Lecture 2

Question 1

General recap of all the past content

Answer 1

In written notes.

But note that I split this by covering what she said in her PP, reading through chapter 15, and doing flashcards where information might be missed!

Question 2

Describe the hybridization of the carbonyl oxygen?

Answer 2

The carbonyl oxygen is also sp2
hybridized. One of its sp2 orbitals forms a s bond with the carbonyl carbon, and each of the other two sp2 orbitals contains a lone pair. The remaining p orbital of the carbonyl oxygen overlaps the remaining p orbital of the carbonyl carbon to form a p bond

Question 3

What are the two resonance contributors for Esters, carboxylic acids, and amides?

Answer 3

Their structure is shown in the notes.

The resonance contributor on the right makes a greater contribution to the hybrid in the amide than in the ester or the carboxylic acid. This is due to the fact that the amide’s resonance contributor is more stable because nitrogen, being less electronegative than oxygen, can better accommodate a positive charge. The resonance contributor on the right makes an insignificant contribution to an acyl chloride, so there is no need to show it.

Question 4

How do carbonyl compounds compare with each other in terms of bp?

Answer 4

Generally

Amides > Carboxylic acids > Alcohols&raquo_space; Esters ∼ acyl chlorides ∼ ketone ∼ aldehydes > ether

The boiling points of an ester, acyl chloride, ketone, and aldehyde of comparable molecular weight are similar and are lower than the boiling points of alcohol, carboxylic acids, and amides of similar molecular weight because only those molecules can form hydrogen bonds with each other.

The boiling points of these four carbonyl compounds are higher than the boiling points of the same-sized ether because of the relatively strong dipole-dipole interactions between the polar carbonyl groups.

Carboxylic acids have relatively high boiling points because each molecule has two groups that can form hydrogen bonds. Amides have the highest boiling points because they have strong dipole-dipole interactions (shown in written notes). This is due to the fact that the resonance contributor with separated charges contributes significantly to the overall structure of the compound. In addition, if the nitrogen of an amide is bonded to hydrogen, hydrogen bonds can form between the molecules.

Question 5

What is the relative solubility of organic molecules?

Answer 5

Carboxylic acid derivatives are soluble in solvents such as ethers, chloroalkanes, and aromatic hydrocarbons. Like alcohols and ethers, carbonyl compounds with fewer than four carbons are soluble in water.

Question 6

What are Esters, N, N-disubstituted amides, and nitriles often used for?

Answer 6

Esters, N, N-disubstituted amides, and nitriles are often used as solvents because they are polar but do not have reactive OH or NH2 groups.

Question 7

What is the general rule about the stability of the carbon tetrahedral intermediate?

Answer 7

Generally, a compound that has an sp3 carbon bonded to an oxygen atom is unstable if the sp3 carbon is bonded to another electronegative atom.

Question 8

Why is the weaker base always the one being eliminated?

Answer 8

when comparing bases of the same type, the weaker base is a better leaving
group. Because a weak base does not share its electrons as well as a strong base does, a weaker base forms a weaker bond—one that is easier to break.

Question 9

Back to the recap

Answer 9

Hallo go back to written notes.

Question 10

How is the general rule that an addition elimination reaction can only form less reactive carboxylic acid depicted and proved by reaction coordinate diagrams?

Answer 10

The reaction coordinate diagram is represented in the digital notes and the explanation is below:

■ To form a more reactive compound from a less reactive compound, the new group in the tetrahedral intermediate has to be a weaker base than the group attached to the acyl group in the reactant. However, the lower energy pathway is for the tetrahedral intermediate (TI) to eliminate the newly added group and re-form the reactants, so no reaction takes place (Figure 15.2a).

■ To form a less reactive compound from a more reactive compound, the new group in the tetrahedral intermediate has to be a stronger base than the group attached to the acyl group in the reactant. The lower energy pathway is for the tetrahedral intermediate (TI) to eliminate the group
attached to the acyl group in the reactant, so a substitution product is formed (Figure 15.2b).

