Tprorgo

Front 1

What is Resolution?

Back 1

The process by which 2 enantiomers are separated from each other.

Front 2

In which of the following solvents should disparlure be most soluble?

Back 2

Ethyl acetate:
Disparlure has 2 long alkyl chains. FOr this reason, it has hydrophobic properties. Water, methanol, and theanol are all hydrophilic solvents because of their small size and ability to hydrogen bond. In fact, water, methanol, and ethanol are all miscible with water. Ethyl acetate, on the other hand cannnot extensively hydrogen bond and is immiscible with water. This indicates hydrophobic properties, so disparlure should be most soluble in ethyl acetate.

Front 3

The most stable conformational isomer of 1,2-dibromoethane is?

Back 3

Staggered, anti:
Conformations in which the single bonds are staggered are more stable than those in which they are eclipsed, due to steric repulsion. Conformations which put the largest substituents in an anti (180 deg) arrangement are more stable than those in a gauche (60 deg) arrangement.

Front 4

Features of SN2 reaction:

Back 4

1) Bimolecular kinetics
2) A single-step mechanism: With the nucleophile attacking opposite the leaving group in a pentacoordinated transition state.
3)A pentacoordinated transition state.
There is NO carbocation intermediate which is a characteristic of E1 and SN1 reactions.

Front 5

Which one will most easily undergo an SN2 reaction, CH3I or CH3F?

Back 5

CH3I:
Since SN2 reactions proceed through a pentacoordinated transition state, they ccur more easily at primary carbons than at secondary or tertiarty carbons. Since iodine is a better leaving group than fluorine due to its larger size, the reaction will proceed most easily with CH3I.

Front 6

E1 reactions are characterized by:

Back 6

1) First-order kinetics: The rate of the reaction is dependent only on the concentration of 1 compound, begins with the breaking of a carbon-leaving group bond to form a carbocation intermediate. In a second step, deprotonation of an alpha-hydrogen by a base results in the formation of a pi bond.
2) Pi-bind formation
3) A 2-step mechanism
4) No nucleophilic attack on a carbocation intermediate: Nucleophilic attack on the carbocation is a characteristic of SN1 reacions.

Front 7

E2 reactions, all of the following occur in the rate determining step:

Back 7

1) Departure of the leaving group.
2) rehybridization of the 2 reacting carbon centers from sp3 to sp2.
3) Formation of a C=C double bond.
4) NO formation of a carbocation.
*The E2 reaction occurs in 1 step, with the dissociation of the leaving group occuring concomitant with deprotonation to yield a C=C double bond. the 2 carbon atoms involved are rehybridized from sp3 to sp2 during this reaction. The formation of a carbocation is characteristic of unimolecular reactions (E1 and SN1)

Front 8

2-Chloro-2-methylbutane more readily undergoes SN1 type reactions than SN2 reactions when in the presence of sodium alkoxide. IF the concetnration of the nucleophile is increased, then the:

Back 8

RATE OF REACTION DOES NOT CHANGE:
The rate law for SN1 reactions is:
Rate=k[substrate]
Since the nucleophile concentration does not affect the rate in SN1 reactions.

Front 9

1)Free-radical Initiation step
2)Free-radical chain propagation step
3)Free-radical chain termination step

Back 9

1)Free-radical Initiation step: The number of radicals increases from 0 to 2.
2)Free-radical chain propagation step: Radical propagation steps do not involve a change in the # of radicals of the reacting species.
2)Free-radical chain termination step: Involves an overall decrease in the # of radicals from 2 to 0.

Front 10

The SN2 reaction that occurs when hydroxide ion is added to an alkyl halide is considerably slowed when the reaction mixture is placed in a solvent. The solvent most probably?

Back 10

SOLVATES THE HYDROXIDE ION:
Solvation of the starting material, thereby stabilizing it, increases the activation barrier and slows the reaction rate.

Front 11

Increasing the size of substituents on alkyl halides decreases the rate of SN2 substitution. This effect occurs because increased size of substituents leads to:

Back 11

INCREASED ENERGY OF TRANSITION STATE:
The rate of SN2 decreases, because it is more difficult for the nucleophile to attack the hindered carbon, and this increases the energy of the transition state.

