Answers to Chem 221 Spectroscopy Problems

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1 nswers to hem 221 Spectroscopy Problems 1. MS: m/z= 156, MW = 156. No info about M+1, no info about # of s. No peak around 1700 cm -1, no carbonyl No broad peaks above 3000 cm -1, no H, NH NMR: Two types of hydrogens (two unique environments) t B q Based on IR this molecules appears at first glance to lack any functional groups, but 1 H NMR δ 3.2 ppm implies some heteroatom. ombination of triplet & quartet is characteristic of -Et (- H 2 H 3 ), implies 2 s & 5H s. This has a mass of 29 and can be seen as a fragment in MS = 127. What has a weight of 127? Iodine. orrect nswer: H 3 H 2 I 2. MS: m/z= 72, MW = 72. No info about M+1, no info about # of s. ν = 1720 cm -1, carbonyl (=) No broad peaks above 3000 cm -1, no H, NH NMR: Three types of hydrogen (three unique environments) 1 3 t B s q We know from IR that molecule must contain (at least one) oxygen (H), i.e. a mass of 16 MW - 16 = 56. What has a mass of 56? Guess: 4 s & 8H s (this is consistent with the total # of H s in NMR). That gives MF: 4 H 8. How many degrees of unsaturation? Fully saturated would be n H 2n+2, or 4 H 10, but we have 4 H 8, which means two H s are missing, or 1 degree of unsaturation. That means 1 ring or one double bond is present. But we already know that it has a carbonyl so that must be the one degree of unsaturation. no other rings or double bonds. How many 4 carbonyl compounds are there? 3: 1 ketone & 2 aldehydes

2 H B H The second one is ruled out because it has different types of H s (4 different environments). The second and third are ruled out because it can not be an aldehyde: aldehyde carbonyls typically occur at larger wavenumbers (ν = = 1740 cm -1, not 1720 cm -1 ), show two unusual H stretches (ν = 2750 & 2850 cm -1 ) and show a very unique signal in the 1 H NMR at δ 9 ppm. It must therefore be the first one. The NMR is consistent with the ketone. Protons near double bonds (= or =) appear ~2 ppm. Generic sp 3 protons appear ~1 ppm. has 2 neighbors and is a triplet, B has no neighbors and is a singlet, has 3 neighbors and is a quartet. The areas are correct. orrect answer: 3. MS: m/z= 60, MW = 60. No info about M+1, no info about # of s. Broad peaks above 3000 cm -1, H, alcohol likely No peak near 1720 cm -1, no carbonyl (=) NMR: Four types of hydrogen (four unique environments).9 3 t B p or sx Multiplicity hard to tell s D t We know from IR that molecule must contain (at least one) oxygen (H), i.e. a mass of 16 MW - 16 = 44. What has a mass of 44? Guess: 3 s & 8H s (this is consistent with the total # of H s in NMR). That gives MF: 3 H 8. How many degrees of unsaturation? Fully saturated would be n H 2n+2, or 3 H 8, so it must be fully saturated. That means no rings or double bonds are present. How many 3 alcohols are there? 2: 1-propanol & 2-propanol H B D H

3 The second one is ruled out because it has only 3 different types of H s (3 different environments). It must therefore be the first one. The NMR is consistent with the 1-propanol. Protons near heteroatoms (, N, halogens) appear ~4 ppm. Generic sp 3 protons appear ~1 ppm. Protons on or N appear as broad singlets. has 2 neighbors and is a triplet, B has 5 neighbors and is a sextet (although this is hard to see), has neighbors but the coupling (splitting) is lost when H is attached to or N, and D has 2 neighbors and is a triplet. The areas are correct - 3:2:2:1. 2-Propanol would have fewer peaks in NMR because by symmetry Me s are equivalent. lso areas would be 6:1:1. lso splitting would give a doublet and a septet. orrect answer: H 4. MF: 6 H 12 2 What could have 2 s? (a) diol (b) di-ether (c) diketone (d) di-aldehyde (e) carboxylic acid (f) ester No broad peaks above 3000 cm -1, no -H. This rules out a & e. ν = 1735 cm -1, carbonyl (=) present. This rules out a & b cm -1 is too high for most ketones. This rules out c. nly d & f are left (dialdehyde & ester). NMR: Two types of hydrogen (two unique environments) s B s The NMR is frustratingly simple and at this point we can not even be certain of the functional group, but... How many degrees of unsaturation are there? Fully saturated would be n H 2n+2, or 6 H 14, so it is missing 2H s, which is one degree of unsaturation. This rules out di-aldehyde (d). It must be an ester. Moreover, there are no other double bonds or rings. How many 6 esters are there? Too many!

