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chem 238 > midterm 1 > Flashcards

midterm 1 Flashcards

ch. 9, 10, 14

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1
Q

nucleophilic substitution

A

-nucleophile swaps places w/ the leaving group
-must be sp^3 carbon for nucleophilic substitution to occur

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2
Q

beta-elimination reactions

A

-strong base grabs H from the beta carbon
-the bond from H form alkene between beta carbon
-beta carbon is the carbon adjacent to the carbon that is attached to the halogen
-alpha carbon is carbon attached to halogen
-two things determine whether we see elimination or substitution: 1) nature of nucleophile/base 2) nature of leaving group compound

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3
Q

nucelophilic substitution equilibria: favors release of weaker base

A

-the side that is favored is the side that has a weaker base forming
-how to determine weaker base: 1) look at the pKa of the conjugate acid 2) if the pKa is low, then it is very acidic meaning that the conjugate base is weak
-if a weaker base is on the reactants side of the equation, the reaction doesn’t proceed forward
-for reactions with conjugate acids that have similar pKa values, you can add something like acetone that precipitates one of the conjugate acids which allows the reaction to proceed forward

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4
Q

leaving group priority

A

-better leaving groups: I (great weak base)>Br>Cl>F (ok weak base and not good LG)

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5
Q

reaction rate depend on rate laws

A

-rate laws that are unimolecular are fast (units are Ms^-1)
-rate laws that are bimolecular are slower (units are Ms^-1)
-one example of bimolecular is SN2 (S means substitution, N means nucleophilic, 2 means bimolecular)

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6
Q

SN2 mechanism: “inversion”

A

-SN2 always has to have OPPOSITE configuration (ex. R changes to S or vice versa) to occur
-backside attack must occur

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7
Q

leaving group as electrophiles

A

-more substituents (alkyl groups) on alpha carbon means SLOWER SN2 rate (called alpha branch)
-more substituents (alkyl groups) on beta carbon also means SLOWER SN2 rate (called beta branch)

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8
Q

nucelophiles in periodic table (row)

A

-across row/across period: as we go left, nucleophilic basicity of the leaving group INCREASES (SN2 rate also increases)
-basicity is proportional to SN2 rule

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9
Q

nucleophiles in periodic table (column)

A

-as you go up the column, nucelophilic basicity of the leaving group INCREASES while SN2 rate DECREASES
-as you go up, H-bonding increases as well
-H-bonding disrupts nucleophilicity (when something is good at H-bonding, it is gonna be a bad leaving group and not as weak a base)

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10
Q

nucleophile overview+ energy

A

-in a row, nucleophilicity is proportional to basicity (increase basicity, increase rate)
- in column, least basic is the BEST nucleophile
-lower energy barrier=faster rate (less energy to get to the top of the hill)

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11
Q

elimination reaction (E2)

A

-mechanism: nucleophile leaves, bond from H that left becomes double bond, nucleophile takes away the H
-in E2, nucleophile and leaving group are 180 degrees apart
-you have to have a beta-H to do elimination
-stereochemistry must be reflected in alkene product (should be opposite face so backside attack)
-regiochemistry: more substituted alkene is favored because its more stable (this will form in the major product)
-magnitude of product isn’t important

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12
Q

SN2 vs. E2

A

-bulkier nucleophiles: have slower SN2 (E2 is MORE likely)
-less basic nucleophiles: SN2 is favored over E2
-primary leaving group: usually SN2 (depends on leaving group ability and branching)
-secondary leaving group: mixture of SN2 and E2
-tertiary leaving group: if beta H is available, E2 is favored over SN2

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13
Q

E1 and SN1

A

-E1 and SN1 are usually neutral or acidic conditions
-however, SN2 and E2 are usually BASIC conditions and have negative charge
-for SN1 and E1, tertiary»secondary»primary (tertiary is most favored bc it forms the most stable carbocation)
-primary alpha carbon DOESN’T UNDERGO SN1 or E1 bc it is TOO slow

