Lecture 3

Question 1

What are the structures of the 20 most common naturally occurring amino acids?

Answer 1

Table 21.2.

We can instantly see that:

• Amino acids differ only in the substituent (R) that is attached to the a-carbon. The wide variation in these substituents (called side chains) is what gives proteins their great structural diversity and, as a consequence, their great functional diversity.
• All amino acids—except proline which contains a secondary amino group incorporated into a five-membered ring—contain a primary amino group.

Finally to better understand the amino acids they have been divided into classes which will be discussed in the upcoming flashcards.

Question 2

What are Imidazole and indole

Answer 2

They are two aromatic compounds whose structures are shown in the notes. They are worth mentioning as Histidine is an imidazole-substituted alanine and Tryptophan is an indole-substituted alanine. It is worth noting that since the lone pair on the nitrogen of indole is needed for the compound’s aromaticity, indole is a very weak base. Therefore, the ring nitrogen in tryptophan is never protonated under physiological conditions.

REMINDER OF AROMACITY CONDITIONS:
1. Cyclic and planar
2. Each of its ring atoms has a p orbital
3. Follows Hickls law 4n+2 (where n is any non zero number)

Question 3

What are the essential amino acids?

Answer 3

They are the amino acids that humans must obtain from their diet because they either cannot synthesize them at all or they cannot synthesize them in adequate amounts. They are represented by an * in Table 21.2

Question 4

What is the stereochemistry of amino acids?

Answer 4

The a-carbon of all the naturally occurring amino acids (except glycine) is an asymmetric centre. Therefore, 19 of the 20 amino acids listed in Table 21.2 can exist as enantiomers with the d and l notation used for monosaccharides also being used for amino acids. All 19 amino acids with chiral centres have an S configuration EXCEPT CYCTINE.

Question 5

How are amino acids generally represented in Ficher projection?

Answer 5

Usually, we have our carboxylic acid group on the top the R group on the bottom, and our amino group on the vertical axis, where if the amino group is on the right we would have a D-amino acid, and if it was on the left we would have a L-amino acid.

Question 6

Which is more abundant in nature L- or D- amino acids?

Answer 6

Unlike monosaccharides, where
the D isomer is the one found in nature, most amino acids found in nature have the L configuration. To date, d-amino acids have been found only in a few peptide antibiotics and in some small peptides
attached to the cell walls of bacteria.

Question 7

Why D-sugars and L-amino acids?

Answer 7

Although it made no difference which isomer nature “selected” to be synthesized, it was important that only one was selected. For example, proteins that contain both D- and L-amino acids do not fold properly, and without proper folding, there can be no catalysis. It was also important that the same isomer was synthesized by all organisms. For
example, because mammals have L-amino acids, L-amino acids must be the isomers synthesized by the organisms that mammals depend on for food.

it is important to mention that in medical applications it was found that the efficiency and safety of the L-amino acid was way greater than the D-amino acid (The D-amino acid quite literally almost caused cancer).

Question 8

What is the relationship introduced between the pH of a solution and the pKa of a compound?

Answer 8

compounds exist primarily in their protonated form (acidic) in solutions that are more acidic than their pKa values (pH < pKa), and primarily in the deprotonated form (basic) in solutions that are more basic than their pKa values (pH > pKa).

Question 9

Applying the relation between pH and pKa, how does an amino acid look like at different pHs?

Answer 9

(this is wrong trust your gut)

The carboxyl groups of the amino acids have pKa values of approximately 2, and the protonated amino groups have pKa values near 9. Both groups, therefore, are deprotonated (acidic form) in a very acidic solution (pH ∼ 0).

At pH = 7, the pH of the solution is greater than the pKa of the carboxyl group but less than the pKa of the protonated amino group; therefore, the carboxyl group is
protonated (basic form) and the amino group deprotonated (acidic form). In a strongly basic solution (pH ∼ 12),
both groups are protonated (basic form). This is shown in the notes.

Imp: Recall from the Henderson–
Hasselbalch equation that half the group is in the acidic form and half is in the basic form at pH = pKa.

Generally, pKa of COOH groups of amino acids is around 1.82-2.63 and 8.84 - 9.8f for amino acids with cysteine having 10.46 and proline having 10.60. This is shown in table 21.3, pay attention also to the pH of side groups.

