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Chemistry
Answers to Additional Questions
O R GA N IC CH EM IS T R Y
UNIT 2
1.
(a)
(b)
T – Br * ; U – HBr; V – C 2 H 5 * ; W – Br * ; X – Br 2 ; Y – C 4 H 10 ;
Z – Br * (any 6 ½)
Endothermic
Energy required to break bonds
3
(1)
(1)
2
(c)
(i)
(ii)
Propagation
Termination
(1)
(1)
2
2.
(d)
Once the Br * radical is produced, the propagation step produces
more of these radicals to keep the reaction going.
1
(8)
(a)
2
(b)
The 2s and 2p orbitals combine to form four hybridised
sp 3 orbitals.
Each occupied by one electron.
1
(1)
(1)
2
(c)
The C=C bond consists of a  bond and a  bond
but the C–C is just a  bond.
(1)
(1)
2
(d)
1
(6)
3.
(a)
Nuclophilic substitution
1
(b)
They have a polar C–X bond.
1
O R GA N IC CH EM IS T R Y
(c)
React the alcohol with an alkali metal (Na).
(d)
(i)
(ii)
Butanoic acid
Propanoic acid
1
(1)
(1)
2
(5)
4.
(a)
CH 3 CHBrCH 3 or full structural formula
1
(b)
(c)
(d)
Nucleophile
An ether (alkoxyalkane)
Any hexanol isomer, e.g. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH
1
1
1
(e)
1
(f)
5.
Use infra-red to detect the presence of C=O in R or its
absence in Q; ether has lower boiling point than ester;
ester can be hydrolysed with NaOH but ether cannot be
(any one).
(g)
Propanone
(a)
Ethoxide ion
(b)
(i)
(ii)
Sodium/alkali metal
PCl 5 /PCl 3 /AlCl 3
1
1
(7)
1
(1)
(1)
2
(c)
1
(4)
O R GA N IC CH EM IS T R Y
6.
(i)
(ii)
(1)
Substitution
HBr/PBr 3 /PBr 5
(2)
Elimination
Alcoholic KOH
(3)
Substitution
(4)
Hydrolysis
(5) Condensation/
esterification
(6)
Neutralisation
(7)
Chlorination
(7  1)
KCN/NaCN/CN –
H + (aq)
CH 3 OH
(concentrated H 2 SO 4 )
NH 3
PCl 5 /PCl 3 /SOCl 2
(7  1)
(14)
7.
(a)
(1 + 1)
2
8.
(b)
Electrophilic (aromatic) attack (substitution) or nitration
1
(3)
(a)
Phenol or hydroxy benzene
1
(b)
(concentrated) H 2 SO 4 + HNO 3
(Note: phenols will react with HNO 3 alone.)
1
(c)
Reduction
1
(d)
C 6 H 4 NH 2 OH  C 6 H 4 NHCOCH 3 OH
111 g
 153 g
 43.6 g  60.1 g
for 80% yield get 60.1  80/100 = 48.1 g
(1)
(1)
(1)
3
O R GA N IC CH EM IS T R Y
(e)
(1 + 1)
2
(8)
9.
(a)
(b)
A reaction in which the carbon atoms of a hydrocarbon
are rearranged to produce a different carbon skeleton
by the action of heat and a catalyst.
1
(Friedel–Crafts) alkylation
using an alkyl halide
in the presence of aluminium chloride.
(1)
(1)
2
(c)
(catalytic) hydrogenation
1
(d)
Detergent
1
(e)
(i)
2
(ii)
10.
(a)
Condensation polymerisation
1
(8)
(i)
OH
(ii) RCHO
(iii) RCOR
(Note: if B were another primary alcohol then it may
be oxidised to a carboxylic acid and D = RCO 2 H.)
(3  1)
3
(b)
acidified dichromate or Tollens’ or Benedict’s
orange to blue-green or silver mirror formed
or blue to orange
Name of a correct reagent
correct change described for C
no change for D
Reagent
Change
(1)
(1)
(1)
O R GA N IC CH EM IS T R Y
or make a derivative
measure melting point
check with literature
(1)
(1)
(1)
3
(c)
3500 to 2500 cm –1 and 1725 to 1700 cm –1
(1+1)
2
(8)
11.
