oxidation of an alkyl side chain of ana alkylbenzene goes through aryl ketone, so an aryl ketone will also oxidized to give the benzoic acid.  We did not go over that reaction

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Q: This is probably not a very important question, but i am reviewing the section on reduction of aryl alkyl ketones, and it says that hydrogenation is limited to aryl alkyl ketones. Is an aryl substituent just one that is connected to an aromatic ring?

A: Aryl usually refers to an aromatic ring substituent. So an aryl alkyl ketone is an aromatic ring directly bonded to the carbonyl carbon of the ketone and an alkyl group on the other side.

As far as carbonyls go, the hydrogenation of a ketone to a CH2 is very specific to an aryl ketone. Non-aryl carbonyl will not reduce. However, C-C pi-bonds are aromatic nitro groups also reduce under these conditions.

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Q: I had a quick question about a question on the practice test – number 15. In your answer key, you add the -NH2 group, then the -Cl group, and then the -NO2 group. This involves putting the -NO2 group in between the -Cl and the -NH2, instead of adding -NH2, then -NO2, and then -Cl, which doesn’t. Why can you do that? I thought it was “bad” to put substituents between other substituents.

A: If two groups are meta to one another, then it is hard to but a third group in between the two existing meta-substituents to give a 1,2,3-trisubstituted benzene. In this case, the NH2 and Cl are para, so putting the third group on isn’t a problem.

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Q: If a benzene ring had two alkyl substituents, neither of which was t-butyl, and it was reacted with KMnO4, would both alkyl groups be oxidized to COOH or would a mixture of products occur?

A: Yes, both will be oxidized

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Q: Am I correct in thinking that section 16.9 (benzyne) is one of the sections you just wanted us to read? I only ask because suggested problem 16.61 involves benzyne and I did not think we were responsible for that reaction.

A: yes, we skipped it

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02-06-2006

Q: If a benzene ring has two alkyl substituents, neither of which are t-butyl, and it is reacted with KMnO4, would both alkyl groups be oxidized to COOH or would a mixture of products occur?

A: Both would expect both to be oxidizied.

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02-05-2006

Q: In question 15 of the practice test, would it be fine if the last two steps were switched? (because NH2 can direct both additional substituents, Cl and NO2, to their respective p and o positions?)?

A: Yes.

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Q: On #2 of the practice test, why does NO2 add para; can’t Br also put it at the ortho position (choice a); is it more likely to go para so as not to produce steric interference?

A: Consider the two aromatic rings.  Would you consider one to be more activated (or deactivated) than the other?

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Q: I am still confused about how deactivating substituents on a benzene ring affect the electrophilic aromatic substitution properties of that compound.  If a benzene with a deactivating group is subjected to F-C alkylation conditions, the reaction will NOT occur?  What if it is subjected to F-C acylation conditions?  Does the reaction not occur or is its rate simply decreased?  What about other electrophilic substitution reactions (e.g., halogenation, nitration, sulfonation)?  Do these reactions not occur in the presence of a deactivating group or are they simply slowed down?

A: The FC reaction will not occur of the aromatic ring is more deactivated than a halobenzene (see fig 16.10, page 541).  There is no reaction because the carbocation or oxacarbenium ion undergoes other reactions or decompositions faster than the FC reaction.  So the reagent is destroyed.  The reactive reagent for other electrophilic aromatic substitions reactions do not suffer this problem, so the reaction proceeds, but at a reduced rate.

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Q: I presume that reaction of a tertiary-alkylbenzene will not undergo reaction with NBS and benzoyl peroxide?

A: If the benzylic carbon is tertiary, there is no reaction.

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Q: Why is is that a NO2 substituent would be reduced to NH2 in the reduction of an aryl alkyl ketone?

A: An aromatic nitro group is very easily reduced.  Catalytic hydrogenation reduces aromatic notro groups at a rate that is competative with an aryl ketone

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Q: How do you determine whether resonance effects outweigh inductive effects or vise versa as an activator or deactivator?

A: Resonace effects will usually outweigh inductive effect.  The exception will be for the halogens because they are not effect electron donors.  However, the group must be able to directly interact, so the position is important.  This is the case for the acidity of substituted phenols.  Groups para to the hydroxy group can directly interact with the phenoxide ion through resonance, so the resonance effect is very powerful (either stabilizing or destabilizing); however, when they are on the meta position they can not directly interact with the phenoxide ion through resonance, so the inductive effect becoemes more important.