Question 10

What is an important note to make about amino acids charge?

Answer 10

Notice that an amino acid can never exist as an uncharged compound, regardless of the pH of the solution. To be uncharged, an amino acid would have to lose a proton from a +NH3 group with a pKa of about 9 before it loses a proton from a COOH group with a pKa of about 2. This is impossible because a weak acid (pKa = 9) cannot lose a proton more easily than a strong acid (pKa = 2) can.

Question 11

What is a zwitterion ion?

Answer 11

A zwitterion is a compound that has a negative charge on one atom and a positive charge on a nonadjacent atom. At physiological pH (7.4), an amino acid exists as a zwitterion.

Question 12

What are the different ways in which we can synthesize amino acids?

Answer 12

1. Hell-Volhard-Zelinski (HVZ) reaction
2. Reductive amination
3. Phthalimidomalonic ester synthesis (combination of Gabriel Synthesis and malon ester synthesis)
4. Strecker Synthesis

Question 13

What is Hell-Volhard-Zelinski (HVZ) reaction?

Answer 13

One of the oldest methods used to synthesize an amino acid is to employ an HVZ reaction to replace an a-hydrogen of a carboxylic acid with a bromine. The resulting a-bromocarboxylic acid can then undergo an SN2 reaction with ammonia to form the amino acid

Look at the wall

Question 14

What is Reductive amination?

Answer 14

Synthesis of amino acids via reductive amination of an a-keto acid:

Look at the wall.

Question 15

What is Phthalimidomalonic ester synthesis (a combination of Gabriel Synthesis and malon ester synthesis)?

Answer 15

Amino acids can be synthesized with better yield using phthalimidomalonic ester synthesis in comparison with the other two methods

Since this mechanism is a combination of Gabriel Synthesis and malon ester synthesis we must first understand both of them before we show the final mechanism. Everything is shown on the wall.

Question 16

What is the Strecker Synthesis?

Answer 16

In the Strecker synthesis, an aldehyde reacts with ammonia to form an imine. A nucleophilic addition reaction with cyanide ions forms an intermediate, which, when hydrolyzed, forms the amino acid. This reaction yields the best for the synthesis of amino acids. Look at the wall

Question 17

What type of mixture is made from the synthesis of amino acids?

Answer 17

When amino acids are synthesized in the laboratory, the product is a racemic mixture of D- and L-amino acids

Question 18

How can we separate the two enantiomers of amino acids and what is kinetic resolution?

Answer 18

The two enantiomers can be separated by an enzyme-catalyzed reaction. This is Because an enzyme is chiral, it reacts at a different rate with each of the enantiomers.

For example, aminoacylase is an enzyme that catalyzes the hydrolysis of N-acetyl-l-amino acids, but not N-acetyl-d-amino acids. Therefore, if the racemic mixture of amino acids is converted to a pair of N-acetylamino acids (by a nucleophilic acyl substitution reaction) and the N-acetylated mixture is hydrolyzed with aminoacylase, the products will be the l-amino acid and unreacted N-acetyl-d-amino acid (as shown in notes), which are easily separated.

because the resolution (separation) of the enantiomers depends on the difference in the rates of reaction of the enzyme with the two N-acetylated compounds, this technique is known as a kinetic resolution.

Note that a racemic mixture of amino acids can also be separated by the enzyme d-amino acid oxidase.

Question 19

What are peptide bonds?

Answer 19

The amide bonds that link amino acids are called peptide bonds. By convention, peptides and proteins are written with the free amino group (of the N-terminal amino acid) on the left and the free carboxyl group (of the C-terminal amino acid) on the right.

Question 20

What are the two cases of representing amino acids in a peptide?

Answer 20

■ When the identities of the amino acids in a peptide are known but their sequence is not known, the amino acids are written separated by commas.
■ When their sequence is known, the amino acids are written connected by hyphens.

Question 21

How are peptides named?

Answer 21

In naming a peptide, adjective names (ending in “yl”) are used for all the amino acids except the C-terminal amino acid.

Each amino acid has the L configuration unless otherwise specified.

Question 22

What are the properties of a peptide bond and the effect of those properties on the peptide?

