midterm 3 Flashcards

1
Q

what is IR spectroscopy?

A

-electromagnetic energy in the infrared region (source) interacting with your molecule
*your molecule has vibrating bonds (stretches, bends, twists)
*absorption happens when the specific wavelength of light being used matches the FREQUENCY of the particular vibration
*most covalent bonds have will a unique signature for that specific functional group
- wavelength=distance between peaks
- electric field and magnetic field propagate with the wavelength
- low energy waves are used for NMR while higher frequency/energy waves are used for IR
- energy is proportional to inverse wavelength (as we increase wavelength, there is decrease in frequency and energy)
- spectroscopy: expose sample with electromagnetic wave, put in detector, and it reads out what wavelength is absorbed
- when a specific frequency matches the frequency of our source, we have an absorption
- most covalent bonds have a unique peak that helps us identify what specific functional groups are present in the molecule

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2
Q

overview of functional groups

A
  • sp/sp2 CHs are usually greater than 3000 cm^-1
  • sp^3 CHs we usually less than 3000 cm^-1
  • C-O alcohol in 1000-1300 cm^-1
  • symmetrical stretch: both stretch in the same way (straight down or straight up)
  • unsymmetrical stretch: one stretches up and one stretches down
  • unsymmetrical bend: both swing to one side
  • symmetrical bend: bond swing towards each other (also called wags and scissors)
  • C—O and O—H stretch are more telling about what functional groups you have
  • below 1400 cm^-1 range is called the fingerprint region (lots of different bending modes and C—C stretches so not lots of useful info except C—O stretch being present or not)
  • you are looking at light transmitted (or light absorbed) as a result of “matching” between the wavelength of light and frequency of a particular version
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3
Q

change in dipole moment is needed

A

why?
*propagating light is a wave that carries a propagating electric field with it
*that electric field will interact with “charges” in different functional groups base on dipoles in certain bonds
*without a change in dipole moment (“moving charges”), the light (and electric field) can’t interact with a particular vibrating bond
- in order to observe a signal in IR spectroscopy, we look at absorption of light or transmittance of light
- to have an IR signal we need a change in dipole moment
- change in dipole moment creates partial positive and partial negative charges that build up on the electronegative or electropositive atom (the change in dipole is what interacts with light)
- unsymmetrical alkenes do have a stretch because it has a change in dipole moment during the stretch (even though its small, it’s still present)
- symmetric C=C bond doesn’t have much of a C=C stretch because there is no change in dipole moment (electropositive and electronegative atoms have same distance to each atom even after you try to stretch it
- the point is to look at bonds to see if extending or compressing them will lead to a change in dipole moment
- C—O bonds usually always have a change in dipole moment
- make sure to check if there is a change in dipole after a potential stretch for symmetrical or unsymmetrical C—C bond

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4
Q

what is mass spectrometry?

A

-x-axis is the mass-to-charge ratio (m/z), z is usually 1 (small molecules typically deal with cations and anions)
-tallest peak is “base peak”
-unfragmented cation of expected molecular weight is “molecular ion” (M+)
-tells relative abundance of ions

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5
Q

mass spectrometer

A

-two types of ionization
*1. electron impact EI: “hard”, produces fragments
*2. chemical ionization CI: “soft”, usually see molecular ion

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6
Q

isotope patterns

A

-good clue for finding Br or Cl in a molecule:
*1. 79Br and 81Br are about the same natural abundance, look for equal sized peaks 2 m/z units apart
* 1. 35Cl and 37Cl has a 3:1 peak ratio 2 m/2 units apart

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7
Q

info obtained

A

-overall, mass spec gives us info about
*molecular weight (using average molecule weight across all isotopes of all atoms in your molecule), “periodic table” mass)
*exact mass (mass of monoisotopic compound, for example, only uses 12C weight instead of averaged weight for all carbon isotopes)= molecular ion (remember, MS can separate all possible isotopes)
*3. molecular formula: the molecular weight can gives us a clue
-highest m/z value peak is usually attributed to the molecular ion
-if every molecule of the molecular ion fragments BEFORE reaching the detector, M+ is not observed
-highest m/z is the assumed mass

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8
Q

what’s a diene?

A
  • diene is two alkenes
  • conjugated diene is a molecule with two adjacent double bonds in one molecule (sp^2 and sp^2)
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9
Q

S-cis vs. S-trans

A
  • having an sp^3 center in a diene (non-conjugated) allows for faster rotation
  • a diene with adjacent carbons that are sp^2 and sp^2 (conjugated) can’t freely rotate
  • since conjugated dienes have delocalized electrons, each pi bond is significantly less stable than pi bonds in unconjugated diene
  • heats of hydrogenation for conjugated diene is less than the heat of hydrogenation for the non-conjugated alkene (bc non-conjugated is less reactive) (conjugated is more reactive)
  • s refers to sigma bond between the adjacent sp^2 centers
  • can convert s-trans to s-cis and vice versa
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10
Q

energy difference is crucial!