■ If the reactant and product have similar reactivities, then both groups in the tetrahedral intermediate will have similar basicities. In this case, the tetrahedral intermediate can eliminate either group
with similar ease, so a mixture of the reactant and the substitution product results (Figure 15.2c). ALSOOOO do the next 2 flash card (they are very easy) and go back to the recap.

Question 11

How does having a weak base attached to the acyl group make the first step of the addition-elimination reaction?

Answer 11

As mentioned before, Weak bases do not share their electrons well, so the weaker the basicity of group, the smaller
the contribution from the resonance contributor with a positive charge on the basic group. In addition, when the basic group is Cl, delocalization of chlorine’s lone pair is minimal due to the poor orbital overlap between the large 3p orbital on chlorine and the smaller 2p orbital on carbon. The less the contribution from the resonance contributor with the positive charge on Y, the greater the contribution from the resonance contributor with the electrophilic carbonyl carbon. Thus, weak bases cause the carbonyl carbon to be more electrophilic and, therefore, more reactive toward nucleophiles.

Question 12

How does a weak base make the second step of addition-elimination reaction easier?

Answer 12

Since, again weak bases are easier to eliminate when the tetrahedral intermediate collapses.

Question 13

What are the possible ways to make an acid?

Answer 13

On the wall

Question 14

How does the reactivity of ketones and aldehydes compare to carboxylic acid derivatives?

Answer 14

Generally, ketones and aldehydes are more reactive than carboxylic acid derivatives due to the lack of electron-donating substituents (R and H are not electron-donating) making the carbonyl carbon more electrophilic, as no stabilizing from resonance contributors will form as no electron will be donated.

Question 15

What is more reactive an aldehyde or a ketone?

Answer 15

An aldehyde is more reactive than ketones since ketones have greater steric crowding in their transition states, so they have less
stable transition states than do aldehydes (as shown in digital notes) and due to The carbonyl carbon of an aldehyde being more accessible to the nucleophile. This is all even though Hyperconjugation stabilizes ketones in comparison with aldehydes.

Also upon looking at additions of aldehyde/ketones, we can further confirm that ketones are much more reactive as shown in written notes)

Question 16

What is the mechanism for the addition of water to aldehydes/ketones?

Answer 16

Look at the wall.

Question 17

What is the mechanism for acid-catalylzed acetal formation?

Answer 17

Look at the wall

Question 18

How can we isolate the acetal formed?

Answer 18

Although the sp3 carbon of an acetal is bonded to two oxygens, which suggests that it is not stable, the acetal can be isolated if the water that is eliminated is removed from the reaction mixture. If water is not available, the only compound the acetal can form is the O-alkylated intermediate, which is even less stable than the acetal. However, if water is available, water can add to the O-alkylated intermediate and the aldehyde (or ketone) will be re-formed. Therefore, the acetal can be hydrolyzed back to the aldehyde or ketone in an acidic aqueous solution.

Question 19

What is the Mechanism for imine formation?

Answer 19

Look at the wall.

Question 20

What are the walls to make aldehydes/ketones?

Answer 20

Look at the wall

Question 21

How are the terms carbohydrate, saccharide, and sugar related?

Answer 21

The terms carbohydrate, saccharide, and sugar are used interchangeably, where saccharide comes from the word for sugar in several early languages:

• Simple carbohydrates are monosaccharides (single sugars).
• complex carbohydrates contain two or more monosaccharides linked together.
• Disaccharides contain two monosaccharides linked together.
• Oligosaccharides contain 3 to 10 (oligos is Greek for “few”).
• polysaccharides contain 10 or more

Question 22

How are Disaccharides, oligosaccharides, and polysaccharides broken down?

Answer 22

Disaccharides, oligosaccharides, and polysaccharides can be broken down to
monosaccharides by hydrolysis

Question 23

How are monosaccharides classified?

Answer 23

A monosaccharide can be a polyhydroxy aldehyde such as glucose or a polyhydroxy ketone such as fructose. Polyhydroxy aldehydes are called aldoses (“ald” is for aldehyde; “ose” is the suffix for a sugar); polyhydroxy ketones are called ketoses.