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14
Q

SN1 stereochemistry

A

-SN1 gives racemic product (50: 50 enantiomers with one as R and other as S configuration)
-SN2 gives inversion
-E1 gives just one product

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15
Q

summary of SN2/E2 and SN1/E1

A

-for poor nucleophile (H2O): no reaction for methyl, no reaction for primary carbocation w/ no branching, no reaction for primary carbocation w/ branching, SN1/E1 mixture for secondary, SN1/E1 mixture for tertiary
-weak base/good nucleophile (I): SN2 for methyl, SN2 for primary w/ no branching, SN2 for primary with brancing, SN2 for secondary, SN1/E1 mixture for tertiary
-stronger base (CH_3O^-): SN2 for methyl, SN2/E2 mixture for primary w/ no branching, E2 for primary w/ branching, E2 for secondary, E2 for tertiary

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16
Q

alkenes in electrophilic addition

A

-mechanism: double bond breaks and attaches to H while Br leaves, Br attaches to more substituted carbon (called Markovnikov rule) while H attaches to less substituted carbon

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17
Q

alkenes in radical addition

A

-mechanism: peroxide breaks double bond and attaches single electron on both sides, bond from Br gives one electron to combine with one electron from RO, you get RO-H + Br with single electron. double bond breaks to give one electron to Br and one electron to adjacent carbon, Br goes to LESS substituted carbon (called anti-markovnikov)
-only need small amount of radical initiator i.e. peroxide here to start the reaction
-see slides for more detail (a lot easier to look at slides for this one)

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18
Q

alkene radical stability

A

-radical stablity: tertiary>secondary>primary>methyl
-tertiary radical i.e. single electron on the tertiary carbon is more stable bc CH_3 bonds help stabilize radical
-bond dissociation energy tells us the strength of the bond
-peroxide RO–OR is an easier bond to break which is why it is the best racial initiator

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19
Q

organolithium(Li)/Organomagnesium (Mg)

A

-when looking at either, assume that the Li and Mg have negative bonds and that Li gets detached and becomes nonbonding pair
-look at slides for this (a lot easier)

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20
Q

organolithium/organomagnesium

A

-they both act as STRONG bases

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21
Q

substitution

A

-substitution is when a nucleophile comes in and then the leaving group comes out
-it goes from a single bond to a single bond
-starts with substrate that reacts with nucleophile

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22
Q

elimination

A

-starts with a single bond
-single bond becomes double bond
-can remove two things: halogen and H

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23
Q

w/ weak bases

A

-in general, with weak bases, we tend to see an SN2 (ex. primary alkyl halide with I^-, secondary alkyl halide reacted with CN^- in aprotic solvent)

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24
Q

w/ strong bases

A

-in general, with strong bases (i.e. RO^- alkoxide), we tend to see SN2 or E2 (tertiary alkyl halide with CH3O^- is an E2, primary alkyl halide with CH3O^- is SN2)

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25
Q

w/ poor nucleophile

A

-in general, with poor nucleophile, we see a very very weak base that is netural or acidic (i.e. solvents)
-poor nucleophiles are not even charged most times and poor nucleophiles are solvents
-with something acidic, it is not very nucleophilic at all (not electron rich and is electron poor)
-we tend to see a combo of SN1 and E1 w/ poor nucleophile

26
Q

small steric hindrance

A

-when there is not much steric hindrance, you get an easy SN2
-methyl halide (least amount of steric hinderane)> primary alkyl halide>secondary alkyl halide>tertiary alkyl halide (no SN2 with tertiary because there is too much steric hindrance and too difficult to do a backside attack)

