MCAT - Organic Chemistry Flashcards

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0
Q

Coordinate Covalent Bond

A

One nucleus donates both of the electrons in the bond.

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1
Q

Electrostatic Forces

A

The attractive force between electrons and the nuclei that is responsible for all molecular bonds.

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2
Q

Hybridization

A

sp - 180° - linear

sp2 - 120° - trigonal planar

sp3 - 109.5° - tetrahedral, pyramidal, or bent

dsp3 - 90°, 120° - trigonal-bipyramidal, seesaw, t-shaped, linear

d2sp3 - 90° - octahedral, square pyramidal, square planar

When one or more shape is possible, it is determined by the number and position of the lone pairs of electrons.

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3
Q

Instantaneous Dipole Moment

A

Exists in an otherwise nonpolar molecule. Occurs because the electrons in a bond move about the orbital and at any moment may not be distributed exactly between the two bonding atoms, even when the atoms are identical.

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4
Q

London Dispersion Forces

A

Occur between two instantaneous dipoles; the weakest dipole-dipole force (vs. hydrogen bonds which are the strongest dipole-dipole forces).

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5
Q

Zwitterion

A

A neutral molecule with a positive and negative electrical charge at different locations within the molecule at a pH of 7.

Ex: amino acids -> amine group deprotonates the carboxylic acid group

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6
Q

Isomers

A

Molecules that have the same molecular formula but are different compounds.

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7
Q

Conformational Isomers (Conformers)

A

Not true isomers - different spatial orientations of the same molecule. At low temperatures, the anti-conformation is the most common.

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8
Q

Structural Isomers

A

Have the same molecular formula but different bond-to-bond connectivity; simplest form of isomer.

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9
Q

Stereoisomers

A

Two molecules with the same molecular formula and same bond-to-bond connectivity that are not the same compound. Like conformational isomers but contain at least one *chiral/stereo center (C bonded to 4 different substituents).

2 types:

  • enantiomers
  • diastereomers
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10
Q

Enantiomers

A

Stereoisomers that have opposite absolute configurations at EACH chiral center (mirror images). Cannot be separated by physical means.

When equal in concentration -> racemic mix

Resolution = the separation of enantiomers.

Same physical and chemical characteristics except in 2 cases:

  1. Reactions with other chiral compounds
  2. Reactions with polarized light
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11
Q

Diastereomers

A

Stereoisomers that are not mirror images (are not the same compound). Can be separated by physical means (crystallization).

Geometric isomer: special type of diastereomer, have different physical properties.

  • Cis: has dipole moment - strong intermolecular forces so has high boiling point; lower symmetry does not form crystals as easily so has lower melting point; has steric hindrance so has higher heats of combustion
  • Trans: no dipole moment - lower boiling point, higher melting point, lower heats of combustion

–> better to use E (opposite) and Z (same)

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12
Q

Epimer

A

Diastereomers that differ at only one chiral center.

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13
Q

Anomers

A

Two diastereomers formed from a ring closure at an epimeric C.

Anomeric carbon: the chiral carbon of the anomer.

Ex: glucose - anomeric C determines if it’s alpha-glucose or beta-glucose.

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14
Q

Meso Compounds

A

Optically inactive (achiral) molecule with two chiral centers; have a plane of symmetry between their centers which divides the molecule into halves that are mirror images of each other.

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15
Q

Absolute Configuration

A

The only way to absolutely describe a chiral molecule.

R = rectus: right (clockwise)
S = sinister: left (counterclockwise)
*for when H (or other 4th priority substituent) is on DASHES.

Mirror images of chiral molecules always have opposite absolute configurations.

*note: retention of configuration does not mean that absolute configuration is retained; it means that there is no inversion.

stereoisomers of a chiral molecule = 2^n where n = # of chiral Cs

Abs config does NOT indicate direction in which a compound rotates plane-polarized light.

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16
Q

Relative Configuration

A

Two molecules have the same relative configuration about a C if they only differ by one substituent and all other substituents are oriented identically around the same C.

–> In SN2 reactions, it is the relative configuration that is inverted.

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17
Q

Polarimeter

A

Screens out photons from a light source to get only photons of a certain orientation of electric field.

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18
Q

Plane-Polarized Light

A

The resulting photons filtered by a polarimeter that all have the same orientation of electric field.

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19
Q

Observed Rotation

A

Gives the direction and degree to which a compound rotates the electric field in plane-polarized light.

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20
Q

Optically Active

A

When a compound does not contain any mirror images (only one stereoisomer present) so when plane-polarized light is projected through the compound, the orientation of the electric field is rotated.

+ or D: compound that rotates plane-polarized light clockwise
- or L: compound that rotates plane-polarized light counterclockwise

vs. racemic mix -> many electric field orientations = optically inactive.

Maximum # of optically active isomers in a compound = 2^n where n = # of chiral centers.

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21
Q

Simple Rotation

A

A standardized form of observed rotation arrived at through calculations using observational rotation and experimental parameters.

