BLOCK 6

Question 1

What are the 5 ways that alcohols can be prepared (inc reagents)

Answer 1

1. RX + OH- (H20) through Nu- sub, Sn1 or SN2 mechanism
2. alkene + H+/H2O for makovnikov addition
3. alkene + 1. B2H6 2. H2O2/OH-
4. reduction of KEA by LiAlH4
5. Nu- addition of R-Mg+X to KA once and EA twice to make 3 alcohol.

Question 2

What is the name for the conjugate base and acid of an alcohol’s OH

Answer 2

O- conj base: alkoxide

OH2 + - conju acid: oxonium

Question 3

Are phenols acidic or basic. Why ?

Answer 3

Phenols are weakly acidic because the phenoxide anion (conjugate base) is resonance stabilized. The O- means that the charge is delocalised across the molecule.

Question 4

How do you name phenols

Answer 4

m/o/p for other substituents- bromo, methyl etc with parent name phenol

Question 5

Which substituents on the phenol increase acidity and why?

Answer 5

Electron withdrawing groups such as CHO, COOR, X are deactivating towards electrophilic aromatic substitution. As they have the electropositive part of the group connected to the ring, they draw e- towards it. It decreases the e- density in the ring making it not as eager to accept a proton, rather it wants to lose it. E- withdrawing groups stabilises the phenoxide anion promoting

Question 6

Which substituents on the phenol decrease acidity and why?

Answer 6

Electron donating groups such as OH, NH2 are activating towards electrophilic aromatic substitution. As they donate e- to the ring, the destabilise the phenoxide anion, increasing its electron density so it is harder for its H to leave.

Question 7

What are 3 reactions that can change the carbon skeleton add more C-C bonds)

Answer 7

Grignard addition, addition of -CN or alkylidine anion C=CR through nu- addition to carbonyl or nu- substitution for akyl halide

Question 8

How do you classify carbohydrates: 2 classifications

Answer 8

Complex carbohydrates are made or two or more simple sugars joined together and can undergo acid hydrolysis to break r-O-r bonds into their monosaccharide units.
Simple carbohydrates are made of a single carbon chain with one carbonyl group (aldehyde/ketone) with hydroxy groups attached to the remaining carbons.

Question 9

How do you name/classify simple carbohydrates

Answer 9

aldo (if aldehyde) or keto (if ketone) + number of carbons in the backbone (tri, tetra, pent, hex) + ose
eg. aldotetrose

Question 10

What is the relationship between the number of stereocentres and the number of stereoisomers

Answer 10

2^n (where n is the number of sterocentres because each stereocentre gives 2 stereoisomers)

Question 11

What are stereoisomers vs enantiomers vs diastereomers

Answer 11

Stereoisomers have the same molecular formula and same connectivity but a different arrangement in space - bonds going in and out of the page). Enantiomers are mirror images of each other. Diastereomers are not mirror images

Question 12

For a compound with multiple stereocentres (monosaccarides) how do you get enantiomer

Answer 12

Have to reverse all stereocentres, not reversing all but just some gives diastereomer.

Question 13

What are the trends in rate of nucleophilic attack depending on what is bonded to the C=O (R1 and R2)

Answer 13

The rate decreases with increasing substitution (less Hs as R1 and more R groups). This is because the more a carbon wants electrons (S+) the easier to attack- and this is greater when there are no e- donating groups attached. And there can be steric hindrance for Nu- attack if the atom groups too big

Question 14

Sugars with stereocentres are optically active. How are they notated and how does it relate to R and S

Answer 14

D/L notation which is determined by the stereochemistry of the sterocentre furthest from the C=O group.
D= R (in clockwise) L= S (in anticlockwise) so they fit the same alphabetically

Question 15

In water, how do monosaccharides exist

Answer 15

They will exist in equilibrium between open chain and cyclic form in six or five membered rings. This is because the OH group can react internally, losing a proton and attacking the C=O to form a cyclic hemiacetal: with RO-C-OH

Question 16

How does the stereochemistry of monosaccharides change when cyclic hemiacetals are formed

Answer 16

The stereochemistry of each stereocentre doesn’t change but a new stereocentre is formed at C-1 where it used to be a sp2 C=O and now its an sp3 RO-C-OH. This can be R or S and are anomers

Question 17

What are the anomers of cyclic sugar

Answer 17

Diastereomers that differ only in configuration at only one symmetric carbon (C1). This happens because the OH can attack on top of the flat C=O or from the bottom of the flat C=O. If it attacks from the top, it pushes the OH group( made from the C=O) down and produces an Alpha hemiacetal. If it attacks from the bottom, it pushes the OH group up and produces a Beta hemiacetal

