Lecture 6 Flashcards

1
Q

What is the easiest way to make nucleophilic carbon?

A

one way to create a nucleophilic carbon is to bond the carbon to a metal (M). This is because metals will always have lower negativities than carbon. This sis shown in digital notes.

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2
Q

What are organometallic compounds?

A

An organometallic compound is a compound that contains a carbon-metal bond. The name of an organometallic compound usually begins with the name of the alkyl group, followed by the name of the metal (E.x: butyllithium)

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3
Q

What are the two most common organometallic compounds?

A

The two most common organometallic compounds are organolithium and organomagnesium (Grigards regents).

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4
Q

How are organolithium metals prepared?

A

Organolithium compounds are prepared by adding lithium metal to an alkyl halide in a nonpolar solvent such as hexane.

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5
Q

How are Organomagnesium compounds prepared (we know)? What is the role of the solvent (new info)?

A

Organomagnesium compounds are prepared by adding an alkyl halide to magnesium metal shavings being stirred
in an ether—usually diethyl ether or tetrahydrofuran (THF)—under anhydrous conditions.

The solvent plays a crucial role in the formation of a Grignard reagent. The magnesium atom is surrounded by only four electrons, so it needs four more to form an octet. Solvent molecules
provide these electrons. Sharing electrons with a metal is called coordination. Coordination of the solvent molecules with the magnesium atom allows the Grignard reagent to dissolve in the solvent, preventing it from coating the magnesium shavings, which would make them unreactive. This is shown in digital notes

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6
Q

What are the different Alkyl halides used to make organolithium and organomagnesium compounds and which is most often used?

A

Alkyl halides, vinylic halides, and aryl halides can all be used to form organolithium and organomagnesium compounds. Alkyl bromides are used most often because they react more readily than alkyl chlorides and are less expensive than alkyl iodides.

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7
Q

Gaining some intuition from the slides

A

Okay, when you choose your alkyl halide you have to keep in mind that Grigard regents aren’t selective like organolithium compounds, y3ani as we saw before regards regents will react with C=O carbonyls so if you use an alkyl halide with a carbonyl group you will form a suicide molecule that always cyclizes! This is shown in digital notes.

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8
Q

What is an important condition of organolithium and organomagnesium compounds?

A

Due to Organomagnesium and organolithium compounds being very strong bases, they will react immediately with any acid present in the reaction mixture, even with very weak acids such as water and alcohols. When this happens, the organometallic compound is converted to an alkane.

This means that Grignard reagents and organolithium compounds cannot be prepared from compounds that contain acidic groups (such as OH, NH2, NHR, SH, C‚CH, or COOH). Because even trace amounts of moisture can convert an organometallic compound into an alkane, it is important that all reagents are dry when organometallic compounds are synthesized and when they react with other reagents. This is shown in digital notes.

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9
Q

What is an application for organometallic compounds given their high basic nature?

A

We can prepare deuterated hydrocarbons by washing the Organometallic compound with D2O. This will allow for the possibility of so much isotope analysis. This is shown in digital notes under flashcard 8

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10
Q

What is the organometallic compound reactivity dependent on?

A

The reactivity of an organometallic compound depends on the polarity of the carbon–metal bond: the greater the polarity of the bond, the more reactive the compound is as a nucleophile. The polarity of the bond depends on the difference in electronegativity between the metal and carbon. As seen in digital notes under Flash card one the most reactive carbon nucleophile is C-Li followed by C-Mg

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11
Q

What is transmetallation and when does an organometallic compound undergo it?

A

Transmetallation is a metal exchange reaction and an organometallic compound transmetallation if it is added to a
metal halide whose metal is more electronegative than the metal in the organometallic compound. In other words, transmetallation occurs if the alkyl group can be transferred to a metal with an electronegativity closer to that of carbon, thereby forming a less polar carbon–metal bond and, therefore, a less reactive nucleophile. (principle of lower reactivity applies)

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12
Q

What are coupling reactions?

A

They are reactions upon which CH-containing groups are joined (coupled) together.

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13
Q

What are the different coupling reactions that involve organometallic compounds?