(a)
Moles C = 69.77/12 = 5.81
Moles H = 11.63/1 = 11.63
Moles O = 18.60/16 = 1.16
Ratio of moles
5
10
1
C 5 H 10 O
(1)
(1)
2
(b)
OH/hydroxyl/alcohol
(c)
C 5 H 10 O
(1)
cyclopentanol
(also accept cyclobutylmethanol or cyclopropylethanol)
(1)
(d)
1
2
Carbonyl/C=O
1
(e)
1
(f)
12.
Compound X or Y or unreacted 2,4-dinitrophenylhydrazine
or water (any one).
(a)
Separate ether layer and distil (evaporate) off the ether.
(b)
Advantage – lack of reactivity/good solvent/low
boiling point
Disadvantage – highly flammable/anaesthetic/toxic/
forms peroxides
1
(8)
1
(1)
(1)
2
O R GA N IC CH EM IS T R Y
(c)
Ethanoic acid/CH 3 COOH
1700 to 1725 cm –1 (C=O stretch)
(1)
(1)
2
(d)
13.
In Grignard reagent C – Mg + Br therefore the C – acts a
nucleophile.
In alkyl halide C + Br  – therefore the C + acts as an
electrophile.
(This difference is due to electronegativity differences.)
(a)
(i)
(ii)
(b)
(i)
(1)
(1)
2
(7)
Carbon to carbon double bond
Geometric isomerism
1
1
1
(ii)
14.
C=C does not allow rotation about this bond
whereas C–C does allow rotation.
1
(c)
If the molecule has changed from cis- to trans- then the enzyme
will not recognise it.
1
(5)
(a)
(i)
2
(ii)
Pentane is a longer molecule and will have more
van der Waals’ forces between the molecules
and thus will have a higher boiling point.
(1)
(1)
2
(b)
(i)
Maleic acid is cis-butenedioic acid; fumaric acid
is trans-butenedioic acid.
(1+1)
2
O R GA N IC CH EM IS T R Y
(ii)
The trans isomer will be able to form hydrogen
bonds with neighbouring molecules but the cis
isomer will form hydrogen bonds with the other
acid group on the same molecule.
The trans isomer will have the higher melting
point.
(1)
(1)
2
(8)
15.
(a)
The two carbon atoms at either end of the molecule, i.e.
1
16.
(b)
Three
1
(2)
(a)
(i)
(ii)
(b)
No
Four of the C atoms have more than one H atom
attached, and two have methyl groups attached therefore
none of them has four different groups attached.
2,3-dimethylbutane
Hexane
1
1
(1)
(1)
2
(c)
1
(5)
17.
(a)
2–hydroxypropanoic acid
1
(b)
2
O R GA N IC CH EM IS T R Y
(c)
18.
It is a racemate (racemic mixture), i.e. it consists of equal
amounts of the two enantiomers.
2
(5)
(a)
1
19.
20.
(b)
(i)
(ii)
No difference
No difference
(c)
+10°.
(d)
It could be a racemate (racemic mixture) of the two
enantiomers.
1
1
1
2
(6)
(a)
Optical isomerism
1
(b)
(i)
(ii)
a C=C
an asymmetric C atom
1
1
(c)
(i)
(ii)
B or H
D or F
(a)
Molecule A reacts with acidified dichromate solution/
hot copper oxide/Benedict’s solution/Tollens’ reagent,
but B does not.
Molecule A decolourises bromine solution but B does
not.
1
1
(5)
(1)
(1)
2
(b)
2
O R GA N IC CH EM IS T R Y
(c)
(i)
(1)
The chiral C has four different groups attached
(1)
2
(ii)
21.
(a)
A racemate (racemic mixture) of the two isomers
will be formed.
(i)
Structural formulae
(iii) C=C
(ii)
(iv)
1
(7)
CH 3 CH 2 CH 2 CH 3
A chiral carbon
6
(b)
1
(c)
1
(d)
1
(9)
O R GA N IC CH EM IS T R Y
22.