Specific rotation = observed rotation after the following experimental factors have been adjusted:

  • length of polarimeter
  • concentration of solution
  • temperature
  • type of wavelength of light used
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22
Q

Physical Properties of Alkanes

A

Increase MW (molec weight) = increase BP (boiling pt); increase MP (melting pt).

Increase branching = decrease BP; increase MP (crystal solids).

Alkanes have the lowest density (think: oil spill -> alkanes float on water).

The first four alkanes are gases at room temperature.

Similar properties in alkenes and alkynes (although alkynes are slightly more polar, slightly more soluble in H2O).

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23
Q

Ring Strain

A

Some ring structures put a strain on the C-C bonds because they bend them away from the normal 109.5° angle of the sp3 C and cause crowding.

–> ring strain is zero for cyclohexane and increases as the number of Cs increases and decreases from there (up until 9, after which it is zero again as Cs are added).

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24
Q

Combustion

A

A (radical) reaction when alkanes are mixed with oxygen and energy (heat) is added.

CH4 + 2O2 –(flame)–> CO2 + 2H2O + Heat

-> exothermic, so heat released is the measure of stability.

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25
Q

Heat of Combustion

A

The change in enthalpy of a combustion reaction. Combustion of isomeric hydrocarbons (same molec formula) requires equal amounts of O2 and produces equal amounts of CO2 and H2O so the heat of combustion is used to compare relative stabilities of isomers.

–> higher heat = more energy = less stable molecule

For cycloalkanes, comparisons can be made of different ring sizes on a “per CH2” basis.
Ex: molar heat of combustion for cyclohexane is almost twice as much as cyclopropane but “per CH2” group heat is much higher for cyclopropane because of ring strain.

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26
Q

Enthalpy (H)

A

A measure of the total energy of a thermodynamic system (a thermodynamic potential). It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.

–> a state function and an extensive quantity. Measured in Joules.

ΔH is positive in endothermic rxns
ΔH is negative in exothermic rxns

-> for an exothermic rxn at constant pressure, the system’s change in enthalpy equals the energy released in the rxn, including the energy retained in the system and lost through expansion against its surroundings.

–> ΔH = Q, where Q is the energy added to the system through heat.

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27
Q

Halogenation (Radical Rxns)

A

Alkanes react with halogens (F, Cl, Br, but not I) in the presence of heat or light to form a free radical.

Chain rxn:

  1. Initiation: the halogen starts as a diatomic molecule. It is homolytically cleaved by heat or UV light, resulting in free radicals.
  2. Propagation: the halogen free radical removes an H from the alkane resulting in an alkyl radical. The alkyl radical may now react with a diatomic halogen creating an alkyl halide and a new halogen radical. Most products are formed here, can continue indefinitely.
  3. Termination: either 2 radicals bond or a radical bonds to the wall of the container to end the chain rxn or propagation.

-> exothermic process.

Alkyl radical stability is the same as carbocation stability: 3° > 2° > 1° > methyl.

Alkyl radicals exhibit trigonal planar geometry.

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28
Q

Halogens

A

Order of reactivity: F, Cl, Br, I (I will not react with alkanes in halogenation).

Order of selectivity: I, Br, Cl, F
–> how selective a radical halogen is when choosing a position on an alkane.

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29
Q

Four Main Reaction Types

A
  1. Addition: A + B –> C
  2. Elimination: A –> B + C, one reactant splitting into two products.
  3. Substitution: A-B + C-D –> A-C + B-D
  4. Rearrangement: A –> B, a reactant undergoing bond reorganization to give an entirely new molecule.
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30
Q

Index of Hydrogen Deficiency (AKA Degrees of Unsaturation)

A

DU = [(2n - 2) - x]/2

n = # carbons
x = # hydrogens
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31
Q

Synthesis of Alkenes

A

Elimination: one or two functional groups are removed to form double bond

  1. Dehydration of an alcohol (E1)
  2. Dehydrohalogenation (E1 or E2 - strong bulky base for deprotonation)
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32
Q

Zaitsev’s (Saytzeff’s) Rule

A

The major product of elimination will be the most substituted alkene.

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33
Q

Reactions of Alkenes

A
  1. Catalytic hydrogenation: adds Hs, makes alkane
  2. Ozonolysis: oxidation (splits double bond and makes carbonyls) using O3
  3. *Electrophilic addition: attraction of electrophiles (often hydrogen halides) to the double bond.
  4. Hydration: addition of H2O in acidic conditions at low T (reverse of dehydration - at high T)
  5. Halogenation: X2 adds anti to either C in the double bond: first X (electrophile) will add to least-substituted C per Markov (with H2O, hydroxide will add second to the most-substituted C instead of other X).
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34
Q

Markovnikov’s Rule

A

The hydrogen will add to the least substituted carbon of the double bond.

-> hydrogen halides (HX) and hydration (H2O) addition to alkenes.

