Practice AAMC exam 1 Flashcards
a stereospecific reaction
- a run in which the stereochemistry of the reactant completely determine the stereochemistry of the product
- look if another stereoisomer of the reactant will give identical products in identical ratios- if so the reactant is NOT stereospecific
stereochemistry of SN1 rxn, vs SN2 Ron
SN1=mixture of retention and inversion (2 steps- unimolecular= only depends on [ ] of one - the alkyl) - only depends on the concentration of the alkyl bc it just depends on how many leaving groups leave)
-carbocation forms
SN2- always an inversion
( one simultaneous step- only a transition state (5 bonds- so cannot do on tertiary carbon)- no carbocation)
SN2 will not happen on?
tertiary alkyls- bc a transition state happens, not a carbocation- so the more substituted the less likely
SN1 favours
tertiary alkyls, and more substituted substrates proceed faster!
speed of SN1 vs speed of SN2 on primary? secondary? tertiary?
SN2 -favours primary- fastest
SN1- favours tertiary - fastest ( most stable carbocation)
SN2 stereochemistry vs SN1
SN2 will always be an inversion bc the nuc attacks on the opposite side of leaving group
Sn1- can be inversion or retention, bc leaving group leaves first- carbocation, which can be attacked on either side
stereospecific vs stereoselective reactions!
A stereospecific reaction is a reaction in which the stereochemistry of the reactant completely determines the stereochemistry of the product without any other option.
A stereoselective reaction is a reaction in which there is a choice of pathway, but the product stereoisomer is formed due to its reaction pathway being more favourable than the others available.
The main difference between stereospecific and stereoselective reactions is that a stereospecific reaction gives one specific product whereas stereoselective reaction gives multiple products.
Which of the following statements does NOT correctly describe the dehydration of malic acid to fumaric acid and maleic acid?
a, The reaction occurs most readily with tertiary alcohols.
B. The reaction involves the loss of a water molecule.
C. The reaction has a carbocation intermediate.
D. The reaction is stereospecific.
in passage it says the dehydration results in the formation of fumaric acid, and its cis isomer, malice acid
therefore - it is NOT stereospecific - bc if it was there would be only one specific product
a- this reaction occurs most readily with tertiary alcohols bc when alcohol acts with protic acids to lose a water molecule and form an alkene–> water must leave= carbocation- the more subsituted the most stable
steps in dehydration of alcohol in presence of protic acid
- protonating alcohol
- water leaves
- carbocation formed (+ charge) - rate-limiting step
- alkene formation -
E1 vs E2
how are they related to sn1 and sn2
E1 - secondary and tertiary
- alkyloxoanium forms, leaves, carbocation, then the water molecule (strongish base) grabs another proton from adjacent carbon- double bond forms
- depending on the hydrogen that gets abstracted- will depend on the placement of the double bond - Zaitsev rule states that the major product will be the most substituted product- and trans is better than cis
SN1 and E1 are grouped together because they always occur together. … Both E1 and SN1 start the same, with the dissociation of a leaving group, forming a trigonal planar molecule with a carbocation. This molecule is then either attacked by a nucleophile for SN1 or a base pulls off a b-hydrogen for E1
E2- proton lost from carbon at the same time as water leaves- no in between formation of unstable acarbocation (if it was a primary alcohol)
dehydration of alcohols yeild
alkenes (undergo E1 or E2) - need strong acid to protonate OH- to for (alkyloxanuim ions -H2O still attached) for leaving group
- and need hella heat !
what type of alcohol needs the most heat for dehydration?
primary
Zaitsev’s Rule
the more substituted alkenes are formed preferential because they are more stable than less substituted alkenes
- furthermore, trans alkenes are more stable than cis alkenes are are also the major product
carbocation rearrangments
very common in run when a carbocation moves to a more stable state through “shifts”
- hydride and alkyl shifts
hydride shift
- This means that the two electron hydrogen from the unimolecular substitution moves over to the neighboring carbon.
- this is typical when alcohol reaches with H-halide
- Whenever a nucleophile attacks some molecules, we typically see two products. However, in most cases, we normally see both a major product and a minor product. The major product is typically the rearranged product that is more substituted (aka more stable). The minor product, in contract, is typically the normal product that is less substituted (aka less stable).
hydride shift mechanism
if SN1 rxn– carbocation forms- but before halide attacks- the hydrogen atom and carbocation can switch places- the Cl can now attack the more stable carbocation (more alkyl substituted)
- bc alkyl groups are electron donating - equalize charge
alkyl shift
Alkyl Shift acts very similarily to that of hydride shift. Instead of the proton (H) that shifts with the nucleophile, we see an alkyl group that shifts with the nucleophile instead. The shifting group carries its electron pair with it to furnish a bond to the neighboring or adjacent carbocation. The shifted alkyl group and the positive charge of the carbocation switch positions
gas-liquid chromatograph
-sample in gaseous form, passes through a liquid or solid stationary form
- mobile= gas carrier
- stationary= liquid
If a solution containing the compounds shown in Figure 4, is injected into a gas-liquid chromatograph, the first peak observed in the gc trace is attributable to which compound?
A.2-Methyl-2-butanol
B.2-Methyl-2-butene
C.2-Chloro-2-methylbutane
D.2-Bromo-2-methylbutane
butene- bc bc lowest molecular weight and also weakest intermolecular forces, so will travel the fastest and be the first peak
GLC seperation principle
-solubility in the stationary liquid phase - higher intermolecular forces
a solution sample that contains organic compounds of interest is injected into the sample port where it will be vaporized. The vaporized samples that are injected are then carried by an inert gas, which is often used by helium or nitrogen. This inert gas goes through a glass column packed with silica that is coated with a liquid. Materials that are less soluble in the liquid will increase the result faster than the material with greater solubility.
The combination of gas chromatography and mass spectrometry is an invaluable tool in the identification of molecules.
what’s D in a molecule
deuterium - represented as a D- it is an isotope of hydrogen with masss number 2
is D higher priority than H ?
yes - they have same atomic number, so break the tie based on mass, which D is higher
SN2 inversion drawing / r–> s
so to draw the molecule- we put the incoming group on the opposite side -> put it on the stick and keep the other substituents exactly as they are
- assuming that the priorities are the same (OH was highest before, and incoming group is now th highest) then if it was S before- now it is R
Isotopic substitution
is a useful technique due to the fact that the normal modes of an isotopically substituted molecule are different than the normal modes of an unsubstituted molecule, leading to different corresponding vibrational frequencies for the substituted atoms.
- adding an isotope to the compound as a tracker
- O-18 (meant it was isotopically labeled, not that it had 18 carbon LOL)
ohms law
V=IR
index of refraction equation
n=c/v
The intensity of the radiation emitted by the oxygen sensor is directly proportional to the:
A.propagation speed of the radiation.
B.wavelength of the radiation.
C.polarization of photons emitted.
D.number of photons emitted.
D - intensity= number of photons
intensity= energy/time
energy proportional to number of photons
E of a single photon = hf
the more photons = the more energy
THz
tera- hertz = 10^12
1/16 in percent
0.06= 6%
1/3, 1/4, 1/5 1/8, 1/9, 1/10
- 33
- 25
- 2 (1/5)
- 16
- 14
- 125 (1/8)
- 111
- 1
1/12,
1/15
1/20
- 08
- 06
- 05
1/20
0.05
bc 1/10= 0.1 —> and 20 is twice as big, so makes this number twice as small