Front 12

E1 reactions are characterized by:

Back 12

1) First-order kinetics: The rate of the reaction is dependent only on the concentration of 1 compound, begins with the breaking of a carbon-leaving group bond to form a carbocation intermediate. In a second step, deprotonation of an alpha-hydrogen by a base results in the formation of a pi bond.
2) Pi-bind formation
3) A 2-step mechanism
4) No nucleophilic attack on a carbocation intermediate: Nucleophilic attack on the carbocation is a characteristic of SN1 reacions.

Front 13

E2 reactions, all of the following occur in the rate determining step:

Back 13

1) Departure of the leaving group.
2) rehybridization of the 2 reacting carbon centers from sp3 to sp2.
3) Formation of a C=C double bond.
4) NO formation of a carbocation.
*The E2 reaction occurs in 1 step, with the dissociation of the leaving group occuring concomitant with deprotonation to yield a C=C double bond. the 2 carbon atoms involved are rehybridized from sp3 to sp2 during this reaction. The formation of a carbocation is characteristic of unimolecular reactions (E1 and SN1)

Front 14

2-Chloro-2-methylbutane more readily undergoes SN1 type reactions than SN2 reactions when in the presence of sodium alkoxide. IF the concetnration of the nucleophile is increased, then the:

Back 14

RATE OF REACTION DOES NOT CHANGE:
The rate law for SN1 reactions is:
Rate=k[substrate]
Since the nucleophile concentration does not affect the rate in SN1 reactions.

Front 15

1)Free-radical Initiation step
2)Free-radical chain propagation step
3)Free-radical chain termination step

Back 15

1)Free-radical Initiation step: The number of radicals increases from 0 to 2.
2)Free-radical chain propagation step: Radical propagation steps do not involve a change in the # of radicals of the reacting species.
2)Free-radical chain termination step: Involves an overall decrease in the # of radicals from 2 to 0.

Front 16

SN2 reaction characteristics;

Back 16

1) 2nd order rate of reaction: Bimolecular and therefore, dependent on concentration of subtrate and nucleophile. Rate=k[substrate][nucleophile]
2) Complete inversion of stereochemistry
3) Reactivity sequence of 1>2>3.
4) Absence of rearrangement.

Front 17

What method would be best to use to differentiate between different isomers?

Back 17

NMR: Only NMR would give the detailed information needed to differentiate the isomeric products.

Front 18

What alcohol would dehydrate the fastest to form alkene?

Back 18

The most substituted alcohol will eliminate the fastest.

Front 19

How many stereoisomers of Camphor which has 2 stereocenter have?

Back 19

Generally the maximum # of possible stereoisomers is 2^n, where n= the # of chiral centers, so we would expect a max of 2^2=4 stereoisomers. However, because of the cyclic nature of the compound, not all the stereocenters of camphor can be inverted. This compound has only one enantiomer and no diastereomers; therefore, there are only 2 stereoisomers of this compound.

Front 20

The acid-catalyzed addition of water across the C=C of propene is characterized by all of the follwing?

Back 20

1) Carbocation formation.
2) The 1st step in the mechanism involves protonation of the C=C double bond.
3) Alcohol product formation.
*The acid-catalyzed addition of water to a C=C double bond begins with protonation of the double bond to form a carbocation. Nucleophilic attack by water results in formation of an alcohol with Markovnikov regiochemistry.

Front 21

The mechanism for addition of HBr to an alkene in the presence of peroxides:

Back 21

Has a planar sp2-hybridized intermediate:
In the presence of peroxides, HBr adds to a C=C double bond via a radical mechanism. Addition of the bromine radical to the double bond produces a planar, sp2-hybridized carbon radical. Note that becaue of the radical mechanism, this reaction, in contrast tomost electrophilic addition reactions, proceeds with anti-Markovnikov regiochemistry (the H is added to the more-substituted carbon of the double bond, while the Br is added to the less-substituted carbon).

Front 22

In electrophilic aromatic substitution, the aromatic ring acts as a:

Back 22

NUCLEOPHILE:
In electrophilic aromatic substituion, the aromatic ring acts as a nucleophile, attacking an electrophile in an electrophilic addition reaction. Deprotonation of the aromatic ring at the site of the attack reforms the double bond (an elimination reaction) and restoresaromaticity . the overall reaction substitutes an electrophile for a hydrogen on the aromatic ring.
-Typically an extremely reactive electrophile is used because the aromatic ring is a very poor nucleophile.

Front 23

Which of the following substituents increases the acidity of a substituted phenol as compared to the unsubstituted form?
I. -CH3
II. -NO2
III. -CN

Back 23

-NO2 and -CN: they are electron-withdrawing groups so they increase pheol acidity.

