AS-LEVEL PAPER 2 PP7 MS

1.     (a)     (i)      enthalpy change when 1 mol of a substance
(or compound) (QL mark)
1
is (completely) burned in oxygen (or reacted in excess oxygen)
1
at 298 K and 100 kPa (or under standard conditions)
1
(ii)     heat produced = mass of water × Sp heat capacity
xΔT (or mcΔT)
1
= 150 × 4.18 × 64 (note if mass = 2.12 lose first 2 marks
then conseq) = 40100 J or = 40.1 kJ (allow 39.9 - 40.2
must have correct units)
1
moles methanol = mass/Mr = 2.12/32 (1)
= 0.0663
1
         ΔH = – 40.1/0.0663 = – 605 kJ (mol–1)
1
(allow –602 to –608 or answer in J)
(note allow conseq marking after all mistakes but note use of 2.12 g loses 2 marks
(b)     (i)      equilibrium shifts to left at high pressure
1
because position of equilibrium moves to favour
fewer moles (of gas)
1
(ii)     at high temperature reaction yield is low (or at low T yield is high)
1
at low temperature reaction is slow (or at high T reaction is fast)
1
therefore use a balance (or compromise) between rate and yield
1
(c)     ΔH = ΣΔHcο(reactants) – ΣΔHcο (products) (or correct cycle)
1
          ΔHcο (CH3OH) = ΔHcο(CO) + 2 × ΔHcο(H2) – ΔH
1
= (–283) + (2 × –286) – (–91) (mark for previous equation or this)
= –764 (kJ mol–1) ( units not essential but lose mark if units wrong) 
(note + 764 scores 1/3)
1
[15]
 
 2.     (a)     (i)      Z (1)
1
(ii)     Collisions (1)
Cause some molecules to slow down or lose energy (1)
2
(b)     Curve starts at origin and is displaced to the right (1)
Curve lower and does not touch energy axis (1)
2
(c)     (i)      Only a small percentage/very few collisions have E >Ea (1)
1
(ii)     Add a catalyst (1)
Lowers Ea (1)
More collisions/molecules have energy > Ea (1)
3
[9]
 
3.      (a)     F2 → 2F•
1
CH4 + F• → •CH3 + HF
1
•CH3 + F2 → CH3F + F• 
1
•CH3 + F• → CH3F
1
OR 2•CH3 → C2H6
(allow credit on this occasion for 2F•→ F2)
(penalise incorrect symbol Fl, once only)
(penalise absence of radical dot once only)
(b)     CH3F + 3F2  → CF4 + 3HF
1
[5]
 
4.      (a)     Electrophile: e– pair / lone pair acceptor or e– deficient species or e–
seeking species (1)
For ‘species’ accept atom, molecule, ion
NOT ‘+’ ion
NOT ‘attracted to ‘‑’ charge’
          Addition: reaction which increases number of substituents or
convert double bond to single bond or where two molecules
form one molecule (1)
2
(b)     (High) e– dense or e– rich C=C or e– rich π bond or 4 e– between the C’s (1)
NOT just ‘C=C’
          causes induced dipole in Br2 (1)
Ignore refs to ‘temporary’
can score M2 from δ+ / δ– on Br2 in (c) unless a contradicting error in (b)
2
(c)     Mechanism: 

If incorrect alkene, lose M3 (wrong cation)
Mark M4 conseq on M3

          Name of product: 1,2-dibromopropane (1)
5
(d)     addition (1)
Not additional
1
[10]

 

5.      (a)     (i)      2-methylpropan-2-ol (1) OR the second one
(ii)     Dehydrating agent:  (1)
Equation: 
Allow C4H9OH in equation provided RHS is correct
if b(i) is blank, b(ii) equation must be full for credit
i.e. NOT C4H9OH
Mark consequential on b(i)
3
(b)     (i)      Isomer: butan-2-ol OR the fourth one
[look at name in table]
wrong isomer = CE
         Structure of the ketone:

(ii)     Isomer: butan-1-ol OR the first one
            OR 2-methylpropan-1-ol OR the third one
[look at name in table]
Wrong isomer = CE
Structure of the aldehyde:
Either 
(iii)     
Reagent	M1	Tollen’s
(AgNO3/NH3)	Fehling’s	
Observation with ketone	M2	Stays colourless
no change	stays blue
no change	
Observation with aldehyde	M3	Silver mirror
black ppt	red solid
orange/red
brown/ red
ppt/solid	
         Other include(*)
K2Cr2O7 / H2SO4
KMnO4/H2SO4 
Schiff’s
Benedict’s 
Wrong reagent R
No reagent = CE
Penalise AgNO3 [Ag(NH3)2] but allow M2 and M3 sequentially. 
 (*)	K2Cr2O7 / H2SO4 acidified	ketone	aldehyde	
 	 	orange
no change	green	
 	KMnO4/H2SO4 acidified	purple
no change	colourless
(v. Pale pink)	

Benedict’s  Fehling’s      ;    Schiff’s colouless → pink with CHO
                                               violet
7

(c)     Equation: CH3CH2CH2CH2OH (or C4H9OH) + 2[O] → CH3CH2CH2COOH
(or C3H7COOH) + H2O (1)
Name of product: butanoic acid (1)
Accept butaneoic acid
2
[12]
 
6.      (a)      Mechanism:

M1 and M2 independent
Curly arrows must be from a bond or a lone pair
Do not penalise sticks
Penalise M1 if  precedes (penalise this once)
Penalise incorrect δ+ δ– for M2
Penalise + on C atom for M2
Only allow M1 for incorrect haloalkane
         Role of the hydroxide ion: nucleophile (1)
                                           electron pair donor/lone pair donor
                                         NOT nucleophilic substitution

(b)     Mechanism:

Only allow M1 and M2 for incorrect haloalkane unless RE on (i)
+ charge on H on molecule, penalise M1
M3 independent
M2 must be to correct C–C
M1 must be correct H atom
Credit M1 and M2 via carbocation mechanism
No marks after any attack of C  by OH–
         Role of the hydroxide ion: base (1)
                                       proton acceptor/accepts H+
7
[10]
 7.	

	

8.	B
[1]
9.	D
[1]
10.	C
[1]
11.	D
[1]
12.	C
[1]
13.	A
[1]
14.	C
[1]
15.	B
[1]
16.	B
[1]
17.	B
[1]
18.	C
[1]
19.	B
[1]
20.	B
[1]
21.	B
[1]