hello everybody my name is Iman welcome back to my YouTube channel today we're going to do a practice problem set that relates to our spectroscopy lecture please make sure to watch the lectures before you tackle the problems let's go ahead and get started problem one says IR spectroscopy is the most useful for distinguishing blank a says double and triple bonds B says carbon and hydrogen bonds C says chirality of molecules and D says relative percentage of inan tumors in mixtures now when we talked about infrared spectroscopy we said it is is the most useful for distinguishing between different functional groups now almost all organic compounds have carbon hydrogen bonds so except for the fingerprint region of a compound these absorptions are not so useful not not little information um about the optical properties of a compound can be obtained by IR spectroscopy so those two facts alone allow us to cancel out a couple of answer choices here all right it is not the most use useful for distinguishing carbon hydrogen bonds if you are concerned about the carbon hydrogen framework of a molecule an unknown molecule NMR would be extremely useful for that it doesn't tell us about anything about the chirality of molecules or the relative percentages of enantiomers in a mixture uh polarimeter might be more useful for something like that especially for relative percentage of enantiomer mixtures this leaves us with an Choice a infrared spectroscopy is the most useful for distinguishing double and triple bonds and if you remember from our lecture we are very concerned about the diagnostic region that is what we worry ourselves for the MCAT and also in most organic chemistry classes as well and you realize that in this diagnostic region you can learn a lot about double bonds and triple bonds so it is the most useful for distinguishing double and triple bonds one is a now two says oxygen does not exib exhibit an IR spectrum because blank a it has no molecular motions B it is not possible to record IR Spectra of a gaseous molecule C molecular vibrations do not result in a change in the dipole moment or D molecular oxygen contains four lone pairs overall now because molecular oxygen so O2 is homo clear it's only composed of one element that's oxygen you guessed it and it's datomic there is no net change in the dipole moment during vibration or rotation so in other words this compound does not absorb in a measurable way in the infrared region now if you remember from our lecture we said IR Spectra is based on the principle that when the molecule vibrates or rotates there is a change in the dipole moment for for O2 we're not going to see that molecular vibrations do not result in a change in the dipole moment so the correct answer for two here is going to be C Choice a is incorrect because oxygen does does have molecular motions they're just not detectable in IR spectroscopy B is also incorrect because it is possible it is possible to record the IR of a gaseous molecule as long as it shows a change in the DI moment when it vibrates and answer Choice D is also incorrect because lone pairs do not have an effect on the ability to generate an IR spectrum of a compound so for two the correct answer is C fantastic 3 says if IR spectroscopy were employed to monitor the oxidation of benzil alcohol to benzal deide which of the the following would provide the best evidence that the reaction was preceding as planned so I have drawn for us Benzel alcohol all right so this R is a Benzene group Benzel alcohol to benzal deide here R is that Benzel group all right now if we look at these two molecules what we notice is that the functional group is changing the functional group is changing from a hydroxy to an aldah this means that a sharp Peak will appear around 1700 1750 which corresponds to this carbonal group based off of our chart that we were supposed that we should memorize about useful signals in the DI diagnostic region we should be familiar with all right so what we're looking for here is that one good evidence the best evidence that this reaction is preceding is that there should be an emergence of a feature around 1750 all right a sharp Peak that will appear around 1750 which corresponds to this carbonal group all right now let's look at these answer choices all right answer Choice a says comparing the fingerprint region of the Spectra of the starting material and product this is not going to be very useful at all all right this is not the answer that we're looking for we're looking for the best evidence that the reaction is going to proceed as planned comparing the fingerprint regions will provide evidence that a reaction is occurring but it's not as useful for knowing that the reaction uh that occurred was indeed the one that was desired all right B says noting the change in intensity of the Peaks corresponding to the Benzene ring this is also not a useful way all right there's Benzene rings in both of these molecules that's what the R Group here is so that's not the dominant thing that you should be looking for to say that your reaction is preceding as planned C says noting the appearance of of a broad absorption peak in the region 31 to 3500 Choice C is actually the opposite of what occurs the reaction is going to be characterized by The Disappearance of the O Peak Fe Peak feature that appears at 31 to 3500 this feature should disappear if the reaction is proceeding forward so C is incorrect the opposite should be happening D says noting the appearance of a strong abs absorption in the region of 1750 this is the correct answer if we begin to see the appearance of a strong absorption in the region of 1750 this indicates this carbonal functionality which is what should happen if this reaction is proceeding forward so three is D four says which of the following chemical shifts would correspond to an alahh protein signal in hnmr Spectra great question so this is this is important for us to know where the alahh feature is going to appear the Peak at 9.5 PPM it corresponds to an alahh proton this signal it lies downfield because the carbonal oxygen is electron withdrawing and it DeShields the proton so the correct answer is a now if we look at the answer choices choice B 7 PPM this corresponds to aromatic protons all right