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35
Q

Oxymercuration/Demercuration

A

A two step process that follows Markovnikov’s rule;

  1. A partially dissociated mercury-containing reagent +Hg(OAc) (as The electrophile) forms a 3-membered ring with the C=C and then adds water in an *anti-addition (from opposite sides of the double bond) to make the organomercurial alcohol.
  2. A reducing agent and base is added for the formation of the alcohol through demercuration.
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36
Q

Anti-Markovnikov Addition

A

In the presence of peroxides (ROOR), Bromine from HBr, not the hydrogen, will add to the least-substituted carbon (does not apply with other halogens - those will still add Markovnikov).

–> hydroboration (BH3) to make alcohols (from the peroxide), syn-addition.

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37
Q

Benzene

A

Undergoes substitution, NOT addition. A flat molecule stabilized by resonance. An EWG (electron-withdrawing group).

-> ortho, meta, para

EWGs (like N+O2, N+R3, CCl3, ketones, aldehydes, esters, SO3H, CN) are meta-directors.
EDGs (electron-donating groups like O-, OH, NR2, OR, R) are ortho/para-directors.
- Halogens are exceptions -> EWGs but ortho/para-directors.

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38
Q

Huckel’s Rule

A

If a compound has planar, monocyclic rings with 4n+2 pi electrons (n being any integer, including 0), it is by definition an aromatic compound.

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39
Q

Substitution Reactions

A

SN1 and SN2 - numbers represent the order of the rate law (unimolecular and bimolecular) and NOT the number of steps.

-> SN1 has 2 steps.
1. LG leaves.
2. Nucleophile attacks.
First step is formation of the carbocation and is the RDS (slowest) -> concentration of the substrate (electrophile) is all that matters (not the nucleophile).
- Has carbocation intermediate so more stable (3°) is better.
- Carbocation intermediate is planar so nucleophile attacks from either side -> gives racemic mix.
- Often has competing elimination (E1) if the nucleophile also acts as a base.

  • > SN2 has 1 step.
    1. Nucleophile attacks the intact substrate from behind the LG and knocks LG off -> rate is dependent on the concentration of the nucleophile and the electrophile.
  • Doesn’t like steric hindrance so 1° is better (with strong base and hindrance in electrophile or nucleophile, will get E2 instead).
  • Inversion of configuration (stereochemistry - not necessarily absolute configuration).
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40
Q

SN1 vs. SN2

A

The nucleophile and the 5 Ss:
Nucleophile: SN2 requires a strong nucleophile while nucleophile strength doesn’t affect SN1.

  1. Substrate: SN2 rxns don’t occur with sterically hindered substrate. SN2 requires a methyl, 1°, or 2° substrate while SN1 requires a 2° or 3° substrate.
  2. Solvent: a highly polar solvent increases the rxn rate of SN1 by stabilizing the carbocation, but slows down SN2 rxns by stabilizing the nucleophile.
  3. Speed: the speed of an SN2 rxn depends upon [substrate] and [nucleophile], while the speed of an SN1 depends only on [substrate].
  4. Stereochemistry: SN2 inverts stereochemistry about chiral center, while SN1 creates a racemic mixture.
  5. Skeleton: SN1 may be accompanied by carbon skeleton rearrangement, but SN2 never rearranges the carbon skeleton.
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41
Q

Physical Properties of Alcohol

A

Follow trends similar to hydrocarbons, but alcohols hydrogen bond, giving them considerably higher BP, MP, and water solubility than similar-weight hydrocarbons.

Alcohols are LESS acidic than water. The order of acidity from strongest to weakest:
Methyl, 1°, 2°, 3°.
–> 3° alcohol has the most methyl groups (EDG to 3°C, prevent it from absorbing excess negative charge of the conjugate); the conjugate base of a 3° alcohol is the least stable so it is the least acidic.

–> EWG and EDG explain many rxns. For alcohol, placing an EWG on the alcohol increases its acidity by reducing the negative charge on the conjugate base.

-> most of the time, if an alcohol is a reactant, it will be acting as the nucleophile (because of lone pairs on O).

42
Q

Synthesis of Alcohols

A
From before (with alkenes):
1. Hydration of alkene
2. Nucleophilic substitution
3. Oxy/deoxymercuration
4. Hydroboration
Also:
5. Grignard synthesis
6. Reduction of a carbonyl: by LiAlH4, NaBH4, H2 + pressure
43
Q

Grignard Reagents

A

React as a nucleophile on C=O, C=N, C(triple)N, S=O, N=O, etc.
Strong enough base to deprotonate O-H, N-H, S-H, C(triple)C-H.
Made in ether, are incompatible with water and acids stronger than water.

44
Q

Why is it more difficult to reduce esters (RCOOR’) and acetates (CH3COOR) than ketones and aldehydes?

A

Because the group attached to the carbonyl of the ester or acetate is a stronger EDG than an alkyl group or hydrogen. By donating electrons more strongly, it reduces the positive charge on the carbonyl carbon making it less attractive to the nucleophile.

–> a hydride (H- from LiAlH4 or NaBH4) prefers the carbonyl with the greatest partial positive charge.