Answer 22

A peptide bond has about 40% double-bond character because of electron delocalization (As shown on the wall). Steric strain causes the configuration that has the a-carbons of adjacent amino acids on the opposite side of the double bond to be more stable.

The partial double-bond character prevents free rotation about the peptide bond, so the carbon and nitrogen atoms of the peptide bond and the two atoms to which each is attached are held rigidly in a
plane (shown in notes). This regional planarity affects the way a chain of amino acids can fold; this has important implications for the three-dimensional shapes of peptides and proteins. Notice that the R groups bonded to the a-carbons are on alternate sides of the peptide backbone.

Question 23

What is a disulfate bond?

Answer 23

When thiols are oxidized under mild conditions, they form a disulfide¬a compound with an S-S bond. (Like C-H bonds, the number of S-H bonds decreases in an oxidation reaction and increases in a reduction reaction.)

Question 24

What is the mechanism for the oxidation of a thiol to disulfate?

Answer 24

The oxidizing agent commonly used for this reaction is Br2 (or I2) in a basic solution. Mechanism on the wall on the same paper as the peptide bond.

Question 25

What are disulfate bridges?

Answer 25

Disulfide bridges are the only covalent bonds that are found between nonadjacent amino acids in peptides and proteins. They contribute to the overall shape of a protein by linking cysteines found in different parts of the peptide backbone.

Note that disulfate bridges aren’t exclusively for cysteines but we will only encounter it with cysteines. An example of two cysteines forming a cystine is shown in digital notes.

Teacher note: If a protein folds then forms s-s bonds it functionalizes, but if a protein forms s-s bonds while (denatured) it will not be allowed to fold and therefore not functionalize.

Question 26

What are the different ways in which an amide bond is made?

Answer 26

This is important to highlight before looking into the peptide bond (amide bond) synthesis strategy. The concept of this is to show that the methods we have learned aren’t suitable for peptide synthesis:

Look at the wall.

Question 27

What problem do the functional groups of the amino acid bring when synthesizing polypeptides?

Answer 27

One difficulty in synthesizing a polypeptide is that the amino acids have two functional groups, enabling them to combine in different ways. Suppose, for example, that you want to make the dipeptide Gly-Ala. That dipeptide is only one of four possible dipeptides that could be formed by heating a mixture of alanine and glycine with the 3 other being Ala-Ala, Ala-Gly, and Gly-Gly.

Question 28

How is the problem with functional groups of amino acids solved?

Answer 28

The problem is solved by

• protecting the amino group of the amino acid that is to be on the N-terminal end, making it not available to form a peptide bond.
• Activating the carboxyl group of the amino acid that is to be on the N-terminal end, before the second amino acid is added, then the amino group of the added amino acid will react with the activated carboxyl group in preference to reacting with a non-activated carboxyl group of another molecule of it self. (y3ani Ala will add to Gly instead of adding to it self)

Question 29

What is the regent used to protect the amino group of an amino acid?

Answer 29

The reagent most often used to protect the amino group of an amino acid is di-tert-butyl dicarbonate. Notice that the amino group rather than the carboxylate group of the amino acid reacts with di-tert-butyl dicarbonate because the amino group is a better nucleophile. When glycine reacts
with di-tert-butyl dicarbonate in a nucleophilic acyl substitution reaction, the anhydride bond breaks, forming CO2 and tert-butyl alcohol. Reaction shown on the wall (not the title not in marker cause i don’t have it with me)

Question 30

What is the regent used to activate the carboxyl group of an amino acid?

Answer 30

The reagent most often used to activate the carboxyl group is dicyclohexylcarbodiimide (DCCD). DCCD activates a carboxyl group by putting a good leaving group on the carbonyl group. (Reaction is shown on the wall with protecting the amino group).

Question 31

What is the reaction of the addition of a second amino acid to our protected and activated amino acid?

Answer 31

On the wall.

Question 32

How can we add more amino acids to the polypeptide being made?

Answer 32

Amino acids can be added to the growing C-terminal end by repeating the same two steps: activating the carboxyl group of the C-terminal amino acid of the peptide by treating it with DCCD and then adding a new amino acid.