this is a good thing for you to know this corresponds to aromatic protons answer Choice T C I'm so sorry answer Choice C corresponds to a carboxy proton and is even further downfill because the acidic proton is D shielded to a greater degree than the AL the alahh proton all right so this is for carboxilic acid and then D is characteristic of an alkal proton on an sp3 hybridized U carbon and so the answer choice for four is going to be a five says the isotope of carbon 12 is not useful for NMR because blank a says it is not abundant in nature c b says its resonance are not sensitive to the presence of neighboring atoms C says it has no Magnetic Moment and D says it the signal to noise ratio in the spectrum is too low all right now five okay the isotope here carbon 12 let's think about this this isotope it has no Magnetic Moment all right and it will therefore not exhibit resonance with an applied magnetic field remember nuclei with odd Mass numbers so hydrogen 1 carbon 13 nitrogen 15 all right and so on or those with an even mass number but an odd atomic number like hydrogen 2 all right will have a nonzero Magnetic Moment but carbon 12 it has no Magnetic Moment it is not that it's not abundant in nature that is the actual abundant isotope of carbon in nature it has nothing to do with resonances not being sensitive to the presence of neighboring atoms or anything to do with signal to noise ratio here it is because simply it has no Magnetic Moment so five is C all right five is C six says in hnmr split splitting of spectral lines is due to blank a says coupling between a carbon atom and protons attached to that carbon atom B says coupling between carbon atom and protons attached to adjacent carbon atoms C says coupling between adjacent carbon atoms and D says coupling between protons on adjacent carbon atoms now spin sping coupling or splitting if you remember from our lecture it's due to the influence on the magnetic environment of one proton by protons on the adjacent atom these protons are about three bonds away from each other and splitting uh and and the splitting is caused by those neighboring protons and their magnetic moments and Fields And so here if we look at the answer choices the answer choice that makes the most sense is going to be D coupling between protons on adjacent carbon atoms that is literally how we defined splitting and multiplicity all right so this this is simply a definition question making sure you understand how we defined splitting of the NMR Peaks and multiplicity so six is D seven says compared to ir and NMR spectroscopy UV spectroscopy is preferred for detecting blank a says aldhy and ketones B says unconjugated alkenes C says conjugated alkenes and D says aliphatic acids and amines now if you remember this this was one of the the selling points for UV is that we said it was for conjugated alkenes most conjugated alkes have an intense ultraviolet absorption all right alahh ketones acids and amines like they're mentioned in answer choices a b and d all absorb in the ultraviolet range however other forms of spectroscopy like ir and NMR are more useful for precise identification and isolated um alkenes which is answer Choice B they can rarely by the way be identified by UV spectroscopy so alahh ketones aliphatic acids and amines they could they absorb an UltraViolet range but again other forms of spectroscopy or more useful and then isolated alkenes altogether here in answer Choice B rarely identified by UV spectroscopy the selling point in short for UV spectroscopy is that it's really useful for conjug ug at molecules like conjugated alkenes so the correct answer for seven is C beautiful 8 says considering only the 0 to 4.5 PPM region of hnmr spectrum how could ethanol and isopropanol be distinguished so I have both molecules drawn out here all right now the region in question often gives information about the types of alkal groups present they are between 0 and 4.5 specifically ethanol ethanol will give a characteristic triplet for the methyl group all right which is coupled to this ch2 and a quartet for the ch2 group which is coupled to the methyl group so what that means is all right we have the ch2 group right here in um ethanol and we have the ch3 group that's right here the splitting for ch2 what is it going to be well how many neighboring hydrogens all right right if we're looking at ch2 how many neighboring hydrogens does it have here at this methyl there's three hydrogens here so what is the splitting the splitting is going to be three hydrogen neighbors plus one so the splitting for this methyl is going to be a for the ch2 is going to be a quartet okay cool let me erase this let's look at the methyl all right let's look at this methyl group now all right here's that methyl group how many neighboring hydrogens does it have well over here is a ch2 there are two hydrogens here what's the splitting for the methyl then if it has two neighboring hydrogens it's going to be 2 + 1 n+ 1 remember our multiplicity Rule and that's equal to three so at the methyl here you're going to have a triplet splitting for that methyl feature and a quartet splitting for the ch2 feature good what about for isopropanol here is isopropanol all right here are the two unique positions for isopropanol all right this is a CH group right here there's only one hydrogen and this is a methyl group there are three hydrogens if we are looking at this position that I am now highlighting in blue what is going to be the splitting all right the splitting is how many neighboring hydrogens does it have it has three on this side and it has another three on this side it has six total hydrogen neighbors plus one because n plus one for multiplicity gives us seven this the peak splitting for the CH feature is going to be a septe all right what about the methyl this methyl group right here what is the splitting for this methyl group well how many hydrogen neighbors does it have only one right here at the CH group so 1 + 1 equal 2 the splitting for this methyl group is going to be a double it all right keeping all this in mind we're going to look for an answer choice that says that there's going to be a triplet and quartet observed for ethanol and a doublet and supplet observed for isopropanol because that's