Monosaccharides are also classified according to the number of carbons they contain: those with three carbons are trioses, those with four carbons are tetroses, those with five carbons are pentoses, and those with six and seven carbons are hexoses and heptoses. Therefore, a six-carbon polyhydroxy aldehyde such as glucose is an aldohexose, whereas a six-carbon polyhydroxy ketone such as fructose is a ketohexose.

Question 24

What is the smallest aldose and how was it used to determine the D,L notation?

Answer 24

The smallest aldose, and the only one whose name does not end in “ose,” is glyceraldehyde, an aldotriose (shown in written notes). Because glyceraldehyde has an asymmetric center, it can exist as a pair of enantiomers (shown in written notes) and rotate the plane of polarized light.

When Emil Fischer and his gang were studying carbohydrates, techniques for determining the configurations of compounds were not available. This made Ficher assign the dextrorotatory isomer of glyceraldehyde to have a D configuration, and its opposite to have an L configuration and from this, he assigned all the other carbohydrates as either D or L.

Later on, we were able to test the configuration of the compounds (the common R and S we use) and we saw that most of the time Emil’s assignment was right, and of course, there were inaccuracies because a Compound with an R configuration can be Levarotatory instead of dextrorotatory.

Due to many research papers have the D and L configuration it was kept until today.

TO SET THINGS STRAIGHT: Like R and S, the symbols d and l indicate the configuration about an asymmetric centre, but do not indicate whether the compound rotates the plane of polarization of plane-polarized light to the right (+) or to the left (-)

Question 25

What are the two rules to follow when drawing the Fischer projection of monosaccharides?

Answer 25

In a Fischer projection of a monosaccharide:

1. the carbonyl group is always placed on top (in the case of aldoses) or as close to the top as possible (in the case of ketoses).
2. If the OH group attached to the bottommost asymmetric center of a Fischer projection (the carbon second from the bottom) is on the right, then the compound is a d-sugar. If that same OH group is on the left, then the compound is an l-sugar. Almost all sugars found in nature are d-sugars. The mirror image of a d-sugar is an l-sugar.

Question 26

What is one more important note:

Answer 26

The common name of the monosaccharide, together with the d or l designation, completely defines its structure, because the configurations of all the asymmetric centers are implicit in the common name. Thus, the structure of l-galactose is obtained by changing the configuration of all the asymmetric centers in d-galactose.

Question 27

What is the configuration of all aldoses and the insights we can gain?

Answer 27

Shown in notes Table 20.1:

Okay, my insights:

• Lowkey, all of the aldoses of the same carbon length are related but just different by the middle carbon configuration, i would not flat out call them diastereomers but they are very similar.
• The number of aldoes per length of the carbon chain is directly related to the number of enantomer pairs it forms. For example, Aldotetroses have two asymmetric centres and, therefore, four stereoisomers (2 enantiomer pairs = the number of Aldotetroses). Other examples, Aldopentoses have three asymmetric centre and, therefore, 8 stereoisomers (four pairs of enantiomers); aldohexoses have four asymmetric centres and 16 stereoisomers (eight pairs of
  enantiomers). This also relates back to how all the compounds are kind of related.
• This is less of an insight and more of a point that d-glucose, d-mannose, and d-galactose are the most common aldohexoses in living systems and knowing their structures is very important not only in Fischer projections (the other projections are coming dw ;) ) An easy way to learn their structures is to memorize the structure of d-glucose and then remember
  that d-mannose is the C-2 epimer of d-glucose and d-galactose is the C-4 epimer of d-glucose. Epimer defined in next flashcard

Question 28

What are epimers?

Answer 28

Monosaccharides that differ in configuration at only one asymmetric center are called epimers. For example, d-ribose and d-arabinose are C-2 epimers because they differ in configuration only at
C-2; d-idose and d-talose are C-3 epimers. An epimer is a particular kind of diastereomer

Question 29

What is the configuration of all ketoses and the insights we can gain?

Answer 29

Shown in table 20.2, the same insights as Aldoses just note that:

• all have a keto group in the 2-position.
• A ketose has one less asymmetric center than an aldose with the same number of carbons. Therefore, a ketose has only half as many stereoisomers as an aldose with the same number of carbons.