Front 24

Cis-2-Butene is less stable than trans-2-butene because cis-2-butene:

Back 24

Has Steric Strain between the 2 bulky substituents:
Cis-Alenes are always less stable than their trans conformers because of steric strain.

Front 25

One would expect the magnitude of the heat of hydrogenation of 1,3-hexadiene to be?

Back 25

SLIGHTLY LESS THAN TWICE THAT OF 1-HEXENE: Because the 2 double bonds are conjugated they are slightly stabilized. The heat of hydrogentation will be slightly less than the value for 2 isolated double bonds.

Front 26

What reagents will react w/ alkenes to give syn-addition products?

Back 26

I. Hydroboration:
1)BH3 2) H2O2 + -OH
II. Catalytic Hydrogenation:
H2, cat. Pd

Front 27

Whats the correct rank of halides in their order of increasing rate of addition to 3-hexene in a nonpolar, aprotic solvent?

Back 27

HCl<HBr<HI:
HI is the strongest acid, so it will protonate the most quickly (followed by HCl). Also, I- is the best nucleophile of the choices, so it will add to the carbocation the most quickly.

Front 28

What 2 steps may be used to synthesize 1-chloro-4-nitrobenzene, the starting material in the reaction from benzene?

Back 28

1) Cl2/FeCl3
2) HNO3/H2SO4
Of the 2 substituents, the Chloro group is para-directing so it should be installed first. While HCl/H2O will add H/Cl across alkenes, these conditions will not substitute Cl in EAS reactions. Also not that Na,NH3 results in the single trans hydrogenation of alkenes but will not install a nitro group in an EAS reaction.

Front 29

Penicillins are relatively nontoxic, because they do not affect human cells. the most likely reason for this is that:

Back 29

HUMAN CELLS DO NOT FORM PEPTIDOGLYCAN BARRIERS TO THE ENVIRONMENT:
Bacterial cells are sensitive to penicillins since penicillins block the synthesis of peptidoglycan cell walls, which bacteria need to survive. Human cells however, lack this structure and are therefore not sensitive to this action.

Front 30

Penicillins can be taken orally if they can survive the low pH of the stomach. THe amide bond between the R group and 6-APA of dicloxacillin and nafcillin is stable at low pH because?

Back 30

AMIDE BONDS DO NOT EASILY UNDERGO HYDROLYSIS:
Since amides are more stable than carboxylic acids, their hydrolysis is uphill in energy.

Front 31

Which of the three compounds:
Methyl Salicylate
Salicylic Acid
Aspirin
Would be soluble in aqueous sodium bicarbonate?

Back 31

Salicylic Acid and Aspirin:
Compounds that have carboxylic acid would be soluble in aqueous sodium bicarbonate. The phenolic proton in Methyl Salicylate is not acidic enough to impart solubility.

Front 32

Treatment of salicylic acid with methanol and dry acid would yield an:

Back 32

ESTER:
A treatment of salicylic acid with methanol and dry acid are standard conditions for making an ester in a process known as FISHER ESTERIFICATION!!

Front 33

What alcohols can be oxidized to a carboxylic acid with KMnO4?

Back 33

ONLY PRIMARY ALCOHOLS CAN BE OXIDIZED TO A CARBOXYLIC ACID!

Front 34

Aldol reaction

Back 34

'Aldol' is an abbreviation of aldehyde and alcohol. When the enolate of an aldehyde or a ketone reacts at the α-carbon with the carbonyl of another molecule under basic or acidic conditions to obtain β-hydroxy aldehyde or ketone, this reaction is called Aldol Reaction.

Front 35

Which one of the acid derivatives can NOT be used to synthesize the Met-Ala-Leu tripeptide?

Back 35

CARBOXYLIC ACIDS:
Recall that the relative reactivity of carboxylic acid derivatives: acid chlorides> acid anhydrides> esters> amides. The peptide (amide) bond may be synthesized by addition-elimination reactions with acid chlorides, esters, and other amides. However, the fast acid-base reaction prevents addition-elimination reactions from occuring between amines and carboxylic acids.

Front 36

An experiment demonstrated that reducing agents broke the 51-residue isulin molecule into 20-residue and 31-residue peptides. This experiment indicates the presence of what in insulin?