4 However only the two circled isomers give two singlets in NMR. t-bu H s are always singlets and Me s are singlets when they are isolated (distant) from other H s. ll other isomers have too many peaks with lots of splitting. But which one these two is it? H s attached to s attached to heteroatoms (such as ) are shifted substantially downfield to δ ~4 ppm. Therefore it must be the second one (methyl pivalate). Note: The NMR integration implies only 4 H s. ctually the integrals are only the relative areas. Since we have the MF, we know the total area (# of H s) should be 12, therefore each integral must be multiplied by 3. This give the correct areas - 3:9. orrect nswer: 5. MS: m/z= 102, MW = 102. No info about M+1, no info about # of s. No broad peaks above 3000 cm -1, no -H, -NH. ν = 1740 cm -1, = present. In the range of aldehyde or ester; too high for most ketones NMR: Three types of hydrogen (three unique environments) q (but see below!!) B q q IR implies ester or aldehyde, but NMR is missing very distinctive aldehyde proton δ 9-10 ppm (aldehydes are the only peaks found in region 9-10 ppm), therefore must be ester. Ester group, has a mass of = 58. What could have mass of 58? Guess: 4 s + 10 H s. Possible MF: 5 H 10 2.

5 How many degrees of unsaturation are there? Fully saturated would be n H 2n+2, or 5 H 10, so it is missing 2H s, which is one degree of unsaturation. But the = is that degree of unsaturation, no other double bonds or rings. How many 5 esters are there? 5: #1 #2 #3 #4 #5 NMR integrals shows a total of 5 H s, but MF implies 10 H s. ctual areas must be 2:2:6. Which of the above choices have 6 identical protons? #2 & #5. But this raises a problem. Those Me H s must be doublets (not quartets), since each Me has one neighbor! We are in fact forced to rule out #2 & #5. #1, #2, #4, and #5 all have isolated Me s, that is, Me s that should appear as singlets, because they have no close neighboring H s. This rules out: #1, #2, #4, and #5. By process of elimination it must be #3! How can we justify the Me s of #3 as being a quartet? fter all, our simple rules tell us it should be a triplet (they have 2 neighbors). In fact those two Me s are not equivalent, one is closer to =, the other closer to. In principle we should see 2 triplets. Why don t we? Because the triplets occur at approximately the same chemical shift (d), they overlap slightly with each other. Two upfield peaks of one triplet exactly overlap with two downfield components of the other triplet, giving rise to an apparent quartet. The fortitous (or unfortuitous) overlap of peaks make this problem tricky. However, such overlap of peaks is not uncommon. The correct answer is ethyl propionate: D B 6. MS: m/z= 136, MW = 136. No info about M+1, no info about # of s. Exceedingly broad peak -H, but not -H of alcohol, that would occur at ν > 3000 cm -1. This H is so broad that it moves into the H stretching region (~ 3000 cm -1 ) and thereby obscuring the H stretches. This is characteristic of carboxylic acid (- 2 H). ν = 1705 cm -1, = present. ould be ketone or carboxylic acid, but in combination with observations above carboxylic acid is more likely. Possibly sharp peaks above 3000 cm -1, sp 2 H bonds (& =) possible, but hard to tell NMR: Three types of hydrogen (three unique environments) s B s s, broad

