Ch 17 Flashcards
Why do benzylic halides undergo SN1 more easily than most alkyl halides?
The carbocation formed at the benzylic position is resonance-stabilized through the aromatic ring.
Why is the first step in Electrophilic Aromatic Substitution (EAS) highly endothermic?
The sigma complex formed does not possess aromaticity because the sp3-hybridized carbon interrupts the ring of p-orbitals.
What is a Lewis Acid? What is the mechanism that shows FeBr3 acting as a Lewis Acid with Br2, and why is FeBr3 used as a catalyst in the bromination of benzene?
A Lewis Acid is a compound that can accept electrons (AlBr3 and AlCl3 are also common Lewis Acids).
Bromine is not a strong enough electrophile (but Br+ is very electron-poor). FeBr3 forms a complex with Br2 that reacts like Br+ (it makes bromine a better electrophile).
What is the rate-limiting step in EAS?
Formation of the sigma complex is rate-limiting because it temporarily destroys aromaticity. The second step is exothermic because aromaticity is restored and a molecule of HX is formed.
What are the common reagents in nitration of a benzene? What type of reaction is nitration of a benzene?
Reagents: Nitric Acid and Sulfuric Acid
Nitration of a benzene is EAS
What happens when you treat nitrobenzene with Zn, Sn, or Fe in dilute acid (aqueous HCl)?
The nitro group is reduced to an amino group.
Even though it is uncharged, why is S03 a strong electrophile? What EAS reaction is it used for and what are the reagents?
Sulfur and oxygen do not share electron density equally (oxygen is more electronegative). Resonance structures of sulfur trioxide help show the positive charge predominately residing on S.
“Fuming sulfuric acid” (SO3 in H2SO4) is used to create sulfonated benzene.
Why is sulfonation one of the MOST important reactions in our aromatic “synthesis toolbag”?
It is REVERSIBLE!
- Concentrated fuming H2SO4 adds -SO3H to a benzene ring.
- -SO3H is a deactivator; therefore, meta-director.
- -SO3H can be removed from the ring by heating in dilute H2SO4 (steam distillation - excess water and heat removes SO3 by hydration into H2SO4).
- After using -SO3H to meta-direct a substituent to an otherwise tricky spot, take it off!
What two factors are considered when determining whether a benzene substituent is activating or deactivating? Which factor is stronger?
Induction and Resonance
Activation means the aromatic ring has more electron density, making it a better nucleophile (electron-rich), thus more reactive. Induction and resonance influence electron density, but resonance is usually a stronger factor than induction.
Example: -OH group is a strong activator. Oxygen is strongly electronegative and withdraws electron density from the ring (counterintuitive of a strong activator). However, nonbonding electrons on oxygen can spread negative charge throughout the ring through resonance.
What factor explains why alkyl groups are activators? What is the strength of their activation and why?
Alkyl groups are EDGs; they donate electron density through the sigma bond to the ring. Alkyl groups are weak activators because the increased electron density on the ring is only due to inductive stabilization. Resonance stabilization, the stronger effect, does not play a role (no pi-bonds in alkyl groups).
Why is a nitro group (-NO2) deactivating? What is its strength?
The nitro group is a strong deactivator because it is electron-withdrawing via resonance and via induction. Resonance structures spread a positive charge throughout the ring (rather than a negative charge, as seen with activators).
What are the directing effects of activators and deactivators? What is the one exception?
- All activators are ortho-para directors.
- All deactivators are meta directors.
- Halogens are weak deactivators, but they are ortho-para directors.
What are the TWO rules for predicting directing effects of multiple substituents?
- Ortho-para directors always beat meta directors.
- Strong activators always beat weak activators.
Remember: First rule trumps second rule. If you have a weak activator against a strong deactivator, the weak activator wins.
Why are the ortho and para sites favored in EAS when the activating substituent has a lone pair next to the ring (like methoxy -OCH3)?
Only in ortho and para attacks, does the sigma complex put a positve charge on the aromatic carbon that joins the substituent. Since the substituent has a lone pair, it can donate electron density through pi-donation and put the positive charge on the atom adjacent to the ring. Meta attack does not allow the LP on the substituent to participate via resonance-donation (pi-donation).