27
Q

how to determine SN1, E1, SN2, E2

A

-if very strong base, it’s either SN2 or E2
*if primary w/ strong base, its SN2
*if primary w/ strong BULKY base, you get an E2 (too much steric hindrance to get SN2)
*if secondary with strong base it’s E2
-for weak base with primary, we tend to see SN2 (bc they’re good enough to do substitution but not good enough to pull an H off)
-note: SN2 ALWAYS does a backside attack
-if there is too much steric hinderance and strong base i.e tertiary alkyl halide, SN2 CAN’T happen so E2 will form
-E2 needs to have strong base to rip an H off of a carbon (needs to be basic enough to pull H off a carbon)
-note: only worry about solvent when looking at secondary alkyl halide in a weak base

28
Q

protic vs. aprotic

A

-protic solvents (H bonding) speeds up SN1/E1 reactions (bc it stabilizes carbocation and minuses which lowers the energy)
*should be weak base
-aprotic solvents (no H bonding) speeds up SN2/E2
*leaves anion alone (can attack right away)
*by leaving the anion alone, it makes the nucleophile more nucleophilic
*should be strong base
-protic solvents (no H bonding) in a strong base is slow bc strong base has to now push all the solvent molecules to do a backside attack

29
Q

strong nucleophiles

A

-strong bases: RO^-, RMgBr, RLi, R triple bond C:^-
-weak bases: I^-, any halogens, CN^-, RC-O double bond O^-, RS^-, HS^-,, NH3 (lone pair is nucleophilic), RNH2 (lone pair is nucleophilic)
*NH3 and NH2 are uncharged but still good nucleophile
most charged nucleophiles here are bimolecular
-OH^- is between strong and weak base (middle)
-bimolecular (SN2/E2) rate=k [R–halogen][nucleophile]
*rate determining step depends on both halogen AND nucleophile
*second order bc R–halogen is first order and nucleophile is another first order to give overall of second order

30
Q

poor nucleophiles

A

-poor nucleophiles are very very weak bases (neutral or acidic)
-ex. of poor nucleophiles: H2O, R–OH, H–halogen, HCN, R–C–Oh double bond O (typically solvents)
-these are so poorly basic that they are not very good nucleophiles (not electron rich so they don’t like to attach that often and wait for R–halogen to lose halogen and form a carbocation)
-halogen MUST LEAVE FIRST before the very very weak bases can attack
-unimolecular reactions (SN1/E1) rate= [R-halogen]
*rate determining step only depends on halogen

31
Q

how SN1 attacks

A

-SN1 can attack from front or back (gives racemic mixture)
*steps that forms carbocation is ALWAYS rate-determining step (slowest step)
*rate-determining step for SN1 only involves halogen so that’s why the rate law depends on the R–halogen only (remember, this is for SN1 and E1 since they both have same rate-determining first step)

32
Q

product of E2

A

-according to Zaitsuve rule, E2’s major product is the MORE SUBSTITUTED alkene

33
Q

chart for SN2, E2, SN1, E1

A

refer to table from notes

34
Q

inversion of configuration (for SN2)

A

-first method is if leaving group is on a wedge, then the nucleophile enters from the dashed side
-second method is what’s on the wedge stays on the wedge and what’s on the dash stays on the dash ( change configuration where R becomes S or vice versa)

35
Q

how heat and cold determine SN1 or E1

A
  • with poor nucleophile (very very weak base) you get a mixture of SN1 and E1
    • however, if it is cold, it favors SN1
    • if it is hot, E2 is favored (heat encourages elimination in general)
    • SN1 can give two products (backside and frontside attack where you’ll have dashed and wedged nucleophile) when there is a chiral center
    • E1 can give many products
    • if you know that a solvent is acting as a nucleophilic, we know that it is not that nucleophilic and is a poor nucleophile (should get SN1 and E1)
    • if water is the nucleophile, make sure to lose one H and let it bond to the halogen in order to get a neutral charge
36
Q

importance of carbocation

A
  • carbocation is the most important intermediate
  • for a carbocation, if there isn’t a tertiary carbocation, the molecule can REARRANGE to become a tertiary carbocation
37
Q

