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Worksheet A
Colorimetry which is known as a scientific technique is used to decide the concentrations of
coloured solutions by the process of the beer lambert law this shows that the concentration
of the solute will be proportional to its absorbance.(class content)
Brass results
Moreover, in the brass one of the errors would’ve been using the colorimeter wrong for
example scanning 700nm first instead of 400nm in the absorbance mode or mixing the
wrong analytes adding too much or too less of what is needed also not filling the solution up
to the marker or going above it these can all causes errors to your method.
Compound A has a C-H peak at 2800-3100 and a C=C at 1700cm -1 which means it’s an alkene
is because it has a double bond and the bonds it doesn’t have are C-O, C=O, and O-H.(class
content)
Compound B has a C=O and a C-O from 1750-3000 which shows that it’s an ester and the
bonds that it doesn’t have are C=C, C-H, O-H.(class content)
Compound C has an O-H, C-H, C-O which makes it an alcohol the O-H bond is between 3000-
3500cm-1 the C-H bond is between 2500-3000cm-1 and the C-O bond is between 1000-
1500cm-1.(class content)
Compound D is an ester because it has a C-H, C=O, and C-O bond and the bonds it doesn’t
have are C=C and O-H the C-H bond is between 2000-2950cm -1 the C=O bond is between
1000-1700cm-1 and the C-O bond is between 1000-1300cm-1.(class content)
Compound E has a C-H, C=O, and C-O bond making it an ester and the bonds it doesn’t have
are C=C and O-H the C-H bond is between 2500-2950cm-1 the C=O is between 1700-2000cm-
1
and the C-O bond is between 1150-1500cm-1.(class content)
Compound F has a C-H bond between 2850-3000cm-1 which makes it an alkane and the
bonds it doesn’t have are C=O, C=C, C-O, and O-H.(class content)
Compound G has a C-H, C-O, C=O making it an ester and the bonds it doesn’t have are C=C
and O-H the C-H is between 2950-3000cm-1 the C-O is between 1150-1300cm-1 and the C=O
is between 1750-2000cm-1.(class content)
Worksheet B
Compound W looking at the carbon Nmr It have two peaks one at 20ppm which is a carbon-
carbon (C-C) and the 2nd peak at 180ppm and it’s shifted because of the electro negative
oxygen it’s a C=O peak out of the four compounds it could be ethanol or ethanoic acid
because It has two carbon peaks.(chemguide.co.uk the background to C-13 NMR
spectroscopy Author unknown, 2022)
Compound X looking at the carbon Nmr It has four peaks one at 5 ppm which is a C-C
another one at 30 ppm which is also a C-C and another one at 38 ppm which is also a C-C
, and the last one is at 210 ppm making it a C=O.(chemguide.co.uk the background to C-13
NMR spectroscopy Author unknown, 2022)
Compound Y looking at the H resonance its showing three separate peaks like a singlet,
triplet, and quartet which straight away tells us that there is a CH2 and CH3 meaning it’s an
alcohol where as on the C resonance it shows two different carbons one below 50ppm and
one above this also indicates there’s a C-O, and C-C bond making compound Y an ethanol.
(chemguide.co.uk the background to C-13 NMR spectroscopy Author unknown, 2022)
Compound Z looking at the H resonance it shows 4 complete separate peaks such as a
quartet two triplets and a peak which consists of six separate pillars on the other hand the C
resonance shows one peak above 200ppm and three peaks below 50ppm which suggests
there is three C-C single bonds and one C=O double bond from this we can conclude that the
compound will be butanal meaning an aldehyde.(chemguide.co.uk the background to C-13
NMR spectroscopy Author unknown, 2022)
Worksheet C
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