Question 30

What are the reactions of monosaccharides in basic solutions?

Answer 30

1. Epimerization
2. The Enediol Rearrangement

Question 31

What is Empimerization?

Answer 31

On the wall, first the mechanism then the proper definition.

Question 32

What is The Enediol Rearrangement?

Answer 32

Basically, it is an aldose is reformed into on or more ketoses. The mechanism is on the wall.

Question 33

What is an important thing to note about enediol rearrangement?

Answer 33

Given the mechanism shown, another enediol rearrangement, initiated by a base removing a proton from C-3 of d-fructose, forms an enediol that can tautomerize to give a ketose with the carbonyl group at C-2 or C-3. Thus, the carbonyl group can be moved up and down the chain.

Question 34

What are the reduction reactions of monosaccharides?

Answer 34

The carbonyl group in aldoses and ketoses can be reduced to an alcohol by NaBH4. The product of the reduction is a polyalcohol, known as an alditol.

The reduction of an aldose forms one alditol and the Reduction of a ketose forms two alditols because the reaction creates a new asymmetric centre in the product (an example is shown in digital notes)

Question 35

How can we determine if a monosaccharide is an aldose or ketose using oxidation reactions?

Answer 35

Br2 is a mild oxidizing agent that easily oxidizes aldehydes but cannot oxidize ketones or alcohols. Therefore, adding a small amount of an aqueous solution of Br2 to an unknown monosaccharide can
tell you whether the sugar is an aldose or a ketose. The reddish-brown colour of Br2 will disappear if the monosaccharide is an aldose because when Br2 oxidizes the aldehyde, Br2 is reduced to Br^-, which is colourless. If the red colour persists, indicating no reaction with Br2, then the monosaccharide is a ketose. The product of the oxidation reaction is an aldonic acid.

The reaction is shown in the notes. Notice that an acid is present.

Question 36

What happens when aldoses and ketoses are exposed to Tollen’s reagent?

Answer 36

Only aldoses are oxidized to aldonic acids by Tollens’ reagent (Ag+, NH3, HO-). The
oxidizing agent in Tollens’ reagent is Ag+. It is reduced to metallic silver that forms a shiny mirror on the inside of the test tube.

Although Tollens’ reagent only oxidizes aldehydes, it cannot be used to distinguish aldoses and ketoses. The oxidation reaction is carried out in a basic solution that will
convert a ketose to an aldose by an enediol rearrangement, and the aldose will then
be oxidized by Tollens’ reagent. Again this is shown the the digital notes

Question 37

What happens when aldoses and ketoses are exposed to Nitric acid?

Answer 37

Again, only the aldose is oxidized with the bonus that nitric acid (being stronger than Br2 and Tollens’ reagent) will oxidize a PRIMARY alcohol.

The product obtained when both the aldehyde and the primary alcohol groups of an aldose are oxidized is called an aldaric acid. This is shown in digital notes

Question 38

What are the different forms that d-Glucose exists in?

Answer 38

d-Glucose exists in three different forms: the open-chain form of d-glucose that we have been discussing, and two cyclic forms—a-d-glucose and b-d-glucose. We know that the two cyclic forms are different because they have different melting points and different specific rotations.

Question 39

How can d-glucose exist in a cyclic form?

Answer 39

we saw that an aldehyde reacts with an
alcohol to form a hemiacetal. The reaction of the alcohol group bonded to C-5 of d-glucose with the aldehyde group forms two cyclic (six-membered ring) hemiacetals.

Question 40

How can we visualize the formation of cyclic hemiacetal from a monosaccharide?

Answer 40

To see that the OH group on C-5 is in the proper position to attack the aldehyde group, we need to convert the Fischer projection of d-glucose to a flat ring structure (a Haworth projection). To do this, draw the primary alcohol group up from the back left-hand corner as shown on the next page. Groups on the right in a Fischer projection are down in the cyclic structure, and groups on the left in a Fischer projection are up in the cyclic structure.