A
  1. Grigards Rregents (Learnt)

Before the rest, it is important to note that new carbon-carbon bonds can be made using an organometallic reagent that has a transition metal as its metal atom:

  1. Gliman regents (Organocuprates) (Discussed next)
  2. Suski reaction (Discussed in this lecture)
  3. Heck reaction (Discussed in this lecture)
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14
Q

What are organocuprtes?

A

The first organometallic compounds used in coupling reactions were copper-containing organocuprates (R2CuLi), also called Gilman reagents. Organocuprates are less reactive than organolithium reagents or Grignard reagents because a carbon–copper bond is less polar than a carbon–lithium or carbon–magnesium bond—that is, Cu is closer in electronegativity to C. Only one of the two alkyl groups in an organocuprate is used as a nucleophile in reactions

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15
Q

How are organocuprates prepared?

A

An organocuprate is prepared by the reaction of an organolithium compound with cuprous iodide in diethyl ether or in THF. Notice that because Cu is more electronegative (1.8) than Li (1.0), transmetallation occurs. This is shown in digital notes.

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16
Q

How are organocuprates used in coupling reactions?

A

The organocuprate reacts by coupling one of its alkyl groups to the alkyl group of an alkyl halide (with the exception of alkyl fluorides, which do not undergo this reaction) and displacing the halogen. This means that an alkane can be formed from two alkyl halides—one alkyl halide is used to form the organocuprate, which then reacts with the second alkyl halide in a coupling reaction. The precise mechanism of the substitution reaction is unknown but is thought to involve radicals. The reaction is shown in digital notes

note that the R groups of the organocuprate and the alkyl halide can be primary alkyl, methyl, aryl, vinylic, or allylic. In other words, any R group except secondary or tertiary alkyl.

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17
Q

What is the “advantage” of using organocuprates?

A
  1. Because organocuprates can react with vinylic halides and aryl halides, they can be used to prepare compounds that cannot be prepared by SN2 reactions with Grignard reagents or organolithium compounds. (Remember that vinylic and aryl halides cannot undergo nucleophilic attack. This is shown in digital notes. Notice that The substitution reaction is stereospecific. In other words, the configuration of the double bond is retained in the product.
  2. Organocuprates can even replace halogens in compounds that contain other functional groups such as compounds with a bromo or chloro substituent attached to a carbon adjacent to a carbonyl group. This is a follow-up from the “initiation form slides” FC
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18
Q

What is the disadvantage of organocuprates?

A

Bad atom efficiency as shown in digital notes under flash card 16.

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19
Q

What are some final notes about organocuprates and some not so final notes about organometallic compounds?

A
  1. When doing retrosynthesis know that the a smaller alkyl chain on the organocuprate is favoured.
  2. Because organocuprates are nucleophiles, they react with electrophiles (no shit). For example (imp!!), in the reaction shown in digital notes), the organocuprate reacts with an epoxide in a nucleophilic substitution reaction to form an alcohol after washing with HCl
  3. tert-Butyllithium (𝑡-BuLi) is more reactive than n-butyllithium (n-BuLi) due to its steric bulk, stronger basicity, and less thermal stability (as tert has the negative carbon surrounded by 3 electron donating alkyl groups making it less stable)
  4. In all organometallic reactions a stoichiometric amount of metal is used which can be nice if a cheap metal but not so nice if an expensive metal
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20
Q

What is the Suzuki reaction?

A

The Suzuki reaction couples the R group of a vinylic or aryl halide with the R′ group of an organoboron compound in a basic solution in the presence of a palladium catalyst (PdL2). The general reaction is shown in the notes. Notice that the carbons that were bonded to the halogen and to the boron are joined by a new C-C bond.

The R′ group of the organoboron compound can be either an alkyl group, an alkenyl group, or an aryl group. When an alkenyl-organoboron compound is used, the new double bond in the product is always trans because an alkenyl-organoboron compound always has a trans configuration. This is shown in digital notes.

Note If the catalysts are modified, halides other than vinylic or aryl can be used

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21
Q

What are some notes about the Suzuki mechanism?