Question 33

How can we remove the protecting group on the N-terminal end?

Answer 33

When the desired number of amino acids has been added to the chain, the protecting group, known by the acronym t-Boc (tert-butyloxycarbonyl; pronounced “tee-bok”), on the N-terminal amino acid is removed with trifluoroacetic acid in dichloromethane, a reagent that does not break any other covalent bonds. The protecting group is removed by an elimination reaction, forming isobutylene in the first step and carbon dioxide in the second step. Look at the wall

Question 34

Why is t-Boc such a good protecting group?

Answer 34

t-Boc is an ideal protecting group because it can be removed easily and, since isobutylene and carbon dioxide are gases, they escape, driving the reaction to completion.

Question 35

What is the yield limitation of synthesizing polypeptides by adding them one by one?

Answer 35

Theoretically, you should be able to make as long a peptide as desired with this technique. Reactions never produce 100% yields, however, and the yields are further decreased during the purification process. (The peptide must be purified after each step of the synthesis to prevent subsequent unwanted reactions with leftover reagents.)

For example, assuming that each amino acid can be added to the growing end of the peptide chain with an 80% yield (a relatively high yield), the overall yield of a nonapeptide such as bradykinin (9 amino acids long) would be only 17% (0.80^8). It is clear that large polypeptides could never be synthesized this way.

NOTE!!! In addition to producing low overall yields, this method of peptide synthesis is extremely time-consuming because the product must be purified at each step
of the synthesis.

Question 36

What is the Merrifield method

Answer 36

Bruce Merrifield described a method (Merrifield method) that revolutionized the synthesis of peptides because it provided a much faster way to produce peptides in much higher yields. Furthermore, because it is automated, the synthesis requires fewer hours of direct attention. With this technique, bradykinin was synthesized in 27 hours with an overall yield of 85%. Subsequent refinements in the technique now allow the synthesis of a peptide containing hundreds of amino acids in a reasonable yield. This synthesis is done on a solid support in a column

Question 37

Important note!!!

Answer 37

okay cute I just realized that with the Merrifield method, the growth happens from the N-terminal y3ani the polypeptide grows from the left while in the first method, the growth happened from the C-terminal y3anin from the right. This means in the Merrifield method the C-terminal side is added to the solid support.

Question 38

What are the steps of the Merrifield automated solid-phase peptide synthesis?

Answer 38

Before you look at the wall note that before the C-terminal amino acid is added to the solid support, its amino group is protected with t-Boc to prevent the amino group from reacting with the solid support. Now you can look (good boy ;)).

Question 39

What is the Advantage of the Merrifield Method?

Answer 39

A huge advantage of the Merrifield method of peptide synthesis is that the growing peptide can be purified by washing the column with an appropriate solvent after each step of the procedure. The impurities are washed out of the column because they are not attached to the solid support. Because the peptide is covalently attached to the resin, none of it is lost in the purification step, leading to high yields of purified product.

Question 40

How can the polypeptide be removed from the resin?

Answer 40

the peptide can be removed from the
resin by treatment with HF under mild conditions that do not break any peptide bonds

Question 41

What is genetic engineering?

Answer 41

Another way to synthesize proteins:

• Inserting DNA into host cells causes them to produce large amounts of a desired protein.
• Genetic engineering techniques have also been useful in synthesizing proteins that differ in one or a few amino acids from a natural protein.

Question 42

What are two notes (from pp) that we should be mindful of when synthesizing via the Merrifield Method?

Answer 42

• necessary, protect reactive groups (NH2, OH, SH, COOH) in the side chains
• Mild conditions ensure that racemization does not occur
• ANOTHER MAJOR MAJOR POINT IS SHOWN ON THE WALLLLL
• make sure to always have peptide bonds! lose the oxygen in the mechanism

Question 43

What are coil conformations?

Answer 43

Generally, less than half of the protein’s backbone is arranged in a defined secondary structure—an a-helix or a b-pleated sheet. Most of the rest of the protein, though highly ordered, is nonrepetitive and, therefore, difficult to describe. Many of these ordered polypeptide fragments are said to be in coil or loop conformations. This is shown in digital notes.

Question 44

What are the criteria for making synthetic polymers?