radical substitution

A
  • only have radical substation for ALKANES
  • stability of radicals: tertiary>secondary>primary>methly (rather remove H from tertiary)
  • need to have H to lose in order to put a halogen where the H is going to leave
  • radical substation can ONLY works for Br2 or Cl2 (Cl2 is much more reactive than Br2)
    • Br only gives most stable product (least reactive and more selective)
    • Cl2 gives multiple different products since it is more reactive
38
Q

radical addition

A
  • radical addition is ANTI MARKOVNIKOV (Br goes to LESS substituted carbon)
  • radical addition ONLY works for H—Br (doesn’t work for H—I or H—Cl)
39
Q

chemically equivalent

A

-all protons (for proton NMR) or all carbons (for carbon NMR) are the same

40
Q

how NMR works

A

-nuclei have a charge and spin of +1/2 or -1/2. in an external magnetic field, these nuclei spins are either parallel or antiparallel to the orientation of the external magnetic field
-there is a slight excess of parallel over antiparallel. we measure the energy absorbed needed to flip a spin from parallel to antiparallel which we call absorption

41
Q

absorption

A

-as the external magnetic field increases, the energy difference increases
-majority of the spins are in the low energy i.e. +1/2 so radiation with energy is added in order to move a low energy spin into the high energy i.e. -1/2
-the exact unique energy required for a spin flip depends on the chemical environment of a given nuclei (i.e. functional group, neighbors)

42
Q

what is a proton NMR spectrum?

A

-the number of peaks is the number of different H’s (if there is symmetry, H’s that are symmetrical to each other are considered the same)
-area under each peak (integration): how many of each type of H (ex. if there is a CH2, it is 2H)
-peak location (chemical shift/resonance): chemical environment of each H
-peak splitting: tells us about the number of nearby NEIGHBORS for each H. how many different H’s are close by?
-if an H is closer to a more electronegative atom, the peak is closer to the left i.e. higher ppm value for the chemical shift

43
Q

some special cases (diasteretopic vs. enantiotopic hydrogens)

A

-2 things that you need to see before considering diastereomeric or enantiotopic: must be an alkene, must have a chiral center
-homotopic have NO chiral centers
-to test if diastereotopic or enantiotopic, switch one H on dash to wedge and compare the two structures. if not mirror images and only one chiral center changes, it’s diastereotopic. if both chiral centers change and are mirror images, it’s enantiotopic

44
Q

chemical shift in an NMR spectrum

A

-chemical shift (or resonance) is DIRECTLY related to energy absorbed during spin flip
-absorption tells us where on the scale the chemical shift will be

45
Q

what’s in a proton NMR spectrum?

A

-number of peaks= number of different H’s
-area under each peak (integration): how many of each specific type of H’s there are?
-peak location (chemical shift/resonance): chemical environment of each H
-peak splitting: tells us about the number of nearby neighbors for each H.. how many different H’s are close by?

46
Q

splitting

A

-non-equivalent H’s on adjacent C atoms “see” each other
-the result is “splitting”
-equivalent H’s DO NOT split each other
-splitting follow the n+1 rule
*the pattern is n is the number of next door H’s+1
*this gives you the type of peak it is
-1 peak is singlet
-2 peaks is a doublet
-3 peaks is a triplet
-4 peaks is a quartet
-more than 4 peaks is a multiplet

47
Q

why do we see splitting?