6 For R 2 H, what is R? 2 H has a mass of = 91. Guess 7 s + 7 H s. Possible MF: 8 H 8 2 How many degrees of unsaturation are there? Fully saturated would be n H 2n+2, or 8 H 18, so it is missing 10H s (!), which is 5 degree of unsaturation. That s a lot for such a small compound. Benzene ring is a good guess, because it has 4 degrees unsaturation (3 = + 1 ring). 2 H adds one degree unsaturation, bringing the total up to 5. Broad peak in NMR at δ 10.6 is characteristic of - 2 H (- 2 H is the only peak found in range δ ppm). onfirms IR interpretation. What is peak so far downfield δ 7.3 ppm? Benzene ring. Protons on aromatic rings are typically the only peaks found in the region δ 7-8 ppm. onfirms guesses above. How many H s on this particular aromatic ring? 5. Why 5 and not 6? Because it is monosubstituted: X What is X? ould it be - 2 H? No. This has the wrong MF, 7 H 6 2. It is short by -H 2 -. NMR also shows peak of area = 2. Because of the rules of valency, there is only one place to put -H 2 -, between the aromatic ring & the - 2 H: B 2 H Why is the -H 2, so far downfield? Usually -H 2 - adjacent to = or = is ~ 2ppm. But this -H 2 - has 2 neighboring double bonds, one on either side & the effects are more-or-less additive, i.e. ~ 4 ppm. Why is there no coupling between H s of aromatic ring? They all have nearly identical chemical shifts (δ s) and identical H s don t split each other. 7. MF: 9 H 11 N 2 (Would its molecular mass be even or odd?) What has 2 s? (a) diol (b) di-ether (c) diketone (d) di-aldehyde (e) carboxylic acid (f) ester (g) nitro How many degrees of unsaturation are there? 9 H 11 N 2 = 9 H 10 Fully saturated would be n H 2n+2, or 9 H 20, so it is missing 10H s (!), which is 5 degree of unsaturation. That s a lot for such a small compound. Benzene ring is a good guess, because it has 4 degrees unsaturation (3 = + 1 ring). What is last degree of unsaturation? The IR spectrum was taken in Hl 3, which gives rise to unwanted 3030, 1220 & 750 cm -1. Whats left? Two weak, broad ν = 3400 & 3500 cm -1, typical of NH 2. ~1700 cm -1, =. This rules out (a) diol & (b) di-ether.

7 NMR: Five types of hydrogen (five unique environments) t B 4.1 B+ = 4 broad s overlaps with quartet 4.2 B+ = 4 q overlaps with singlet D d E d IR = rules out diol & di-ether. lso no intense 1500 & 1600 cm -1 ; rules out nitro. Exceeding broad - 2 H peak absent. Rules out carboxylic acid. an not be dialdehyde because NMR shows no peak δ 9-10 ppm. NMR clearly shows aromatic peaks (benzene ring); that uses up 4 degrees unsaturation. Therefore can not be diketone (or dialdehyde) because they would bring total degrees of unsaturation to 6. nly the ester has one degree of unsaturation. But ester = typically occurs at 1740 cm -1. Why is this one at 1700 cm -1? It must be conjugated with =. In fact this is indirect evidence for =. (Unfortunately, any sp 2 H IR stretch is obscured by Hl 3 solvent.) ll evidence points to an aromatic ester. NMR integration shows 4 H s on ring. onclusion: it must be disubstituted. H s are identical by symmetry. H B s are also identical. lthough both H & H B are aromatic H s, they are in slightly different environments (one is closer to X, the other to Y), so two different δ s are observed. H & H B couple with each other to give doublets (one neighbor each). H H B X Y H H B But what are X & Y? We know it must have a conjugated ester: aromatic- 2 R. What is R? We have used up 7 s, leaving 2 s, so guess R = Et. Et shows characteristic triplet of area 3 & quartet of area 2 in 1 H NMR. The -H 2 - has 3 neighbors (thus quartet), and is downfield because it is on oxygen. What is left? From MF only NH 2 remains. This is consistent with IR at ν = 3400 and 3500 cm -1. orrect answer: D E NH 2 B H s directly N (or ) are typically broad singlets. In this case it overlaps with the quartet. 8. MF: 4 H 8 2 (Would its molecular mass be even or odd?) No broad peaks above 3000 cm -1, no -H, -NH. ν = 1740 cm -1, = present. In the range of aldehyde or ester; too high for most ketones 13 NMR: Four unique carbons are seen (ignore peak from carbon in solvent, Dl 3 ). ne of them is characteristic of = (δ 171 ppm) and one is adjacent to heteroatom (δ 60 ppm).