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Q: If you want only a mono-substituted products from the F-C alkylation, is the best way to acylate and then reduce? When should you use alkylation over acylation?

A: Yes, the FC-acylation followed by reduction of the benzylic carbonyl group to a CH2 can be equivalent to the FC-alkylation.  The real value of the F-C acylation is:  1) it has a different directing effect than a alkyl group (m vs o,p) and 2) the bigger problem with the FC-alkylation is the alkyl halide is converted to a carbocation which can undergo rearrangements (slides 49-50).  The F-C acylation does not do that.

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Q: Do you also get tri-substituted products in the F-C alkylationas or only mono and di?

A: The book makes a big deal about this but it really isn’t a problem.  The F-C of benzene can give mono-, di- and tri-alkylated products.  However, if you use benzene is large excess, only the mono-alylated product is obtained because of stoicheometry.  So you can assume that you can get a mono-alkylate product from an F-C reaction.

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Q: When a resonance structure places a positive charge on oxygen, is this stable because the oxygen donates electrons to the ring and increases its stability?

A: Yes, oxygen is a good electron pair donor and can handle a positive charge. This it can stabilize a positive charge through resoancne. It will destabiolizes a negative charge.

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02-04-2006

Q: For a Friedel-Crafts alkylation of a phenol, does the alkyl group ever add to the C with the OH?

A: No. If that occurs, the system could not regain aromaticity. The OH of a phenol directs the FC alkylation to the ortho or para positions

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Q: Is sulfonation followed by the alkai fusion reaction the fastest way to put an –OH group on a benzene ring?

A: Yes. That is the only way that we have learned.

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01-22-2006

Q: For Problem 16.31, why are some rings deactivated and others aren’t? What determines if a ring is deactivated?

A: A substituted benzene is considered deactivated if the rate of electrophilic substitution is slower than for benzene. Activated is just the opposite, the rate of electrophilic substitution is faster than for benzene. The change in rate is attributed to the influence of the substituent on the pi-electrons of the aromatic ring. In general, if the net effect of the substituent is that it is electron withdrawing, the ring is deactivated. An electron withdrawing pulls electron density away from the aromatic ring making it electron deficient and consequently less nucleophilic. Electron donating groups push electron density toward the aromatic ring making it more electron rich and more nucleophilic.

The two components for determining if a substituent activates or deactivates the aromatic ring are its inductive and resonance effect. This is summarized on Table Table 16.2 and Figure 16.10. There is related discussion on these issue below.

Problem 16.31 asks you to predict where FC alkylation will occur on the given substituted benzene. FC alkylation is not compatible for all substituted benzenes and these limitations are noted in Figure 16.8. FC alkylation works for benzene, alkyl benzenes, phenols, anisoles, and halobenzenes. If does not work for other deactivated benzenes and for anilines. There is related discussion on this issue below.

(a) bromobenzene is slightly deactivated and there are no incompatible functional groups. It will undergo FC alkylation at the ortho or para-positions.
(b) m-bromophenol has a modest deactivating group and a strong activating group so overall it is activated. There are no incompatible functional groups. The directing effect for the two groups reinforce so you would expect FC alkylation to occur at the 4- and 6-positions
(c) p-chloroaniline, although aniline is an activating group it is incompatible with the FC alkylation reaction and therefore gives no reaction.
(d) 2,4 dichloronitrobenzene, this compound has three deactivating groups on it so it is strongly deactivated. A single nitro group is enough to prevent FC alkylation- no reaction
(e) 2,4-dichlorophenol. The two chloro groups are each modestly deactivating groups but the phenol is strongly activating. Overall, this is still an activated aromatic. There are no incompatible functional groups. The directing effect of the two chlorine atoms reinforce each other but are oppose the directing effect of the phenol. The phenol wins because it is the more powerful activator. FC alkylation occurs at the 6- position.
(f) benzoic acid is deactivated so this does not react in an FC alkylation
(g) p-methylsulfonic acid, the methyl group is a modest activator, the sulfonic acid group is a strong deactivator. The next effect is that the aromatic ring is deactivated and does not undergo FC alkylation.
(h) 2,5-dibromotoluene, The two bromine substituents are modestly deactivating while the methyl group is modest activator. The net effect is that the ring is probably not much different than benzene itself, perhaps very modestly activated, perhaps very modestly deactivated. It certainly is not strongly deactivated and there are no incompatible function groups. The directing effect of the two bromine oppose each other. The directing effect of the methyl group reinforces with the 5-bromo but is opposed to the 2-bromo. The methyl group is the stronger activating group and will direct the FC-alkylation to the 4-position.