A
  • s-trans is MORE stable than s-cis because alkene are far apart from each other while s-cis has the alkenes right next to each other
  • mostly work with s-cis
    -s-cis is important for more downfield reactivity
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11
Q

“seeing” a difference in extended polyenes

A
  • longer alkene chain has longer maximum wavelength
  • as energy decreases, the wavelength goes up so the longer chain absorbs into a higher wavelength
  • as you have more alkenes in a row, the energy gap between LUMO and HOMO becomes higher and alkene becomes more reactive
    -energy is proportional to the inverse of the wavelength
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12
Q

diene+dieneophile

A

-we are making a six membered ring
-reaction involves a diene (electron rich) and dienophile (electron poor)
-it’s a concerted mechanism (happens all at once, arrows move together)
-4+2-> gives a Diels-Alder
-to get the most electron poor dienophile, we want electron withdrawing groups

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13
Q

Diels-Alder “bicycle”

A

-a diels-alder occurs when there is withdrawing groups on the dienophile
-with no withdrawing groups, the ring is locked into place and can’t react with diels alder
-acyclic

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14
Q

back to conformation

A

-there is no s-trans option because s-trans is NOT reactive (even though s-trans is more stable, it is not reactive)
-must have an s-cis so it’s locked in place

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15
Q

stereochemistry

A

-stereochemistry stays the same from the alkene dienophile to the product
*if originally cis, groups must be syn
*if originally trans, groups must be anti

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16
Q

more complex

A

-it is preferred when the R’s are tucked under (endo)
-it is less preferred when the R’s are pointing away (exo)
-try the cube trick to figure out the way the R’s are pointed

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17
Q

alkene+H–X

A
  • addition of HX (1 equivalent) to a diene looks a lot like addition of HX to a regular alkene w/ a slight twist (resonance can occur)
  • what is the ratio of our products (trying to find the answer to this)
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18
Q

product ratio depends on carbocations stability

A
  • we know primary<secondary<tertiary (tertiary is the most stable bc it is the more substituted carbocation) (more R groups=more stable) -diene make things more interesting
    • secondary allylic carbocation is about as stable as a tertiary cation bc there is resonance in the allylic carbocation
    • tertiary allylic carbocation is the most stable of them all (even more stable than a tertiary cation)
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19
Q

diene+H–X product

A
  • at cold temps, we favor external alkene with MORE substituted halide
  • at higher temps, we favor internal alkene with LESS substituted halide
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20
Q

product classification

A
  • kinetic product: formed FASTER, favored at lower temps (products with terminal alkenes)
  • thermodynamic product: more stable product, favored with heat (products with internal alkenes)
  • less stable carbocation has to go through HIGHER transition barrier to get to more stable thermodynamic product
  • more stable carbocation has to go through LOWER transition barrier to get to less stable kinetic product
  • kinetic forms irreversibly
  • thermodynamic equilibrates to some ratio of the 2 products and is reversible
  • thermodynamic control: favors thermodynamic product
  • kinetic control: favors kinetic product
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21
Q

visualizing kinetic vs. thermodynamic product

A

-kinetic control:
*1. kinetic product is favored (can still form thermo product but small)
*2. forms irreversibly
-thermodynamic control:
*1. reversible (equilibrate)
*2. thermodynamic product is favored (can still form kinetic product but only small)
- the more stable carbocation forms really fast but it leads to less stable kinetic product (just in this example)
- less stable carbocation forms slower but it leads to the more stable thermodynamic product (just in this example)

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22
Q

when do we get major of each type of product?

A

-kinetic product: lower temp leads to whatever intermediate forms fastests (non-reversible), doesn’t always give most stable product
-thermodynamic product: higher temp allows for reversibility; via high barrier, ultimately gives most stable product

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23
Q

benzene is unique!