Back 36

DISULFIDE BONDS:
2 Cysteine residues may form disulfide bonds within proteins to stabilize its tertiary or quaternary structure. Since disulfide bonds are an oxidized form of cysteine, they are broken by reducing agents. While disulfide bonds commonly stabilize the tertiary structure of proteins, it si apparent that the disulfide bonds of insulin bridge 2 distinct polypeptides. Therefore, they support the quarernary structure of insulin.

Front 37

During acid hydrolysis of proteins, which amino acid side chain might also be hydrolyzed?

Back 37

GLUTAMINE:
Since acid hydrolysis cleaves the peptide amide bond between amino acid residues, it can also hydrolyze amide side chains. This could convert glutamine to glutamic acid and asparagine to aspartic acid.

Front 38

All amino acids-except glycine- are naturally optically active. However, during acid hydrolysis, amino acids may lose their optical activity. Formation of which of the following species may account for this phenomenon?

Back 38

ENOLS:
Amino acids contain carbonyls, and under certain conditions may reversibly tautomerize to an enol-containing compound. Upon reversal of tautomerism, the C alpha mayb be protonated in such a way as to form the D amino acid or the L-amino acid. Thus a racemic misture is formed and the amino acid loses its optical activity.

Front 39

Of the following, which would be the best choice of solvent for an aldol addition?
A) Ethanol
B) Diethyl ether
C) Acetone
D) Ethyl acetate

Back 39

DIETHYL ETHER:
Since the aldol rection involves the use of strong base, usable solvents must not be either acidic or electrophilic; this eliminates ethanol, acetone, and ethyl acetate. Therefore, of the choices given, only diethyl ether is suitable.

Front 40

TLC and Rf

Back 40

The less polar a compound is, the faster it will travel during TLC, and the larger Rf value it will have.

Front 41

Which of the following molecules will have an infrared stretch nearst to 2250 cm-1?
A) CH3OH
B)(CH3)2CO
C)H2C=CH2
D) CH3CN

Back 41

CH3CN:
THe region from 2100-2400 cm-1 of the infrared spectrum shows the characteristic absorptions of C/C and C/N triple bonds. OF the molecules listed only CH3CN contains a triple bond.

Front 42

When an aldehyde is added to dilute acid it undergoes?

Back 42

ALDOL CONDENSATION, FORMING A HYDROXYLATED ALDEHYDE:
ALDOL REACTIONS ARE FAST AND OCCUR IN EITHER ACIDIC OR BASIC MEDIA.

Front 43

The acid-catalyzed hydrolysis products were identified by spectral analysis and qualitative analysis. It was observed that tert-butyl alcohol was more soluble in water than n-butyl alcohol. Which of the following statements support this conclusion?

Back 43

1) The tert-butyl group is smaller and requires less room than the n-butyl group.
2) The tert-butyl group breaks fewer water hydrogen bonds.
3) A more complex orderly solvation shell is formed around n-butyl alcohol.
4) Solubility of an alcohol decreases as the hydrocarbon chain length increases.
-Branched hydrocarbons take up less space in aqueous solution tahn their unbranched equivalents. This means they disrupt fewer water hydrogen bonds and require a less complex solvation shell (orderly solvation shells are unfavorable entropically)

Front 44

Acid and base hydrolysis of esters are different in the following ways?

Back 44

1) Acid-catalyzed hydrolysis is more readily reversible.
2) Base-catalyzed hydrolysis has a larger equilibrium constant. Less readily reversible reactions have larger Ks.

Front 45

The reaction of ehtyl 2-hydroxypropanaoate with ammonia yields:

Back 45

2-hydrocypropanamide and ethanol:
Amides (carboxylic acid and amine) are more stable than esters (carboxylic acid and alcohol). Therefore, the ester ("oate") will be converted to an amide ("-amide") Note that the alcohol part of the original ester will be a byproduct of the reaction.

Front 46

Which amino acid would most likely be present in the hydrophobic binding region of a protein?

Back 46

VALINE:
Valine is the only one with a hydrophobic side chain.

Front 47

If an amino acid is at its isoelectric point, then a decrease in pH will most likely cause the amino acid to become?

Back 47

POSITIVELY CHARGED:
Decreasing the pH (making the solution more acidic) will tend to protonate basic groups. This will make the amino acid positively charged.

Front 48

Naturally-occuring sugars may have which of the following configurations?