A

-recall, nuclei have spin (+1/2 or -1/2)
-magnetic field of each neighbor can slightly shield or deshield the proton you are observing

48
Q

pascal’s triangle

A

-non equivalent neighbors are “additive” ex. triplet+quartet are triplet of quartets

49
Q

alkene splitting

A

-cis has J value of 6-10 Hz
-trans has J value of 11-18 Hz
-geminal has J value of less than 3.5 Hz
-must memorize these values
-only looking at J values of ALKENES
-J is the spacing between the lines in the peak

50
Q

proton exchange/hydrogen bonds

A

-OH is not seen on an NMR when D_2O is added

51
Q

carbon NMR

A

-gyromagnetic ratio: relationship between magnetic moment and angular momentum
-weaker signal and no carbon-carbon splitting observed
-smaller gyromagnetic ratio than 1H so less intense signal
-13C i.e. carbon NMR has larger range of chemical shifts

52
Q

distinguishing structures

A

-there is no splitting here so we say spectra are proton decoupled (you only see single lines)
-there is also no integration

53
Q

DEPT

A

-special technique used to determine attached protons

54
Q

3 major components of NMR

A

-chemical shift: location on the spectrum which tells us the type of proton like sp^3, sp^2, or sp H (can also tell us neighboring functional groups; deshielding or shielding)
-integration tells us the number of H’s in each signal; must be relative ratio so must divide to get the SMALLEST ratio)
-splitting/coupling (or multiplicity is name of that specific split): share the same J value
*splitting is caused by the number of NEIGHBORING H’s within 3-bond rule
*n=number of neighboring H’s
*n+1=number of peaks you see

55
Q

symmetry for proton NMR

A

-number of different H’s=number of signals (based on symmetry)
-if there is symmetry, each proton is considered same even if theres two CH_3, they are considered same H’s
-if there is no symmetry, each proton is considered different even if theres two CH_3, they are two different H’s
-note: when there are alkenes, draw out the branched double bond

56
Q

symmetry for carbon NMR

A

-no symmetry means each C is different
-symmetry means the C’s that are symmetrical to each other are considered the same C

57
Q

splitting/coupling/multiplicity

A
  • N is neighboring H’s
  • N+1 is the number of peaks
  • neighbors of 4 and up are considered multiplet
  • height of the peaks are based on the pascal triangle
  • if H is within 3 bonds, molecule is close enough to split
  • note: common thing on test we see: quartet and triplet (ethyl), septet and doublet (isopropyl)
58
Q

integration

A
  • integration is number of H’s it contains
  • area under the curve represents the number of H’s it containts (if there are more H’s, the area under that curve is greater)
  • multiply by lowest factor to get the smallest whole number
  • count total number of H’s to make sure it matches the NMR that is given
  • when symmetrical, adds both A’s (2 CH$_3$) that are similar but their integrals act as if there is only ONE CH$_3$
  • if there is no splitting for a part of the molecule, it is isolated so automatically a singlet
  • can’t split with IDENTICAL neighbors
59
Q

chemical shift

A
  • don’t need to memorize chemical shifts
  • chemical shift tells us location in spectrum
  • ppm increases as you go from RIGHT to LEFT
  • TMS marks the zero point of the chemical shift
  • most sp_3 show up around 0-2 ppm where there are NO neighboring functional groups
  • as you go to higher ppm, you are deshielding (downfield shift)
  • as you go to lower ppm, you are shielding (upfield shift)
  • if there are extra functional groups, there is an additive effect
  • downshield effects are additive
  • electron withrawing groups cause neighborings H’s to experience DOWNFIELD SHIFT (or deshielded)
  • electron donating groups cause neighborings H’s to experience UPFIELD SHIFT (or upshielded)
  • only worry about electron donating and electron withdrawing groups when you have a double bond ring like benzene
60
Q

special cases

A
  1. diastereotopic protons
    a. two different times that you’ll see this: CH$_2$ next to chiral center and CH$_2$ on an alkene (asymmetrical alkene)
    b. when two carbons are on the same carbon, you have things like doublet of a doublet (bc you have very different neighbors where one is like 2 bonds away and the other is like 1 bond away); split them separately and then multiply them)
    c. 2 H’s on the same carbon are usually homotopic
    d. when near by a chiral center, H’s on the same carbon are NOT identical

chem 238 (4 decks)

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