Answer 44

In order to synthesise a polymer you need a superefficient organic reaction, where ideally no side reactions occur, no mistakes are made, and each step occurs very fast.

Note that since the polypeptides we synthesis are considered to be synthetic polymers, then they must also follow that criteria!

Question 45

What are the two major classes of synthetic polymers?

Answer 45

1. Chain-growth polymers: polymers made by chain reactions—the addition of monomers to the end of a growing chain. The end of a growing chain is reactive because it is a radical, a cation, or an anion. There are three mechanisms: radical polymerization, cationic polymerization, or anionic polymerization.
2. Step-growth polymers: Polymers made by linking monomers as a result of removing (in most cases) a small molecule, generally water or an alcohol. The monomers have reactive functional groups at each end.

Question 46

What are the monomers that most commonly go through chain polymerization?

Answer 46

The monomers used most commonly in chain-growth polymerization are ethylene (ethene) and substituted ethylene
CH2=CHR). Polymers formed from ethylene or substituted ethylenes are called vinyl polymers.

Question 47

What is the mechanism for radical polymerization?

Answer 47

It is in digital notes.

Note that Molecule XY can be a solvent, a
radical initiator, or any molecule with a bond that can readily be cleaved homolytically. Chain transfer can be used to control the molecular weight of the polymer.

Question 48

If a polymer has a large Mw what determines its physical properties?

Answer 48

As long as the polymer has a high molecular weight, the groups (RO) at the ends of the polymer chains—arising from initiation and chain transfer—are relatively unimportant in determining its physical properties and are generally not even specified; it is the rest of the molecule that determines the properties of the polymer.

Question 49

Why is head-to-tail addition favoured?

Answer 49

1. First, there is less steric hindrance at the unsubstituted sp2 carbon of the alkene; therefore, the propagating site attacks it preferentially.
2. Second, radicals formed by addition to the unsubstituted sp2 carbon can be stabilized by the substituent attached to the other sp2 carbon. For example, when Z is a phenyl substituent, the benzene
   ring stabilizes the radical by electron delocalization. (This is shown in digital notes!)

Question 50

What are the cases in which head-to-tail addition doesn’t occur?

Answer 50

In cases where the substituent is small (so that steric considerations are less important) and is not able to stabilize the growing end of the chain by electron delocalization, some head-to-head addition and some tail-to-tail addition occur. This has been observed primarily when the substituent is fluorine. This abnormal
mode of addition, however, has never been found to constitute MORE THAN 10% of the overall chain. So y3ano even then do head-to-tail addition but also mention that tail-to-tail or head-to-tail can occur.

Question 51

What are the monomers that undergo radical polymerization?

Answer 51

Monomers that most readily undergo chain-growth polymerization by a radical mechanism are those in which the substituent Z is a group able to stabilize the growing radical species by electron delocalization or by inductive electron withdrawal. Examples shown in notes

(note that structures of acrylate, methacrylate, styrene and its polymers are required to be memorized!!! Acrylate structure is not their so search it up)

(GOOD BOY)

Question 52

What are the two factors that must be taken into account when choosing a radical

Answer 52

The first is the solubility of the initiator. For example, potassium persulfate is often used if the initiator needs to be soluble in water, whereas an initiator with several carbons is chosen if the initiator needs to be soluble in a nonpolar solvent. The second factor is the temperature at which the polymerization reaction is to be carried out. For example, a tert-butoxy radical is relatively stable, so an initiator that forms a tert-butoxy radical is used for polymerizations carried out at relatively high temperatures.

note: radicals can be any compound with a weak bond that readily undergoes homolytic cleavage by heat or ultraviolet light to form radicals sufficiently energetic to convert an alkene into a radical.

Question 53

How can branching occur in radical polymerization?

Answer 53

If the propagating site removes a hydrogen atom from a chain, a branch can grow off the chain at that point. The propagating site can remove a hydrogen atom from a different polymer chain or from the same polymer chain (This is shown in digital notes).

Removing a hydrogen atom from a carbon near the end of a chain leads to short branches, whereas removing a hydrogen atom from a carbon near the middle of a chain results in long branches. Branched polymers are more flexible.

Question 54

What is the initiator for cationic polymerizations?

