Emphasis Topics for the Third Exam in CH610B:Bauld

CHAPTER 20: REACTIONS OF BENZENE AND ITS DERIVATIVES

Reaction Mechanisms

• You should be able to write the detailed, general mechanism for aromatic electrophilic substitution, using a general electrophile (E+) and benzene as an aromatic substrate. You should know that the first step is usually the rds and that it leads to the formation of a special carbocationic intermediate called an arenium ion.
• You should be able to draw the three canonical structures for the arenium ion and the DL/PC summary structure which shows partial positive charge at the ortho and para positions, but none at the meta position.
• You should also be able to write the detailed, specific mechanisms for bromination, nitration, and Friedel-Crafts acylation and alkylation.You should know two different mechanisms for Friedel-Crafts alkylation, depending upon whether the alkyl halide is secondary/tertiary (SN1) or primary (SN2).
• You should also know that because of the strong resonance stabilization of benzene,bromine itself is not reactive enough to substitute benzene, although it reacts readily in addition to alkenes. Therefore, an electrophilic catalyst is needed to generate a more reactive electrophile.
• You should be able to derive a TS model for the rds of a general electrophilic substitution reaction.
• You should know why, in the last step of the mechanism, the arenium ion prefers to eliminate a beta proton rather that to react with the nucleophile at carbon, yielding electrophilic addition. (Elimination restores aromaticity, but reaction at carbon does not).

Substitutent Effects on the Rates and Orientation of Electrophilic Substitution of Monosubstituted Arenes

• You should know that substitutents present on a benzene ring (i.e., a monosubstituted arene) which stabilize a carbocation center typically accelerate the rate of electrophilic aromatic substitution. These are called activating substituents. These substituents are also referred to as electron donating groups (EDG's). They stabilize the TS by stabilizing the carbocation character in the arenium ion-like TS.
• These EDG's also typically exert a positional preference upon the entering electrophile (nitronium ion or whatever), causing the electrophile to prefer the ortho and para positions over the meta position. These groups are called ortho-para directing groups.Thus, most activating substituents are also ortho-para directing groups.
• You should be able to use the Method of Competing TS's to explain the preference for the para TS over the meta TS (the same things which apply to the para also apply to the ortho, so just use the para case). Note especially that the entering electrophile (E+) generates carbocation character only at the ortho and para positions in the arenium ion-like TS. Therefore a carbocation stabilizing substituent can only stabilize this character if it is attached to the position ortho or para to the entering electrophile.
• For a simple case, you should be able to show how methoxy and amino groups stabilize a carbocation center (resonance effect). You should know already how methyl and alkyl groups stabilize positive charge. Additionally, you should know that the order of activation is amino>methoxy>alkyl and be able to explain why in terms of the resonance effects involved in each case.
• Using these principles, you should be able to predict the major products of any of the familiar electrophilic reactions of toluene, anisole, and aniline.
• You should know that groups which destabilize a carbocation center (or carbocation character) retard the rate of electrophilic substitution. They are said to be deactivating groups. In general, such groups are called electron withdrawing groups (EWG's). They act by destabilizing the carbocation character of the arenium ion-like TS.
• You should know that EWG's are, in contrast to EDG's,generally meta directing, preferring the electrophile to enter meta to the electron withdrawing substituent on the benzene ring.
• You should be able to use the Method of Competing TS's to explain this preferred meta direction. Essentially, the EWG, when present on the ortho or para position destabilizes the carbocation character at these two positions, but since there is no carbocation character at the meta position there is less destabilization.
• The deactivating effect of EWG's arises from the destabilization of the positive charge by the substituent dipole, which is oriented with the positive end of the dipole nearest to the charge on the ring. This inductive destabilization affects even the meta transition state.
• You should be able to explain the deactivation effect by applying the Method of Competing TS's to the TS's for substitution of an electrophile (E+)meta to an EWG such as nitro with that for the substitution of the same electrophile upon benzene, itself, where there is no substituent present to destabilize the positive charge.

