Lecture 7

Question 1

How does the substitutant on substituted phenols, protonated anilines, and benzoic acid affect acidity?

Answer 1

Electron-donating substituents destabilize a base (i.e the base of phenol, protonated aniline, and benzoic acid) therefore, decrease the strength of its conjugate acid; electron-withdrawing substituents stabilize a base, which increases the strength of its conjugate acid. Remember: the stronger the acid, the more stable (weaker) its conjugate base.

An example is shown in digital notes.

Question 2

What are the different ways in which a substitutant can donate and withdraw electrons?

Answer 2

• Electron withdrawal:

1. Inductive Electron Withdrawal: If a substituent that is bonded to a benzene ring is more electron withdrawing than a hydrogen, then it will draw the sigm electrons away from the benzene ring more strongly than a hydrogen will. Withdrawal of electrons through a s bond is called inductive electron withdrawal. The +NH3 group is an example of a substituent that withdraws electrons inductively
2. Electron Withdrawal by Resonance: If a substituent is attached to a benzene ring by an atom that is doubly or triply bonded to a more electronegative atom, then the electrons of the ring can be delocalized onto the substituent; these substituents are said to withdraw electrons by resonance. Substituents such as C=O, C≡N, SO3H, and NO2 withdraw electrons by resonance. These substituents also withdraw electrons inductively because the atom attached to the benzene ring has a full or partial positive charge and is, therefore, more electronegative than a hydrogen. Example is shown in notes.

• Electron donation:

1. Electron Donation by Hyperconjugation: We saw that an alkyl substituent (such as CH3) stabilizes alkenes and carbocations by hyperconjugation—that is, by donating electrons to a p orbital.
2. If a substituent has a lone pair on the atom directly attached to a benzene ring, then the lone pair can be delocalized into the ring. These substituents are said to donate electrons by resonance. Substituents such as NH2, OH, OR, and Cl donate electrons by resonance. These substituents also withdraw electrons inductively because the atom attached to the benzene ring is more electronegative than hydrogen. Example shown in notes

Question 3

Important note

Answer 3

As we said in flashcard 2 atoms such as Cl can donate electrons by resonance but they also withdraw electrons inductively due to differences in electronegativity. Sometimes (such as in the case of Cl) the withdraw of electrons is greater than the donation, so the ring is overall stabilized and thus more acidic. This is not weird as we talked about it before and we should know more from the last lecture about which atoms are activating and which are deactivating.

Question 4

Why is this concept important to remember?

Answer 4

Okay, now we will start to talk about the main topic of this lecture, which is heterocyclic compounds. We will apply the knowledge learned above to analyze the acidity of a heterocyclic compound depending on the atom incorporated into the ring and the aromaticity (if applicable).

Question 5

What are Heterocyclic compounds?

Answer 5

Heterocyclic compounds (or heterocycles) are cyclic compounds in which one or more of the atoms of the ring are heteroatoms. In this chapter, we will consider the most prevalent heterocyclic compounds—the ones containing the heteroatom N, O, or S.

Question 6

What are some important notes about amines before we move forward?

Answer 6

■ The nitrogen in amines is sp3 hybridized with the lone pair residing in an sp3 orbital.

■ Amines invert rapidly at room temperature through a transition state in which the sp3 nitrogen becomes sp2 nitrogen. Shown in digital notes

■ The lone-pair electrons of the nitrogen atom cause amines to react as bases and as nucleophiles. Shown in digital notes

Question 7

How do we name nitrogen-, oxygen-, and sulfur-containing saturated heterocycles?

Answer 7

A saturated cyclic amine, can be named as a cycloalkane, using the prefix aza to denote the nitrogen atom, For example, azacyclopropane. Saturated heterocycles with oxygen and sulfur heteroatoms are named similarly. The prefix for oxygen is oxa and the prefix for sulfur is thia.

Note that heterocyclic rings are numbered to give the heteroatom the lowest possible number, For example, 2-methylazacyclohexane

Question 8

What are the relative acidities of ammonium and anilinium ions?