45
Q

Reactions of Alcohols

A

From before (with alkenes):
1. Dehydration
Also, alcohols like to be the nucleophile!
2. Nucleophilic addition: to alkene
3. Nucleophilic substitution: to LG to make alkyl halides, mesylates and tosylates (last 2 retain configuration).
4. Oxidation: for 1° or 2° alcohols –> (aldehydes), carboxylic acids or ketones by K2Cr2O7, KMnO4, H2CrO4, O2, Br2
5. Pinacol Rearrangement: dehydration of vicinal diol that results in an aldehyde or ketone b/c of methyl rearrangement.

46
Q

Ethers

A

Relatively non-reactive (other than epoxides), polar, can’t H bond with themselves but can H bond with other compounds that contain an H attached to N, O, or F.

–>*Ether is almost always the answer to solvent questions! Roughly as soluble in H2O as alcohols of similar MW but organic compounds tend to be much more soluble in ethers than alcohols because no H bonds need to be broken.
Have relatively low BPs like alkanes with similar MWs.

–> cleaved by the halo-acids HI and HBr to form the corresponding alcohol and alkyl halide.

Epoxide is a 3-membered cyclic ether. More reactive b/c of ring strain.

47
Q

Acidities of the Functional Groups

A
From weakest to strongest acid:
H3C-CH3 alkane < 
H2C=CH2 alkene < 
H2 dihydrogen < 
NH3 ammonia < 
HC(triple)CH alkyne < 
H3C-COH aldehyde < 
H3C-CH2-OH alcohol < 
H2O water < 
H3C-COOH carboxylic acid
48
Q

Stereoselective Reaction

A

A certain stereoisomer or set of stereoisomers predominate as products.

49
Q

Stereospecific Reaction

A

Different isomers lead to isomerically opposite products.

–> any reaction that is stereospecific is also stereoselective, but the converse is not true!

50
Q

Enol

A

Molecule with both an alkene and an alcohol at the same carbon.

51
Q

Boiling Point

A

Strong intermolecular bonds increase boiling point. Hydrogen bonds are the strongest intermolecular bonds so atoms that are better able to hydrogen bond will have higher BPs.

Ex: oxygen is more EN than nitrogen (makes stronger hydrogen bonds) so the BP of methanol will be higher than ammonia.

Carbonyls raise boiling point (higher than just hydrocarbons).

52
Q

Aldehydes and Ketones

A

Prefer Nucleophilic ADDITION (vs. carboxylic acids -> Nucleophilic Substitution).

Know: aldehyde (RCOH), formaldehyde/methanal (HCOH), ketone (RCOR), acetone/2-propanone (CH3COCH3)

More polar and have higher BP than alkanes and alkenes of similar MW, but cannot H bond with each other so lower BP than corresponding alcohols. Aldehydes and ketones with up to 4 Cs are soluble in water (and readily form hydrate). Aldehydes are slightly more acidic than ketones (alkyl groups are EDG).

–> accept H bonds - makes H2O, etc excellent solvents for these substances.

Mostly act as the substrate in Nucleophilic Addition or as a Brownsted-Lowry acid by donating an alpha-H (corresponding ion is the enolate with negative charge).

Adding an alcohol (ROH) to an aldehyde (R’COH) or ketone (R”COR”’) forms the hemiacetal R’HC(OH)OR and hemiketal R”R”‘C(OH)OR respectively.

–> aldehydes can be oxidized to carboxylic acids while ketones cannot.

53
Q

Tautomerization

A

Due to properties of the alpha-H and carbonyl, ketones and aldehydes exist at room temperature as tautomers, with the ketone or aldehyde tautomer favored.

NOT resonance -> hydride shift.

54
Q

Aldol Condensation

A

Aldehyde + Aldehyde
Ketone + Ketone
Aldehyde + Ketone
Catalyzed by an acid or base (here).
1. Deprotonation: base deprotonates alpha-H, leaving enolate ion.
2. Nucleophilic attack: enolate ion as the nucleophile attacks the carbonyl carbon to form an alkoxide ion.
3. Protonation: alkoxide ion is a stronger base than a hydroxide ion (b/c EDG alkyl attached to O) and deprotonates water to complete the aldol.
4. Condensation: strong base (OH-) does E2 dehydration rxn by deprotonation of the alpha-C to make the enal (subsequent double bond and original carbonyl).

  • > Aldol is unstable and is easily dehydrated by heat or a base to an enal which is stabilized by conjugated double bonds.
  • VERY important on MCAT
55
Q

Halogenation of Ketones

A

Halogen adds to the alpha-C in the presence of acid or base.