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01-22-2006

Q: I do not understand the mechanism whereby NH3 adds to a benzyne to form aniline. Is this similar to the addition of BH3 to an alkene to form an alcohol? Aan other strong bases be used to form a benzyne and subsequently add to the benzyne? For example, could a deprotonated terminal alkyne (acetylide anion) react with benzene to form an alkyne-substituted benzene?

A: We are skiping the section on benzyne (Ch. 16.9). The triple bond of benzyne is pretty strained. The reactivity of benzyne can be thought of as going through a zwitter ion rather than a triple bond. See the attached pict. You can draw resonance structures if benzyne in which there is a positive and negative charge on carbons instead of a triple bond. You can also draw benzyne as a di radical. so the nucleophile (ammonia) reacts with the cation to give a new zwitter ion in which an NH3+ group is now covalently bonded to the benzene and C2 is still deprotonated and has a negative charge. A proton is then transfered from the NH3 group to the carbon to give the product.

The only C-C bond forming reaction I have seen with benzyne is a Diels-Alder, which is mentioned in the book. I suspect that acetylide reacts with benzyne is a much more complex manner, but I don’t know for sure.

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Q: I do not understand 16.54 (b). The last step in the mechanism, they treat the CH2Br groups with NaOH. How does result in conversion to CH2OH groups?A: This is a nucleophilic substitution reaction, most likely an SN2.

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Q: On the bottom of slide 44, several condtions for the reduction of nitro groups are listed. Could you explain the two involving tin?

A: The mechanism of these reaction has not been extensively studied. Many zero valent metals are reducing agents, Fe(0) and Sn(0) are very mild ones. The mechanism is similar to a dissolving metal reaction that you learned last semester for the reduction of alkynes to trans double bonds with Li(0) in liq. NH3 and involves a series of single electron transfers from the metal to the nitro group. The oxygens of the nitro group are ultimately lost as water and the oxygens are eventually replaced by hydrogens on the nitrogen. Thus, acid is necessary. Aromatic nitro groups are among the easy functional groups to reduced

The reductions with SnCl2 or with Fe(0) or Sn(0) in acid are important because they selectively reduce aromatic nitro groups to anilines. They do not reduced ketones or C-C pi-bonds. This is opposed to H2, Pd/C which will reduce aryl ketones, and C-C pi bonds

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Q: Related to the previous question below, does the limitation of F-C alkylation reactions regarding the nature of the substituent (p. 536 of McMurry, strong deactivating groups and amino groups are incompatible) also apply to the F-C acylation?

A: FC acylation of nitrobenzene does not occur. In fact nitrobenzene is often used as the solvent for the reaction. Benzenes with other strong deactivators also do not undergo FC acylation.FC acylation with anilines works poorly. If the aniline is converted to an amide, the reaction proceeds. Amides of anilines are o,p-directors and still activated, although less than aniline itself. I think is covered in Chapter 24. Phenols and anisoles are fine for the F-C acylation reaction.

So it appears as though the substituent effect for FC alkylation and acylation are the same.

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Q: If the directing effects of the two groups are antagonistic, how does one decide whether two products will form or only one? This requires knowing the relative reactivity of all of the substituents, I imagine. Take for example problem 16.32 (c). The book gives two products (and identifies neither as major/minor) and rationalizes this by saying that both groups deactivate to a “similar extent” even though the relative reactivity chart provided on page 541 shows that this is actually not true (the nitro substituent is significantly more deactiving than the carboxylic acid substituent). On other problems, the book will simply give one possible product as the “correct” answer without indicating the minor product. How then, does one go about determining whether two products will form in roughly equal amounts, two products will form in different amounts (a major and a minor product), or one product will form to a significantly greater to extent than the other, thereby allowing us to neglect contributions from the other potential product?