Answer 8

ammonium ions have pKa values of about 10 )nd anilinium ions have pKa values of about 5 (Sections 8.9 and 8.10). The greater acidity of anilinium ions compared with ammonium ions is due to the greater stability of the conjugate bases of anilinium ions as a result of electron delocalization and inductive electron withdrawal.

Question 9

How does the acidity of heterocyclic amines compare with acyclic amines?

Answer 9

Saturated amine heterocycles containing five or more atoms have physical and chemical properties like those of acyclic amines. This is shown in digital notes.

Question 10

What are the different aromatic aromatic heterocyclic compounds?

Answer 10

pyridine, pyrrole, furan, and thiophene. The structures are shown in written notes.

Question 11

What is the electronic structure of pyridine?

Answer 11

Pyridine is an aromatic heterocyclic compound. Each of the six ring atoms of pyridine is sp2 hybridized, which means that each has a p orbital, and the molecule contains three pairs of p electrons. Do not be confused by the lone-pair electrons on the nitrogen—they are not p electrons. Recall that lone-pair electrons are p electrons only if they can be used to form a p bond in the ring of a resonance contributor. Because nitrogen is sp2
hybridized, it has three sp2 orbitals and
a p orbital. The p orbital is used to form the p bond. Two of nitrogen’s sp2 orbitals overlap the sp2 orbitals of adjacent carbons, and its third sp2 orbital (which is perpendicular to nitrogen p orbital) contains the lone pair. The structure is shown in digital notes.

Question 12

What are the resonance contributors of pyridine?

Answer 12

Shown in digital notes

Question 13

What is the electronic structure of pyrrole?

Answer 13

The nitrogen atom of pyrrole is sp2
hybridized. Thus, it has three sp2 orbitals and a p orbital. It uses its three sp2 orbitals to bond to two carbons and one hydrogen. The lone-pair electrons are in the p orbital that overlaps the p orbitals of adjacent carbons. Pyrrole, therefore, has three pairs of p electrons and is aromatic. This is shown in digital notes

Question 14

What are the resonance contributors of pyrrole and what do they intell

Answer 14

The resonance contributors are shown in digital notes and they intell that the lone-pair electrons of pyrrole form a pi bond in the ring of a resonance contributor; thus, they are pi electrons.

Question 15

What is the electronic structure of furan and thiophene?

Answer 15

Similar to pyridine and pyrrole, furan and thiophene are aromatic compounds. Both the oxygen in furan and the sulfur in thiophene are sp2 hybridized and have one lone pair in an sp2 orbital. The orbital picture of furan in digital notes (under flashcard 13) shows that the second lone pair is in a p orbital that overlaps the p orbitals of adjacent carbons, forming a pi bond (as shown from the resonance contributors shown in digital notes). Thus, they are pi electrons.

Question 16

What are other heterocyclic aromatic compounds?

Answer 16

Shown in digital notes, maybe know imidazole by heart, but other than that just look at them long enough to be able to recognize them in the exam and recognize that they are aromatic.

Question 17

Break and review

Answer 17

Understanding the above heterocyclic aromatic compounds is very important, and memorizing their resonance contributors is even more important and they intell so much information about the compound (such as how they react) as will be seen in the upcoming flashcards

Question 18

Structure of how I’m going to split this

Answer 18

Okay, I find it best to start with the five-membered rings (Pyrrole, furan, and thiophene) and then go the the six-membered pyridine. With each will discuss their electron density map, acid-base properties, and relevant reactions ;).

Question 19

What is pyrrolidine?

Answer 19

A nonaromatic heterocyclic amine, structure is shown in the notes (under Flash card 20), it is very important!!

Question 19

What are the electron density maps of pyrrole and pyrrolidine?

Answer 19

Pyrrole has a dipole moment of 1.80 D and pyrrolidine has a slightly smaller dipole moment of 1.57 D, but as we see from the electrostatic potential maps shown in digital notes, the two dipole moments are in opposite directions. the dipole moment in pyrrolidine is due to inductive electron withdrawal by the nitrogen. Apparently, the ability of pyrrole’s nitrogen to donate electrons into the ring by resonance more than makes up for its inductive electron withdrawal. Note that a better understanding of pyrrole’s electron density map can be made from looking at its resonance contributor and resonance hybrid (flashcard 14)

Question 20

How do the delocalization energies of the aromatic five-membered heterocyclic compounds compare with benzene and cyclic enolate ion?

