Mgc Org

Front 1

What does solvent extraction exploit?

Back 1

exploits the particular solubilities of organic compounds in various solvents. Allows the chemist to separate a substance froma mixture of substances by adding a solvent that the cmopound of interest is highly soluble in.

Front 2

distribution (or partition) coefficient

Back 2

Ratio of the substance's solubilities in the two solvents is called the distribution (or partition) coefficient: Kd=Ca/Cb

Front 3

Explain what the simplest liquid extraction is.
What is it good at removing?

Back 3

simplest liquid-liquid extraction is accomplished when an organic compound is extracted with water b/c it removes substrances that are highly polar or charged like inorganic salts, strong acids and bases, and polar, low MW compounds (less than 5 carbons) such as alcohols, amines, and carboxylic acids.

Front 4

What can organic compounds that are basic (sa. amines) be extracted from?

Back 4

Organic compounds that are basic (ie. Amines) can be extracted from mixtures of organic compounds upon treatment with dilute acid (usually 5-10% HCl).

Front 5

Explain the process of extraction of organic amines.

Back 5

- This treatment will protonate the basic functional group, forming a positively charged ion.
- Resulting cationic salts of these basic compounds are usually freely soluble in aqueous solution and can be removed from the organic compounds that remain dissolve in the organic phase.

Front 6

Isomer

Back 6

same molecular formula, different structure

Front 7

Constitutional isomers

Back 7

different bond connectivity

Front 8

stereoisomers

Back 8

same bond connectivity, different spatial arrangement of atoms

Front 9

Enantiomers

Back 9

Non-identical mirror images (chiral molecules)

Front 10

Meso

Back 10

contains chiral centers and an internal plane of symmetry

Front 11

diastereomers

Back 11

stereoisomers that aren't enantiomers

Front 12

geometric isomers

Back 12

diastereomers that differ in orientation of substituents around a ring or double bond

Front 13

epimers

Back 13

diastereomers that differ in absolute configuration at only one chiral center

Front 14

anomers

Back 14

epimers that form as a result of ring cluser.

Front 15

Compare the reactivity of R-X's for E1 and E2 reaction and reasoning.

Back 15

Both E1 and E2 prefer tertiary the most.
E1: 3 > 2 >> 1 (b/c of carbocation stability)
E2: 3 > 2 > 1 (due to product stability)

Front 16

What is the role of solvent in E1 and E2 reactions?

Back 16

- E1 reactions are favored by protic (hydrogen bonding) solvents. (This stabilizes the carbocation).
- E2 reactions are favored by polar, aprotic (non-hydrogen bonding) solvents.

Front 17

Compare the product compositions of E1 and E2 reactions.

Back 17

- E1 reaction: more substituted alkenes are favored over less substituted ones.
- E2 reaction: Alkene stereochemistry is determined by the anti arrangement of the beta-proton and the leaving group.

Front 18

What are the intricacies of the E2 reaction in terms of which proton is removed?

Back 18

E2 rxns works best if the proton to be removed is anti to the leaving group, similar to the way the nucleophile must approach from the backside in an SN2 rxn. Since the rxn occurs in a single step and with the beta-proton anti to the LG, the stereochemistry of the dbl bond is determined by the rxn transition state.

Front 19

What are the kinetics of an E1 and E2 rxn.

Back 19

E1: reaction rate=k[haloalkane]
E2: reaction rate=k[base][haloalkane]

Front 20

What are the conditions of an E1 and E2 rxn?

Back 20

E1: 1st order elimination rxns are favored over substition rxns by the use of weak bases and high temperatures.

E2 conditions: Elimination rxns are favored over substition rxns by the use of strong bases and high temperatures.

Generallyh speaking, strong bases such as -OH and -OR yield exclusively the elimination product(s).

Front 21

Compare the reactivity of R-OH's in an SN1 and SN2 rxn.

Back 21

SN1: 3 > 2 >> 1 (due to stabilization of the carbocation)

SN2: 1 > 2 >> 3 (b/c of steric hindrance)

Front 22

Compare the rearrangements of an SN1 and SN2 rxn.