  • > Base: difficult to prevent halogenation at more than one of the alpha positions and also is consumed by the rxn with H2O as a by-product.
  • > Acid: more control of X addition and acts as a true catalyst (not consumed)
56
Q

Haloform Reaction

A
  1. Complete halogenation of alpha-C of a methyl ketone in basic conditions.
  2. Trihalo pdt reacts further with base to produce a carboxylic acid and a haloform (chloroform CHCl3, bromoform CHBr3, or iodoform CHI3).
57
Q

The Wittig Reaction

A

Converts a ketone (C=O) to an alkene (C=C).
1. A phosphorous ylide (a neutral molecule with a negatively charged carbanion) acts as the nucleophile toward the carbonyl C forming a betaine (beta-ine) with a P+ and an O-.
2. Betaine is unstable so cyclization quickly gives the alkene (C=C) and a triphenylphosphine oxide (P=O).
When possible, a mixture of cis and trans isomers is formed.

58
Q

The Tollens Test

A

Gives a “silver mirror” product for reducing sugars. Reducing sugars are hemiacetals in their ring form and either aldehydes or ketones in their straight-chain form. Acetals do not open easily because they contain blocking groups (in “formation of acetals” in book). Promotes enediol rearrangement of ketones and aldehydes.

59
Q

Carboxylic Acids

A

Prefer Nucleophilic SUBSTITUTION (vs. aldehydes/ketones -> Nucleophilic Addition).

Know: formic acid/methanoic acid (HCOOH), acetic acid/ethanoic acid (H3CCOOH), benzoic acid (PhCOOH)
-> salt of acetic acid, acetate (H3CCOO-), is often abbreviated -OAc.

Behaves as an acid or substrate in a *nucleophilic substitution rxn. Makes strong double H bonds to form a dimer - significantly increases BP by effectively doubling the MW. Double bonds in unsaturated carboxylic acids impede crystal lattice and lower MP. With 4 Cs or less they are miscible with water. Also soluble in most nonpolar solvents because of the dimer form.

60
Q

Transesterification

A

An alcohol reacts with an ester and one alkoxy group is substituted for another.

RCOOR’ + R”OH –(H+)–> RCOOR” + R’OH

61
Q

Amines

A

Considerations when dealing with nitrogen-containing compounds:

  1. They may act as a (weak) Lewis Base donating their lone pair of electrons.
  2. They may act as a Nu- where the lone pair of electrons attacks a positive charge.
  3. Nitrogen can take on a fourth bond (becoming positively charged).
Amine basicity from highest to lowest when the functional groups are EDG: 2°, 1°, ammonia.
Aromatic amines (attached to benzene) are much weaker bases than nonaromatic because the electron pair can delocalize. Other EWG substituents on benzene further weaken the amine.
Can also be weakened by large sterically-hindering substituents.

Hydrogen bonds with each other and water, increasing BP and solubility.

62
Q

Amine Condensation with Carbonyls

A

1° and 2° amines react with aldehydes and ketones losing water to produce imines (C=N-R) and enamines (C-NR2).

  1. Amine acts as the nucleophile attacking the carbonyl C,
  2. H2O forms as a LG,
  3. Dehydration (E2) of the alpha-C for 2° or of the N for 1°.

Imines and enamines exist as tautomers.

63
Q

Wolff-Kishner Reduction

A

Reduces a ketone or aldehyde by removing the oxygen and replacing it with two hydrogens (can do this by adding hot acid but some ketones and aldehydes can’t survive the conditions).

  1. Same mechanism of imine but uses hydrazine (H2N-NH2) to add to ketone or aldehyde.
  2. Produces a hydrazone (C=N-NH2) by nucleophilic addition.
  3. Hot strong base is added to deprotonate the second N and give the desired product with H2O and N2(g) as by-products.
64
Q

Amine Alkylation

A

A nucleophilic substitution rxn with amine acting as the nucleophile and an alkyl halide (RX) as the electrophile. Makes a poor LG (NH2) into a good LG (C-NR3+).

65
Q

Hoffman Elimination

A

E2 rxn of a 4° ammonium salt like [RN(CH3)3]+. Hot hydroxide (OH-) removes H from the C that will form the LEAST stable alkene (anti-Zaitsev) as the major product -> Hoffman product.

66
Q

Diazotization of an Amine

A

Primary aromatic amines (H2NAr) and nitrous acid (HNO2) react to form the diazonium ion (ArN+(triple)N:).

Diazonium group can be easily replaced by a variety of groups because it’s a good LG.

67
Q

Amides

A

Can behave as a weak acid or weak base. Less basic than amines due to EWG properties of the carbonyl. Hydrolyzed by either strong acids or strong bases.
Amides with an H attached to the N can hydrogen bond to each other (N-H–O).

Primary amides are named by replacing the -ic in the corresponding acid with -amide.
Ex: acetic acid (H3CCOOH) –> acetamide (H3CCONH2)

Substituents on the nitrogen are prefaced by “N-“.
Ex: N-ethylacetamide

68
Q

Lactams

A

Cyclic amides. Amides are the most stable of the carboxylic acid derivatives but beta-lactams are highly reactive due to large ring strain. Good electrophiles. Found in several antibiotics.

Beta: 4-membered ring (N attached to carbonyl C and beta-C)
Gamma: 5-membered ring
Delta: 6-membered ring

69
Q

Hofmann Degradation (Rearrangement)

A

1° amiDes react with strong basic solutions of chlorine or bromine to form 1° amiNes with CO2 as a by-pdt.