Answer 20

we saw that a compound’s delocalization energy increases as the resonance contributors become more stable and more nearly equivalent (3yani similar in structure). The delocalization energies of pyrrole, furan, and thiophene are not as great as the delocalization energies of benzene or the cyclopentadienyl anion, each a compound for which all the resonance contributors are equivalent.

Thiophene, with the least electronegative heteroatom, has the greatest delocalization energy of the three, and furan, with the most electronegative heteroatom, has the smallest delocalization energy. This is what we would expect, because the resonance contributors with a positive charge on the heteroatom are the most stable for the compound with the least electronegative heteroatom and the least stable for the compound with the most electronegative heteroatom*. Recall that the more stable the resonance contributors, the greater the delocalization energy. A summary is shown in digital notes.

• A Connection is made here, where we can assume that the resonance contributors of thiophene are the same as pyrrole and furan and that they have similar electron density maps and properties with differences in extremity (this is not always the case thou in the exam really think about it please)

Question 21

What are the acid-base properties of pyrrole?

Answer 21

Pyrrole is an extremely weak base because the electrons shown as a lone pair in the structure are part of the pi cloud. In other words, the nitrogen donates the lone pair into the five-membered ring. Therefore, protonating pyrrole destroys its aromaticity. As a result, the conjugate acid of pyrrole is a very strong acid (pKa = -3.8).

Question 22

Where is pyrrole protonated?

Answer 22

Building up on the previous flashcard, , pyrrole is protonated on a carbon rather than on the nitrogen since the resonance hybrid of pyrrole (flash card 14) shows that there is a partial positive charge on the nitrogen and a partial negative charge on each of the carbons. The carbon it attaches to is C-2 because a proton is an electrophile (as shown in digital notes), and we will later see that electrophiles attach preferentially to the C-2 position of pyrrole.

Question 23

Why is pyrrole unstable in strongly acidic solutions?

Answer 23

Pyrrole is unstable in strongly acidic solutions because, once protonated, it can readily polymerize as shown in digital notes.

Question 24

Which is more acidic pyrrole or pyrrolidine?

Answer 24

pyrrole due to the partial positive charge on its nitrogen atom (flashcard 14*) which makes it donate protons easier (acts as less of a nucleophile), and the sp2 nitrogen in pyrrole is more electronegative than the sp3 nitrogen in the saturated amine, contributing to pyrrole’s increased acidity, and finally, if it acts as a base it will lose its aromaticity as shown in the above flashcards!! * You can see how important resonance contributors are i keep referring to them

Question 25

what are the pKa Values of different Nitrogen-Containing Heterocycles?

Answer 25

Shown in digital notes. Note that pyrrolidine has a pka of 36 and that Indole is protonated at C-3, because protonation at C-2 would disturb the aromaticity of the benzene ring.

Question 26

What reaction do the aromatic five-membered heterocyclic compounds undergo?

Answer 26

Because pyrrole, furan, and thiophene are aromatic, they undergo electrophilic aromatic substitution reactions. Notice that the mechanism of the reaction is the same as the mechanism for electrophilic aromatic substitution of benzene as shown on the wall.

Question 27

Why is C-2 substitution preferred in electrophilic aromatic substitution?

Answer 27

Substitution occurs preferentially at C-2 because the intermediate obtained by adding a substituent to this position is more stable than the intermediate obtained by adding a substituent to C-3. Both intermediates have a relatively stable resonance contributor in which all the atoms (except H) have complete octets. The intermediate resulting from C-2 substitution of pyrrole has two additional resonance contributors, whereas the intermediate resulting from C-3 substitution has only one additional resonance contributor. This is shown in digital notes. If both positions adjacent to the heteroatom are occupied, electrophilic substitution will take place at C-3.

Question 28

What are the relative reactivities of the five-membered heterocyclic compounds?

