• link on course website when studying for final exam
• choose the highest energy - deep UV. highly energetic wave lengths induces
core electron transitions.
• moving into the UV vis region, lower energies. valence electrons (HOMO into
• infrared, lower energies (longer wavelengths) not enough to promote
electrons from one orbital to another. induced molecular vibration
• infrared is often called vibrational spec, exchange in dipole moment. in order
to have IR absorption, molecules must have change in dipole moment. still
have non polar molecule, change in dipole must occur in the course of the
◦ bond stretching
◦ bond bending (change of angles)
• N number of atoms before vibrational motion, 3N-6. exception are linear
molecules, 1 degree of freedom more in vibrational modes, hereford they
have 3N-5 vibrational states.
• ex. NO2 is a bent structure (3N -6) CO2 is a linear molecular (3N-5) therefore
CO2 has more vibrational modes that NO2.
• oscillation of one part of the molecule will have an effect on the other part of
the molecule. ex. CO2. imagine it as 2 carbonyl stretches that are stuck
together where the 2 carbonyls share the carbon atom. effect of oscillation
will affect oscillation on the other.
• u is for dipole moment (not reduced mass) change in dipole moment is equal
to zero; no change in dipole moment for symmetric stretch which is why it's
inactive in IR. hook's law, calculated value is 1715 cm-1. if it were to appear,
it would appear at 1715. can confirm indirectly using ramen spec.
asymmetric. stretches out of phase, not in phase. should have the same
calculated value as symmetric. change in dipole moment is not equal to zero
therefore it is IR active. doesn't actually appear at 1715, it occurs are 2350
cm-1. why? because of the strong degree of coping between the two
oscillators. length of one carbonyl is directly related to the other car bony; as
one gets longer the other one gets shorter. can change the position, higher or
lower in wave number. can also add extra stretching/bands. can change
position of bands and can see the appearance of extra bands in the IR
spectrum. where do they come form? coupling between 2 strong fundamental vibration in which we can add vibrations to make combination
bands or get the difference and gives a difference band. ex. 2000 cm-1 and
750 cm-1 (fundamental). difference band = 2750 (combination) and 1250
(difference). can definitely see the first 2, you can have 2,3 or 4.
• ex. amines, (primary) 2 NH stretches. mehtyl and methylene. if you have ch2
group in the middle of a molecule, will give a doublet instead of a singlet
because of CH stretching. or two overlapping peaks. because of the degree of
coupling, can be difficult to assign individual peaks. look at bands instead of
• harmonic overtones. intense bands, abroptions are so probable that you
might get multiples of the individual absorptions.
• tertirary interacts whihc is coupling of coupling. ex. fundamental coupling
with combined bands (fermi resonance) observed when you have intense
fundamental absorptions. ex carbonyl groups. practically impossible to
assign every single peak in IR spectrum. loooookk at bands.
• slde 25: coupling. fermi resonance in between. 2 alpha CHs, fermi resonance
betweenen the fundamental ibration along witht he combination band
between the 2 CH streches. because this is a dipole, the stitch is going to be
dependien on the environment around the dipole. when you change the
polarity, you change the stretch. solvent dependent.
• slide 26: H bonding. between X-H and Y. XH is the hydrogen bond donor. Y is
the acceptor. between he amine of one molecule and the carbonyl of the 2nd
molecule. amine is the H bond donor and carbonyl is the H bond acceptor.
intermolecular interaction. if we establish thehese H bonds, affect the
oscillation of the 2 individual oscillator.s for H bond donor, will shift in lower
frequencies (right) decrease in energy. will brooding because of range of H
bonds, distribution of H bonds perfectly aligned to imperfectly aligned.
intensity will increase, improve IR absorption for H bond donor. for H bond
acceptor, won't have same shane in intensity, but will shift it to lower
frequencies. weakening the bond for the acceptor aka shifting to lower
energies, easier to stretch the carbonyl. look t difference betweenn free H
bond donor and bonded H bond donor and look before and after and look at
difference between the two.
• intermolecular bonding is concentration dependent. intramolecular bonding
is solvent dependent, can dilute.
• linear alkanes: stretching modes on left hand side, bending modes on the
right. key thing is the CH stretch. draw an imaginary line at 3000. absorptions
to the right, those are the CH stretches associated with H attached to spa
hybridizes carbons aka saturated carbon. absorptions to the left is something
different. wide range of symmetric nd asymmetric. bending modes in
fingerprint region (1000 and 1500) most bending modes appear in that window, all overlap. last peak on 720725, unique peak to long chain linear
alkanes. if you have at least 4 CH2s in a row, you get a small peak where
nothing is absorbing at around 720-725. diagnostic for a saturated long
• branched alkanes: peak at 720 is gone because there are no 4 CH2s in a row.
rest of spectrum looks the same-ish. below 3000, draw line. all overlapping
one other producing a fat band, CH stretch for spa hybridized carbons,
• cyclic alkanes: difficult to discern from loin ear or branches. general rule,
cyclopentage, pentane, etc. as the number of CH2s in the ring decrease, the
stretch associated with Chs will go up, more difficult to stretch CHs but easier
to bend. frequencies associated with bending modes go down. widening in
the placements of the peaks. only diagnostic tool..