THE SPECIAL CASE OF THE HALOGENS

• The halogens are a special case: they are deactivating (EWG's), but unlike other deactivating groups, they are o,p-directing.
• The deactivating nature of chlorine can be seen, for example,by using the Method of Competing TS's and considering the TS for meta substitution upon chlorobenzene relative to that for substitution upon any of the six equivalent positions of benzene (same electrophile in each case).The C-Cl bond has a large dipole with the positive end on the ring (ipso)carbon to which it is attached.. This gives an inductive destabilization of the positive charge on the ring positions ortho and para to the electrophile (especially on those ortho to the chlorine substituent, because these are closest). Therefore, chlorine is a deactivating group at the meta position.
• Now,use the Method of Competing TS's and compare this meta TS with the TS for para substitution. Now the chlorine is attached to a position with carbocation character, and inductively it should destabilize this carbocation character and result in a deactivating effect. Indeed, chlorine is deactivating even at the para position. The anomaly is that it is actually less deactivating that when at the meta position, even though the para TS has carbocation character on the carbon to which the chlorine is attached. The inductive effect should be largest here, therefore, and the destabilization the greatest, inductively. Undoubtedly, this is the case. However, there is another effect which operates only at the para (and ortho) position, and not at all at the meta, and this effect is a carbocation stabilizing effect.
• This additional effect is a resonance effect. You should be able to show how a chlorine substitutent present at a carbocation center stabilizes a carbocation site via resonance (at the same time it destabilizes it inductively; the net effect is a dominant inductive effect, but the resonance effect counteracts a large part of the inductive effect, leaving only a rather weakly EWG).
• You should also know that methoxy has both of these same effects, but since the electron pair on oxygen is more available for delocalizaiton than is that on chlorine, the resonance effect is large in the oxygen case, giving a net dominant resonance stabilization. Recall that alkyl groups, like methyl, are electron donating both inductively and via resonance (hyperconjugation).

CHAPTER 21: AMINES

You should be able to name amines using two two types of nomenclature. The older approach names the alkyl or aryl group first and then attaches the word amine (without a separating space). The newer IUPAC mechanism deletes the final -e of the parent chain and attaches the suffix "amine", e.g., CH3NH2 is methanamine in the latter system, but methylamine in the older system.
• You should know and be able to recognize the sub-classes of amines, viz., primary, secondary , and tertiary amines, and you should be able to name these: e.g., dimethylamine, a secondary amine, in the newer IUPAC nomenclature is N-methylmethanamine; trimethylamine, a tertiary amine, would be called N,N-dimethylmethanamine.
• You should know that the nitrogen atom of amines is tetrahedral (or pyramidal, if you prefer) and that the unshared electron pair is in an sp3 type orbital. Also, a tertiary amine with three different alkyl substituents is chiral, but rapidly racemized by an inversion process.
• You should know that all amines are hydrogen bond acceptors and are substantially basic, both via the unshared electron pairs. You should also know that all amines except tertiary amines are hydrogen bond donors. Therefore, in terms of water solubility, amines are like alcohols and ethers, rather than like alkanes and alkyl halides, since water can hydrogen bond to them. In terms of boiling points, amines are also like alcohols and ethers in having higher bp's, since they can hydrogen bond to themselves. The exception to the latter is tertiary amines, which have no oacidic proton and are not H-bond donors.
• You should be able to write the equation for the reaction of an amine with water to given a substituted ammonium ion and hydroxide ion and to formulate Kb and pKb. You should also be sure to know the relationship between pKb and pKa of the conjugate acid of the amine.
• You should know that the stronger the basicity of the amine, the lower the pKb and the higher the pKa of its conjugate acid. That is, the stronger the amine is as a base, the weaker the ammonium ion conjugate acid is as an acid.
• You should know that amines are comparable in basicity to ammonia, but slightly stronger in general. You should know, e.g., why methylamine is a stronger base than ammonia, and that resonance plays no part in this(why?) ,unlike the stabilization of a carbocation by an alkyl group.
• You should know that tertiary amines are weaker bases than primary or secondary amines and why (steric hindrance to solvation).
• You should know that amines themselves (not their conjugate acids) are acidic unless they are tertiary amines. You should also be able to write the equation for the dissocation of an N-H bond of an amine in water, giving an amide type ion. You should know that the approximate pKa of amines is 40, and how to form these ions quantitatively (e.g., butyl lithium, as in the preparation of LDA).

