1.
(a)
(i)
NO2
+ H 2O
+ HNO3
(1)
(ii)
1
HNO3 + 2H2SO4 → H3O+ + NO2+ + 2HSO4– (1)
or
HNO3 + H2SO4 → H2O + NO2+ + HSO4-
1
(iii)
(HSO4- )
(1)
NO2+
+
H
NO2
m
NO2
(HSO4- )
(1)
H
+
NO2
NO2
ve
NO2+
.co
(+H SO4)
or H +
(b)
(c)
(i)
em
ac
ti
(+H SO4)
or H +
1 (1)
1 (1)
3
2
(ii)
Rate = k[benzene] [NO2+] or k [benzene] [electrophile] (1)
1
(iii)
zero or 0 (1)
1
(i)
360 (kj mol–1)
ch
(ii)
1
ww
w.
Delocalisation
energy
Relative positions of cyclohexatriene and benzene (1)
Some comment that the difference is due to `extra' stability from the delocalisation
energy of benzene. (1)
2
(iii)
Delocalisation is due to overlap of p–orbitals (1)
or π bonds
Overlap of orbitals to give delocalised system like that in benzene only possible if
molecule is flat. (1)
Could be explained via a diagram.
2
[14]
NT Exampro
1
(b)
Substance A is CH3CH2COCH3 (1)
Substance B and C are CH3CH2CH2CHO (1)
and CH3CH(CH3)CHO (1)
(i)
C2H5Br + Mg → C2H5MgBr
(1)
(ii)
dry
ether solvent
(2)
CH3CH2C(OH)(CH3) C2H5
(1)
(iii)
(c)
(d)
3
step 1
step 2
(i)
PCl5 /PCl3 SOCl2
NH3
4
2
two isomers drawn with an attempt at a 3-d diagram (1) each = 2
CH 3
C2 H5
C2H5
C
C
OH
HO
H
H
m
(a)
.co
2.
CH 3
rotate the plane of plane polarised light in opposite directions
(1)
(i)
butanone
CH3CH2COCH3
(2)
(ii)
(a)
(b)
(1)
CH3CH(NH2)COOH + HCl → CH3CH(NH3+Cl−)COOH (1)
Cl− can be separated but then changes must be shown
1
(ii)
CH3CH(NH2)COOH + NaOH → CH3CH(NH2)COO−Na+ + H2O
charges optional but if shown must be correct
1
• Exists as zwitterion (or diagram) (1)
• Strong attraction between oppositely charged ions (1)
(i)
2 unambiguous 3-D diagrams (2)
COOH
COOH
C
CH 3
H
NH 2
H
5
3
[17]
(i)
ww
(c)
green
ch
3.
em
ac
ti
(ii)
w.
(e)
ve
They have an asymmetric carbon atom (4 different groups on a carbon) (1)
mirror image non-superimposable (1)
(4)
2
2
C
NH 2
CH 3
Must show attachment to correct atoms
(ii)
NT Exampro
Rotates the plane of (plane) polarised (monochromatic) light (1)
in opposite directions (1) consequential on mention of polarised light
or
Use polarimeter (1)
measure rotation (of plane of polarised light) in opposite
directions (1)
2
2
(d)
(i)
diagram of but-1-ene (1)
CH3CH2CH==CH2 or
H
H
H
H
H
C
C
C
C
H
H
H
or C2H5CH=CH2
diagram of but-2-ene (1)
H
H
H
H
H
C
C
C
C
H
(a)
CH 3
H
C
CH 3
CH2
C
H
H
C
.co
CH 3
em
ac
ti
4.
geometric (or cis-trans) (1)
H3C
CH 3
H3 C
C C
C
H
H
H
diagram of cis- but-2-ene (1)
diagram of trans-but-2-ene (1)
H
4
COOH is present anyway – absorbs at app 280and 1700
If A is present – nothing else
If B – peak at 3600
or
A spectrum will include 1700 but not 3600
B will include 3600
Marking points
identification of groups to note (1)
link to peaks in spectrum (1)
how to distinguish (1)
3
Correct test (1)
PCl5
or
2,4 dinitrophenylhydrazine
or dilute sulphuric acid plus potassium dichromate(VI)
or sodium hydroxide and iodine
2 correct observations (1)
2
(ii)
NT Exampro
ww
w.
(i)
[13]
ch
CH 3
CO 2 H
• Identification of chiral centre (1)
2 diagrams to show two isomers of non-superimposable molecules
• 3D diagram (1)
mirror image (1)
Distinguish by rotation of the plane of (plane)-polarised light (1)
NB could identify their own real molecule.- correctly shown.
Not necessarily Ibuprofen
(b)
3
ve
(ii)
H
m
CH3CH ==CHCH3 or
2
3
(iii)
(i)
(ii)
5.
(a)
0.0485/50 × 6.023 × 1023 (1) = 5.84 × 1020 (1)
2
−COOH (1)
warm with a little conc sulphuric acid (1)
ester (1)
3
(i)
.co
m
4
1
Rotation (of the plane of polarisation) of plane-polarised
(monochromatic) light
1
bromine (1) allow Br2, not Br, not bromine water
and sodium hydroxide (1)
heat or warm (1) conditional on the correct reagents
(i)
CH3CH2CH(MgBr)CH3 (1)
CH3CH2CH(COOH)CH3 (1)
CH3CH2CH(COCl)CH3 (1)
If substituents are on the end max. (2)
ww
(d)
3
3
(ii)
Grignard reagents react with water (1)
1
(iii)
ammonia (1)
room temperature (1)
2
(i)
CH3
CH
CH 2 CH 3
OH
(ii)
(iii)
(iv)
NT Exampro
[25]
A molecule that is non-superimposable on its mirror image (1)
or
has no plane of symmetry (1)
or has a single asymmetric carbon atom
w.
(c)
5
• At pH=3 equilibrium pushed back in favour of undissociated
acid molecules (1)
• So acid is insoluble because large benzene carboxylic acid
molecules are insoluble (1)
At pH=8 salt is formed (1)
Ionic bonding results in increase solubility (1)
(ii)
(b)
25.75 × 2/1000 mol of NaOH left (1) so
(100 − 51.50)/1000 mol reacted (1)
0.0485 mol of acid in 50 tablets (1)
0.0485 × 206/50 (1) = 0.1998 g per tablet (0.200)
199.8 mg / 200 mg (1) 3 or 4 sig figs
ve
(e)
2
em
ac
ti
(d)
1 correct COOH group scores 1 mark
2 correct COOH groups score 2 marks
ch
(c)
COOH or sodium salt
HOOC
contains C=O so reacts with 2,4 dnp (1)
but cannot be oxidised so no reaction with Fehlings’ solution (1)
CH3 C
CH 3 CH
if
included then
O
OH
zero
(1)
1
2
1
CHI3 (1)
4
CH3CH2COO– (1) Or CH3CH2COONa
2
ww
w.
ch
em
ac
ti
ve
.co
m
[17]
NT Exampro
5