A: This is a nuance (and perhaps a nuisance) that you are not expected to have a feel for at this stage. I agree completely with you regarding 16.32c. With the information you are given, I would expect the major product to be 3-chloro-4-nitro-benzoic acid. The opposing directing effect “rule” that the stronger activating group will dominate works best when the two groups are activating or if one is activating and the other deactivating. When both are deactivating, mixtures are much more common.

It is difficult to know for sure when to expect mixtures and what will be the major product and by how much in every case. The book gives “rules” and mechanistic rationale for the rules and in general these are pretty good predictor. So we do the best we can with these rather than trying to know some general rules and all the specific exceptions for each rules, as this would become overwhelming. However, it would be nice if the problems and answers were more self consistent or did not use a rationale that obviously isn’t well supported by the text. I can understand how this could be a source of frustration. I suggest that you work with the “rules” that are given in the text (and lecture) the best you can and you seem to be doing fine with them. If your answer differs from the answer book, it does not necessarily mean that you are wrong. You are doing the correct thing by asking when an answer in the book confuses you.

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Q: How does one determine whether an electrophilic aromatic substitution reaction will or will not occur. Do strongly deactivating groups (i.e., the deactivators mentioned in the book excluding halogens) cause a reaction not to occur only for Friedels-Craft alkylation reactions? That is, do these compounds generally only decrease the rate of a reaction but in the specific case of F-C alkylation, they actually prevent the reaction from occuring?

A: This is a complex questions. The short answer is yes, electron withdrawing groups only slow down most electrophilic aromatic substitution reactions but in the case of the FC alkylations, they prevent the reaction from occurring.

The long winded answer: Most electron-withdrawing groups will only affects the rate of the reaction because nothing else can happen to the electophile or the aromatic ring. For F-C alkylation, the reactive electrophile is a carbocation, which can undergo other reactions. For instance, they can lose a proton to give alkenes. Alkenes under strongly acidic conditions polymerize. So because of the decreased rate of F-C alkylation of a deactivated benzene, the reactive electrophile (carbocation) actually undergoes other reactions fasters than electrophilic substitution. This is the case for a nitro group and a quarternary ammonium group (Ar-NR3+).

The F-C alkylation also works poorly for benzenes with some electron donating (activating) groups for other of reasons. Substituents such as carbonyl groups, hydroxyl groups or amino groups, have lone pairs of electrons and can act as a Lewis base. Aniline (amino groups) will react with AlCl3 to form a Lewis acid-Lewis base complex- a lone pair of electrons from N is donated to a vacant p-orbital of Al. The result is that the substituent now has a formal positive charge which strongly deactivates the ring toward electrophilc substitution. So a strong activating group is converted to a strong deactiving groups because of the acid-base chemistry. While aromatic ketones or esters are electron-withdrawing groups, the lone pair of electrong on the carbonyl oxygen interact readily with the Lewis acid catalysts and it is this interaction that causes aromatic carbonyl compounds (aldhydes, ketones, esters) from undergoing electrophilic substitution.

However, phenols and anisoles (methoxybenzene) are fine for F-C alkylations reactions. So here is another wrinkle. The difference between phenols or anisoles and anilines is that alcohols and ethers are weaker Lewis bases. So the complex between them and the catalyst is not as strong. In this case there will a reasonable concentration of free AlCl3 and phenol (or anisole) to react in the F-C reaction. In the case of an aniline or a carbonyl, the Lewis acid- Lewis base complex with AlCl3 is so strong that all of the aniline and AlCl3 is associated and consequently there is no reaction
The nature of the catalyst (Lewis acid) is central to all of this. The Lewis acids catalysts for electrophilic aromatic substitution are not necessarily interchangeable and there is a continuum of reactivities. FeX3 is strong enough to activate X2 for electrophilic substitution with a variety of substituted benzenes and does not have the drawbacks noted above. However, FeX3 is not a strong enough LA catalyst to generate a carbocation from an alkyl halide in most cases. So FeX3 is not a common FC alkylation catalyst.