A

Okay before we start note that this mechanism is asked in the exam, and a nice FC coming up (“What are the several common features of the Suzuki and Heck reactions?”) shows almost everything, buttt i still want to highlight some things (the mech is also in notes nice to look at):

  • The mechanism for the Suzuki reaction is under active investigation; it has been found to depend on the ligand (L) and the base
  • Pd is an active species that allows R and Br to bind to it in OXIDATIVE ADDITION.
  • The base substitutes the Br
  • Finally, the formation of the B-OH bond is the driving force of this reaction.
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22
Q

What is oxidative addition and reductive elimination?

A

The insertion of a metal between two atoms is called oxidative addition—two new groups are added to the metal; it is an oxidation because the oxidation state of palladium is 0 in L2Pd and +2 in L2Pd(X)R. (Recall that oxidation is loss of electrons; reduction is gain of electrons.) Reductive
elimination eliminates two groups from the metal. Thus, the first step in a Suzuki reaction is an oxidative addition, and the last step is a reductive elimination.

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23
Q

How are Organoboron Compounds for Suzuki reaction prepared?

A

The alkyl- or alkenyl-organoboron compound used in a Suzuki reaction is prepared by hydroboration of a terminal alkene or a terminal alkyne, respectively. Often the boron-containing compound is catecholborane. The boron in an the alkenyl-organoboron compound is always trans to the substituent on the adjacent sp2
carbon (Recall that boron adds to the triply bonded carbon that is bonded to the hydrogen and that B and H add to the same side of the triple bond). This is shown in digital notes.

An aryl-organoboron compound is prepared from an organolithium compound and trimethylborate. also shown in digital notes

Note that these compounds are generally hard to make which is a disadvantage of the Suzuki reaction. (STILL BETTER THAN HECK)

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24
Q

What is The Heck reaction?

A

The Heck reaction couples a vinylic or an aryl halide with an alkene in the presence of a base (such as triethylamine) and a palladium catalyst (L2Pd). Like the Suzuki reaction, the Heck reaction is a substitution reaction: the R group of the halide replaces a vinylic hydrogen of an alkene. If there is a substituent attached to the alkene (Z), the R group will be trans to that substituent in the product. Notice in the reactions shown in digital notes that the new C¬C bond joins two sp2 carbons. When the reactant is a vinylic halide,the configuration of its double bond is retained in the product.

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25
Q

What is the Mechanism of Heck’s reaction?

A

Shown in digital notes

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26
Q

How do you obtain a high yield of a single product in a heck reaction?

A

The nucleophilic R group can add to either sp2 carbon of the alkene. Therefore, the reaction leads to a high yield of a single product only in the following situations:

■ if the alkene is symmetrical.

■ if one of the sp2 carbons is sterically hindered to the addition of the nucleophile. Therefore, internal alkenes are much less reactive than less sterically hindered terminal alkenes, so internal alkenes are generally not employed in Heck reactions.

■ if one of the sp2 carbons is bonded to a group that can withdraw electrons by resonance (such as C“O or C‚N), causing the other sp2 carbon to have a partial positive charge that makes it more susceptible to nucleophilic addition. This is shown in the digital notes

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27
Q

What are the several common features of the Suzuki and Heck reactions? long srry also take some things with a grain of salt

A

■ Both use a palladium catalyst (L2Pd)!!

■ Both reactions form a new C-C bond by replacing a halogen of a vinylic halide or an aryl halide (only vinylic and aryl halide explained why next FC) (bromides and iodides work best) with a carbon-containing group (R). Thus, they are substitution reactions. Shown in digital notes

■ The palladium atom of the catalyst is coordinated with ligands (L2Pd). Several different ligands (L) can be used; a common ligand is triphenylphosphine [P(C6H5)3].

■ Both reactions occur via a catalytic cycle in which the palladium atom participates in breaking bonds in the reactants and forming the new C¬C bond in the product (this might not fully be accurate).

■ The first step of each cycle is the insertion of palladium between the carbon and halogen to form an organopalladium compound. This is shown in digital notes

■ The reactions can be carried out if the reactants have other functional groups.