Synthesis of Amines

• You should know that amines cannot be efficiently prepared, generally, by the reaction of ammonia with an alkyl halide , and why not (the primary amine formed initially can also react with the alkyl halide to give a secondary amine, and on to a tertiary amine,etc.Thus a mixture is obtained. One way of getting around this is to use a huge excess of ammonia, if this is feasible.
• You should know that primary amines can be synthesized by the reduction of alkyl azides by lithium aluminum hydride (LAH), which ,in turn, can be obtained by the SN2 reaction of azide ion (N3-) with alkyl halides. This works, in contrast to the reaction of the previous part(involving ammonia), because the alkyl azide is not a good enough nucleophile to react with an alkyl halide.
• You should know that primary, secondary, or tertiary amines can be prepared by LAH reduction of amides.

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The Hoffman Elimination of Quaternary Ammonium Hydroxides

• You should be able to write the equation for the overall reaction in a general case, and understand that this is a concerted (E2) elimination, which requires a beta hydrogen and at least a resonably good leaving group. As usual in an E2 elimination , the preferred stereochemistry is anti (the beta H and the leaving group are on opposite faces, with a dihedral angle of 180 degrees). The leaving group is an amine, almost always trimethylamine.
• You should know that the Hoffman elimination is very different from the Zaitsev elimination of alkyl halides or tosylates with base. The latter tends to produce the more stable alkene as the major product because, as you will recall, the predominant TS character is alkene character. The Hoffman elimination tends to produce the less stable, less highly alkyl substituted alkene as the major product, e.g., 1-pentene over 2-pentene. Therefore these two types of elimination, both of which are E2, are synthetic complements for each other.
• You should review the TS model for the E2 reaction of an alkyl halide and its alkene character. Now, you should be able to derive(!) a more refined TS model for E2 TS's in general, and note that in addition to alkene character, the TS can also have some carbanion character,specifically at the beta carbon (i.e., C-beta carbanion character).
• You should be aware that, since alkyl groups are EDG's, they destabilize a carbanion, in sharp contrast to their effect on a carbocation, so that the stability order for carbanions is CH3 carbanion>primary carbanion (ethyl,e.g.) >secondary carbanion (isopropyl)>tertiary carbanion (tert-butyl).
• You should be able to use the Method of Competing TS's to compare the TS characters which lead, respectively, to 1- and 2-pentene, noting that the former has primary carbanion character and the latter secondary carbanion character.
• You should carefully note that the position adopted here is not that there is now no alkene character, but that both characters are quite generally present in all E2 eliminations, but that carbanion character dominates in the case of an alkylammonium ion (Hoffmann) elimination and alkene character dominates in the case of an alkyl halide or tosylate.
• You should know that the blend of carbanion and alkene characters changes to emphasize carbanion character in the Hoffmann elimination because the trimethyl amine leaving group is a poorer leaving group than the halide ion, thus leaving more charge on the beta carbon.
• Finally, you should understand why the leaving group is almost always trimethylamine (this amine has no beta hydrogens, so no competing eliminations exist in this case).

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Arylamines

• Arylamines and phenols (which are aryl alcohols) are considered separately from their corresponding aliphatic counterparts for a number of reasons, including the fact that the acidities (phenols) and basicities (arylamines) are grossly different from those of the aliphatic amines and alcohols.Also, aryl amines and phenols feature much of their reactivity on the aryl ring, since the -NH2 and -OH groups are powerful EDG's when attached to such rings,and especially the -NH2 group.
• Arylamines are much less basic than aliphatic amines because the arylamine is resonance stabilized and the anilinium ion (the conjugate acid) is not. This disfavors the side of the equation which generates the hydroxide ion (the product side).
• you should be able to write resonance structures for aniline which show the development of much partial negative charge on the o,p positions. You should also be able to explain why the anilinium ion has no resonance stabilization.
• You should know that a pKa for the anilinium ion of 4.63 means that it is much more acidic than the methylammonium ion, which has a pKa of 10.66 ( you need not memorize these pKa values, of course). The point is that a lower pKa means a stronger acid. In this case the factor is about a millionfold. You should know therefore that aniline is a stonger base than methylamine (the stronger the acid, the weaker the corresponding conjugate base). You should be able to calculate the pKb of aniline from the pKa of the anilinium ion, and you should know that a lower pKb means a stronger base , e.g., 9.37 for aniline vs. 3.34 for methylamine means that methylamine is the stronger base.
• You should know that anilines (arylamines) can be prepared by catalytic hydrogenation of nitroarenes or by Fe/HCl reduction of nitro groups.

