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Substitution Reactions of Aromatic Compounds Simple alkenes tend to undergo addition reactions:

The elements of the reagent (HBr or Br2) are simply added to the starting material. This is called, unsurprisingly, an addition reaction. Aromatic compounds do not react in this manner; and it usually takes a catalyst to initiate reaction with halogens:

What we have done in this case is substitute a bromine atom for a hydrogen atom. Hence the reaction is termed an “aromatic substitution.” Because the benzene ring is quite electron-rich, it almost always behaves as the nucleophile in a reaction - which means that the substitution on benzene occurs by the addition of an electrophile to benzene; thus, the reactions are termed “electrophilic aromatic substitution”:

There is basically one simple mechanism for all electrophilic aromatic substitutions:

The benzene acts as a nucleophile, attacking the electrophile with a pair of its π-electrons. This initial step destroys the aromaticity of the molecule! The resulting positive charge is delocalized over the ortho and para positions. The conjugate base of the initial electrophile then assists in removing the now extraneous proton, and restores aromaticity. Note that addition of the conjugate base to the cyclohexadienyl cation (an addition reaction) does not occur because the addition reaction is much less exothermic than the rearomatization reaction and cannot compete with it. Because all electrophilic aromatic substitutions proceed in this way, the only thing that matters is the preparation of a “reactive’ electrophile. Why a “reactive” electrophile? As you can see, the first step of the reaction involves destroying aromaticity. In order to do this, there must be a

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significant energetic driving force. This driving force comes in the form of a very reactive (unhappy) electrophile. How are such electrophiles generated? Halogenation As you can imagine, halogens bearing a positive charge are particularly reactive. We will focus on preparing halogen electrophiles from Br, Cl and I. Bromine: Allowing bromine to react with iron metal first generates FeBr3, which then interacts with the remaining Br2 to form a highly Lewis acidic system that is capable of reacting with weakly nucleophilic aromatic compounds:

The reaction proceeds by the mechanism shown below to give brominated benzene:

Chlorine: The same chemistry shown for bromine also works with chlorine togenerate “Cl+.” A mixture of benzene, chlorine and iron(III)chloride yields the chlorobenzene:

Iodine: It is a little more difficult to make iodine sufficiently electrophilic. For relatively activated compounds, where a mild source of “I+” is required, copper salts are often used as a catalyst:

Reaction then proceeds by the standard mechanism, with I+ as the electrophile, to give iodinated benzenes. What good are aromatic halides? The halogens are excellent synthetic “handles” – they can be easily converted into other functional groups. For

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example, bromobenzenes can be turned into Grignard reagents, and then reacted with aldehydes, ketones, etc....

Nitration We can also make a highly electrophilic form of NO2 (from HNO3/H2SO4), the nitronium ion:

Which can then react with aromatic compounds via the standard mechanism to give nitrated aromatics...

Why nitrate aromatics? There are a couple of good reasons to nitrate things. The first is in the manufacture of explosives – highly nitrated organic molecules are frequently used as explosives (trinitrotoluene (TNT), nitroglycerine, etc.). The second reason is that nitro-groups are generally easy to reduce to amines. And since it is nearly impossible to make an amine electrophilic (in order to add it to an aromatic ring under electrophilic aromatic substitution conditions), aromatic nitro compounds are about the only precursors to aromatic amines:

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Sulfonation Just as with nitration, it is easy to make a solution of highly electrophilic SO3

(SO3/H2SO4). This solution can be used to sulfonate aromatic compounds:

Why is this a useful reaction? First of all, the reaction is REVERSIBLE, especially at high temperature! Cook it up in hot AQUEOUS acid, and the SO3

group falls right off again. Second, aromatic sulfonic acids were used as the first antibiotics - the so-called “sulfa” drugs, such as sulfanilamide.

Friedel-Crafts Reactions: This reaction comes in two flavors - alkylation and acylation. Alkylation first: The basic premise of this reaction is the electrophilic addition of alkyl groups to an aromatic ring. The general scheme:

whether a true carbocation is formed depends on the nature of the R group A simple example:

There are a number of drawbacks to this reaction: 1) Does not work at all on aromatic rings with de-activating groups (nitro, any carbonyl, cyano) attached. 2) Because alkyl groups are activating, over-alkylation is a significant problem. 3) Because a carbocationic intermediate is involved, rearrangements tend to take place. For example,

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However, Friedel & Crafts came up with another reaction without so many drawbacks. The Friedel-Crafts acylation generally proceeds without complication (through an acylium ion):

For example:

Note: the only significant restriction is that we still can’t have any deactivating groups on the ring. Over-acylation is not a problem because an acyl group is de-activating. Therefore only one acyl group can add! There is also no problem with rearrangements. This is a very efficient reaction! Substituent Effects Most of what we’ve talked about so-far has involve the addition of a compound to an unsubstituted benzene ring. What of additions to a substituted ring? Let’s summarize: 1. Activating groups Because the benzene ring’s electrons are acting as the nucleophile in all of the above reactions, rings substituted with strong electron-donating groups

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(particularly π-electron donating) are considered “activated” - they often will react without a catalyst! Some examples of activating groups are: OH, OR, NH2, NR2, Alkyl. Oxygen and nitrogen-based activating groups increase reactivity by a resonance effect:

As you can see, the very nature of the activation requires ortho-para direction! Alkyl groups work somewhat differently. They are not as strong at activating the ring, and their main function is stabilizing the positive charge formed after the attack on the electrophile:

2. The Halogens: The halogens are in a class by themselves, and behave in an unusual manner. They are VERY electron withdrawing, and thus DE-ACTIVATE the ring towards electrophilic substitution. However, their multiple lone-pairs are able to stabilize the cation formed after electrophilic addition, and thus direct ortho-para:

3. Meta-Directing De-activators. Strongly electron-withdrawing groups de-activate the ring towards electrophilic substitution. Examples of such groups are: Ester (COOR), Acid (COOH), Aldehyde (CHO), Nitro (NO2), Ketone (or acyl, COR), Cyano (CN). However, with a strong enough electrophile, the rings often will react.

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However, in order to avoid putting the charge that develops (after nucleophilic attack on an electrophile) on the carbon attached to the electron-withdrawing group, the incoming electrophile must attach to the meta carbons (see below). Note that the charge cannot be delocalized onto the carbon containing the acyl group (also note - the charge delocalizes over all of the ortho and para positions!)

Other considerations: 1. Sterics is particularly a problem with bulky ortho-para directors - the ortho positions are blocked, forcing addition to the para positions:

Multi-functional compounds: READ Sections 18.8-12 carefully What if the ring has several substituents? Your text has a rather detailed description of what to do, so here I’ll just supplement that data. 1) If the effects of two substituents point to reactivity at one particular carbon, you’re in luck! That’s where the electrophile will go (providing the sterics are not horrible). 2) The strongest directing group almost always prevails (e.g. OH over alkyl, NR2 over Br, etc). The order of precedence is: a) Strong o,p directors (OR, NR2) b) Alkyl groups and halogen c) All meta-directors. 4) And of course, keep a close eye on the steric environment. Birch reduction. a) Li, liquid NH3 (or Na/NH3) converts benzene into 1,4-cyclohexadiene:

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Na/ liq. NH3

EtOH•

••

••

•

etc

EtO-H

•

H H

Na

H H

••

etc

etcEtO-H

H H

HH b) Alkyl and alkoxy groups prefer to end up on the C(sp2) of the cyclohexadiene product. c) CO2H group prefers to end up on the C(sp3) of the cyclohexadiene product. Reactions on the Aromatic Side Chain: Alkyl groups attached to a benzene ring are surprisingly reactive. The hydrogens on a carbon atom attached to the benzene ring are called benzylic. Notice a similarity to allylic systems?

Benzylic hydrogens show a reactivity similar to allylic hydrogens: They can be brominated with NBS, for example:

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Aromatic compounds with benzylic hydrogens are also susceptible to oxidative degradation: Potassium permanganate (KMnO4) decomposes these to the benzoic acid (caution -you loose the WHOLE side-chain on this one!):

Note - if there are NO benzylic hydrogens (last example), the benzylic position is not oxidized (in this case, only the alcohol on the side chain gets oxidized. Hydrogenation: With a sufficiently active catalyst, it is possible to hydrogenate the aromatic system to a completely saturated system. Because aromatic chemistry is so rich & diverse, this is one method to prepare highly substituted cyclohexanes:

It is also extremely easy to hydrogenate a side-chain. As a simple example, take ethyl cinnamate:

Along with hydrogenation, we can also do a different kind of reduction – a

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deoxygenation! This reaction ONLY works on Aryl Ketones (R ≠ H), and is a PERFECT way to turn a Friedel-Crafts Acylation into an Alkylation:

NOTE ALSO THAT a Clemmenson reduction using Zn(Hg), HCl or Wolf-Kishner reduction using N2H4, KOH, will accomplished the same thing.