■ The reactions are stereospecific: the configuration of the double bond in a vinylic halide is retained in the product.

■ The reactions are very efficient, often requiring less than 0.1% of the palladium catalyst (compared to the reactants) and giving high yields of products (80–98%).

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28
Q

Why do Suzuki and Heck’s reactions only happen with vinylic or aryl halide? (Remember than if PdL2 is modified Suzuki can react with other compounds bas out of our scope)

A

If the organopalladium compound (firmed after the first step of the cycle after insertion of pd) has a b-hydrogen on an sp3 carbon, the organopalladium compound will rapidly undergo an elimination reaction before the coupling reaction has a chance to occur. This is shown in digital notes

This explains why vinylic and aryl halides are the reactants in these reactions—they cannot undergo an elimination reaction under the conditions used to carry out the coupling reactions.

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29
Q

Final notes about Suzuki and Heck?

A
  • Heck reaction needs to be carried out in very clean environments to maintain the activity and efficiency of the palladium catalyst cause yk Heck isn’t as good as Suzuki.
  • Suzuki can be used to make condensation-like polymers.
  • Heck can be used to make conjugated structures.
30
Q

What is alkene metathesis? (Discussed lecture 1 i think)

A

Alkene metathesis, also called olefin metathesis, is a reaction that breaks the strongest bond in an alkene (the double bond) and then rejoins the fragments. When the fragments are joined, each new double bond is formed between two sp2
carbons that were not previously bonded. this is shown in the notes.

31
Q

How can we obtain the best yeild for a single alkene in metathesis?

A

Terminal alkenes give the best yield of a single alkene product because one of the new alkene products is ethene. Ethene is a gas, so it is removed from the reaction mixture as it is formed. This shifts the equilibrium that exists between the two pathways for joining the fragments toward the pathway that forms ethene and the other new alkene.

32
Q

What is the catalyst used for metathesis?

A

Alkene metathesis requires a transition metal catalyst. If the catalyst does not affect other functional groups that are in the starting alkene, then a large variety of alkenes can be used for metathesis. Grubbs catalysts (in digital notes)—catalysts that contain ruthenium center—have been found to be the ones most tolerant of other functional groups. The ligands (L) in the Grubbs catalyst shown in digital notes are not identified because there are several generations of Grubbs catalysts, each with different ligands.

Note that the ligands determine the reactivity of metathesis.

33
Q

What are some examples of alkene metathesis?

A

DIGITAL NOTES READ CAREFULLY!!!

34
Q

What are the 5 most common electrophilic aromatic substitutions considered in this course?

A
  1. Halogenation: a bromine (Br), a chlorine (Cl), or an iodine (I) substitutes for hydrogen.
  2. Nitration: a nitro group (NO2) substitutes for hydrogen.
  3. Sulfonation: a sulfonic acid group (SO3H) substitutes for hydrogen.
  4. Friedel–Crafts acylation: an acyl group (RC=O) substitutes for hydrogen.
  5. Friedel–Crafts alkylation: an alkyl group (R) substitutes for hydrogen.
35
Q

What is the general mechanism for electrophilic aromatic substitutions?

A

It is shown in digital notes. steps are explained below:

Step 1: The electrophile (Y+) adds to the nucleophilic benzene ring, thereby forming a carbocation intermediate. The structure of the carbocation intermediate can be approximated by three resonance contributors. Notice this step is in Eq.

Step 2: A base in the reaction mixture (:B) removes a proton from the carbocation intermediate and the electrons that held the proton move into the ring to reestablish its aromaticity. Notice that the proton is removed from the carbon that has formed the bond with the electrophile. Notice this step isn’t in Eq.

The aromatic reaction we are going to go through only differs in how the Y is made, but the general mechanism stays the same.

36
Q

What is the reaction coordinate diagram for electrophilic aromatic substitution?

A

It is shown in digital notes. The first step is relatively slow and endergonic because an aromatic compound is being converted to a much less stable nonaromatic intermediate The second step is fast and strongly exergonic because this step restores the stability-enhancing aromaticity

37
Q

Why does the reaction of benzene with Br2 or Cl2 require a catalyst when the reaction of an alkene with these reagents does not require a catalyst?