A

-it is not like a normal alkene (can’t brominate benzene)
-it is not unique just because of it’s cyclic structure (8-membered rings an be brominated while benzene can’t be)

24
Q

aromaticity also refers to fragrance

A

-some rings smell good and others smell bad

25
Q

bonding

A

-in a benzene ring, there are 6 pi electrons (3 bonds in a ring)
-3 C–C bonds are shared among 6 atoms

26
Q

added stability of benzene

A

-measure via heat of hydrogenation
-more negative delta H= exothermic (favorable)
-more positive delta H=endothermic (unfavorable); also stable

27
Q

aromaticity rule pt.1

A

-must be planar molecule (flat)
-all atoms should be fully conjugated (all pi electrons must be involved in the resonance) (closed cycle)
*lone pair can be involved, but only if that atom with the lone pair doesn’t already have a double bond
-4n+2 pi electrons (n can be any integer)

28
Q

aromaticity rule pt.2

A

-antiaromatic: doesn’t have 4n+2 pi electrons but is planar and fully conjugated (less stable)
*if it has 4n pi electrons, it is antiaromatic
-if 8 atoms or bigger, it can distort and not be planar/flat
-how to determine it’s flat:
*if it’s a big molecule, it’s smart and selfish
*if it’s a big molecule but looks antiaromatic, it’ll distort and become non-aromatic
*small rings (7 or below) can’t distort so can’t be non-aromatic
-aromatic: most stable
-nonaromatic: middle
-antiaromatic: unstable

29
Q

frost circle

A

-draw out with vertex at 6-o-clock
-fill orbitals from the bottom up and analyze
-6 pi electrons is stable
-4 pi electrons is unstable (2 unpaired electrons)

30
Q

what about heterocycles?

A

-rings with heteroatoms (N,O,S)= heterocycle
-if one lone pair is resonance, one lone pair is orthagonal
-don’t count lone pairs if a double bond is connected to that atom
-an atom with lone pairs can be put into resonance and therefore could be aromatic (all pi bonds are conjugated in the ring i.e. all move around the ring)

31
Q

aromatic ions

A

-not limited neutral molecules

32
Q

benzene substitution patterns (configuration of substituents in a benzene ring)

A
  • ortho (o): two substituents right next to each other
  • meta (m): two substituents separated by one carbon
  • para (p): two substituents directly across the ring from each other
33
Q

IR/NMR

A
  • IR: absorptions around 1500-1600 c^=1 (C=C stretch)
  • recall NMR shifts (1H) around 6.5-8.0 ppm)
34
Q

benzene and 1H NMR

A
  • recall our discussion of alkenes and NMR
  • for alkenes in a magnetic field, the space on the outside that protons occupy are deshielded (higher frequency is needed and you’ll see higher chemical shifts for protons to resonate in our NMR spectrometer)
  • benzene rings align perpendicularly to the applied magnetic field B_naught
  • in the deshielding religion (outside), we are reinforcing B_0 which leads to a higher frequency and higher chemical shift (deshielding)
35
Q

expanding on coupling constants

A
  • previously only discussed coupling between hydrogens on adjacent carbons in benzene ring
    • in reality, ortho meta and para coupling are possible
    • J ortho is 6-10 Hz
    • J meta is 1-3 Hz
    • J para is 0-1 Hz (always assume its 0 Hz in our class so ignore J para)
36
Q

addition vs. substitution

A

-for benzene, we must have Br2 and Lewis acid (accepts electrons) in order to turn Br2 into a good electrophile
-Br2 alone does nothing to benzene
-replaces H on benzene ring w/ a Br after turning Br into a good electrophile
-it’s not a simple SN2 but we will see features of an SN2

37
Q

EAS mechanism

A

-after turning into a good electrophile, dobule bond from the nucleophile (i.e. benzene ring) will attack E and halogen leaves as leaving group
*we use heat for this reaction
-E is now bonded and we have broken aromaticity
-we have an E1-like reaction occur where H bonds to carbocation to make a double bond while halogen takes the H
-aromaticity is now restored and we have an H–X by-product

38
Q

example: halogenation

A

-Br2 is a poor electrophile so we have to activate it by adding a lewis acid
-once the lewis acid is bonded to the Br2, we have a good leaving group
-Br–FeBr3 now acts as a leaving group while the double bond from the nucleophile bonds to the other Br
-E1 reaction like occurs again and Br from Br-FeBr3 takes the H to give a by-product of H–Br and FeBr3

39
Q

sulfonation

A

-to make an activated electrophile, you react H2SO4 and SO3 together
-SO3 takes an H from H2SO4 and the bond that was attached to H in the H2SO4 now moves to the oxygen
-after this, you’ve created a new HSO4^- and the new SO3H now gives one of the bonds of the double-bonded O to the O to give +SO2OH
-repeat the EAS mechanism steps to finish the rest of the reaction
-activated electrophile: +^SO2OH

40
Q

nitration

A

-to make an activated electrophile, react HONO2 and H2SO4
-the OH in HONO2 takes the H from H2SO4 and becomes HSO4^-
-the water in the HONO2 leaves as a leaving group to give +^NO2 (nitronium ion)
-repeat rest of EAS mechanism to complete the reaction
-activated electrophile: +^NO2

