a compound containing ion height: - intec chemistry · web viewnext to ch3 and ch2 each of...

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A compound containing ion height:

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 and 1 for 1H and 2H respectively similar reason why 13C doesnt 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 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.

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.

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)










radio wave



visible light.


i) Electron transitions from bonding to non-bonding orbirtals

ii) Molecular Rotation

iii) Electron transitions from bonding to bonding orbitals

iv) Nuclear spin

v) Bond vibration

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)

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 Hs 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.

ii) What effect must the C=C have in making H(A) and H(B) be pulled downfield compared to CH2s on alkanes? (1)

It must be exerting an electron withdrawing effect which deshields the CHs 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.

Q10 Answer:

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 = 3

SDBS 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=3

Chem 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 2

A 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.

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 = C3H8O

Mass 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?


#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=6

Most 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.

Peak C, 2.2 ppm, no split, int = 2/1 = 1

This 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.

sdbs 673 spectra:

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 fro