A

Benzene’s aromaticity makes it much more
stable and, therefore, much less reactive than an alkene, so, benzene requires a better electrophile.

38
Q

What is the mechanism for bromination of benzene?

A

Found in the old notes

39
Q

What is the mechanism for chlorination of benzene?

A

Found in the old notes

40
Q

What is the problem with using ferric bromide and ferric chloride catalysts and how is it avoided?

A

The ferric bromide and ferric chloride catalysts react readily with moisture in the air during handling, which inactivates them. To avoid this problem, they are generated in situ (in the reaction mixture) from iron filings and bromine (or chlorine). As a result, the halogen component of the Lewis acid is the same as the halogen used in the substitution reaction. The reaction for which the catalyst are formed is shown in digital notes

41
Q

What is the mechanism for iodination?

A

Iodobenzene can be prepared using I2 and an oxidizing agent under acidic conditions. Hydrogen peroxide is commonly used as an oxidizing agent. The mechanism is shown in the old notes

42
Q

What is the mechanism for nitration?

A

Look at old notes

43
Q

What is the mechanism for sulfonation and desulfonation?

A

Look at old notes. Also note that the SO3H group is often used as “dummy” to guide regioselectivity in further substitution reactions of benzene, as a benzene ring can be desulfonated with relative ease

44
Q

What is the mechanism for the Friedel–Crafts acylation of benzene?

A

An acyl chloride or an acid anhydride is used as the source of the acyl group needed for a Friedel–Crafts acylation. The mechanism is in the old notes.

45
Q

What is the problem with the product of Friedel-crafts acylation?

A

The product of a Friedel–Crafts acylation contains a carbonyl group that can complex with, Friedel–Crafts acylations must be carried out with more than one equivalent of AlCl3. When the reaction is over, water is added to the reaction mixture to liberate the product from the complex. The reaction is shown in old notes

46
Q

What problem does the synthesis of benzaldehyde from benzene pose?

A

The synthesis of benzaldehyde from benzene poses a problem because formyl chloride, the acyl halide required for the reaction, is unstable and must be generated in situ. The Gatterman–Koch reaction uses a high-pressure mixture of carbon monoxide and HCl to generate formyl chloride and uses an aluminum chloride–cuprous chloride catalyst for the acylation. The reaction is shown in old notes

47
Q

What is the mechanism for Friedel-crafts alkylation?

A

It is shown in the old notes. From the intuition I gained if the alkylation leads to a secondary or tertiary carbon then it is Friedel-crafts alkylation, this is because we encounter a lot of other coupling reactions such as the ones discussed in this lecture and also because of the ways to reduce the Friedel-crafts acylation product to from alkyl benzene

48
Q

What are the ways to alkylate benzene using acylation-reduction reactions?

A

Look at old notes. I love myself so much for this. (Look at reading the notes) An extra note from the slides is that the Wolff-Kishner and Clemmensen reduction is used because the reduction of CO with H2/Pd-C is extremely slow and requires very high H2 pressure. But at the end of the day, choice depends on (1) Safety (toxicity, pressure) (2) Ease of work up (3) Costs

49
Q

How can some substituents on the benzene ring be changed chemically?

A

Benzene rings with substituents other than those placed on a benzene ring by the five reactions listed above can be prepared by first synthesizing one of those substituted benzenes and then chemically changing the substituent. Several of these reactions will be familiar to you. I decided to split them into 2 categories:

  1. Substitution and elimination reactions
  2. Oxidation and reduction reactions
50
Q

How can we change the substituents on the benzene ring chemically using substitution and elimination reactions?

A

We can use a radical substitution reaction to substitute a bromine for a benzylic hydrogen (the reaction is shown in digital notes). Once a halogen has been placed in the benzylic position, it can be replaced by a nucleophile via an SN2 or SN1 reaction. A wide variety of substituted benzenes can be prepared this way (The reaction is shown in notes).