  1. Base deprotonates N.
  2. N picks up a halogen ion, making N more acidic than original amide.
  3. R group on amide carbonyl transfers to N (forms isocyanate).
  4. Isocyanate reacts with H2O forming a carboxylic acid group.
  5. Decarboxylation of N - CO2 is given off as by-pdt, leaving 1° amine.

–> useful because other methods of producing amines rely on the SN2 mechanism which prevents the production of amines on a 3° carbon. This can produce amines with 1°, 2°, or 3° alkyl positions.

70
Q

Phosphoric Acids

A

H3PO4. Forms phosphoric anhydrides (two acyl groups - phosphorous here - on the same oxygen) when heated. Forms esters when reacted with alcohols. Negative charge at pH 7 gives stability (less susceptible to nucleophilic attack).

ATP -> triphosphate with ester and anhydride linkages.

71
Q

Fatty Acids

A

Long carbon chains with a carboxylic acid end. Have three basic functions in the human body:

  1. Serve as hormones and intracellular messengers.
  2. Components of phospholipids and glycolipids in cell membranes.
  3. Act as fuel for the body (stored as triglycerols in adipose cells).

Nonpolar even though they’re amphipathic (hydrophobic and hydrophilic ends) because the hydrophobic carbon chain predominates.

Saturated fatty acids have the greatest energy storage potential.

72
Q

Lipolysis

A

Triacylglycerols are hydrolyzed to form glycerol and the corresponding fatty acids (reverse of esterification). Takes place in adipose cells when blood levels of epinephrine, norepinephrine, glucose, or ACTH are elevated. The resulting fatty acid products are then exported to other cells for utilization of energy.

Once inside the cell, the fatty acid is linked to Coenzyme A and carried into the mitochondrial matrix by g-amino acid L-carnitine. The fatty acid is then oxidized two carbons at a time with each oxidation yielding a NADH, FADH2, and an acetyl CoA. Each acetyl CoA enters the Krebs cycle for further oxidation by condensation with oxaloacetate.

73
Q

Saponification

A

In the lab, triacylglycerols are cleaved into glycerol and fatty acids by the addition of NaOH. Use in the production of soap.

74
Q

Amino Acid

A

The building blocks of proteins. Has a carboxylic acid end and an amine end. Peptide bond forms amide group -> resonance of the N and the O allow for a partial double peptide bond that prevents rotation and contributes to the secondary and, to some degree, the tertiary structure.

The body has 20 nonessential (alpha) amino acids and requires 10 essential amino acids from the diet.
The form of the amino acid depends on the pH –> dipolar ion at pH = 7, fully saturated in acidic conditions and unsaturated in basic conditions. Carboxylic acid is the first to lose its proton.

The number of possible polypeptides that contain one each of n amino acids is “n!”.
Ex: for n = 3, n! = 3x2x1 = 6 possible combinations.

75
Q

Side Chain

A

The R groups; have different chemical properties which are divided into four categories:

  1. Acidic - also polar (carboxylic acids)
  2. Basic - also polar (amines)
  3. Polar
  4. Nonpolar

The first three will increase the amino acid’s solubility in aqueous solutions.

76
Q

Isoelectric Point, pI

A

The point in the titration of an amino acid with a strong base where all of the carboxylic acid functional groups have been deprotonated - molecule has no net charge between the (R-COO)- and the (R-NH3)+. Maximum number of zwitterions.

Dictated by the side group of the amino acid. The more acidic the side group, the lower the pI.

Ex: only amino acids with an acidic isoelectric point (all except histidine, arginine, and lysine) will be negatively charged at a basic pH (8) - for electrophoresis where the negatively charged amino acids will run towards the anode.

77
Q

Carbohydrate

A

Think: carbon and water; Cn(H2O)n.

78
Q

Hexoses

A

Six carbon carbohydrates - glucose (aldehyde) and fructose (ketone). May appear as Fischer projections or ring structures.

Five C: pentose
Four C: tetrose
Three C: triose
Etc. 
-> names are often combined with aldose (aldehyde) and ketose (ketone).

Ex: glucose = “aldohexose”

79
Q

Aldose

A

A monosaccharide (simple sugar) that contains only one aldehyde (AKA polyhydroxyaldehyde), such as glucose.

80
Q

Ketose

A

A monosaccharide (simple sugar) that contains only one ketone (AKA polyhydroxyketone), such as fructose.

81
Q

Carbohydrate Chirality

A

In a Fischer projection, the carbohydrate can be:

D: if the hydroxyl group on the highest numbered chiral carbon points to the right.

L: if the hydroxyl group on the highest numbered chiral carbon points to the left.

82
Q

Anomeric Carbon

A

The only carbon attached to two oxygens and its hydroxyl group can point down or up, giving the alpha (axial) or beta (equatorial) anomer.