Diazonium Salts

• You should know that aniline and other arylamines are converted, in the presence of nitrous acid (HNO2) and sulfuric acid at 0 degrees C to diazonium salts--ArN2+/HSO4-.
• You should be able to write two main canonical structures for the diazonium salts which show that it is a resonance stabilized cation and that the charge is delocalized upon both nitrogens.
• You should know that, as an electrophile, the diazonium ion reacts only at the terminal nitrogen, even though there is positive charge on both nitrogens and why this is the case.
• You should know that these salts are stable at 0 degrees, but decompose when warmed to room temperature, to give dinitrogen and the phenyl carbocation.
• You should know that dinitrogen is perhaps the very best leaving group known.
• You should know that aliphatic diazonium ions can be generated in the same way, but they are unstable, at 0 degrees or even far below this temperature, because nitrogen is such a good leaving group. This is, however, a good method for generating carbocations.
• You should know that aryldiazonium ions are much more stable than this because the diazonium group is in resonance with the benzene ring. You should be able to draw appropriate structures which show this stabilization and the development of double bond character between the N and the ring carbon attached to it.

Synthetic Uses of Aromatic Diazonium Salts

• You should know that when an aryldiazonium salt in aqueous solution is warmed to room temperature in the presence of various nucleophiles, the nitrogen molecule is lost and the aryl carbocation which results reacts with the nucleophile. In this way, CuX ( or just halide ion) gives halogen attached to the ring, CuCN ( or just cyanide ion), gives the nitrile, and water gives the phenol as a replacement for the diazonium group.
• You should review the synthesis of diazonium ions from anilines,and of anilines from nitroarenes, and of the latter from arenes by nitration. That is, you should be able to start with benzene and sketch a synthesis of a diazonium salt.
• You should know that diazonium ions are mild electrophiles which react only with the most highly activated aromatic rings, especially arylamines .You should know that the products are azo compounds, highly colored materials that are used as dyes (azo dyes). You should also be able to write the mechanism for the reaction of diazonium ions with such reactive aromatics.

HETEROCYCLIC AMINES

• You should know the structures of pyridine and pyrrole and that both are less basic than typical aliphatic amines.
• You should know that the unshared electron pair of pyridine is in an sp2 orbital (not included in the aromatic sextet). Since this orbital is sp2 it is lower in energy than the sp3 AO which the unshared pair of an aliphatic amine occupies. The electron pair is thus more stable in pyrdine and is more difficult to protonate.
• You should know that pyrrole is less basic than aliphatic amines because its unshared pair is part of the aromatic sextet. To protonate pyrrole removes its aromaticity, which requires a large amount of energy.
• You should also know the structure of guanidine and that it is the strongest base among neutral compounds. The pKz of its conjugate acid is 13.6, which makes it only a little weaker than water as an acid. The reason for the great ease of protonation of guanidine is that the conjugate acid is highly resonance stabilized. It has three equivalent resonance structures.

CHAPTER 20 (THE LATTER PORTION, NOT COVERED ON THE SECOND EXAM)

• Be sure to study the resonance stabilization of benzylic radicals and cations, and be able to write the resonance structures (four of them in each case) and summarize as a DL/PC structure for the carbocation and as a dotted line/partial radical character structure for the benzyl radical.

• You should know which resonance structure is the lowest energy one and why, and the consequences of this for the distribution of charge in the cation and of radical character in the radical.
• You should also know what a benzylic carbon and a benzylic hydrogen are and that benzylic hydrogens are readily induced to undergo radical chain halogenation to give benzylic halides.
• You should know that benzyl halides can undergo SN1 reaction mechanisms because of the stability of the benzyl carbocation, but also SN2 mechanisms because the benzylic carbon is primary.
• You should know that the choice of mechanisms depends primarily upon the strength of the nucleophile. Weak nucleophiles give the SN1 mechanism preference, but good nucleophiles prefer the SN2 mechanism.

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