Remember that alkyl halides can also undergo elimination reactions. In some reaction, the elimination product predominates because the double bond is stabilized by conjugation with the benzene ring. This is also shown in digital notes

51
Q

How can we change the substituents on the benzene ring chemically using oxidation and reduction reactions?

A

2 types of reactions are mentioned:

  1. catalytic hydrogenation: Substituents with double and triple bonds can be reduced by catalytic hydrogenation. Because benzene is an unusually stable compound, it can be reduced only at high temperature and pressure. Therefore, only the substituents are reduced. IMP!! A nitro substituent can be reduced to an amino substituent. Catalytic hydrogenation is commonly used to carry out this reaction
  2. Oxidation using chromic acid: An alkyl group attached to a benzene ring can be oxidized to a carboxyl group by an oxidizing
    agent such as chromic acid. Notice that the benzene ring is too stable to be oxidized—only the alkyl group is oxidized. Regardless of the length of the alkyl substituent, it is oxidized to a COOH group, provided that a
    hydrogen is bonded to the benzylic carbon. This is shown in digital notes. If the alkyl group lacks a benzylic hydrogen, the oxidation reaction will not occur because the first step in the oxidation reaction is removal of a hydrogen from the benzylic carbon.
52
Q

Okay break a moment to check we are on the same page

A

Okay, this was a fast recap, I hope I covered everything. You will find this amazing mind map in the digital notes hat shows everything you need to know!!! Look at it, do mechanisms you arent sure of and memorize it like your LIFE depends on it. Everything you have learned. Recognize that the two reactions above the red highlighted one are Suzuki and reaction with organocuprates, make sure you understand them!!! Okay now we go to the ortho para meta shiz ;)

53
Q

Everything you need to know about para ortho shiz

A

literally, go to lecture 6 of organic 1 Alhamdulillah i finally have my notes showing me in real life how important they are!!! I will maybe add some notes i think i messed in the next FC

54
Q

What are the rules for synthesizing trisubstituted benzenes?

A

When a disubstituted benzene undergoes an electrophilic aromatic substitution reaction, the directing effects of both substituents have to be considered:

  • If both substituents direct the incoming
    substituent to the same position (only one), the product of the reaction is easily predicted.
  • If the substitutes both activate the same positions (multiple) the major one will be the least sterically hindered one
  • If the two substituents direct the new substituent to different positions, a strongly activating substituent will win out over a weakly activating substituent or a deactivating substituent
  • If the two substituents have similar activating properties, neither will dominate and a mixture of products will be obtained.
55
Q

Break moment

A

Sorry for so many breaks. Okay, now we are going to look into how we can consider the Benzene can act as an electrophile. This is basically Diazonium salts for Sandmeyer reactions and aryl halides with strongly withdrawing electrons. First, we consider Arenediasonium salts.

56
Q

How can we use arenediazonium salts?

A

Okay so we learnt only how to place a limited number of different substituents on a benzene ring but by using arenediazonium salts The list of substituents that can be placed on a benzene ring can be expanded. The structure is shown in digital notes!

57
Q

why do arenediazonium salts have the ability to expand the list of substituents that can be placed on a benzene ring?

A

This is because Displacing the leaving group of a diazonium ion by a nucleophile occurs readily as the leaving group is nitrogen gas (indicated by an upward pointing arrow in digital notes), which is very stable.

note that Some displacements involve phenyl cations, whereas others involve radicals—the actual mechanism depends on the particular nucleophile.

58
Q

How can we form arenediazonium salts?

A

Aniline can be converted to an arenediazonium salt by treatment with nitrous acid (HNO2). Because nitrous acid is unstable, it must be formed in situ, using an aqueous solution of sodium nitrite and an acid. N2 is such a good leaving group that the diazonium salt must be synthesized at 0 °C and used immediately without isolation. The mechanism for this reaction is shown in written notes.

59
Q

Now the important question is, Why do arenediazonium salts make benzene act as an electrophile?

A

Simple, as already mentioned the nitrogen is such a good leaving group also I’m pretty sure it deactivates the benzene ring a lot, this forces the benzene to react as an electrophile instead of a nucleophile.

