10 Marking Scheme: End-Of-Chapter Test

10 Marking scheme: End-of-chapter test

10 Marking scheme: End-of-chapter test

Element / Carbon / Hydrogen / Oxygen
Percentage / 62.1 / 10.3 / 27.6
Mass in 100g / 62.1g / 10.3g / 27.6g
Number of moles / = 5.18 / = 10.3 / = 1.73 / [1]
Relative number of atoms / = 3 / = 6 / = 1

empirical formula = C3H6O[1]

bmolecular mass is 58 (from the m/e value for the penultimate peak).[1]
empirical formula mass = molecular mass, therefore the molecular formula = C3H6O [1]

[1]

iiIt is propanal.[1]

iiiThere are three peaks present in the 1H NMR spectrum;[1]
propan-2-one has only one type of proton [1]
and therefore it would have only one peak.[1]

dThree types of proton and therefore three peaks.[1]
The peak for the –CH3 protons at δ = 1.1ppm is split into a triplet because of the
two chemically different protons on the adjacent carbon.[1]
The peak for the –CH2– protons at δ = 2.5 ppm is split into a quadruplet because of the
three chemically different protons on the adjacent carbon.[1]
The peak for the –CHO proton at δ = 9.8ppmis split into a triplet because of the
two chemically different protons on the adjacent carbon.[1]

eThe strong absorption at 1750cm–1[1]
which is due to the presence of the >C=O group. [1]

Element / Carbon / Hydrogen
Percentage / 90.6 / 9.4
Mass in 100g / 90.6g / 9.4g
Number of moles in 100 g / = 7.55 / = 9.4 / [1]
Relative number of atoms / = 1 / = 1.25
Whole numbers / 4 / 5

empirical formula = C4H5[1]

bThe m/e value for the penultimate peak on the mass spectrum = 106 [1]
this is twice the empirical formula mass, so the molecular formula is C8H10[1]

1 mark for each structure[4]

[1]

This is because when chlorinated it forms only the following compound:[1]

The other possible isomers form several different compounds. [1]
The 1H NMR spectrum shows two peaks as C has only two different types of proton; [1]
the other possible isomers have larger numbers of different types of proton.[1]

dAt δ = 7ppm [1]
there is the peak from the benzene ring protons;[1]
at δ = 2.3ppm [1]
the peak corresponds to the –CH3 protons.[1]

Element / Carbon / Hydrogen / Oxygen
Percentage / 77.8 / 7.41 / 14.8
Mass in 100 g / 77.8g / 7.41g / 14.8g
Number of moles in 100g / = 6.48 / = 7.41 / = 0.925 / [1]
Relative number of atoms / = 7 / = 8 / = 1

empirical formula = C7H8O [1]

bThere is a strong absorption at ~3300cm–1[1]
corresponding to the –OH group in alcohols. [1]

cThe m/e value for the molecular-ion peak is 108;[1]
this corresponds to the mass of the empirical formula, so
molecular formula = empirical formula = C7H8O [1]

1 mark for each structure[5]

ii[1]

iiiThe relative area of the peak at δ = 7.3 ppm is 5, so there are five protons on the
benzene ring; thereforeD cannot be a di-substituted compound.[1]
The other mono-substituted compound has only two types of proton[1]
while the 1H NMR spectrum of D has three peaks, corresponding to the three different types of proton in D. [1]

eThere are three peaks because of the three chemically different types of proton in D:[1]
the peak for five protons at δ = 7.3ppm is caused by the five benzene protons;[1]
the peak given by two protons at δ = 4.6ppm is caused by the –CH2– protons;[1]
the peak at δ = 2.4ppm is caused by the –OH proton. [1]

4a / Element / Carbon / Hydrogen / Oxygen
Percentage / 69.8 / 11.6 / 18.6
Mass in 100 g / 69.8 g / 11.6 g / 18.6 g
Number of moles in 100 g / = 5.82 / = 11.6 / = 1.16 / [1]
Relative number of atoms / = 5 / = 10 / = 1

empirical formula = C5H10O [1]

bThe m/e value for the molecular-ion peak is 86so its relative molecular mass is 86;[1]
this is the same as the empirical formula mass, so the molecular formula is also C5H10O.[1]

1 mark for each structure

1 mark for each structure[7]

dIt must have a carbonyl group, because of the positive result with 2,4-DNPH.[1]
It is an aldehyde, because of the positive result from the silver mirror test. [1]
On the NMR spectrum there is an absorption at δ = 9.5ppm, characteristic of the
aldehyde –CHO proton, and on the IR spectrum there is an absorption at 1720cm–1, characteristic of a C=O group. [1]

eOn the 1H NMR spectrum there is an absorption at δ = 9.5ppm, characteristic of the
aldehyde –CHO proton;[1]
it has only two distinct peaks and therefore only two types of proton;[1]
therefore E must be[1]

There is no splitting of the peaks because there are no adjacent carbons with chemically different protons on them. [1]

5aCH3CH2CH2COOH[1]
CH3CH(CH3)COOH[1]
CH3COOCH2CH3[1]
CH3CH2COOCH3[1]
HCOOCH2CH2CH3[1]
HCOOCH(CH3)2[1]

bF is butanoic acid, CH3CH2CH2COOH[1]
because it gives effervescence with Na2CO3 and has four types of carbon atom. [1]
G is 2-methylpropanoic acid, CH3CH(CH3)COOH [1]
because it gives effervescence with Na2CO3 and has three types of carbon atom. [1]
H is 2-methylethyl methanoate, HCOOCH(CH3)2[1]
because it does not give effervescence with Na2CO3 and has three types of carbon atom. [1]