1 1) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound. First, your molecular ion peak is 112 and you have a M+2 peak at 114. Therefore, you have a halogen. Now, your molecular ion peak and M+2 peak are in a 3 to 1 ratio. This means chlorine. So, =77 # C s 77/12=6 carbons so C 6 H 5 Cl. DOUS (2(6)+2-5-1)/2=4 Cl
2 2) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug containing a cloudy liquid (bp C), marked only "STUDENT PREP". She ran a quick mass spectrum, which is shown below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, "I know what it is." What compound is the "student prep"?
3 So molecular ion peak at 136 and M+2 peak at 138, so halogen present. They are in a 1:1 ratio so Br. So =57/12 = 4x12= =9, C 4 H 9 Br (2(4)+2-9-1)/2=0 Br The peaks at 107 (C 2 H 5 ) and 93(C 3 H 7 ) tell us it is a linear chain instead of a branched one.
4 3) A laboratory student added 1-bromobutane to a flask containing dry ether and Mg turnings. An exothermic reaction resulted, and the ether boiled vigorously for several minutes. Then she added acetone to the reaction mixture, and the ether boiled even more vigorously. She added dilute acid to the mixture, and separated the layers. She evaporated the ether layer, and distilled a liquid that boiled at 143 C. GC MS analysis of the distillate showed one major product with a few minor impurities. The mass spectrum of the major product is shown below. Show the structure of this major product.
5 BrMg/etherMgBrOO-H+OH The molecular ion peak should be at 116, but the loss of a carbon from the quatenary C forms a stable carbocation at 101.
6 1) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound. 1-butene, 1-butanol, 4-hydroxy-1-butene, methyl propyl ether, butanoic acid.
7 First thing to notice is the presence of an alcohol at cm -1, which narrows our choices down to 1-butanol, 4-hydroxy-1-butene and butanoic acid.. OHO OH OH Only 1-butanol works, because there are no peaks corresponding to C=C and C=O.
8 2) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound. phenylacetone, benzoic acid, acetophenone, benzyl alcohol, benzaldehyde
9 First you can eliminate benzoic acid and benzyl alcohol because there is no OH peak. Second you can eliminate benzaldehyde because there is no peak at 2740 cm -1 (aldehyde peak). That leaves phenylacetone and acetophenone. OO Acetophenone is the answer because the carbonyl peak is at 1700 cm -1 and a simple ketone like that on phenylacetone would absorb at a higher energy.
10 3) One of the following compounds is responsible for the IR spectrum shown. Draw the structure of the responsible compound. 2-ethynylcyclohexanone, 2-methyl-2-cyclohexenone, acetophenone, cyclohexylmethyl alcohol, 4-ethylcyclohexanone.
11 First, 2-ethynylcyclohexanone can be eliminated because there is no peak for a carbon carbon triple bond. Second, cyclohexylmethyl alcohol is eliminated because there is no OH peak present. Acetopheone can be eliminated next because there is no peak for aromatic C-H stretches. Also the carbonyl peak would be at a higher energy like around 1800 cm methyl-2-cyclohexenone can be eliminated because there is no O carbon carbon double bond peak. Leaving 4-ethylcyclohexanone.
12 (4) Determine the structure of the compound that gives rise to the following mass and IR spectra.
13 The molecular ion peak is at 162 and the M+2 peak is at 164, and they are in a 1:1 ratio, therefore there is a bromine atom =83 83/12=6 carbons and 6x12= 72 so 83-72=11 C 6 H 11 Br So 2(6) =2/2=1, which means 1 double bond or 1 ring. Looking at the IR, there is no C=C peak so that means a ring. Br
14 (5) Determine the structure of the compound that gives rise to the following mass and IR spectra.
15 So the molecular ion peak is 72. So 72/12= 6 carbons 2(6)+2=14/2=7 a bit high. So subtract 1 C and replace with 12 Hs. C 5 H 12. 2(5)+2-12=0 Not pentane, there is a carbonyl stretch.
16 So to add an O, subtract a methyl group C 4 H 8 O. 2(4)+2-8=2/2=1 Lastly there is a peak at 2740 which tells us that the carbonyl si due to an aldehyde. O
17 (6) Draw a structure consistent with the following data: The MS shows a molecular ion at 59 amu. The IR spectrum shows a double-humped strong absorbance at around 3300 cm 1 (the only absorbance in the functional group region) and a single absorbance at about 1385 cm 1. Odd molecular ion peak tells you there is a nitrogen =45 45/12=3 carbons 45-36=9 hydrogens C 3 H 9 N. 2(3)+2-9+1=0 The peak at 3300 cm 1 tells us that the N is part of an amine. NH2
18 (1) The following 1 H NMR spectrum is of a compound of molecular formula C 3 H 8 O. Propose a structure for this compound.
19 First you have a septet that integrates to 1 H and a doublet that integrates to 6 Hs. This is typical of an isopropyl group. Then the peak at 2.5ppm. Is a singlet and represents a H on an OH group. OH
20 2) Draw the structure of the compound with the 1H NMR and IR spectra shown and the formula C5H12O.
21 2(5)+2-12=0 so no double bonds or rings. Also there is no OH or C=O peaks in the IR, so it has to be an ether. Looking downfield you have a triplet and a singlet. For there to be a singlet there must be only a methyl on one side of the ether. Thus giving us the following structure. O Looking at this structure, it explains the presence of the pentet and sextet for the middle two CH 2 s. And finally the triplet upfield is for the terminal methyl group.
22 3) Draw the structure of the compound with the 1H NMR and IR spectra shown and the formula C6H12O2.
23 First 2(6)+2-12=2/2=1 And based on the carbonyl peak in the IR we know this is our degree of unsaturation. Also we know that there must also be an ether since there is no OH peak in the IR. And based on the proton NMR we have two types of protons. One has to be connected to the ether. And for the rest to be all the same, there must be an isobutyl group. OO
24 (4) Determine the structure of the following compound based on its mass, IR, and 1 H NMR spectra.
25 114/12=9 carbons =6 hydrogens C9H6 2(9)+2-6= 7 degrees of unsaturation Based on the IR we know there is a carbonyl So CH4 add O C8H2O 2(8)+2-2=16/2=8 Lets take off a C and add 12 Hs C7H14O 2(7)+2-14=2/2=1 Which accounts for the C=O
26 And now look at the proton NMR and there are 3 types of protons, this indicates a symmetrical ketone. There should be two triplets and one sextet. O
27 (5) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1- nitropropane, nitroethane, or 2-bromopropane is responsible for the 1 H NMR spectrum shown. Draw the structure of the responsible compound.
28 There are two signals in the spectra so we can eliminate 2- butanone, 2-methyl-2-nitropropane and 1-nitropropane because they have either more or less than 2 types of protons. Next you can eliminate 3-pentanone because it would have a triplet and a quartet which is not seen in the spectra. Also nitroethane can be eliminated because it would have a doublet and a quartet. While the answer is 2-bromopropane which has a doublet and a heptet. Br
29 6) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1- nitropropane, nitroethane, or 2-bromopropane is responsible for the 1 H NMR spectrum shown. Draw the structure of the responsible compound.
30 Because there are 3 signals it is either 2-butanone or 1- nitropropane. 1-nitropropane should have a doublet, a quartet and a triplet. While our answer, 2-butanone should have a singlet, a quartet and a triplet.. O
31 7) The molecular formula of a compound is C 6 H 12 O. Determine the structure of the compound based on its molecular formula and its 13 C NMR spectrum.
32 First 2(6)+2-12=2/2=1 so either a ring or double bond. No peak shows up in the double bond region, C=C or C=O. So that leaves a ring. Four peaks and with this structure we have 4 different types of carbon. OH
33 8) Identify the compound with molecular formula C 3 H 5 Cl 3 that gives the following 13 C NMR spectrum. (The resonance at 0 ppm is due to the TMS standard, not the unknown.)
34 First, 2(3)+2-5-3=0 so no double bonds or rings. Secondly there are 3 peaks, so 3 different kinds fo carbon. So that leaves two choices that are correct. ClClClClClCl
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