This is shown in written notes.

Question 41

What is (formally Haworth projections?

Answer 41

Haworth projections is when the six-membered ring is represented as flat and is viewed edge-on. The ring oxygen is always
placed in the back right-hand corner of the ring, with C-1 on the right-hand side, and the primary alcohol group attached to C-5 is drawn up from the back left-hand corner as shown in written notes under flash card 40.

Question 42

Why is there two different cylic hemiacetals?

Answer 42

There are two different cyclic hemiacetals because the carbonyl carbon of the open-chain aldehyde becomes a new asymmetric centre in the cyclic hemiacetal.

If the OH group bonded to the new asymmetric center points down (is trans to the primary alcohol group at C-5), the hemiacetal is a-d-glucose; if the OH group points up (is cis to the primary alcohol group at C-5), the hemiacetal is b-d-glucose.

The mechanism for cyclic hemiacetal formation is the same as the mechanism for hemiacetal formation between individual aldehyde and alcohol molecule

Question 43

What are anomers and anomeric carbon?

Answer 43

a-d-Glucose and b-d-glucose are anomers. Anomers are two sugars that differ in configuration only at the carbon that was the carbonyl carbon in the open-chain form. This carbon is called the anomeric carbon. The prefixes a- and b- denote the configuration about the anomeric carbon.
Because anomers, like epimers, differ in configuration at only one carbon, they too are a particular kind of diastereomer. Notice that the anomeric carbon is the only carbon in the molecule that is bonded to two oxygens.

Question 44

How does the equilibrium system between the open form of the glucose and the two cyclic forms operate?

Answer 44

In an aqueous solution, the open-chain form of monosaccharide is in equilibrium with the two cyclic hemiacetals (As long as 5 or 6-membered ring). Because formation of the cyclic hemiacetals proceeds nearly to completion (unlike formation of acyclic hemiacetals), very little glucose is in the open-chain form (about 0.02%). Even so, the sugar still undergoes the reactions discussed in previous sections (oxidation, reduction, imine formation, and so on) because the reagents react with the small amount of open-chain aldehyde that is present. As the open-chain aldehyde reacts, the equilibrium shifts to produce more open-chain aldehyde, which can then undergo a reaction. Eventually, all the glucose molecules react by way of the open-chain form.

Question 45

What is an example of an aldose that forms a 5-membered ring?

Answer 45

d-Ribose is an example of an aldose that forms five-membered-ring hemiacetals:
a-d-ribose and b-d-ribose. Shown in written notes.

Question 46

What are Pyranoses and Furanoses?

Answer 46

Six-membered-ring sugars are called pyranoses, and five-membered-ring sugars are called furanoses. Consequently, a-d-glucose is also called a-d-glucopyranose, and a-d-ribose is also called a-d-ribofuranose

Question 47

What are the most important things to note when it comes to cyclic hemiacetals from ketoses?

Answer 47

• When Ketose forms a 5-membered ring the carbon that forms the anomeric centre is C-2 not C-2 as in aldoses.
• To decide whether a five-membered ring has alpha or beta configuration always always look at the OH, if down it is alpha. Do not be tricked by the other group attached to the anomeric centre.
• Specific to d-Fructose: The pyranose form predominates in the monosaccharide, whereas the furanose form predominates when the sugar is part of a disaccharide.

Question 48

How accurate is the Haworth projection?

Answer 48

Haworth projections are useful because they show clearly whether the OH groups on the ring are cis or trans to each other. Five-membered rings are nearly planar, so furanoses are represented fairly
accurately by Haworth projections. Haworth projections, however, are structurally misleading for pyranoses because a six-membered ring is not flat—it exists preferentially in a chair conformation

Question 49

What is mutarotation and give an example of it?

Answer 49

mutarotation is a slow change in optical rotation to an equilibrium value. An example is when crystals of pure a-d-glucose are dissolved in water, the specific rotation gradually changes from +112.2 to +52.7. When crystals of pure b-d-glucose are dissolved in water, the specific rotation gradually changes from +18.7 to +52.7.