Created when the hydroxyl group on the chiral carbon furthest from the carbonyl acts as a nucleophile and attacks the carbonyl. This nucleophilic addition to an aldehyde or ketone results in the corresponding hemiacetal (C attached to an OH and an OR group) ring structure.

83
Q

Carbohydrate Ring Structure Names

A
  • Reducing sugars: named according to the number of ring members (including O). End in -ose.

5-membered: furanose
6-membered: pyranose

Ex: glucose ring = pyranose

  • Nonreducing sugars: sugars that are acetals (C attached to two OR groups), not hemiacetals, are called glycosides. End in -oside.

Ex: if the hydroxyl group on the anomeric C of glucose was replaced with an O-methyl group, it would form methyl glucopyranoside.

84
Q

Disaccharides and Polysaccharides

A

Glycosides where the group on the anomeric C that is not OH is another sugar. The linkages are glycosidic linkages.

85
Q

Glycosidic Linkages

A

The linkages between the anomeric C of one sugar with any of the hydroxyl groups of another sugar. Three common bonding arrangements:
1,4’ link
1,6’ link
1,1’ link
Where the numbers refer to the carbon numbers on the sugars (carbonyl C is 1).

Broken via hydrolysis. Without enzyme, broken down slowly by water.

  • > animals do not possess the enzyme to break down the beta-1,4’ glucosidic linkage in cellulose.
  • > some adult humans lack the enzyme to break down the beta-1,4’ galactosidic linkage in lactose.
86
Q

Spectroscopy

A
  1. Nuclear Magnetic Resonance (NMR)
  2. Infrared Spectroscopy (IR)
  3. Ultraviolet Spectroscopy (UV)
87
Q

NMR

A

Typically studies the H nucleus (HNMR), but done with C-13 and others as well (only H on MCAT). Needs nucleus with odd atomic or mass number for nuclear spin to register.
Observes the magnetic field that nuclear spin creates around the nucleus. When placed in an external magnetic field, the nucleus aligns its own field with or against the external field. Nuclei aligned with the field have a lower energy state than those aligned against the field.
Stronger magnetic field = greater difference between these two energy states.

88
Q

Resonance (NMR)

A

Photons (electromagnetic radiation) of just the right frequency (energy state) are shone on the nuclei. Those nuclei whose magnetic fields are oriented with the external magnetic field can absorb the energy of the photon and flip to face against the external field.

  • > a nucleus subjected to this perfect combination of magnetic field strength and electromagnetic radiation frequency is said to be in resonance.
  • > an NMR spectrometer can detect the energy absorption of a nucleus in resonance.

In NMR, the frequency of the electromagnetic radiation is held constant while the magnetic field strength is varied.

89
Q

Chemical Shift (HNMR)

A

The difference between the resonance frequency of the chemically shifted hydrogens and the resonance frequency of hydrogens on a reference compound (such as tetramethylsilane Si(CH3)4 - most common because it contains many hydrogens that are all symmetrical and all very well shielded; gives one large peak all the way upfield/right).

EWGs tend to lower shielding so they decrease the magnetic field strength at which resonance takes place -> will be further left.
EDGs tend to increase shielding and increase required field strength for resonance -> will be further right.

90
Q

Integral Trace

A

A line drawn above the peaks on an HNMR graph that rises each time it goes over a peak. The rise is in proportion to the number of chemically equivalent hydrogens in the previous peak. *Only determines a RATIO of the hydrogens from one peak to another!

91
Q

Splitting

A

AKA spin-spin splitting. Results from neighboring hydrogens that are not chemically equivalent. Use “n + 1” when examining the number of peaks due to splitting where n is the number of neighboring hydrogens that are not chemically equivalent.

92
Q

Solving HNMR Spectroscopy

A
  1. Identify chemically equivalent Hs
  2. Identify and count neighboring Hs that are not chemically equivalent. Use “n + 1” and the splitting of the peaks of neighboring hydrogens to determine the number of neighboring hydrogens that are not chemically equivalent.
  3. If necessary, identify EWGs/EDGs near the chemically equivalent Hs (EWGs move their signal left).

*aldehyde Hs have a very distinctive shift at 9.5ppm!!

93
Q

CNMR

A

Rarely on MCAT. Just remember that the nucleus must have an odd atomic or mass number to register, so carbon-13 is the only isotope used. Treat the same as HNMR, except ignore splitting.

94
Q

IR Spectroscopy

A

In IR Spect: an infrared spectrometer slowly changes the frequency of IR light shining upon a compound and records the frequencies of absorption in number of cycles per cm (cm^-1).

Uses dipoles - the centers of positive and negative charge do not coincide. When exposed to an electric field, the oppositely charged centers will move in opposite directions - either towards or away from each other.

Infrared radiation: the direction of the electric field oscillates, causing the positive and negative centers within polar bonds to move towards and then away from each other (stretch/contract in vibrating motion).

Different bonds vibrate at different frequencies. When the resonance frequency of the oscillating bond is matched by the frequency of infrared radiation, the IR energy is absorbed.