41
Q

EAS summary

A

-1. activate the electrophile (ex. are Br2 w/ FeBr3, suflonation, and nitration
-2. nucleophilic attack of the aromatic ring
-3. E1 occurs to regain aromaticity

42
Q

friedel-crafts reaction

A

-when you add an alkyl group (ex. methyl, ethyl) or an acyl group (ex. C w/ double bond O and R group and any other group)

43
Q

methods we used so far to create new carbon-carbon bonds

A

-organometallic (R-MgX2 or R–Li) with an epoxide
-EAS where a benzene ring nucleophile and alkyl electrophile react

44
Q

alkylation and acylation overview

A

-all friedel crafts are EAS but NOT all EAS are fridel-crafts
-we still want to make an active electrophile with a good leaving group that will react with a benzene ring (simply different electrophiles but pretty much the same process as EAS)
-for alkylation, it’s a benzene ring with R–X with AlX3 reagent to give benzene with R group attached and H–X by product
-for acylation, it’s benzene ring with C double bond O and R and Cl with reagent AlCl3 and H2O to give a benzene ring with a C double O and R group attached and H–Cl by product

45
Q

alkylation mechanism

A

-1.
*a. react alkyl halide with AlCl3 (i.e. a lewis acid) to make the alkyl halide into a great electrophile
*b. if secondary or tertiary R–X, we have AlCl4 leave and a carbocation form; if primary R–X, rearrange to get a more stable carbocation
-2. double bond from benzene reacts with carbocation and then aromaticity is restored when beat H forms a double bond at the carbocation and Alcl3 takes H

46
Q

watch out for rearrangements in alkylation!

A

-if a more stable carbocation can form, rearrangement will occur with either a hydride or alkyl shift

47
Q

acylation mechanism

A

-1. acid chloride reacts with AlCl3 to make an acylium ion (look at notes for details)
-2. double bond from benzene ring reacts with C in acylium ion and one bond goes to O as nonbonding electrions; E1 occurs where bond to H creates a double bond and AlCl4 takes an H
-note: there is H2O is this reaction in order to mop u the rest of the excess Lewis acid

48
Q

intramolecular friedel-crafts

A

-remember, this is a reaction within the same molecule
-alkylation happens faster
-look at notes for this one

49
Q

other carbocations can also be useful

A

-ex. shows OH in a 5-membered ring reactiong with H2SO4 to give a carbocation where the OH used to be and then binding the 5-membered ring to the benzene ring (look at notes)

50
Q

any substituent can “activate” or “deactivate” the ring

A

-occurs via induction and/or resonance
-electron donating groups are donors
-if we have donor substituent, we get ortho/para position
-electron withdrawing groups are acceptors
-if we have acceptor substituent, we ONLY get the meta position
-ex. NO2 is an electron withdrawing group bc it inductively pulls electrons away from the benzene ring so we only see NO2 in the meta position

51
Q

induction (inductive effects)

A

-effects that don’t involve “curved arrow pushing”
-donors activate and are in the ortho/para position (ex. alkyl groups)
-acceptors deactivate and are in the meta position (ex. carbonyl, nitrile, nitro, sulfonic acid, halide)

52
Q

resonance effects

A

-involves curved arrow pushing
- donors are amines, ether, halides cus it has lone pairs)
*ortho/para builds up negative charge which increases nucelophilicity
-acceptors are carbonyl, nitrile, nitro, sulfonic acid
*ortho/para here builds up positive charge which decreases nucleophilicity

53
Q

ex. 1: why does alkyl direct ortho/para?

A

-look at carbocation stability
*you get the best possible carbocation intermediate with ortho and para while meta doesn’t have this
-para is more accesible and there is usually a mix of ortho/para because they are in competition

54
Q

list of strong, moderate, and weak activators (electron donating groups)

A

-strong activators: –NR2, –OH
-moderate activators: –OCH3, –NH–C (w/ double bond O)–CH3
-weakly activator: –CH3, –R
*usually lone pairs are activating
-ortho/para directors are strong, moderate, and weak activator AND weak deactivators
-reactivity: more electron donating groups are faster

55
Q

list of strong, moderate, and weak deactivators (electron withdrawing groups)

A

-strong deactivator: –NO2, –CF3
-moderate deactivator: –C (w/ double bond O) and other group, C triple bond N, –SO2OH
-weak deactivator: –X (halogen)
*usually pi bonds are deactivating
-meta directors are strong and moderate deactivators