Why is an amine substituent an ortho, para director?
EAS attack at the ortho or para position allows nitrogen’s lone pair (nonbonding electrons) to provide resonance stabilization. A pi-bond with the adjacent carbon can delocalize the positive charge onto the nitrogen.
What do strong activators have in common?
Lone pair of electrons on the atom bonded to the aromatic ring.
It provides resonance stabilization to the sigma complex.
What are the three categories of activators? What is one common feature in each category that most species within the group share?
Strong Activators: Lone Pair of electrons on atom next to ring.
Moderate activators: Lone Pair on atom next to the ring, but LP also tied up in resonance outside of ring.
Weak Activators: Alkyl groups.
Take-away: the stronger activators are pi-donors (resonance effect stronger); moderate activators are also pi-donors but have to share their love outside of the ring too. Weak activators only contribute electron density through the inductive effect (sigma donors).
Why are meta-directors also referred to as “meta-allowing” substituents?
Meta-directors are always deactivators. The sigma complex formed by meta attack is the least unstable complex. Ortho and para attacks give rise to sigma complexes that are particularly unstable because the carbon joining the substituent can bear a positive charge. This puts a positive charge adjacent to an atom that either has a positive formal charge or partial positive charge (not a stable situation).
Remember a common feature of moderate and strong deactivating groups: the atom bonded to the ring has a positive formal charge or partial positive charge.
Why are alkyl groups ortho, para directors?
Ortho and para attacks on an alkylbenzene produce sigma complexes that form tertiary carbocations. This improved stability of the sigma complex increases the reactivity of benzene rings with alkyl substituents; therefore, they are weak activators (ortho, para directors).
Compare and contrast the ortho, para positions during “EAS with an activator” and “EAS with a deactivator”
The ortho and para positions are the most heavily influenced in EAS. A__ctivators are ortho, para-directors - they turn on the ortho and para positions. D__eactivators are meta-directors - they turn off the ortho and para positions, thus making the meta position the least unfavorable (meta-allowing).
Specifically, the ortho and para positions with an activator form a sigma complex that either 1) creates a tertiary carbocation (meta only creates secondary carbocations), or 2) places a positive charge on the carbon adjacent to a substituent atom with a lone pair (creating pi-donation and better resonance stabilization).
However, the ortho and para positions with a deactivator form sigma complexes that places a positive charge on a carbon that is adjacent to a substituent atom that either has a postive formal charge or a positive partial charge.
Why are halogens deactivating but also ortho, para-directors?
Bottom-line Up Front: for halogens, inductive effect trumps resonance effect making them weak deactivators, but resonance effect still directs o, p.
Halogens are strongly electronegative, so they withdraw electron density through the sigma bond with the benzene ring, making it a less effective nucleophile, deactivating.
However, halogens have nonbonding electrons that are capable of pi-bonding. When EAS occurs at the ortho or para positions, a positive charge in the sigma complex is shared by the carbon bearing the halogen. This allows for further delocalization of the positive charge onto the halogen. This is resonance stabilization through pi-donation, but halogens do not want a positive charge on them, so the resonance effect plays a smaller role.
Inductive effect alone is weakly deactivating, but for halogens it is strong enough to outcompete resonance (remember periodic trends for electronegativity).
What are the three categories of deactivators? In each category, what common feature do most species within the category share?
Weak Deactivators: Halogens
Moderate Deactivators: Atom next to ring has a pi-bond to an electronegative atom.
Strong Deactivators: Very powerful electron-wtihdrawing effects (either through resonance or induction)
Summarize the common features of strong, moderate, and weak activators, provide examples. Summarize the common features of strong, moderate, and weak deactivators, provide examples.
What do deactivating meta-directors have in common?
Positive charge or partial positive charge is present on the atom bonded to the aromatic ring.
Note: resonance forms help make this apparent.
How do you get a nitro group (-NO2) ortho to the methyl group on toluene without getting the para product?
Block the para position by first sulfonating the ring, then adding the nitro group, and then desulfonating the ring.
What type of reacton is Friedel-Crafts alkylation? What is the nucleophile, what is the electrophile, and how is the electrophile created?
- Friedel-Crafts alkylation is EAS.