95
Q

Important IR Spectra

A
C=O: a sharp dip around 1700cm^-1
O-H: a broad dip around 3200-3600cm^-1
(note: together = carboxylic acid)
"Fingerprint region": 600-1400cm^-1 
-> Two compounds are very unlikely to have exactly the same IR spectrum; most of the complex vibrations that distinguish one compound from a similar compound are found here - like the identifying fingerprint of a compound.
96
Q

Ultraviolet Spectroscopy

A

Detects conjugated double bonds by comparing the intensities of two beams of light from the same monochromatic light source. One beam is shone through a sample cell, s (with sample compound dissolved in solvent) and the other is shone through a reference cell, r (with solvent only).

  • > the sample cell will absorb more energy from the light beam than the reference cell. The difference in radiant energy is recorded as a UV spectrum of the sample compound.
  • > conjugated pi systems have vacant orbitals (LUMO: lowest unoccupied molecular orbital) at energy levels close to their HOMO (highest occupied molecular orbital) energy levels. UV photons can momentarily displace electrons to the LUMO and the energy is absorbed (typically pi -> pi*).

The wavelength of UV light is between 200 and 400 nm (much shorter than IR light and at much higher energy level). Absorption starts at around 217 nm with butadiene.

*Generally, there’s a 30-40nm increase for each additional conjugated double bond and a 5nm increase for each additional alkyl group. Isolated (not conjugated) double bonds do not increase absorption wavelength.

Lacks detail. Samples must be extremely pure or the spectrum is obscured.

Carbonyls also absorb light in the UV region. Here, n -> pi* (recall: n = lone pair in molecular orbital diagrams).

97
Q

Visible Spectrum

A

The absorbance of compounds with eight or more double bonds moves from UV into the visible spectrum.

“Beta-carotene”: precursor of vitamin A, has 11 conjugated double bonds and a maximum absorbance at 497nm.
-> the electromagnetic radiation of 497nm has blue-green color. Carrots, having beta-carotene, absorb blue-green light, giving them the “complementary color” of red-orange.

98
Q

Mass Spectrometry

A

Gives the MW and (in high resolution) the molecular formula by % abundance. The molecules of a sample are bombarded with electrons, causing them to break apart and ionize. These ions are accelerated through a magnetic field and the resulting force deflects the ions around a curved path.
-> radius of curvature of their path depends on the “mass to charge ratio” (m/z) of the ion.
Most ions have a +1 charge. The magnetic field strength is altered to allow the passage of different sized ions through the flight tube and a computer records the amount of ions passing through at a given strength.
Base peak: largest peak
“Parent peak”: peak made by the molecular ions (original molecules that did not fragment, missing one electron -> +1) should be on the right of the spectrum with heavy isotopes.

99
Q

Separations

A
  1. Chromatography
  2. Distillation
  3. Crystallization
  4. Extraction
100
Q

Chromatography

A

The resolution (separation) of a mixture (usually dissolved into a solution) by passing it over or through a matrix (usually a solid) that adsorbs different compounds with different affinities, ultimately altering the rate at which they lose contact with the resolving matrix.
-> More polar compounds elute more slowly because they have a greater affinity for adsorption with the matrix.
Result: establishment of separate and distinct layers, one pertaining to each component of the mixture.

Recall: paper chromatography - sample to be separated is spotted onto paper, one end of paper is placed into the solvent which moves up the paper via capillary action and dissolves the sample as it passes over. More polar components move slower because they are attracted to the polar paper.

101
Q

Distillation

A

Separation based upon vapor pressure.
A solution of two volatile liquids with boiling point differences of approx 20°C or more may be separated by slow boiling.
-> compound with lower BP (higher vapor pressure) will boil off and can be captured and condensed in a cool tube.

Note: if a solution of two volatile liquids exhibits a positive deviation to Raoult’s law, the solution will boil at a lower temperature than either pure compound (can also have a solution with a higher BP than either pure substance). Cannot be separated by distillation if it brings the BPs of the two liquids closer than a 20°C difference.

Fractional distillation: just more precise. Vapor is run through glass beads allowing the compound with the higher BP to condense and fall back into the solution.

102
Q

Crystallization

A

Based upon the principle that pure substances form crystals more easily than impure substances.

Ex: an iceberg is formed from the ocean but is made of pure water, not salt water. Pure water forms crystals more easily.

Inefficient method of separation.
An exothermic process for most salts.

103
Q

Extraction

A

Based upon solubility due to similar polarities. Like dissolves like.
Start with an organic mixture on top of an aqueous layer - different polarities, don’t mix.
1. Add strong acid and shake. The acid protonates bases like amines in the organic layer, making them polar. The polar amines dissolve in the aqueous layer and are drained off.
2. Add a weak base. The base deprotonates only the strong acids like carboxylic acids, making them more polar. The polar carboxylic acids dissolve in the aqueous layer and are drained off.
3. Add a strong base. The strong base reacts with the rest of the acids (weak acids like phenol). These acids dissolve in the aqueous layer and are drained off.