This change in rotation occurs because, in water, the hemiacetal opens to form the aldehyde, and both a-d-glucose and b-d-glucose are formed when the aldehyde recyclizes. Eventually, the three forms of glucose reach equilibrium concentrations. The specific rotation of the equilibrium mixture is +52.7. This is why the same specific rotation results whether the crystals originally dissolved in water are a-d-glucose or b-d-glucose or any mixture of the two.

Question 50

How can you convert Haworth projection to chair conformer?

Answer 50

To convert the Haworth projection to a chair conformer, start by drawing the chair so that the backrest is on the left and the footrest is on the right. Then place the ring oxygen at the back right-hand corner and the primary alcohol group in the equatorial position. Note that the primary alcohol group is the largest of all the substituents, and, therefore, it is more stable in the equatorial position where there is less steric strain.

From then go to the next carbon in the ring and see if it is cis or tans with the primary alcohol group if cis put it in the axial position (opposite to the alcohol ) if trans put it in the eq position (same as alcohol), then go to the next carbon and compare the position of the OH to the previous carbon fulling the general rule that if cis flip position if trans don’t change position.

Question 51

Why is d-Glucose the most stable aldohexose?

Answer 51

As you move around the ring, you will find that all the OH groups in b-d-glucose are in
equatorial positions and the axial positions are all occupied by hydrogens, which require little space and, therefore, experience little steric strain. No other aldohexose exists in such a strain-free conformation. This means that d-glucose is the most stable of all the aldohexoses; therefore, we should not be surprised that it is the most prevalent aldohexose in nature.

Note that the OH group bonded to the anomeric carbon is in the equatorial position in b-d-glucose, whereas it is in the axial position in a-d-glucose. Therefore, b-d-glucose is more stable than a-d-glucose, so b-d-glucose predominates at equilibrium in an aqueous solution.

Question 52

How do you draw the chair conformer of l-pyranose

Answer 52

To draw an l-pyranose, draw the d-pyranose first, and then draw its mirror image.

Question 53

What is a glycoside and a glycosidic bond?

Answer 53

In the same way that a hemiacetal reacts with an alcohol to form an acetal. The cyclic
hemiacetal formed by a monosaccharide can react with an alcohol to form two acetals (alpha and beta).

The acetal of a sugar is called a glycoside, and the bond between the anomeric carbon and the alkoxy oxygen is called a glycosidic bond

Question 54

How are glycosides named?

Answer 54

Glycosides are named by replacing the “e” ending of the sugar’s name with “ide.” Thus, a glycoside of glucose is a glucoside, a glycoside of galactose is a galactoside,
and so on. If the pyranose or furanose name is used, the acetal is called a pyranoside or a furanoside.

Question 55

What is the Mechanism for Glycoside Formation?

Answer 55

On the wall.

Question 56

What are N-glycosides?

Answer 56

The reaction of a monosaccharide with an amine is similar to the reaction of a monosaccharide with an alcohol. The product of the reaction is an N-glycoside. An N-glycoside has a nitrogen in place of the oxygen at the glycosidic linkage. The subunits of DNA and RNA are b-N-glycosides. This is shown in the notes

Question 57

What is the anomeric effect?

Answer 57

The preference of certain substituents bonded to the anomeric carbon for the axial position is called the anomeric effect. This is the cause for beta cyclic hemiacetals being only a little beta more stable than alpha, when usually they are much much much more stable. Alos, this is the cause for a-glucoside being the major product When glucose reacts with an alcohol.

Question 58

What is responsible for the anomeric effect?

Answer 58

If the substituent is axial, one of the ring oxygen’s lone pairs is in an orbital that is parallel to the s* antibonding orbital of the C–Z bond. The molecule then can be stabilized by hyperconjugation—some of the electron density moves from the sp3 orbital of oxygen into the s* antibonding orbital. If the substituent is equatorial, neither of the orbitals that contain a lone pair is aligned correctly for overlap. This is shown in digital notes.

Question 59

What are Disaccharides?