The FC acylation suffers from many of the same drawbacks. Most oxacarbenium ions, although highly reactive, are not as fragile as alkyl carbocation. However, they react with amines and alcohols (anilines and phenols) to give amides and esters rather than the aromatic ring to give electrophilic substitution products. For F-C acylation, a stoichiometric amount of the Lewis acid catalyst is required since after electrophilic aromatic substitution the aryl ketone product will form a complex with the catalyst and prevent it from turning over.

So overall, we see that there is a narrow window of substituted benzenes for FC reactions: benzene itself, simple alkyl benzenes, halobenzenes, phenols and anisoles (or any ether of phenol). This needs to be taken into account in planning the order of reactions for a synthesis problem along with the directing effects.

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01-20-2006

Q: When a substituent is o/p directing, does it electrophile add to both the o and p positions or one over the other?

A: We will assume that a single addition occurs. For ortho-para directors you usually get a mixture of ortho and para disubstituted benzenes as the product. The ratio is dependent upont the substituent and the electrophil and is not necessarily statictical, equal or easily predictable. So we will not worry about these details.

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Q: Why is there a difference between o/p directing activators and deactivators since addition of the electrophile occurs at the o or p position?

A: These are two separate effects (see question below). The o,p part is the orientation of the reaction relative to the substituent. The other has to do with the rate of the reaction. An o,p-directing activator, directs electrophilic aromatic substitution to the o- and p-positions AND the reaction is faster than electrophilic substitutuion for benzene. o,p-deactivators have the same directing effect, but the rate of the reaction is slower than for benzene.

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Q: Is halobenzene the same as phenol?

A: No, a halobenzene has a halogen as a substituent. Phenol has a hydroxyl group (-OH).

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Q: Are the two classifications for substituents: activating/deactivating and o,m,p directing?

A: The substituent has two effects:

a) the substituent is either activating or deactivating- does the substituent make the benzene more reactive toward electrophilic aromatic substitution or less reactive (relative to benzene), and

b) orientation of further electrophilic aromatic substitution- does electrophilc aromatic substitution occur ortho, meta or para to the substituent.

Combining the two effects, substitutent are classified as either :
1) o,p-directing activators

2) m-directing deactivators, or

3) o,p-directing activators.

There are no m-directing activators (see question below and Chapter 16.6).

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Q: Is a substituent with an electron withdrawing inductive effects a deactivating? (and if that substituent is not a halogen, then it is a meta-directing deactivator?)

A: The overall effect is a combination of inductive and resonances effects. All groups other than an alkyl group, will be more electronegative than C, so most substitunts will have an electron withdrawing inductive effect. Resonance effects can play a large and sometime even a dominant role the reactivity.

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Q: Do substituents have either a inductive OR resonance effect?

A: All substituents have an inductive effect; not all substituents have a resonance effect. Many substituents have both an inductive and resonance effect. The combination of the inductive and resonance effects determine if the substituent is an activator or deactivator, and if it is an ortho-para or meta director

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Q: Is there a relation between reactivity and orientation? i.e. do meta directors tend to be deactivating (electron withdrawing) and o/p directors tend to be activating except for the halogens which are deactivating?

A: All activators are ortho-para directors. This is usually due to an electron-donating resonance effect. The exception here is an alkyl group which has an electron donating inductive effect and no resonance effect.

All meta-directors are deactivators; however, not all deactivators are meta-directors. The halogens are othro-para directing deactivators because they have an electron donating resonance effect and a strong electron wothdrawing inductive effect.

The oxygen of phenol has an electron withdrawing inductive effect, but the lone pairs of electrons of the oxygen interact very strongly with the pi-system of the benzene ring. Therefore it is overall electron donating, activating and ortho-para directing

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Q: From slide 54, what is the difference between activators and deactivators? Why can’t you have meta-activators?

A: A substituent on a benzene ring is regarded as activating when the substituted benzene is more reactive toward electrophilic aromatic substitution than benzene itself. Another way to look at it is that rate of electrophilic aromatic substitution for toluene, phenol, aniline, anisole, etc., is faster than the rate of electrophilic aromatic substitution for benzene. A deactivator is the opposite. The rate of electrophilic aromatic substitution for chlorobenzene, nitrobenzene, benzaldehyde, etc. is slower than the rate of electrophilic aromatic substitution for benzene.