- The aromatic ring is the nucleophile (a deactivated ring will not undergo Friedel-Crafts alkylation).
- The alkyl cation acts as the electrophile.
- Lewis Acid catalyst forms the alkyl cation electrophile, and the catalyst is regenerated during the last step (deprotonation). AlCl3 or FeBr3 often used.
What is the product of this reaction?
No reaction - nitrobenzene is a deactivating group.
What reagents are used in Friedel-Crafts acylation? How is the electrophile formed?
- Acyl halid is used as the electrophile.
- Lewis Acid catalyst forms the acyl cation electrophile, and the catalyst is regenerated during the last step (deprotonation). AlCl3 or FeBr3 often used.
Which substitution position dominates in Friedel-Crafts acylation when the aromatic substrate contains an ortho, para-director?
Para position prevails - steric effect partially blocks the ortho position.
How can we overcome possible rearrangments in carbocations when considering Friedel-Crafts alkylation to alkylbenzenes?
Don’t use it! Use Friedel-Crafts acylation, instead, then reduce the acylbenzene to the alkylbenzene via Clemmensen reduction.
Clemmensen reduction treats acylbenzenes with aqueous HCl and Zn(Hg).
What is the product?
Conditions for Clemmensen reduction are similar to conditions used for nitro reduction to amine. When reducing an acylbenzene to an alkylbenzene, nitro groups will also be reduced to amine groups.
What are the three criteria for Nucleophilic Aromatic Substitution (NAS)?
- There must be an EWG on the ring.
- There must be a LG on the ring.
- The LG must be ortho or para to the EWG.
Without strong resonance-withdrawing groups (like -NO2) in the ortho or para positions, formation of the negatively charged sigma complex is unlikely.
How do nitro groups (-NO2) in the ortho and para positions (relative to the LG) help stabilize the “sigma complex” in NAS addition-elimination?
Nitro groups in the ortho and para positions (relative to the LG) allow the sigma complex to further delocalize the negative charge onto the EWGs.
In other words, the negative charge is spread out over THREE carbon atoms and TWO oxygen atoms (in the figure). The ortho and para positions allow the ring (during a nucleophilic attack) to kick the negative charge up into “negative-charge-loving reservoirs”, increasing resonance-stabilization.
Note: if the LG is meta relative to the electron “reservoir”, there is no way to place the negative charge up in the reservoir; therefore, the intermediate is not stabilized.
Why is an acid added at the end of NAS (addition-elimination) when placing an -OH group on the aromatic ring?
The product contains a phenolic proton, which is mildly acidic and cannot survive the strongly basic conditions used (hydroxide is a strong base and is present in the reaction flask). Under basic conditions, the product is deprotonated. To regenerate the desired product, a proton source (acid) must be added to reprotonate the phenoxide ion, giving phenol.
Why is the benzyne mechanism also called elimination-addition?
The benzyne mechanism is essentially an elimination follwed by an addition.
Step 1: Nucleophile acts as a base to abstract a proton and expel the LG to form another “pi-bond” - benzyne. Elimination
Step 2: Nucleophile attacks either side of benzyne and adds to the ring. Addition
How do you determine which mechanism is operating: EAS, NAS: Addition-Elimination (aka SNAr), or NAS: Benzyne Mechanism (aka Elimination-Addition)?
Check the reagents:
EAS: Electrophile Adds to Smelly (aromatic ring). E+ in solution, check for a Lewis Acid catalyzing the formation of an E+.
NAS: Nucleophile Adds to Smelly. Nuc- in solution.
Remember the three criteria for NAS: Addition-Elimination:
- There must be an EWG on the ring.
- There must be an LG on the ring.
- The LG must be ortho or para to the EWG.
What type of reagent is this reaction creating? What is this reagent used for?
Gilman Reagent (lithium dialkylcuprate)
- Used to couple two groups together
- R group on Gilman reagent can couple with alkyl halides, vinyl halides (R2C==CRX), and aryl halides.
- Reactions of lithium dialkylcuprates with vinyl halides usually preserves the setereochemistry of the vinyl halide (trans stays trans, cis stays cis, E stays E, Z stays Z)
How is the reactivity of Gilman reagants (lithium dialkylcuprates R2CuLi) different from that of Grignard reagents?