Answer 54

In cationic polymerization, the initiator is an electrophile (generally a proton) that adds to the monomer, causing it to become a carbocation. The initiator cannot be an acid such as HCl because its conjugate base (Cl−) will be able to react with the carbocation. Thus, the initiator most often used in cationic polymerization is a Lewis acid, such as BF3, together with a proton-donating Lewis base, such as water.

Question 55

What is the mechanism for cationic polymerization?

Answer 55

Notice that the reaction follows the rule that governs electrophilic addition reactions—that is, the electrophile (the initiator) adds to the sp2 carbon bonded to the most hydrogens.

Ofc mechanism is shown in the digital notes :).

Question 56

What is a very important thing to note about cationic polymerization?

Answer 56

That carbocation rearrangements can occur by either a 1,2-hydride shift or a 1,2-methyl shift if rearrangement leads to a more stable carbocation. This means (READ CAREFULLY) that you will have a polymer that will propagate by both the unarranged and the rearranged form, where most of the time the cation formed will be in the rearranged and a few times it will be in the unarranged form. An example is shown in digital notes.

Question 57

What are the monomers that undergo cationic polymerization?

Answer 57

Monomers that are best able to undergo polymerization by a cationic mechanism are those with substituents that can stabilize the positive charge at the propagating site either by hyperconjugation
(the first compound in Table 27.4) or by donating electrons by resonance (the other two compounds in the table). The table is ofc in digital notes

Question 58

What is the initiator for anionic polymerization?

Answer 58

In anionic polymerization, the initiator is a nucleophile that reacts with the monomer to form a propagating site that is an anion. Nucleophilic attack on an alkene does not occur readily because alkenes are themselves electron-rich. Therefore, the initiator must be a very good nucleophile, such as sodium amide or butyllithium (BuLi), and the alkene must contain a substituent that can withdraw electrons by resonance, which decreases the electron density of the double bond.

Question 59

What is the mechanism for anionic polymerization?

Answer 59

Check digital notes.

Question 60

What is termination in anionic polymerizations and what does it entail about the nature of anionic polymerization?

Answer 60

The chain in anionic polymerization can be terminated by a reaction with an impurity in the reaction mixture (like a proton). If all impurities are rigorously excluded, chain propagation will continue until all the monomers have been consumed. At this point, the propagating site will still be active, so the polymerization reaction will continue if more monomer is added to the system. Such nonterminated chains of anionic polymerization are called living polymers because the chains remain active until they are terminated (“killed”).

Question 61

What are the monomers that undergo anionic polymerization?

Answer 61

Alkenes that undergo polymerization by an anionic mechanism are those that can stabilize the negatively charged propagating site by resonance electron withdrawal as shown in digital notes.

Question 62

What is an application of anionic polymerization (mentioned in the lecture)?

Answer 62

Super glue: A polymer made by anionic polymerization of methyl a-cyanoacrylate (see digital notes). Because the monomer has two electron-withdrawing groups, it requires only a moderately good nucleophile to initiate anionic polymerization, such as surface-absorbed water. The ability to form covalent bonds with groups on the surfaces of the objects to be glued together is what gives Super Glue its amazing strength.

Question 63

What determines the mechanism of chain growth polymerizations?

Answer 63

We saw that the substituent attached to the alkene determines the best mechanism for chain-growth polymerization. Alkenes with substituents that can stabilize radicals readily undergo radical polymerization, alkenes with electron-donating substituents that can stabilize cations undergo cationic
polymerization and alkenes with electron-withdrawing substituents that can stabilize anions undergo anionic polymerizations (Lovely table summarizing this is in notes).

note that some alkenes undergo polymerization by more than one mechanism. For example, styrene can undergo polymerization by radical, cationic, and anionic mechanisms because the phenyl group can stabilize benzylic radicals, benzylic cations, and benzylic anions. The mechanism followed for its polymerization depends on the nature of the initiator chosen to start the reaction.

Question 64

What is ring-opening polymerization (reoccurring exam question)

Answer 64

Here we specifically focus on epoxides, which can undergo CHAIN-GROWTH polymerization via two mechanisms:

1. Anionic mechanism: This happens when the initiator is a nucleophile, where the nucleophile attacks the LESS sterically hindered carbon of the epoxide. The reaction is shown in the notes.
2. Cationic mechanism: This happens when the initiator is acidic. The reaction is shown in the notes. Notice that under acidic conditions, the nucleophile attacks the MORE substituted carbon of the epoxide.