The activator or deactivator designation therefore pertains to relative reactivity of the substituted benzene. Things that donate electrons make the benzene ring more electron rich and therefore more nucleophilic (faster reaction with electrophiles). Substituents that withdraw electrons make the benzene ring electron deficient, and therefore less nucleophilic because the electron are less available (slower rate of reaction with electrophiles).

The question of ortho-para versus meta directing pertains to how the substituent influences the relative stability of the intermediate carbocations from electrophilic addition. For electrophilc addition at the ortho or para positions, the positive charge of the carbocation resonances structures are spread out on the same carbons of the benzene ring (see Figures 16.12, 16.13, 16.14, and 16.15). The charge is on a different set of carbons for electrophilic addition to the meta position. The directing effect can be rationalized by analyzing the influence of the substituent on the carbocation resonance structures. We find that a carbocation resonance structure that arises from either ortho or para addition of the electrophile, places the positive charge on the carbon bearing the substituent. If the substituent is electron donating (either by inductive effects or resonance), the change can be stabilized by the substituent (see the boxed structures in Figure 16.12, 16.13 and 16.14). Thus, ortho or para addition is favored. If the substituent has an electron withdrawing resonance effect, the charge is destablized for this same carbocation resoanance structure from ortho or para addition (see the boxed structure in Figure 16.15). In this case, ortho or para addition is disfavored, leading to meta addition. While the carbocation for meta addition, is still detabilized by the elctron withdrawing group, it is less destabilzed than for ortho or para addition because none of the resonance structures place the charge next to the electron withdrawing group.

So the reason there are no meta directing activators is because the carbocation intermediate from ortho or para addition will always be more stable for an activating group.

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Q: When you say “electron withdrawing substituents” on slide 47, do you mean a substituent with very electronegative atoms like O, S, and N?

A: Those are electron withdrawing through inductive effects, which is one part of the overall effect. You also have to consider resonance effects, which can sometimes outweigh the inductive effect. This is the case for the -OH of phenol, -OCH3 of anisole, -NH2 for aniline, etc. These groups have a more electronegative atom (than C) attached to the benzene ring and therefore have an electron withdrawing inductive effect. However, these groups have a very strong electron donating resonance effect which outweighs the inductive effect, and therefore the overall effect for these substituents is that they are strong electron donating groups.

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Q: You said that the alkyl halide does not act as the electrophile but on slide 46 you have termed it that.

A: Alkyl halides are generally regarded as electrophiles, however they are not reactive enough to react with a benzene ring, which is a very weak nucleophile. So for the FC alkylation the actual electrophilic species is the alkyl carbocation. When we write out the reaction and identify the participants, we would generally call the alkyl halide the electrophile since it directly gives rise to the reactive electrophilic species.

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Q: On slide 44, do all reagants/catalysts from (a) or all from (b) reduce aromatic nitro groups to the aniline? Is one combination preferred over the other? What is unique about these reagents ?

A: For the reduction of nitro group, each line has the complete set of reagents needed to carry out the reduction to the aniline. So, one set of conditions in SnCl2, H3O+ followed by hydroxide. A second set of reaction conditions is Fe(0) in acid, etc. SnCl2 is useful because it will only reduce aromatic nitro groups to anilines; C=C double bonds, etc are not reduced. H2, Pd/C on the other hand, will reduce aromatic notro groups, carbon-carbon pi-bonds, aromatic carbonyl groups with no functional groups selectivity. (see Chapter 16.11 and slides 70-72)

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Q: From slide 43, would an example of a base that removes the proton from the carbocation to give the electrophilic substitution product be Cl- (left over from Cl2)?

A: Yes. Electophilic addition by X2 and FeX3 to benzene gives a resonances stabilized carbocation intermediate. The counter ion is FeX4- (anion), which is in equilibrium with FeX3 and X-. The X- acts as a base and removes a proton from the carbocation, thus restoring aromaticity and giving the substitution product. So in all cases, HX is a byproduct from electrophilic aromatic substitution. In the case of electrophilic iodination, the base is from the Cu(II) ligands.

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Q: Is there an actual 3D complex made between the Br2 and the FeBr3 that allows Br to be attacked by the benzene ring?

A: Yes. There is a physical association between the FeBr3 and the Br2. This polarizes the Br-Br bond so that the Br is more electrophilic.

Questions from Spring 1999

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