Lithium dialkylcuprates are not as reactive towards carbonyl groups as Grignard reagents. Gilman reagents react much slower with ketones than with acid chlorides, so ketones can be formed from acid chlorides and the reaction can be stopped, preventing the Gilman reagent from reacting again with the ketone (because it is very slow).
What type of reaction is this? What is the product?
Coupling using Gilman reagent (reactants: alkyl halide and aryl cuprate)
What type of reaction is this? What is the product?
Coupling using Gilman reagent (reactants: vinyl halide and aryl cuprate)
stereochemistry preserved
What type of reaction is this? What is the product?
Coupling using Gilman reagent (reactants: aryl halide and vinyl cuprate)
What type of reaction is this? What is the product?
Coupling using Gilman reagent (reactants: acyl halide and aryl cuprate)
the ketone formed is preserved
What does the Birch reduction achieve? What reagents are typically used?
The Birch reduction converts benzene derivatives into cyclohexa-1,4-diene derivatives. The product loses aromaticity and is nonconjugated.
Birch reduction reagents form solvated electrons in ammonia solution that add to benzene, creating a radical anion. The anion abstracts a proton (2x) to reduce benzene.
Reagents:
- liquid ammonia (NH3)
- Na, Li, or K (solvated electron source)
- alcohol (typically ethanol or tert-butanol)
What are the two empirical rules for Birch reduction regioselectivity?
- EWGs stabilize carbanions, so reduction takes place on carbon atoms bearing EWGs.
- EDGs (or electron-releasing substituents) destablize carbanions, so reduction does not take place on carbon atoms bearing EDGs.
Another way to state the two rules:
- EDGs seek (end up on) the double bond.
- EWGs avoid the double bond.
What is the product?
What benzene functional groups are not affected by permanganate oxidation?
Permanganate oxidation is harsh, and only one carbon atom on the side chain survives (forming the carbon in the carbonyl of carboxylic acid).
Functional groups that typically survive:
- Nitro (-NO2)
- Halogens
- Carboxylic Acid (-COOH)
- Sulfonic Acid (-SO3H)
- Groups where the benzylic position is completely substituted (tert-butyl), i.e. no hydrogen attached to the carbon joining the aromatic ring.
Why does free-radical chloronation at the benzylic position produce the major product in free-radical chloronation of ethylbenzene?
The benzylic radical is resonance stablized.
The radical is too reactive to give entirely benzylic substitution, but it is favored.
Br2 is much cheaper than N-bromosuccinimide (NBS). Why is NBS used in allylic bromination, but it is usually fine to use Br2 in aromatic side-chain bromination at the benzylic position? When can the use of Br2 for side-chain bromination produce undesired products?
NBS is preferred in allylic bromination because Br2 in high concentrations chooses to add across the double bond, forming vicinal dibromides (not desired allylic bromide). NBS provides a constant low concentration of Br2.
When side-chain halogenation is desired, Br2 will not mess with the relatively unreactive benzene ring, so it will favor free-radical bromination at the benzylic position (this is a good thing). Remember, benzylic radicals are resonance stablized. However, powerful activating groups on the ring can sometime enable bromination of the benzene ring without at catalyst. In this instance, the preference for bromination at the benzylic position is lowered and regioselectivity is reduced. A mixture of products is likely.
Why do benzylic halides undergo SN1 reactions more easily than most alkyl halides?
The carbocation intermediate is resonance-stabilized.
Why are benzylic halides more reactive than primary alkyl halides in SN2 displacement?
The energy of the tranisition state is lower during SN2 reactions of benzyl halides; therefore, the reaction rate increases.
During SN2 displacement of the benzylic halide, the carbon p-orbital partially bonded with the Nuc- and LG overlaps with the pi-electrons of the aromatic ring, thus stabilizing the transition state.
Propose a synthesis pathway to put a nitrile on the methyl functional group.
The benzene ring does not have a strongly activating functional group, so bromination will take place exclusively at the benzylic position. Bromination on the ring would require a Lewis Acid catalyst (FeBr3).
- Free-radical brominate the benzylic carbon to add a LG.
- SN2 that sucka.