Question 18

How do you differentiate between alpha and beta anomers

Answer 18

Alpha has trans (down) C1 OH group and C5 CH2OH (up) while Beta has cis (up and up) C1 OH and C5 CH2OH

Question 19

How can you evaluate the stability of anomers in equilibrium

Answer 19

It is not a racemic mixture because the stablility of conformer of anomers are different with more stable favoured. Each anomer has two different chair conformers (O replacing one C) with the most stable having big groups equatorial

Question 20

What is Mutorotation and how does it come about

Answer 20

Spontaneous change in optical rotation observed when a pure anomer of a sugar is dissolved in water and equilibrates to an equilibrium mixture of anomers. This equilibration occurs by reversible ring opening to the open chain form followed by reclosure. The change in optical rotation is due to the uneven ratio of B and A anomers that rotate plane polarised light in specific ways .

Question 21

What are glycosides and what are their properties

Answer 21

Glycosides are formed by the anomeric OH group (OH group attached to C1 that is bonded to 2 Os) only being replaced by an OR group. This acetal is called a glycoside and it is stable to water and are not in equilibrium with open chain form: Therefore NO MUTOROTATION.

Question 22

How are polysaccharides formed and why are there so many able to be formed

Answer 22

When the OH of a second sugar is used as the alcohol used to form an acetal, it forms a dissaccharide. This bond takes water out: OH from one sugar and H from the OH group of another to make a R-O-R’ bond. These glycoside bonds can form between any OH group on the second sugar so there are infinite number able to be formed

Question 23

What is the reason for carboxylic acids acidity

Answer 23

The deprotanation of carboxylic acids is more favourable because if the H is taken off the OH of COOH, the remaining negative charge on O can be delocalised through resonance stabilised carboxylate anion and this means that its carboxylate anion (conj base) is pretty stable

Question 24

Compare acidity of alcohol, phenol and carboxylic acid

Answer 24

Carboxylic acid and phenols both form anions (conj base) that are resonance stabilised which means they are both more acidic than akyl alcohols which can’t do this. As the stabilisation of carboxylate anion is greater than the phenoxide anion ( stabilised through the adjacent aromatic ring) , carboxylic acid is more acidic.

Question 25

Which type of electron ... group attached o/p to COOH in aromatic or directly attached to COOH in aliphatic acids increases acidity

Answer 25

Electron withdrawing groups like X, help to stabilise the carboxylate anion formed whereas Electron donating groups destabilise the carboxylate anion formed

Question 26

Are acid chlorides good nucleophiles? why

Answer 26

They have two e withdrawing groups (O and Cl) attached to carbonyl carbon which makes the S+ on carbon larger and therefore will react with Nu- more rapidly -> it is an electrophile no nucleophile

Question 27

What are the 3 ways carboxylic acids can be prepared

Answer 27

Oxidation of alcohols, grignard addition to CO2 (2 steps 1. CO2 and 2. H3O+ because the MgX group goes onto that O-) and Nucleophilic substitution of Akylhalides via nitrile

Question 28

Describe the process of getting a carboxylic acid from a akyl halide

Answer 28

First do nu- substitution by SN2 mechanism with NC- to make RCN : nitrile. Then through addition of acidified water (H+/H2O) you can break one of the pi bonds to add the electrophile OH and this changes to its tautomer which has the C=O instead of C=N (amide). Then the NH2 group is replaced with OH in amide hydrolysis using acidified water

Question 29

What is the reactivity order for nu- acyl substitution and what determines reactivity

Answer 29

Acid chlorides most reactive then acid anhydrides (OCOR), esters (OR) and amides least (NH2). This is determined at the rate of nucleophilic attack of carbonyl carbon in first step. Electron withdrawing Y group (Cl-) makes the carbonyl carbon more positive, therefore reacting more rapidly than Electron donating Y group (OR or NR2) which decreases the polarity of the bond. Acid anhydride is in the middle of Ester and acid chloride reactivity because the OCOR group is donating but its being pulled to both adjacent carbons so not as much. Donation of e- would result in unfavourable resonance hybrid. For amide, lone pairs on nitrogen readily donated making resonance hybrid with C=N so e- density is high.