Answer 28

Pyrrole, furan, and thiophene are all more reactive than benzene toward electrophilic aromatic substitution because a lone pair on the heteroatom can donate electrons into the ring by resonance. Therefore, they are more nucleophilic than benzene and they are better able to stabilize the positive charge on the intermediate. Furan is not as reactive as pyrrole in electrophilic aromatic substitution reactions. The oxygen of furan is more electronegative than the nitrogen of pyrrole, making oxygen less effective than nitrogen in increasing the nucleophilicity of the ring (less electron donation) and in stabilizing the carbocation (cant handle positive charge. Thiophene is less reactive than furan because the 3p orbital of sulfur overlaps less effectively than the 2p orbital of nitrogen or oxygen with the 2p orbital of carbon. The electrostatic potential maps in the notes (IMP) illustrate the different electron densities of the three rings. Again you can use the intuition about activating and deactivating substituted learned in the last lecture here!

Question 29

What is underway to measure the relative reactivities of the five-membered heterocyclic compounds?

Answer 29

The relative reactivities of pyrrole, furan, and thiophene are also reflected in the Lewis acid required to catalyze a Friedel–Crafts acylation as shown in digital notes. Try to attempt these mechanisms!! Also now we will move on to 6-membered rings (basically pyridine)

Question 30

What is the electron denisty map of pyridine?

Answer 30

The electron denisty map of pyridine is shown in digital notes. This is supported by the resonance contributors of pyridine (Shown under flashcard 12) and due to the dipole moment of the compound pointing in the direction of the electron-withdrawing Nitrogen (shown in digital notes). Again here you can make the connection with what was learned in the previous lecture. Notice how the electron density map shows the centre to be less electron-dense than benzene.

Question 31

What are the acid-base properties of pyridine?

Answer 31

The pyridinium ion is a stronger acid than a typical ammonium ion because its proton is attached to an sp2 nitrogen, which is more electronegative than an sp3 nitrogen. This is shown in digital notes It is important to note here that the bond made with hydrogen will not disrupt the aromaticity of pyridine, this is opposite to what was seen with pyrrole. This is because pyridine nitrogen has a lone pair of electrons in the sp^2 orbital that bonds with the hydrogen while the nitrogen in pyrrole has its only lone pair incorporated into the ring's pi electrons. Furthermore, looking into each ring electron density map (flashcard 25ii and flashcard 31i) it can be seen that the most nucleophilic position in pyridine is at the nitrogen while in pyrrole it is within the ring (explained by the respective resonance contributors), this explains why when pyrrole acts as a base it adds the hydrogen to one of its double bonds while pyridine just straight up adds it to the nitrogen. CONNECTIONS PEOPLE CONNECTIONS!!!

Question 32

What are the reactions pyridine is involved in?

Answer 32

1. Nucleophilic Reactions 2. Electrophilic Aromatic Substitution Reactions 3. Nucleophilic Aromatic Substitution Reactions 4. Reactions of Substituted Pyridines

Question 33

Pyridine and Nucleophilic Reactions

Answer 33

Pyridine undergoes reactions characteristic of tertiary amines (due to the available lone pair). For example, pyridine undergoes an SN2 reaction with an alkyl halide as shown in digital notes. Note that pyrrole cant act as a nucleophile for obvious reasons (BOHO)

Question 34

Pyridine and Electrophilic Aromatic Substitution Reactions

Answer 34

Because it is aromatic, pyridine (like benzene) undergoes electrophilic aromatic substitution reactions and by the same mechanism as shown on the wall.

Question 35

Why does electrophilic aromatic substitution of pyridine occur at the C-3 position

Answer 35

Electrophilic aromatic substitution of pyridine takes place at C-3 because the most stable intermediate is obtained by placing an electrophilic substituent at that position (Figure 19.2). When the substituent is placed at C-2 or C-4, one of the resulting resonance contributors is particularly unstable because its nitrogen atom has an incomplete octet and a positive charge. This is shown in digital notes.

Question 36

What is the relative reactivity of pyridine towards electrophilic aromatic substitution?