End of Material for Exam 2 Palladium-mediated Reactions of Aromatic Compounds Palladium chemistry is dominated by two oxidation states 0 and +2. Typical Pd(0) sources include:

Heck reaction

The Heck Reaction is one of the most synthetically useful palladium-catalyzed reactions and is very efficient; it couples together a halide or triflate compound with an alkene to form a new alkene. Pd(0) is electron rich and will undergo what is termed “oxidative addition” reaction with suitable R’–X substrates, such as aryl halides (Ar–X) or triflates (Ar–OTf; TfO- = CF3SO3-), to form a Pd(II) complex. Note that in this reaction R’-X can be aryl, vinyl, or any alkyl group without β hydrogens on an sp3

carbon (we’ll see why in a moment); and X = halide or triflate. The Pd-C sigma bond is very reactive, especially toward carbon-carbon π bonds:

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Note that the choice of (R’X) substrates is limited to aryl, heteroaryl, vinylic, and benzylic halides or triflates. This is because the presence of an sp3

carbon in the β position carrying a hydrogen rapidly results in β-hydride elimination. The alkene can be mono- or disubstituted and can be electron rich, -poor, or neutral; the base need not be strong and can be Et3N, NaOAc, or aqueous Na2CO3. For example,

Electron withdrawing groups such as esters promote excellent regioselectivity in favor of terminal attack:

In contrast, electron donating groups, such as ethers, lead to attack at the end of the alkene substituted by oxygen to produce in this case the 1,1- disubstituted product

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By the way, Pd(0) is generated in-situ in the above reaction by:

Another common route to Pd(0) is:

Et3N + PdL2Cl2NEt2

PdL2Cl

H

Cl

HPdL2Cl

Pd0L2

ligand exchange

!-H elimination

reductive elimination

More Metal-Catalyzed Cross-Coupling Reactions i) C–C bond-forming reactions The overall reaction is:

R1–X: R1 must not have β-hydrogens that can eliminate hence aryl, vinyl, allyl, benzyl, and polyfluoroalkyl halides and triflates are typical. R2–M: R2

can be almost anything including examples with β-hydrogens; this is because reductive coupling is faster than β-H elimination. M = MgX, ZnX, Cu, SnR2, AlMe2, B(OR)2, etc.

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R1–X

a) ArCl + RMgCl + cat. (Ph3P)2NiCl2 → ArR (Kumada coupling) b) ArX + RM + cat. Pd(0) + phosphine ligand → ArR (M = SnBu3 or SnMe3, B(OH)2, ZnCl) (Stille, Suzuki, Negishi couplings) c) ArX + CO + MeOH + base + cat. Pd(0) + phosphine ligand → ArCO2Me d) ArX + RC≡CH + base + cat. CuI + cat. Pd(0) + phosphine ligand → ArC≡CR (Sonogashira coupling). In general, use C(sp2) with B(OH)2, C(sp2) or C(sp3) with Sn or Zn. C(sp3) can be primary, secondary, or tertiary — no rearrangements, unlike Friedel–Crafts chemistry! These are the most widely used methods these days for attaching C(sp3) and C(sp2) to aromatic rings. Disconnection: Ar–R ⇒ Ar–Br + R–M. (1) If R is C≡C, M = H is best. (2) If R is C=C, M = B(OH)2 is best. (3) If R is C–C, M = ZnCl or SnMe3 is best. Friedel–Crafts alkylation is also an option (but watch for rearrangements when working forward!). (4) If in doubt, use M = SnMe3, which works with all kinds of C. (5) If R is C=O, use Friedel–Crafts acylation! more on Stille Coupling Reaction the moist widely used cross-coupling reaction. For example,

- the Stille reaction maybe coupled with cabonylation. For example, when the normal Stille reaction is carried out in a CO saturated solution, under 1 atm CO pressure, excellent yield of the carbonylated product can be obtained. For example,

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ii) C–N and C–O bond-forming reactions (Buchwald–Hartwig couplings) a) ArX + ROH + base + cat. Pd(0) + phosphine ligand → ArOR b) ArX + R2NH + base + cat. Pd(0) + phosphine ligand → ArNR2

Note: Cannot do these reactions by SN2 or SN1! Now, we have many ways to make ArCO2H! a) ArH + X2 + Lewis acid → ArX + Mg → ArMgX + CO2 → ArCO2H. b) ArH + X2 + Lewis acid → ArX + CuCN → ArCN + aq. NaOH → ArCO2H. c) ArH + X2 + Lewis acid → ArX + CO + MeOH + cat. Pd(0) + phosphine ligand → ArCO2Me ArCOOH (ester hydrolysis). d) ArH + CH3Cl + AlCl3 → ArCH3 + KMnO4 → ArCO2H.