(a)
Mass of C = 12/44  0.630 = 0.172 g
Mass of H = 2/18  0.258 = 0.0287 g
Mass of O = 0.315 – (0.172 + 0.0287) = 0.1143 g
(–1 for each error)
Moles of C = 0.172/12 = 0.0143
Moles of H = 0.0287/1 = 0.0287
Moles of O = 0.1143/16 = 0.00714
The empirical formula is C 2 H 4 O
Mole ratio
2
4
1
2
(1)
(1)
2
(4)
23.
(a)
Mass of water = 0.610 g  moles of water = 0.61/18 = 0.034 
moles of hydrogen = 2  0.034 = 0.068
 mass of hydrogen = 1  0.068 = 0.068g (1)
Volume of CO 2 = 610 cm 3  moles CO 2 = 610/24000
= 0.0254
 moles of carbon = 0.0254
 mass of carbon = 12  0.0254 = 0.305 g(1)
Mass of oxygen = 0.508 – (0.068 + 0.305) = 0.135 g
(1)
Mole ratio
Moles of C 0.305/12 = 0.0254
3
Moles of H 0.068/1 = 0.068
8
Moles of O 0.135/16 = 0.0084
1
Empirical formula is C 3 H 8 O, CH 3 CH 2 CH 2 OH
24.
(1)
(1)
5
1
(6)
(b)
To improve the reliability of the results
(a)
CO + 28; COH + 29; CH 3 O + 31; CH 3 OH + 32.
2
(b)
Relative abundance of each ion formed.
1
(c)
At m/z = 14; 14.5; 15.5; 16
1
(4)
O R GA N IC CH EM IS T R Y
25.
(a)
2
(b)
A
Molecule B cannot be fragmented into all of these
parts. (1)
(1)
2
(4)
26.
(a)
1
(b)
(c)
The peaks at 112 and 114 correspond to the two iso topes
of chlorine, 35 Cl and 37 Cl, being present.
A molecular ion is formed from the molecule with the
loss of only one electron.
C 6 H 5 Cl +
1
(1)
(1)
2
(d)
(i)
(ii)
It is part of the molecule with a positive charge.
There is only one peak at 77 therefore the two
isotopes of Cl are not present.
(iii) C 6 H 5 + has a mass of 77.
1
1
1
(7)
27.
(a)
Vibration in the molecule, i.e. specific bonds lengthen
and shorten rapidly.
1
(b)
The presence of different types of bonds, e.g. C=O, O–H, etc.
1
(c)
Compare the infra-red spectrum of the distilled sample
with that of a pure sample of the expected distillate.
They should be identical.
(1)
(1)
2
(4)
O R GA N IC CH EM IS T R Y
28.
29.
(a)
The sample is not used up nor changed during analy sis
and can be recovered and used again.
1
(b)
X–ray or infra-red of liquid sample
1
(c)
No, because the sample is used up.
1
(3)
NMR – radio; infra-red – infra-red; colorimetry – visible
(3  1)
(3)
30.
(a)
It produces a magnetic field.
1
(b)
(i)
(ii)
1
1
(c)
The nucleus changes orientation from high to low spin,
i.e. changes from being aligned against to being aligned
with the magnetic field.
(d)
(e)
Absorption as energy is required to bring about the
change.
(i)
(ii)
31.
Lined up with or against the magnetic field.
The one that is lined up against the magnetic field.
1
(1+1) 2
The chemical environment, i.e. the other atoms close
to the hydrogen atom.
The number (proportion) of hydrogen atoms in the
same chemical environment.
(a)
12
(b)
All the H atoms are in CH 3 groups attached to the central
1
1
(8)
1
silicon.
32.
(c)
0.0
(a)
Tetramethylsilane
It is used as a reference and is given the chemical shift
value of 0.
1
1
(3)
(1)
(1)
2
O R GA N IC CH EM IS T R Y
(b)
The chemical shift of the peak is where the CH 3 peak is
found.
The area under the curve/integral shows three H atoms.
(1)
(1)
2
(c)
Peak A has a chemical shift of 3.6 and has an integral of 2,
therefore it corresponds to CH 2 .
1
Peak B has a chemical shift of 2.8 and has an integral of 1,
therefore it corresponds to the H in the OH group.
33.
1
(6)
(a)
Peak D is the reference peak (TMS).
1
(b)
(i)
(ii)
1
Integral
It gives the relative number of hydrogen atoms in
the same environment.
1
(c)
The phenyl group has five H atoms; the CH 2 has two H
atoms; the CH 3 has three H atoms and this agrees with
the integral values of 5; 2; 3.
Also, the chemical shift values from the Data Booklet
agree.
(1)
(1)
3
(6)
34.
(a)
CH 3 CH 2 CH 2 OH
CH 3 CH(OH)CH 3
(1)
(1)
2
(b)
(i)
CH 3 CH 2 CH 2 OH
(ii) 3
(iii) 6 (2  CH 3 ): 1 (CH): 1 (OH)
1
1
1
(c)
CH 3 CH 2 OCH 3 , therefore expect three peaks.
1
(6)
O R GA N IC CH EM IS T R Y
35.
(a)
(i)
CH 4 ; C 2 H 6
(1+1)
2
(ii)
All the H atoms are in the same chemical
environment for CH 4 and for CH 3 CH 3 .
For the CH 3 CH 2 CH 3 and the CH 3 CH 2 CH 2 CH 3
expect two peaks from the CH 3 and the CH 2
groups.
(1)
(1)
2
(b)
CH 3 CH(CH 3 )CH 3
2 peaks – one for the CH 3 groups
and one for the CH group
CH 3 CH 2 CH 2 CH 3
2 peaks – one for the CH 3 groups
and one for the CH 2 groups
(1)
(1)
2
(c)
2
(8)
36.
(a)
The atom has only one electron.
1
(b)
1
(c)
1
(3)
O R GA N IC CH EM IS T R Y
37.
(a)
X–ray crystallography
1
(b)
Helps to find mode of action or allows vitamin B12 or
related compounds to be synthesised
1
(c)
(i)
(ii)
(d)
A carbon atom has fewer electrons and would
appear smaller.
They have only one electron and so their electron
density is too small for them to be easily seen.
1
1
(i)
1
(ii)
At the centre of the concentric circles.
1
(iii)
1
(7)
38.
(a)
(b)
(c)
(d)
A receptor is a large protein molecule situated in the cell
membrane.
1
Receptors bind the body’s chemical messengers, e.g.
neurotransmitters or hormones, and this produces a response in
the cell.
1
An agonist has the same effect on the cell as the body’s
own chemical messengers and produces a biological
effect.
1
An antagonist has no effect on the cell when it binds to
the receptor.
(1)
O R GA N IC CH EM IS T R Y
However, by binding to the receptor, it blocks the
body’s own chemical messengers from binding and thus prevents
them having their usual effect on the cell
(this could then lead to a biological effect).
(1)
2
(5)
39.
(2)
40.
(a)
Pharmacologically active – a substance that alters the
biochemical processes of the body.
Derivative – another molecule with a significant part of
its structure the same as the chosen molecule, and
which is found to be pharmacologically active.
(1)
(1)
2
(b)
The pharmacophore is the structural fragment of the molecule,
that confers pharmacological activity
(alters the biological processes in the body).
(1)
Look for a common structural fragment.
(1)
2
(c)
41.
Many tests are carried out in the laboratory on animals
then on humans – checking toxicity, side effects, checking
all isomeric forms of the derivative, test effectiveness
of medicine, etc. (some appreciation of testing shown).
(a)
Moulds, land plants and seaweeds, etc.
(b)
(i)
(ii)
1
(5)
1
A pharmacophore is the structural fragment of the
molecule that confers pharmacological activity
(alters biological processes in the body).
The structure of a (lead) compound can be altered
to give a large number of derivatives.
As long as the pharmacophore is retained, there is
a good chance these derivatives will be biologically
active.
1
(1)
(1)
2
(4)
O R GA N IC CH EM IS T R Y
42.
(a)
1
43.
(b)
Tests on both optical isomers are carried out.
(a)
X–ray crystallography
1
(b)
(i)
(ii)
1
(c)
44.
(a)
(b)
1
(2)
Pharmacophore
It has to fit into the receptor site in the cell
membrane.
1
Computers can have a large database of molecules. They
can examine many molecular structures quickly looking
for a possible pharmacophore.
1
(4)
They will only work as an enzyme for one specific
biological reaction.
(i)
The surface shape at the active site of the enzyme
is such that only a specific molecule with a
definite shape can fit into the active site.
1
(1)
(1)
2
(ii)
Once the reaction has taken place at the active sites,
the products leave the site. The enzyme is now
available, leaving it free for another reactant to fill
the site.
1
(4)
EX T RA QU E S T IO N S
28.
(a)
Rate = k [RCl] or rate  [RCl]
1
(b)
Rate = k [R 3 CCl ][I – ] or rate  [(CH 3 ) 3 CCl ][I – ]
1
(2)
29.
(a)
The order with respect to HgCl 2 (aq) is first.
The order with respect to C 2 O 4 2– is second.
So rate = k [HgCl 2 ] 1 [C 2 O 4 2– ] 2
(½)
(½)
(1)
2
(b)
k = 1.82  10 –4  (0.128  (0.304 ) 2 )
(½)
(applying units to this formula: mol l –1 min –1  mol l –1
mol 2 l –2 )
(½)
= 0.0154 mol –2 l 2 min –1
(½ + ½)
2
(c)
Rate = 0.0154  0.1  0.1  0.1
= 1.54  10 –5 mol l –1 min –1
(remember the units!)
(½ + ½)
1
EX T RA QU E S T IO N S
(d)
The first mark is for a line graph that curves generally as
shown, representing a decreasing rate of change of
concentration.
The second mark is for a tangent to the curve at time
zero.
(1)
(1)
2
(7)
30.
(a)
(1)
Neither C atom has four different groups attached.
(1)
2
(b)
The methyl group could be R or any group.
(c)
1
It will exist in two enantiomeric forms that are optically active. 1
(4)
EX T RA QU E S T IO N S
31.
(a)
(i)
(ii)
By an addition reaction with hydrogen using a finely
divided Ni catalyst.
1
To lower the melting point of the triglycerides in the
oil so that the margarine can have the desired properties,
e.g. spread straight from the fridge.
1
(b)
(½ + ½ for each cis-arrangement)
(½ + ½ for double bond correctly placed on C 9 and C 12 )
2
(c)
1
(d)
trans-octadec-9-enoic acid
1
(e)
There is no chiral centre. or There is no optically active
carbon
so it is not optically active.
(½)
(½)
1
(7)
32.
(a)
Mass of C = 12/44  0.905
Mass of H = 2/18  0.369
Total mass of C + H
So mass of O
=
=
=
=
0.247
0.041
0.288
0.507
g
g
g
– 0.288 = 0.219 g
(½)
(½)
(1)
EX T RA QU E S T IO N S
Elements
Masses
C
H
O
0.247
0.041
0.219
12
1
16
0.021
0.041
0.014
21
41
14
3
6
2
Relative atomic mass
Ratio
Allowing for experimental
error
(–1 for each error)
(2)
empirical formula = C 3 H 6 O 2
(1)
5
(b)
C 2 H 5 COOH
1
(6)
33.
(a)
Mass = 60 therefore isomers are C 3 H 6 O
(i)
(b)
CH 3 CH(OH)CH 3 and
(ii)
CH 3 CH 2 CH 2 OH
2
(i)
4
(c)
Isomer B the has biggest intensity at m/z = 31. This corresponds
to CH 2 OH and is only possible with the primary alcohol,
structure (ii).
2
(8)
EX T RA QU E S T IO N S
34.
(a)
CH 3 –CH 2 –OH (1)
(b)
(i)
(ii)
CH 3 –CH 2 –OH
0 is TMS; 1.0 is CH 3 ; 3.5 is CH 2 ; 4.0 is OH (4  ½)
(c)
(i)
(ii)
Integrals
The length of the lines indicates the relative areas
under the peaks.
In this case relative areas are 1:2:3 (A:B:C)
(d)
CH 3 –O–CH 3 (1)
2
1
2
(1)
(1)
2
The molecule is symmetrical with all the hydrogen atoms
in the same environment. There will only be one peak.
1
(8)
35.
(a)
The NMR operator sets the value to zero as it is the internal
standard.
1
(b)
There are only two ‘types’ of protons in compound G.
1
(c)
From the integral ratio of the peaks of 2:1 no reasonable structure
can be drawn for CH 3 . Thus, try C 2 H 6 O but molecule must be
symmetrical.
The structure must be HOCH 2 CH 2 OH – it is the only one
to fit the data.
3
(d)
A broad peak due to OH stretch between 3600 and
3300 cm –1
1
(6)
EX T RA QU E S T IO N S
36.
(a)
(b)
CH 3 CH 2 CH 2 CH 2 OH
five peaks; five different hydrogen environments
butan-1-ol
CH 3 CHCH 2 CH 3
five peaks; five different hydrogen environments
OH
butan-2-ol
CH 3 CHCH 2 OH
four peaks; four different hydrogen environments
2  CH 3 groups are identical
CH 3
methyl propan-l-ol
CH 3
CH 3
C
OH
two peaks; two different hydrogen environments
3  CH 3 groups are identical
CH 3
methyl propan-2-ol
(structures and names ½ each)
(correct number of peaks ½ each;
correct explanation ½ each)
(6)
37.
(a)
1
(b)
(i)
Cyclohexane has one peak
Cyclohexane-1,4-diol has three peaks
(1)
(1)
2
EX T RA QU E S T IO N S
(ii)
All hydrogen atoms in cyclohexane are in the same
environment.
In cyclohexane-1,4-diol there are three different
environments and three peaks, as shown.
(1)
(2) (–1 per error)
3
(6)
38.
(a)
2
(b)
(i)
(ii)
CH 3 CH 2 OH
CH 3 OCH 3
(2  ½ )
1
EX T RA QU E S T IO N S
(c)
Isomer
Group
Approximate 
value
Integral
CH 3 CH 2 OH
CH 3
0.8–1.3
3
CH 2
3.5–4.0
2
OH
3.0–6.0
1
Isomer
Group
Approximate 
value
Integral
CH 3 OCH 3
2  CH 3 O
3.0–4.0
6
(8  ½)
(½ for approximate  value and ½ for integral)
4
(d)


39.
(a)
(b)
Both peaks in correct position, i.e. at correct approximate 
 value (ignore integral)
C 70.59/12 = 5.88
H 5.88/1 = 5.88
(divide by 1.47)
C 5.88/1.47 = 4
H 5.88/1.47 = 4
Therefore empirical formula = C 4 H 4 O
2
(9)
O 23.53/16 = 1.47 (1)
O 1.47/1.47 = 1
(1)
(which equates to mass of 68)
2
Molecular mass = 136 so molecular formula = C 8 H 8 O 2
1
EX T RA QU E S T IO N S
(c)
The infra-red absorption at 1710 indicates the presence
of a carbonyl group.
Two oxygens present so ester or acid
but acid ruled out because no large peak between
3500 and 2500 cm –1 .
NMR spectrum peak at 6–7 (integral 5) indicates a
monosubstituted benzene ring.
Peak around 2 (integral 3) indicates a CH 3 attached
directly to C=O.
(1)
(1)
(1)
(1)
(1)
5
(8)
40.
(a)
(i)
(ii)
(b)
Agonist interacts with receptor to give response
similar to the natural compound.
1
Antagonist interacts with receptor to produce no
response and it prevents action by a natural
compound.
1
Agonist acts like a good copy of a car key and is able to
switch the car engine on
but an antagonist is a poor copy of the key which will go
in the lock but will not turn and so cannot switch the
engine on.
It can also prevent the real key from being used if it
sticks in the lock.
(1)
(1)
(1)
3
(c)
There are three chiral carbons at atom numbers 3, 5 and 6
(–1 for each wrong answer.)
2
(7)