8 1 H NMR: Three types of hydrogen (three unique environments) t B s q What has 2 s? (a) diol (b) di-ether (c) diketone (d) di-aldehyde (e) carboxylic acid (f) ester How many degrees of unsaturation are there? Fully saturated would be n H 2n+2, or 4 H 10, so it is missing 2H s, which is 1 degree of unsaturation. nly 1 ring or 1 double bond. That rules out diketone and dialdehyde. bsence of H peak rules out diol and carboxylic acid. Diol and di-ether are ruled out because they have no =. nly ester is left. Its = is the one degree of saturation. The IR = is typical of a generic, unconjugated ester. There are only 2 four esters: They would have virtually identical 13 NMR. Their 1 H NMR would give identical integrals - 3:3:2. They have identical splitting patterns: the typical triplet/quartet for Et and a sharp singlet for Me. They differ only in the chemical shifts. In one case the Me is attached to a double bond (expected δ ~2 ppm), in the other case it is attached to an oxygen (expected δ ~4 ppm). This Me is closer to 2 ppm, therefore it must be the first one, ethyl acetate: B 9. MF: 8 H 10 (Would its molecular mass be even or odd?) No broad peaks above 3000 cm -1, no -H, -NH. No peak at ~ 1700 cm -1, no = present. Sharp peaks above just 3000 cm -1, sp 2 H bonds and = 13 NMR: Five unique carbons signals (ignore peak from carbon in solvent, Dl 3 ). Four of them is characteristic of = (δ ppm). 1 H NMR: Two types of hydrogen (two unique environments) s B ~7 4 m

9 How many degrees of unsaturation are there? Fully saturated would be n H 2n+2, or 8 H 18, so it is missing 8H s, which is 4 degrees of unsaturation. The MF & absence of functional groups in IR indicate that this is a hydrocarbon. hemical shifts in both 1 H & 13 indicate aromatic ring. The IR (sp 2 H) and relatively large degrees of unsaturation hint at this as well. Why does the 1 H NMR show 4 H s in the aromatic region instead of 6? Beause the aromatic ring is disubstituted. What are the two substituents? MF ( 8 H 10 ) - 6 H 4 fragment = 2 s & 6H s. Based on the sharp singlet of area =1 in the 1H NMR, 2 Me s are likely. Me s must be attached to directly to the ring; the Me s chemical shift (~2 ppm) is typical of H s near double bonds. But what is the relative orientation of the Me s? rtho, meta or para? ortho meta para Why does the 13 NMR show only five peaks if the MF indicates 8 s? The molecule must have some symmetry. The ortho would show 4 unique signal and can thus be ruled out. The meta and para would both show 5 peaks. How can the meta and para be distinguished? By the 1H NMR coupling pattern. For the para, all 4 aromatic hydrogens are in the same environment; all 4 are identical by symmetry and identical H s don t couple (split). Thus it would appear as a singlet. Since it is a complicated multiplet, we conclude it must be the meta isomer. meta The 4 meta aromatic H s are in 3 different environments: between Me s, adjacent to Me s and remote from Me s, w/ coupling of singlet, doublet, and triplet (if you look closely you can pick out this coupling). (Meta substituted aromatics show unique peak in IR fingerprint region at ~900, ~800, & ~ 700 cm -1 )