• alkenes: C=C has at leads on CH group. not tetra substituted. carbon carbon
double bond stretch in the 1620 to 1680 region. C=C region is around 16 and
17. for a vinyl group, appear around 1640 +/-. for trans alkenes, 1670
because they are disubstitued. cis alkenes, added sterics between 2 R groups,
1650 cm-1. absorptions to the left of the 3000, you automatically have an
• conjugated alkenes: saturated carbonds in the molecule hence around 2900.
unsaturated Chs around 3100. 2 oscillators that are connected with a single
bond, result is coupling, 2 C-C stretching bands, symmetric and asymmetric
in C=C region (16-17)
• in between 2800 and 3200 is the CH region. in between 1600 and 1700 is the
C=C region. in between 1000 and 1500 is the fingerprint region. next is
between ~650 to 900 is the OOP region (out of plane). bending modes which
correspond to out of plane bending for planing molecules. imagine the C=C
plane, 2 hydrogens that are in trans, hydrogen binding in and out of the plane.
can be very useful when dealing with planar molecules determining their
regiospecificity. for alkenes, can use it to see if its trans or cis. for trans, the
out of plane bending is found around 900 cm-1. for cis, it's around 700 cm-1.
• alkynes: draw line at 3000. shift even further to the left, CH stretching mode
for sp hybridized carbon. from sp3 to sp2 to sp. terminal alkyne. internal
alkyne, no hydrogen peak dissapeares. region of alcohol's and amines.
terminal alkyne and alcohol, alcohol will definitely absorb over the CH
stretch. C-C triple bond stretch is in the middle of the spectrum. between
2000 and 2250.
• monocycluc aromatic hydrocarbons: absorption to right and left of 3000. left
due to H atoms on aromatic ring itself. saturated CHs due to methyl group on
benzene ring. C-C stretch bond orders are no single or double, they're 1/2. get low value for carbon carbon bond at 1600. close to the border of C=C
zone. between 1000 and 1500, intense band due to aromatic C-C stretching.
entire ring expanding and contracting. ring breathing. OOP region, tell us the
substriution of the aromatic ring.
• ex. 1: absorptions to the right, CH stretch, exclusively sp3. CH stretching
that's sp3. saturated carbons. do not have ch tratch for spa or sp. do not have
any double or triple bonds. triple bond region is empty. for C-C double bond
is empty. do have strong peak at 1712 (indicates a carbonyl ketone). some
sort of alkyl ketone. actual molecule is 4-heptanone.
• alcohols and phenones: intense bands because of condensed media (soltuo
phase and gel phase). OH stretch. if you were to dilute it (gas phase).
unhydrogen hounded OH group. nice sharp peak between 3500 and
3700. shift to lower wavelengths and more broad.
• intermolecular and intra to solve internal h bonding. for intra, no way you
can make it not h bonded.
• alcohols, see some saturated CHs, and unsaturated. see OH stretch
dominating spectrum. ring breathing in fingerprint region. can look at OOp
region to determine it's a monosubstiruted molecule.
• phenol: same, but saturated sp3 CH stretches. mountain range bumps is the
#1 hint for phenols. result of secondary interactions. CO stretch (carbon and
oxygen), normally in fingerprint region but sometimes it's very intense
• ethers: the only unique bond in the either is the C-O bond. alcohol is gone,
can use the CO. hunt for it in fingerprint region. tricky.
• cabronyl region: overlaps with c-c double bond region, about 1650-1850.
always going to see nice intense band, not too wide. relative position of the
band (where is the band in the area) tells what kind of carbonyl it is. base
values on slide 49. electronic effect a runs carbonyl are going to affect
stretcgth of C-O doubleb one, as you strengthnbond, shift to the left. as you
weaken, easier to stretch so you shift peak to the right. get varies positions.
ex. acid vs. acid chloride by switching OH to Cl (inductive effect) which
strengthens CO bond versus carboxylic acid. memorize 1715 is the calculated
value for a carbonyl in isolation.
• CH of 2700, might be able to see CH stretch for aldehyde hydrogen.
• i you raise energy, making it more difficult to stretch carbon bond. base
values at 1715 for a saturated symmetric ketone, as you introduce increase
degrees of ring strain, higher wave values
• acetophenone: carbony stretch in IR spec, between 1600 and 1800 is
carbonyl region relatively narrow, value of 1686 lower than predicted value. why? added resonance, created by substation of aromatic ring and carbonyl
group, weaking double bond, easier to stretch and band displease towards
• inductive effects: G=group. electron withdrawing group, withdraws electron
density away form carbonyl itself. 2 dipoles: pulled away form carbonyl and
toward the electronegative element. ex