Back 48

1)D:Most biologically active sugars have the D configuration. Thjis means that the last chiral carbon has the same configuration as D-glyceraldehyde.
2)L: There are a few examples of biologically-active L sugars. For example, L-galactose is a constituent of some polysaccaride structures, and L-arabinose, a pentose, is an important part of cell walls and plant glycoproteins.
3)Alpha and Beta: Alpha and beta refer to the configuration of carbon #1, the anomeric carbon, when a sugar is in the furanose or pyranose (ring) form. the alpha and beta forms are said to be anomers of each other.

Front 49

What is correct regarding classification of carbohydrate isomers?

Back 49

EACH D-ALDOHEXOSE HAS EXACTLY 2 ANOMERS:
When 6 Carbon sugar cyclizes, yielding a pyranose, a bond known as a hemiacetal is formed. This is a linkage b/w the OH at C #5 and the anomeric carbon, C-1. The carbonyl of C-1 is converted to an alcohol. The resulting C-1 hydroxyl may be in the axial or equatorial position on the ring. These 2 possible ring forms are called anomers. If the OH at carbon #1 is in the down position (which is axial in the case of D-glucopyranose, or glucose), the ring is referred to as the alpha anomer; when it is in the up position, the structure is called the beta anomer. A monosaccharide in solution freely converts between the open chain form and the 2 anomeric forms; this is known as mutarotation. It is necessary to specify the anomeric form b/c the acetal linkage (glycosidic bond) b/w monosaccharides in di- and polysaccharides prevents mutarotation. Enzymes accept only 1 of the 2 anomers in the formation and hydrolysis of glycosidic linkages. Hence all the linkages in starch and glycogen are alpha-1,4 glycosidic linkages, whereas all the bonds in cellulose are beta-1,4. Even though all 3 macromolecules are simple glycose polymers, our cells can digest only the former 2, since we lack enzymes for the hydrolysis of Beta-glycosidic linkages.

Front 50

Phosphoric acid esters of nucleosides are known as:

Back 50

NUCLEOTIDES:
A nucleotide is a phosphoric acid ester of a nucleoside.
Nucleotide:Any of various compounds consisting of a nucleoside combined with a phosphate group and forming the basic constituent of DNA and RNA.
Nucleoside: Any of various compounds consisting of a sugar, usually ribose or deoxyribose, and a purine or pyrimidine base, especially a compound obtained by hydrolysis of a nucleic acid, such as adenosine or guanine.

Front 51

The reaction conditions typically used to hydrolyze a protein are:

Back 51

CONCENTRATED ACID, HEAT:
The standard conditions for peptide hydrolysis are concentrated HCl and several hours worth of reflux. The reaction time depends on whether partial or complete hydrolysis is desired.

Front 52

2,4-dinitrofluorobenzene is used in protein analysis to determine?
AND
The reaction mechanism by which 2,4-dinitrofluorobenzene reacts with a protein is?

Back 52

THE AMINO ACID AT THE N-TERMINUS:
2,4-dinitrofluorobenzene is used to identify the N-terminus of a peptide. It reacts with the N-terminal amino acid, and remains attached even after complete acid hydrolysis. This allows isolation of the N-terminal amino acid from the rest of the residues.
NUCLEOPHILIC AROMATIC SUBSTITUTION: The reaction mechanism by which 2,4-dinitrofluorobenzene reacts with an amine is called nucleophilic aromatic substitution. The nitro groups are so electron withdrawing that they make the benzene ring quite electrophilic.

Front 53

Mutarotation

Back 53

Mutarotation is the interconversion between 2 anomers. If the anomeric carbon are stable as acetals they will not mutarotate. However, having a free hemiacetal, will mutarotate between the alpha and the beta anomers.

Front 54

Purifications Methods

Back 54

1) Extraction: Separates dissolved substances based on differential solubility in aqueous versus organic solvents.
2) Filtration: Separates solids from liquids.
3) Recrystallization: Separates solids based on differential solubility; temperature is important here.
4) Sublimation: Separates solids based on their ability to sublime.
5) Centrifugation: Separates large things (like cells, organelles and macromolecules) based on mass and density.
6) Distillation: separates liquids based on boiling point, which in turn depends on intermolecular forces.
7) Chromatography: Uses a stationary phase and a mobile phase to separate compounds based on how tightly they adhere (generally due to polarity but sometimes size as well).
8) Electrophoresis: Used to separate biological macromolecules (such as proteins or nucleic acids)based on size and sometimes charge.