60
Q

What are Sandmeyer’s reactions?

A

They are found in old notes, start with the loose paper, then go to the iodo subsistent etc etc ;)

61
Q

Replacing a Diazonium Group with a Hydrogen?

A

Found in old notes ;) Same title sooo

62
Q

What other reactions can arenediazonium be used in?

A

In addition to being used to synthesize substituted benzene via nucleophilic attack from other species, arenediazonium ions can be used as electrophiles in electrophilic aromatic substitution reactions (that is when the aromatic ring is the nucleophile). Because an arenediazonium ion is unstable at room temperature, it can be used as an electrophile only in reactions that can be carried out well below room temperature. In other words, only highly activated benzene rings (phenols, anilines, and N-alkylanilines) can undergo electrophilic aromatic substitution reactions with arenediazonium ion electrophiles. ‘The product of the reaction is an azobenzene. The N=N linkage is called an azo linkage.

63
Q

Where does the arenediazonium ion get placed in electrophilic aromatic substitution reactions?

A

Okay, first we need to note that the benzene ring is activated so it will be ortho-para directing. Because the electrophile (arenediazonium ion) is so large, substitution takes place preferentially at the less sterically hindered para position. However, if the para position already has a substituent, then the substitution will
occur at an ortho position.

64
Q

What is the mechanism for electrophilic aromatic substitution reaction with arenediazonium electrophile?

A

Look at the wall.

65
Q

WALL TOLD ME TO COME HERE

A

Diazobenzenes are

  • extremely important as dyes
  • Switching elements to change material properties by light
  • Important in photopharmacology: Prontosil is a prodrug, a compound that becomes an active drug only after it undergoes a reaction in the body. Used to treat bacterial infections. Cis and trans one is active. Shown in digital notes
66
Q

Okay to make sure we are on the same page

A

Okay, so arenediazonium ion makes the benzene act as an electrophile due to the very good leaving group it has so it undergoes a different substitution reaction where the N2 group is removed. Also, the arenediazonium ion itself can act as an electrophile but only with an activated benzene ring and low temperatures.

67
Q

How can an aryl halide make the benzene act as an electrophile?

A

We saw that aryl halides do not react with nucleophiles because the p electron cloud repels the approach of a nucleophile. If, however, the aryl halide has one or more substituents that strongly withdraw electrons from the ring by resonance, a nucleophilic aromatic substitution reaction can take place. The electron-withdrawing groups must be positioned ortho or para to the halogen. The greater the number of
electron-withdrawing substituents ortho and para to the halogen, the more easily the nucleophilic aromatic substitution reaction occurs. Notice the conditions required for each reaction found in the digital notes

68
Q

Why must the electron-withdrawing substituent be in the ortho or para position?

A

The electron-withdrawing substituent must be ortho or para to the site of nucleophilic attack because the electrons of the attacking nucleophile can be delocalized onto the substituent only if the substituent is in one of those positions. This is shown in digital notes and in the upcoming mechanism

69
Q

What is the mechanism for nucleophilic aromatic substitution?

A

Nucleophilic aromatic substitution takes place by a two-step mechanism—nucleophilic addition, followed by elimination. The reaction is called an SNAr reaction (substitution nucleophilic aromatic). It is shown on the wall.

70
Q

What are some examples of nucleophilic aromatic substitution?

A

A variety of substituents can be placed on a benzene ring by means of a nucleophilic aromatic substitution reaction. The only requirement is that the incoming group be a stronger base than the group that is being replaced. Examples are shown in digital notes.

71
Q

Imp note

A

An electron-withdrawing substituent (Like NO2) increases the reactivity of the benzene ring toward nucleophilic substitution and decreases the reactivity
of the benzene ring toward electrophilic substitution.

72
Q

Final notes

A
  • Okay so the diazonium Group can act as dummy molecules!!! Similar to how the SO3H group does. This is because the Diazonium Group can easily be replaced with hydrogen when washing with H3PO2 as we have seen already in the previous flashcards.
  • Also look at Sanger’s regent in digital notes. Enjoy holidayssss you better be doing these in Jordan.