Question 30

For base or acid promoted hydrolysis of esters - this is acyl substitution. Which mechanism does each one do

Answer 30

Base promoted has OH- a strong nucleophile so it can directly attack the C=O and swap OH for OR group. It then results in the base form of Carboxylic acid. For Acid promoted, the O=C needs to be activated by getting a H+ to make C=OH+ so it can make the S+ on C=O big enough so it can be attacked by H-OH.

Question 31

What do you call compounds with 4 groups attached to N

Answer 31

quarternary ammonium salts

Question 32

what do you call a benzene ring with an amine group attached

Answer 32

phenylamine

Question 33

What is the chirality of amino acids = stereochemistry

Answer 33

They have a stereogenic centre at the carbon attached to the amine group, called the alpha carbon with the R group varying for most amino acids except glycine. There are two enantiomeric forms possible D= R and S =L. L more predominantly found in nature.

Question 34

Amine can act as both an lewis base = attacking proton and nucleophile (attacking electron deficient things). What determines basicity for amines

Answer 34

Higher pka for conj acid= more basic. By changing N-H bonds to N-C bonds (adding an electron donating group, it means that N has more electron density so can act as a base more easily. This effect is limited though as there is some steric hindrance when adding 3 methyls to N making it harder for lone pairs to attack. Amine attached to aromatic ring is much less basic as delocalisation of N lone pairs (into the ring) make them less available to act as a base. Adding electron donating or electron withdrawing groups can mediate this effect making it slightly more basic or slightly less basic.

Question 35

what is the order of basicity for arylamines with e withdrawing groups, donating groups, no groups and aliphatic amines

Answer 35

aliphatic amines most basic, then arylamine with e- donating group, without groups and then e- withdrawing group.

Question 36

How do you name dipeptides

Answer 36

The first name is from the N-terminal Amino acid (free NH2) on the left and the last name is from the C terminal carboxyllic acid (free COOH) on the right.

Question 37

What is the difference between protein and peptide

Answer 37

protein is >50 amino acids, peptide is <50 amino acids

Question 38

Describe the reactivity of the amide bond (O=C-N-H) and the relative position of the bond- angle and arrangement

Answer 38

The amide bond has low reactivity as the lone pairs of nitrogen are donating to the carbonyl group (making a resonance hybrid thingy). This results in a double bond character of C-N bond where amide bond is planar with the N-H orientated 180 degrees to C=O.

Question 39

What are the 4 levels of 3D protein structure

Answer 39

primary: amino sequence in peptide chain secondary: h bonding between N-H and O of the C=O below alters the local peptide geometry producing alpha helix or beta sheets. Tertiary: disulfide bonds alter entire protein shape - S - S bonds. Quartenary: different proteins aggregate to form new structures

Question 40

What is a radical and how does it arise

Answer 40

An atom, molecule ro ion that has one or more unpaired valence electrons. These are generated by homolytic bond cleavage where each partner gets just one unpaired electron. So you will only use fish hook arrows

Question 41

What are the three steps of Radical reaction

Answer 41

1. initiation: homolytic bond cleavage of a weak bond generates two radicals. This is facilitated by light (hv) and heat 2. Propagation: a free radical reacts with a molecule with no unpaired electrons to generate a new radical and new covalent bond. 3. Propagation is a repeating cycle that does not change the net number of radical species present and will continue until all the starting material is consumed or until Termination 4. Termination: when two radicals combine to generate a species with no unpaired electrons.

Question 42

What is more likely: termination or propogation in general and relating to the stability of radicals

Answer 42

Propogation because the concentration of radical species is low in radical reactions so more likely to interact with non radical rather than second radical. However if radicals are more stable they are more likely to survive long enough to find a second radical-> therefore termination more likely than unstable radicals

Question 43

What makes a radical more stable

Answer 43

generally radicals are highly reactive but delocalisation of charge through the ability to form resonance hybrids is one factors that increases radical stability.

Question 44

What are common radicals in the body and what damage do they cause if they build up : Oxidative stress

Answer 44

Hydroxyl OH with one unpaired e- and super oxide O2-. because they are highly reactive they can cause DNA damage, denature proteins, or disrupt lipid membranes. Radicals like to steal the H adjacent to an unsaturated fatty acid as the resulting radical is resonance stabilised and further attack leads to decomposition of fatty acid

Question 45

How does the body protect itself from radical damage: oxidative stress

Answer 45

It uses enzymes and antioxidants: Vitamin E is antioxidant and forms a more stable radical when it reacts with radical that doesn't damage the cell further. It is fat soluble so protects the lipid membranes.