Answer 36

The electron-withdrawing nitrogen atom makes pyridine less nucleophilic than benzene and the intermediate obtained from electrophilic aromatic substitution of benzene. Pyridine, therefore, is less reactive than benzene. Indeed, it is even less reactive than nitrobenzene. (Recall from the last lecture that an electron-withdrawing nitro group strongly deactivates a benzene ring toward electrophilic aromatic substitution.) QUICK CONNECTION ALSO IN THE ELECTRON DENSITY MAP WE CAN SEE HOW DEACTIVATED THE RING IS. Pyridine, therefore, undergoes electrophilic aromatic substitution reactions only under vigorous conditions, and the yields of these reactions are often quite low. If the nitrogen becomes protonated under the reaction conditions, the reactivity decreases further because a positively charged nitrogen will make the PYRIDINE RING (NOT THE NIRTOGNE THE RING) less nucleophile and the carbocation intermediate even less stable. This is shown in digital notes. (Notice the relative temperatures of the reactions)

Question 37

Pyridine and nucleophilic aromatic substitution reactions?

Answer 37

Because pyridine is less reactive than benzene in electrophilic aromatic substitution reactions, it should not be surprising that pyridine is more reactive than benzene in nucleophilic aromatic substitution reactions. The electron-withdrawing nitrogen atom that destabilizes the intermediate in electrophilic aromatic substitution stabilizes the intermediate in nucleophilic aromatic substitution. The mechanism is shown on the wall (on the same paper as the electrophilic aromatic substitution).

Question 38

Why does nucleophilic aromatic substitution of pyridine take place at the C2 and C4 positions?

Answer 38

Nucleophilic aromatic substitution of pyridine takes place at C-2 or C-4 because addition to these positions leads to the most stable intermediate. Only when addition occurs to these positions is a resonance contributor obtained that has the greatest electron density on nitrogen (3yani the negative charge is on nitrogen), the most electronegative of the ring atoms. This is depicted in digital notes. Note that If the leaving groups at C-2 and C-4 are different, the incoming nucleophile will preferentially substitute for the weaker base (the better leaving group). THis is also depicted in digital notes ;).

Question 39

Reactions of Substituted Pyridines

Answer 39

Substituted pyridines undergo many of the side-chain reactions that substituted benzenes undergo: 1. alkyl-substituted pyridines can be brominated and oxidized. This is shown in digital notes 2. When an aminopyridine is diazotized, a pyridone is formed. The diazonium salt reacts immediately with water to form a hydroxypyridine, which is in equilibrium with its keto form. Digital notes 3. The electron-withdrawing nitrogen and the ability to delocalize the negative charge cause the hydrogens on carbons attached to the 2- and 4-positions of the pyridine ring to have about the same acidity as the hydrogens attached to the a-carbons of ketones This is shown on digital notes. Consequently, the hydrogens can be removed by a base, and the resulting carbanions can react as nucleophiles. This is shown on digital notes.

Question 40

What is everything we need to know about Imidazole? (The last important heterocyclic aromatic compound)

Answer 40

Imidazole, the heterocyclic ring of histidine, is an aromatic compound because it is cyclic and planar, every atom in the ring has a p orbital, and the p cloud contains three pairs of p electrons. The lone-pair electrons on N-1 are part of the p cloud because they are in a p orbital, whereas the lone-pair electrons on N-3 are not part of the p cloud because they are in an sp2 orbital that is perpendicular (KEY WORD REMEMBER TO MENTION WHEN APPROPRIATE) to the p orbitals as shown in digital notes. Because the lone-pair electrons in the sp2 orbital are not part of the p cloud, imidazole is protonated in acidic solutions. The conjugate acid of imidazole has a pKa = 6.8. Therefore, imidazole exists in both the protonated and unprotonated forms at physiological pH (7.4). This is one of the reasons that histidine, the imidazole-containing amino acid, is a catalytic component of many enzymes. Notice that both protonated imidazole and the imidazole anion have two equivalent resonance contributors. Thus, the two nitrogens become equivalent when imidazole is either protonated or deprotonated. This is shown in the notes.

Question 41

The retrosynthesis bit

Answer 41

Okay go to the notes and look its straight forward!! MEMORIZE LIKE YOUR NAME ;)