going to be what distinguishes the two in this small region of 0 to 4.5 PPM and the answer choice that exactly says that is going to be answer Choice B so 8 is B fantastic nine says before absorbing an UltraViolet Photon electrons can be found in blank homo lumo both or neither well the homo is going to be the highest occupied molecular orbital only after absorbing ultraviolet light is an electron um excited from the homo to the L lumo which is the lowest unoccupied molecular orbital all right so before absorbing an UltraViolet Photon electrons can be found all right in the highest occupied molecular orbital that's going to be answer Choice a and only after excitation do they go into the lumo the lowest occupied molecular orbital fantastic so nine is a 10 says in an IR spectrum how does extended conjugation of double bonds affect the absorbance band of carbonal stretches compared with normal absorption this is a really good problem let's think about this carbonal groups in conjugation with double bonds they tend to absorb at lower wave numbers because the delocalization of Pi electrons causes that carbonal bond to lose double bond character Shifting The Frequency closer to a single Co stretch remember that high order bonds tend to have higher absorption frequencies so loss of double bond character should decrease the absorption frequency of the group so if we're looking at the answer choices here a says the absorbance band will occur at lower wavelengths this is true all right that is how extended conjugation of double bonds that's how it's going to affect the absorbance bond of a carbonal it will not occur at higher wave numbers and it will not occur at the same wave number and it sure as hell won't just disappear all right carbonal groups in conjugation with double bonds tend to absorb at lower wave numbers because of the delocalization of Pi electrons that cause the carbonal bond to lose double bond character and shift the stretching frequency closer to the single carbon oxygen stretch so lower wave numbers the correct answer for 10 is a beautiful 11 says wave number is directly proportional to Blink wave number is directly proportional to Blink now wave number all right we talked about this elure is equal to 1 over wavelength all right wave number is directly proportional to frequency frequency is speed of light over wavelength okay that means the answer is B it is not directly proportional to wavelength it is inversely proportional to wavelength because wave number is one over wavelength all right it is directly proportional to frequency though all right and C and D have nothing to do with wave number there's no proportionality to percent transmittance or absorbance those are throwaway answers and so 11 is B 12 says two enantiomers will blank are they going to have identical IR Spectra because they they have the same functional groups are they going to have identical IR Spectra because they have the same specific rotation are they going to have different IR Spectra because they are structurally different or are they they going to have different IR Spectra because they have different specific rotations now enan humors are going to have identical IR Spectra because guess what they're going to have the same functional groups and therefore have the same exact absorption frequencies and antier do have opposite specific rotations but you're not going to be able to detect that or distinguish that on an IR spectrum all right so the correct answer here is going to be a they're going to have identical IR Spectra because they have the same functional groups all right especially when you're looking at your diagnostic region fantastic 13 in a molecule containing a carboxilic acid group what would we expect in hnmr Spectra all right the oxygen of the hydroxy group all right it's going to dshield the hydroxy hydrogen what what does that mean it means it's going to shift it downfield or left and so hydrogens and carboxilic acids can have some of the most down field absorbances around 10.5 to 12 PPM that's what we want to think about when we're looking at these answer choices all right we're looking for dshield hydrogen Peak for the hydroxy hydrogen and that's going to shift it downfield or left and the answer that corresponds to that for 13 is going to be a a down shielded hydrogen Peak for the hydroxy hydrogen shifted left 14 the coupling constant J is blank a says the value of n + 1 when determining splitting in NMR Spectra B says measured in parts per million C says corrected for by calibration with TMS and D says a measure of the degree of splitting caused by other atoms in the molecule now the coupling constant is a measure is is a measured of the the degree of splitting introduced by other atoms in a molecule and is the frequency of the distance between sub Peaks all right so that's going to align exactly with answer Choice D now if you look at answer Choice B measured in PPM uh all right that's not true it's measured in hertz all right the coupling constant is independent of the value of N1 of n + one and also it's not changed by the calibration of TMS which is why A and B are also incorrect 14 is D all right the coupling constant is a measure of the degree of splitting caused by other atoms in the molecule 15 says the IR spectrum of a fully protonated amino acid would likely contain which of the following Peaks all right a sharp Peak at 1750 a sharp Peak at 3,300 wave numbers or a broad Peak at 3,300 wave numbers now amino acids in their fully proteinated form form contain all three of the Peaks that should be memorized for testing they're going to have Co NH H and o h now while statements one and two correctly give the Peaks for the carbonal bond all right sharp Peak at 1750 and the NH Bond sharp Peak at 3,300 wave numbers the peak for the O bond is in the wrong place so in a carboxilic acid the carbonal Bond withdraws electron density from the O Bond Shifting the absorption frequency down to 3,000 wave numbers and so the only correct statements are going to be one and two not three all right so 15 is going to be B and with that we've completed the practice problem set that relates to our spectroscopy lesson I really hope this was helpful let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors