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Tutorial (B) Spectroscopy & Chromatography - ANSWERS Q1) Suggest why 1 H gives peaks on a 1 H NMR spectrum where as is 2 D does not. (1) (They are both “NMR” active, however) signals from 2 H appear at quite separate parts of the NMR spectrum (1) or The energies/frequencies of resonance for 2 H are significantly different from that of 1 H. {Teachers comment: This due to their spin ½ and 1 for 1 H and 2 H respectively similar reason why 13 C doesn’t appear on 1 H spectra and vice versa} Q2) a) Describe what information can be obtained from the following four aspects of a 1 H NMR spectrum. (2) i) integration height: ii) chemical shift of the peak iii) number of peaks iv) peak splitting i) Relative number of protons in a particular environment ii) The protons chemical environment. iii) The number of proton environments iv) The number of adjacent protons (all 4 = 2 marks, 3 correct =1 mark) b) which of the above, If any, is most relevant for functional group determination. No ii), the protons chemical environment. Q3) Give the symbol for chemical shift and state its units: Symbol = units = parts per million (ppm) Q4) Discuss why peaks appear with different chemical shifts. Page 1 of 24

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Page 1: A compound containing ion height: - INTEC Chemistry … · Web viewNext to CH3 and CH2 each of which cause this CH2 group to split into a complex pattern. Molecule = CH3-CH2-CH2-Cl

Tutorial (B)

Spectroscopy & Chromatography - ANSWERS

Q1) Suggest why 1H gives peaks on a 1H NMR spectrum where as is 2D does not. (1)

(They are both “NMR” active, however) signals from 2H appear at quite separate parts of the NMR spectrum (1) or The energies/frequencies of resonance for 2H are significantly different from that of 1H. {Teachers comment: This due to their spin

½ and 1 for 1H and 2H respectively similar reason why 13C doesn’t appear on 1H spectra and vice versa}

Q2) a) Describe what information can be obtained from the following four aspects of a 1H NMR spectrum. (2)

i) integration height:ii) chemical shift of the peakiii) number of peaksiv) peak splitting

i) Relative number of protons in a particular environmentii) The protons chemical environment.iii) The number of proton environmentsiv) The number of adjacent protons (all 4 = 2 marks, 3 correct =1 mark)

b) which of the above, If any, is most relevant for functional group determination.No ii), the protons chemical environment.

Q3) Give the symbol for chemical shift and state its units:Symbol = units = parts per million (ppm)

Q4) Discuss why peaks appear with different chemical shifts.

Different chemical environments cause different degrees of e- withdrawal from nearby protons, causing different degrees of deshielding/shielding on those protons which therefore experience the effect of the applied magnetic field B0

to different extents., each requiring slightly different radio wave frequencies/energies to make the protons flip, hence peaks appear at different frequencies/energies/chemical shifts on the spectrum.

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Q5) How does low resolution NMR differ from high resolution NMR and what benefit does this bring? (4)

High resolution NMR reveals splitting patterns(1) which are useful in structure elucidation/deduction(1) by helping determine local/adjacent proton count(1) Low res NMR does not ‘reveal splitting patterns’/show splitting patterns clearly/in sufficient detail(1).

Q6) a) To absorb microwave radiation, what molecular condition must be present.

The molecule must have a permanent dipole

b) What is necessary for a molecule absorbs microwave radiation, what internal and external effect does this have? (2)

The internal molecular rotation increases.(1) This leads to more energetic collisions (with other molecules) producing heat.(1)

c) Rank the spectroscopic frequencies: Radio, micro, infra-red waves in terms of increasing energy (1)

Radio < micro < infra-red Correct order (1) Q7) Using the table of different types of radiation, identify which is associated with the following quantum phenomenon. (3)

Table:A B C D E F

X-ray Infra-red radio wave microwave UV visible light.

Phenomenon:

i) Electron transitions from bonding to non-bonding orbirtalsii) Molecular Rotationiii) Electron transitions from bonding to bonding orbitalsiv) Nuclear spinv) Bond vibration

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i) = E (UV)ii) = D (microwave)iii) = F (visible light)iv)= C (radio wave)v) B (IR) (all 5 = 3 marks, 4 correct = 2 marks, 3 correct – 1 mark)

Q 8) a) Give one advantage of spectroscopic techniques such as NMR or IR compared to techniques like elemental analysis and mass spectroscopy.

NMR and IR are non-destructive i.e. the compound is not destroyed, unlike elemental analysis or mass spec.(1)

b) Give one disadvantage of NMR as a spectroscopic technique.(1)

The initial cost of NMR machines is high OR the liquid helium currently necessary in the operation of NMR machines is expensive to purchase (approx USD 5 per dm-3)

Q9) Account for the fact that N2 is not a greenhouse gas, yet H2O is. (3)N2 (because of its symmetry) experiences no change in dipole moment(1) as it vibrates (1) and cannot produce heat from rotation as it has no permanent dipole and is unresponsive to microwave radiation (1 for correct consideration of microwaves)

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Q 9) Using the 1H NMR spectrum of propene is given below.

a) Identify the peak likely to come from the methyl group and justify your choice (2)The peak at 1.7 ppm(1). It is the only group near to ppm 1 where methyl groups frequently are found(1).

b) As indicated, there are four peaks present. What is the significance of the four peaks? (1)

Four peaks signify 4 H environments(1)

c) Read the following analysis suggest answers for the corresponding questions.

Compared to CH2 in alkanes, the peaks H(A) and H(B) are very much more downfield. In terms of chemical environment, the two H’s on the Left hand siede, H(A) and H(B), are in a more similar to each other than the single hydrogen, H(C), on the RHS. The peaks at 4.88 and 4.96 are likely to be from H(A) and H(B). Of these two, the peak at 4.96 is likely to be the proton H(B) and H(A) the one with chemical shift of 4.88.

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ii) What effect must the C=C have in making H(A) and H(B) be pulled downfield compared to CH2’s on alkanes? (1)

It must be exerting an electron withdrawing effect which deshields the CH’s on the C in the C=C.

i) Why will H(B) to be shifted slightly more downfield compared to H(A)? (3)

There are more electrons in the upper part of the rigidly held molecule than the lower part.(1) So H(A)’s local environment contains more electrons than H(B),(1) hence H(B) experiences less shielding than H(A) and is pulled/shifted downfield.(1) Last mark reverse argument can be made.

Q10) Using the spectra below, suggest a structure for the molecule whose ms, 1H NMR and IR spectra are given below (SDBS spectra 3362)

Hint: the peaks at 49 & 51, 63 & 65 and 78 & 80 all have relative intensities of 3:1.

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Q10 Answer:

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Multiple occurrences of peaks separate my m/z = 2 units with 3:1 ratios suggest evidence for 1 Cl atom. M+. (35Cl isotope) = 78 residual mass = 78 - 35 = 43 maximum No of C's = 43/12 = 3SDBS 3362 has no evidence of an N or O functional group. Has a peak between 600 and 800 cm-1 so halogenalkane confirmed. C3H8Cl Mass spec: base peak = 42 3C fragment 3xC = 36 with 6 H's [C3H6]+

Peak at peak at 29 = ethyl? [CH3CH2]

NMR (building on MS and IR info e.g. absolute no of H's = 8 determined from the mass spec)

#peaks = 3 so 3 H environments

Peak C, 0.855 ppm, triplet, int=3Chem shift suggests methyl, and not next to an e- withdrawing group.Spliiting reveals 2 adjacent and equivalent protons CH2 group?

Peak A, 3.3 ppm, triplet, int=2 (typo used to say 3)splitting suggests next to the previously identified CH2 group. Corrected integration height suggests itself contains 2 H.Shifted far downfield therefore beside an e- withdrawing group. This CH2 probably holds the Cl

Peak B, 1.6 ppm, multiplet, int 2A CH2 group from int height. Next to CH3 and CH2 each of which cause this CH2 group to split into a complex pattern.

Molecule = CH3-CH2-CH2-Cl

Q11) Molecule “SDBS 2149” con contains % C, H and O by mass of 59.9, H 13.42 and O 26.61 respectively.The MS, IR and 1H NMR are given below. Deduce the structure of this molecule.

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Q11 ANSWER: % by mass gives impirical formula C3H8O has mass 60. IR shows OH group.

Mass Spec = M+. = 50 or 60. Must be 60 due to empirical formula.m/z 60 - 17(OH gp) = 43, corresponds to C3H7. Suggested Molec formula = C3H8OMass spec has base peak at 45 = C3H9 (imposs) so must contain one O.45-16 = 29 = [C2H5]+ so base peak = [C2H5O]+Missing mass from 45 to 60 = 15 i.e. methyl group. Molecule = propan-2-ol?

NMR:#peaks = 3 so 3 H environments.Total integration height = 12+2+2=16. Impossible. Molecule has 8H's so divide all integration ratios by 2 to get the absolute no of H's Peak C, 1.2 ppm, doublet, int 12/2=6Most likely a methyl and this methyl must be nect to 1 proton due to doublet. Must have 2 CH3's to account for the 6 integration ratio.

Peak A, 4ppm, heptet, int = 2/1 = 1.This must be the must be the CH next to the two methyl groups. It's shofted quite far downfield so probably has the OH bonded to it.

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Peak C, 2.2 ppm, no split, int = 2/1 = 1This is going to be the OH.

So (CH3)2CH(OH) confirmed {2-methyl propan-2- ol)

Q12) The following spectra are of two interconvert able compounds “sbds 1460” and “sbds 673”

Using the spectra, determine the structure of each molecule and identify the reaction taking place from “sbds 1460” to “sbds 673”. The spectra for “sbds 1460” are given first.

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“sdbs 673” spectra:

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Answer to Q12) SDBS=1460.................Look at IR first to get a flavour of the functional groups. IR: spectrum was taken in a CCl4 solution hence may appears slightly different from the liquid based IR. But C=O is clearly shown about 1720 cm-1. Are there peaks at >3000cm-1?? look too weak to distinguish from noise {They are actually there but CCl4 obscures it} and peaks at around about 1600 (C=C) were physically absent from the spectrum (there was originally a 'gap' in the specrum itself, which I was forced to fill in with a straight line!) Probably aldehyde, ketone or ester

Mass spec: Large mass therefore expect (M+1)+. and (M+2)+. peaks. Therefore choosing peak at 150 to be the molecular ion. Peak at 77 classic monosubstituted benzene ring [c6H5]+. Peak at 122. Has lost 27 mass units from M+. is not less than or equal to (150-77) so it must contain the ring and the rest of mass = 122-77 = 45 units. i.e. ring + 45 units = peak at 122. [CH3-CH2-O]+ = 45 Peak at 51 = hard to assign.

Ar and C=O present possible ethoxy group.

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NMR: Appears to show 5 peaks. 3 of them in the Arene region and all are complex so treat all as Ar-H's mass spec supports this (m/z=77) Integration height for these comes to 5 so int. height is probably the absolute no of H.

Peak E, 1.39 ppm, triplet, int = 3,sugggests a methyl group. Skhew to LHS (and seems to have a smaller triplet embedded)

Peak D, 4.4 ppm quartet, int = 2Suggests CH2 group. Chem shift highly downfield could be next to electron withdrawing group say -O- (as suggested by mass spec) MS supports this and

Suggests Ar-COOEt. Ethylbenzene. Mass = 150.

Spectrs SDBS 673.............

IR: IR shows characteristic COOH and C=O. The mass of COOH is 45. So carboxylic acid is suspected.

Now the molecular ion is 122, assuming due to the high mass that 123 is the (M+1)+, peak. Again it has a strong peak at 77 indicating a monosubstituted benzene ring. This leaves 122 - 77 mass units = 45 which accounts for COOH. Benzoic acid suspected.

NMR: # peaks=2 so two H environments Peak A, 12ppm, no split, int = 1This is a 'magic number' for carboxylic acids.The peak is unsplit and quite broad.

Peaks B,C&D, 7-8 pmm, complex splitting, sum of int=5Each peak assumed to be individual H's in the ring.

Going back to MS, base peak at 105 = [Ar-C=O]+ and expected for carbox acids. Peak at 51 - difficult to assign, probably a McLaferty rearrangement fragment???

Spectra SDBS 1460 = benzoic acidSpectra SDBS 673 = benzoic acid

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The chemical reaction therefore was hydrolysis.

Q12 (second Q12!): With reference the spectra for propan-1-ol and propanone given below, discuss briefly how the progress of a chemical reaction could be followed.

By following the disappearance of the OH peak at about 3300cm-1 or the appearance of the C=O peak at about 1700cm-1.

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Q13) Which technique (give all if more than one is applicable) would be suitable to detect (3)a) E-Z isomers - NMR (best)b) optical isomers - NMR (only)c) positional isomers - NMR (perhaps IR)

Q13) a) What purpose does HPLC serve? (1)

To separate(or purify) mixtures (in small quantities) (1)

b) is used to In HPLC, why is a high pressure used? (2)

To speed up the passage of the eluent through the column (1) as retention time is long due to fine particle size in the column.(1)

c) Identify the mobile phase and the stationary phase in a HPLC experiment.

The stationary phase is the material inside the column,(1) the mobile phase is the solvent (and eluent) (1)

d) Other than using high pressure, what three other factor could shorten retention time? And in general, are shorter times favourable in terms of giving good separations? (2)

Use a shorter column OR use larger particle size OR a solvent more similar in polarity to the eluent than the material in the column. (1) Shorter retention times generally give poorer separations. (1)

e) Give one suitable and one unsuitable chromatographic method for separating volatile compounds.(2)

Suitable = HPLC (1) , unsuitable = GC (2)

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--------------

SDBS 2149

--------------------

DBS=1460 MS.M+, in the region of 150 so (M+1)+. peaks expected. So C12 version has a mass of 150. Magic number of 77 suggests monosubstituted benzene ring (a phenyl group) leaving rest of mass = 150 - 77 = 73. Max other carbons(outside ring=6).

SDBS=1460 IR was done in a CCl4 solution hence may appear slightly different, but C=O is clearly shown. "Peaks" at >3000cm-1 too small to see. peaks at around about 1600 (C=C) were physicall absent from the spectrum (there was originally a gaps in the specrum itself!)

So Ar and C=O present. 77+12+16 = 105 so max C now = 150-105 = 3 Ar(5 H) + 3C leaves other H. ImpossibleNMR:

Appears to show 5 peaks. 3 of them in the Arene region and all are complex so treat these as Ar-H's mass spec supports this. MS supports this and beside these two methyls. It

CH3 and CH obtained.

Cannot have 12 H's in molecule, must be

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