Back 22

SN1: carbocation rearrangement is possible. if the carbocation can rearrange to one that is more stable, it will.

SN2: Not possible, because carbocations aren't present in the reaction

Front 23

Compare the role of the solvent in SN1 and SN2 rxns.

Back 23

SN1: SN1 rxns are favored by protic (hydrogen bonding) solvents (this stabilizes the carbocation.)

SN2: Sn2 rxns are favored by polar, aprotic (non-hydrogen bonding) solvents.

Front 24

Compare the role of the stereochemistry in SN1 and SN2 rxns.

Back 24

SN1: Almost complete racemization due to nucleophilic attack on either side of p orbital. However, usually a little net inversion due to ransient sheilding of LG. Net inversion increases with increasing solvent polarity.
SN2 rxns: Complete stereochemical inversion of the carbon that is being attacked by the nucleophile.

Front 25

Compare SN1 and SN2 kinetics.

Back 25

SN1: rxn rate = k[electrophile]

SN2: rxn rate = k[nucleophile][electrophile]

Front 26

If the resonance contributors are similar in that they are equally stable, what would you expect?

Back 26

That they produce equal % products.

Front 27

If a product has a greater portion of the products produced, then how does this relate to its activation energy?

Back 27

Its activation energy should be lower than that of the other products.

Front 28

- what is a conjugated system? - which is more stable, conjguated or unconjugated

Back 28

- molecules having alternating double bonds - more stable than unconjugated system because it has resonance

Front 29

- Heat of hydrogenation; - Is something with greater heat of hydrogenation more stable or less stable?

Back 29

- Heat released upon the addition of H2 to one mole of a compound (to generate the corresponding alkane) - this will usually be an exothermic process - both starting material and products are organic compounds - Greater heat of hydrogenation for same molecular formula, it is LESS STABLE

Front 30

Endothermic reaction

Back 30

Heat is absorbed during this process (heat taken in)

Front 31

Exothermic reaction

Back 31

heat released (heat give out)

Front 32

Heat of Combustion

Back 32

Heat released when one mole of a compound undergoes complete combustion in O2. - combustion (released CO2 and H2O) usually exothermic process - the organic compound is a starting material for teh reaction

Front 33

Heat of Formation

Back 33

- heat released if one mole of a compound were formed from its component elements in their standard state - can be either endothermic or exothermic process - important to make sure they are balanced when drawing diagrams. - the organic compound is a product for the reaction

Front 34

Is something with greater heat of hydrogenation more stable or less stable?

Back 34

- Greater heat of hydrogenation for same molecular formula, it is LESS STABLE

Front 35

(T/F) Iodine in the presence of UV light will break up into radicals that can abstract hydrogens.

Back 35

FALSE. It will break up into radicals but it has great difficult abstracting hydrogens. - Iodine is often used as a free-radical inhibitor b/c it will readily react with other radicals. IODINE INHIBITS REACTIONS, IT RARELY INITATES THEM

Front 36

After radical intiaitoin step by UV light, what would slow down or stop the halogenation reaction?

Back 36

water b/c it prevents propagation by mopping up the bromine radical

Front 37

What is not an example of a conjugated system: a. benzene b. 1,3-cyclohexadiene c. 1,2-butadiene d. cyclobutadiene

Back 37

1,2 butadiene. (draw it out)

Front 38

What happens when an alkene is reacted with H2O, H+/heat?

Back 38

The OH goes to the more substituted markovnikov.

Front 39

Will alkanes be substitued by a bromide from hydrogen bromide?