Answer 59

If the hemiacetal group of a monosaccharide forms an acetal by reacting with an alcohol group of
another monosaccharide, the glycoside that is formed is a disaccharide. Disaccharides are compounds that consist of two monosaccharide subunits hooked together by a glycosidic linkage.

Question 60

What are the most common disaccharides mentioned?

Answer 60

All structures are shown in the notes

1.Maltose:
- a disaccharide obtained from the hydrolysis of starch, contains two d-glucose subunits connected by a glycosidic linkage
- This particular linkage is called an a-1,4′-glycosidic linkage. An α-1,4′-glycosidic linkage joins the C-1 of one sugar to the C-4 of another, with the oxygen on the anomeric C-1 in the α-position, and the prime indicates C-4 is in a different ring.
- Notice that the structure of maltose does not specify the configuration of the anomeric carbon that is not an acetal.
- Because maltose can exist in both a and b forms, mutarotation occurs when crystals of one form are dissolved in water

2. Cellobiose:
   - A disaccharide obtained from the hydrolysis of cellulose, also contains two d-glucose subunits.
   - The two glucose subunits are hooked together by a b-1,4′-glycosidic linkage
   -mutarotation occurs when crystals of one form are dissolved in water.
3. Lactose:
   - Lactose is a disaccharide found in milk.
   - The subunits of lactose are d-galactose and d-glucose
   - The d-galactose subunit is an acetal,
   and the d-glucose subunit is a hemiacetal
   - The subunits are joined by a b-1,4′-glycosidic linkage
   - mutarotation occurs when crystals of one form are dissolved in water.
4. Sucrose:
   -The most common disaccharide, the substance we know as table sugar
   - Sucrose consists of a d-glucose subunit and a d-fructose subunit linked by a glycosidic bond between C-1 of glucose (in the a-position) and C-2 of fructose(in the b-position).
   - Does not exhibit mutarotation because its glycosidic bond is between the anomeric carbon

Question 61

How do you confirm that the d-glucose subunit is a hemiacetal in lactose?

Answer 61

A simple experiment can prove that the hemiacetal group in lactose belongs to the glucose residue and not to the galactose residue. The disaccharide is treated with excess methyl iodide in the presence of Ag2O, reagents that methylate all the OH groups via SN2 reactions. Because the OH group is a relatively poor nucleophile, silver oxide is used to increase the leaving propensity of the iodide ion. The product is then hydrolyzed under acidic conditions.

The mechanism is shown in the notes.

This treatment hydrolyzes the two acetal groups, but the ethers, formed by methylating the OH groups, are untouched. Identification of the products shows that the glucose residue contained the hemiacetal group in the disaccharide because its C-4 OH group was not able to react with methyl iodide (because it was an acetal linkage with galactose). The C-4 OH of galactose, on the other hand, was able to react with methyl iodide. Allowing us to identify between them

Question 62

What are Polysaccharides?

Answer 62

Polysaccharides contain as few as 10 or as many as several thousand monosaccharide units joined together by glycosidic linkages

Question 63

What are the most common polysaccharides mentioned?

Answer 63

All structures are shown in the notes:

1. Starch:
   -It is a mixture of two different polysaccharides: amylose (∼20%) and amylopectin (∼80%)
   -Amylose is composed of unbranched chains of d-glucose units joined by a-1,4′-glycosidic linkages.
   -Amylopectin is a branched polysaccharide composed of chains of d-glucose units joined by a-1,4′-glycosidic linkages and a-1,6′-glycosidic linkages.
2. Cellulose:
   - Cellulose is composed of unbranched
   chains of d-glucose units.

• The glucose units in cellulose are joined by b-1,4′-glycosidic linkages.

3. Chitin (not as common)
   - Chitin has b-1,4′-glycosidic linkages
   - Chitin has an N-acetylamino group instead of an OH group at the C-2 position
   -The b-1,4′-glycosidic linkages give chitin its structural rigidity.

Question 64

What is Saccharin?

Answer 64

Saccharin (Sweet’N Low), the first artificial sweetener, Structure shown in notes

Question 65

Okay, I’m burning out and the reading material is thick and it’s not that deep. Look at the pp for the one slide I think is important.