Question 64

What are the two reactions that can be undergone via step-polymerization?

Answer 64

1. The reaction of a single bifunctional compound with two different functional groups, A and B
2. The reaction of two different bifunctional compounds. One contains two A functional groups and the other contains two B functional groups.

Question 64

What are some trends we see in polymer configuration (isotactic, syntactic, and so on)

Answer 64

• Radical polymerization leads primarily to branched polymers in the atactic configuration.
• Anionic polymerization can produce polymers with the most stereoregularity (syn or iso).
• The percentage of chains in the isotactic or syndiotactic configuration increases as the polymerization temperature decreases and the solvent polarity decreases

Question 64

What are the issues of step-growth polymerization?

Answer 64

1. Because the growing polymer will have the structure A—[polymer]—B it can cyclize (an intramolecular reaction) thereby terminating polymerization.
2. Difficult to attain high molecular weights, since the viscosity will increase causing the removal of the small group (H2O, etc) to become more demanding, and making the diffusion of the end groups (and thereby them meeting) even harder. This means that if step-growth polymerization is to lead to long-chain polymers, very high yields must be achieved.
3. Difficult to control the nature of the end groups, especially with AA / BB polymerizations. (Here she ment what the final end group of the polymer will be, it could be A or B)
4. Often catalysts are needed to enhance the reaction rate, which leads to higher mw making catalyst removal hard
5. Reactive groups are needed for high conversions

Question 65

How can we reduce the issue of cyclization?

Answer 65

The intramolecular reaction can be minimized by using a high concentration of the monomer. Because large rings are harder to form than smaller ones, once the polymer chain has more than about 15 atoms, the tendency for cyclization decreases.

Question 66

What are the different classes of step-growth polymerizations?

Answer 66

Examples of all these are found in notes (I want to highlight nylon (6 and 66), Kodel, Lexan, polyurethane)

1. Polyamides: s. The acyl chloride group of one monomer reacts with the amino group of another monomer to form an amide
2. Aramides: aromatic polyamides! The incorporation of aromatic rings into polymers results in polymers with great physical strength. The chains are hydrogen bonded, forming a sheet-like structure.
3. Polyesters: Made in two ways. Via transesterification (acid or base catalysed with alcohol and ester). You lose alcohol here as the ester group is replaced. Via direct esterification (acid catalysed with alcohol and acid). You lose water here.
4. Polycarbonates: Polyesters with two OR groups bonded to the same carbonyl carbon (RO-(C=O)-OR)
5. Epoxy Resins: are the strongest adhesives known; they are extensively cross-linked systems. They can adhere to almost any surface and are resistant to solvents and extremes of temperature.
   Epoxy cement is purchased as a kit consisting of a low-molecular-weight prepolymer and a hardener that reacts when mixed to form a cross-linked polymer.
6. Polyurethanes: a compound that has an OR group and an NHR group bonded to the same carbonyl carbon. Urethanes can be prepared by treating an isocyanate with an
   alcohol, in the presence of a catalyst such as a tertiary amine. n. Notice that polyurethanes prepared from diisocyanates and diols are the only step-growth polymers we have seen in which a small molecule is not lost during polymerization.

Question 67

What are biodegradable polymers?

Answer 67

Polymers that can be degraded by microorganisms such as bacteria,
fungi, or algae. An example is Polylactide (PLA), a biodegradable polymer of lactic acid.

Question 68

What are the ways that lactic acid can be polymerized to form PLA? (the ROMP one is a reoccurring exam question)

Answer 68

1. by step-growth polymerization with itself releasing a molecule of water that can hydrolyze the new ester bond, reforming lactic acid again. Shown in notes
2. ROMP: If the lactic acid is converted to a cyclic dimer, the dimer can form a polymer without the loss of water by ring-opening polymerization initiated by a Nu. Shown in notes

Note: Because lactic acid has an asymmetric centre, there are several different forms of the polymer.

Question 69

Leftover

Answer 69

LOL