Back 39

No alkanes are hihgly unreactive molcuels alkanes which will only undergo substitution when extreme conditions are employed -=- like UV light for instance. UV light will initiate highly reactive free radicals which will now react. (Br2 will also work with hv to get radical rxn with alkanes)

Front 40

glucoside

Back 40

a carbohydrate in its acetal or ketal form, in which the anomeric hydroxyl group is replaced with an alkoxy group. (oxygen attached to another monosaccharide or a simple alkyl group)

Front 41

- glucoside bond
- alpha-glucoside linkage

Back 41

- one in which the glucose subunit is attached to an alkoxy group
- alpha correponds to the downwards or axial orientation of the hydroxyl group on the anomeric carbon. In summary: look for a linkage btw glucose and its acetal group which is oriented axially or downward on the page.

Front 42

Benedict's reagent

Back 42

used to test for reducing sugars; in other words, those sugars that contain hemiacetal or hemiketal groups.
- when it reacts with the reducing sugar, a brick-red copper oxide precipitate is formed -- characteristic of a postive test.
- reaction only occurs when himacetals and hemiketals exist in equilbrium with their open chain forms.

Front 43

carbonyl group

Back 43

C=O

Front 44

Does glucose have a carbonyl group?

Back 44

Yes.

Front 45

Do some dissacharides possess free hemiacetal and himketal groups?

Back 45

Yes, maltose consists of two glucose residues, one of which is present in the hemiacetal form.

Front 46

alpha-anomers

Back 46

defined as those in which the hydroxyl group on the anomeric carbon in the ring is oriented AXIALLY. (straight up or down)

Front 47

Mutarotation

Back 47

- the change in specific rotation that occurs when an α (alpha) or β (beta) hemiacetal form of carbohydrate is converted into an equilibrium mixture of the two forms.
- also refers to the interconversion between the α and β forms.

Front 48

- Is sucrose a reducing sugar?
- Is fructose, glucose, glyceraldehyde, lactose, arabinose, and maltose reducing surgars?

Back 48

- No.
- Yes.

Front 49

What is a reducing sugar?

Back 49

- A reducing sugar occurs when its anomeric carbon is free, which is the chiral carbon.
- ANy sugar that contains a hemi-acetal wil lbe a reducing sugar
- Reducing agent: element or a compound in a redox reaction that reduces another species. In doing so, it becomes oxidized and is therefore the electron donor in the redox.

Front 50

Reducing agent;

Back 50

- Reducing agent: element or a compound in a redox reaction that reduces another species. In doing so, it becomes oxidized and is therefore the electron donor in the redox.
- 2Mg(s) + O2 => 2Mg2+(s) + 2O2-
here Mg(s) is the reducing agent... O2 is being reduced.

Front 51

What makes a strong reducing agent?

Back 51

- A strong reducing agent is one which loses or donates electrons easily.
- Nucleus in an atom is the main force pulling the electrons inwards, keeping them from going anywhere. ==> if atomic radius is large, infludence of the nucleus is not sufficient.
==> Elements with large atomic radii are strong reducing agents since no significant force keeps them from releasing electrons easily.
- Reducing agents can be ranked by increasing strength by ranking their oxidation potentials: strongest when it has a more postiive oxidation potential and will be week reducing agent whenever it has a negative oxidation potential. (in order to tell, change sign if in reduction potential)

Front 52

Nucleophile

Back 52

describes the tendency of an electron rich species to be attracted to the positive nuclear charge of an electron poor species, the electrophile .

Front 53

Nucleophilicity trends (compared with basicity)

Back 53

1) Across a row in the periodic table nucleophilicity (lone pair donation) C- > N- > O- > F- since increasing electronegativity decreases the lone pair availability. This is the same order as for basicity.
2. If one is comparing the same central atom, higher electron density will increase the nucleophilicity,
e.g. an anion will be a better Nu (lone pair donor) than a neutral atom such as HO- > H2O. This is the same order as for basicity.
3. Within a group in the periodic table, increasing polarisation of the nucleophile as you go down a group enhances the ability to form the new C-X bond and increases the nucleophilicity, so I- > Br- > Cl- > F-. The electron density of larger atoms is more readily distorted i.e. polarised, since the electrons are further from the nucleus.
Note that is the opposite order to basicity (acidity increases down a group) where polarisability is much less important for bond formation to the very small proton.