OCR Chemistry F322~ 4th June 2013~ AS Chemistry

I've attempted one, but considering it's my own answers simple-d down even more it's going to be heavily flawed.

Please list improvements and I'll change it

I'll write it up all fancy-pancy once the improvements are finished tomorrow after my darn physics exam.

P.S I had no idea what I was doing with the Ozone stuff

I put exactly that however in the heat of the moment I may have got my left and right mixed up :-( hope I didn't lol

I've attempted one, but considering it's my own answers simple-d down even more it's going to be heavily flawed.

Please list improvements and I'll change it

I'll write it up all fancy-pancy once the improvements are finished tomorrow after my darn physics exam.

P.S I had no idea what I was doing with the Ozone stuff

For the functional group would I get marks for put c-o and c double bond o?


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Found it ok . Just some weird questioms

Hopefully did good!

Wbu? Best of luck

I put ketones and alkenes, and I am not even sure if thats right!

Do you think they could be any lower?

https://dl.dropboxusercontent.com/s/1ic6a1a8nyttotd/full%20f322%20june%202013%20paper_merged.pdf?token_hash=aahkil7xalll9wtllwvp6rtll2wus9opynpjhrncq5urwq&dl=1

here you go!
This is the full paper that was another one uploading earlier...
That will 10x better!
:d


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In box 3 and 4 didn't it want the two possible isomers?

UNOFFICIAL MARKSHEME - TO BE CONTRIBUTED...

Q1.
a) i) C10H22 (1)
ii) 2-methylnonane (2)
iii) Fewer points of contact, less VDWs, less energy required to break them. (2)

b) i) C15H32 à C10H22 + C5H10 (1)
ii) Carbon to carbon bond can be broken in a variety of places. (1)

c) i) Draw an octagon (1)
ii) Cyclic hydrocarbons burn more efficiently as fuels (1)

Q2.
a) i) E & H (1)
ii) H (1)
iii) F (1)

b) i) C5H9OH (1)
ii) 2-methylpentan-3-ol (1)

c) A series of organic compounds with the same functional group, each differing by CH2. (2)

d) 1st box -> Ketone, same as diagram but change –OH to =O (4)

2nd box -> Ester, same as diagram minus the H on –OH, add ethanoic acid but minus H on the carboxyl.

4th & 3rd box -> Alkene, same as diagram minus –OH and put a double bond in the ring. H20 in other box due to dehydration.
Q3.
a) Alkene and Ester (2)

b) Double carbon bond within chain. (1)

c) Orange to Colourless (1)

d) i) Same structural formula, but different arrangement of atoms in space. (1)

ii) Ester group on opposite side of the double carbon bond (pointing downwards). (1)

e) i) The enthalpy change when 1 mol of a substance reacts completely with excess oxygen, under standard conditions. (2)

ii) Q = mcT, (50x4.18x(20.2-54))/1000 = (-)7.06kJ (2)

iii) n = m/Mr 1.34/(12x17)+32+32 = 5x10^-3mol (2)

iv) 5x10^-3 mol 7.06kJ
1mol 7.06 x 200 = 1412kJ (depending on rounding) (3)

v) Incomplete combustion, carbon particulates were created. (1)

f) C6H12O6 à 2C2H5OH + 2CO2
Yeast + 37 degrees Celsius. (3)

Q4.
a) i) Curly arrows from bond in IBr to Br. From double bone to I.
Next molecule: I bonded to one carbon, leaving Br- ion with lone pair, and a carbocation. Arrow from lone pair to carbocation. (3)

ii) Electrophilic Addition (1)

iii) Same as last molecule, but switch I and Br around. (1)

b) i) Ultraviolet radiation (1)





ii) Free Radical Substitution (7)

Initiation
IBr I + Br (2 radicals formed due to UV, bond breaks homolytically)

Propagation
I + CH4 CH3I + H
Br + CH4 HBr + Ch3

Termination
CH3 + H CH4
Br + H HBr
I + CH3 CH3
Q5.
a) Products – Reactants
(4 x +54) (4 x -394)
(5 x -20) (12 x -242)

+116 – (-4480) = 4596 kJmol-1 (3)

b) i) n = 240/24 (N2O) = 10 mol
1 mol à 164kJ
10 mol à 1640kJ (2)

ii) For 2N2O = 164
for N20 = 82kJmol-1 (1)

iii) 447 – 164 = +283kJmol-1 (1)

c) O3 split by UV (2)
O3 à O2 + O
O2 + O à O3

d) NO + O3 à NO2 + O2 (2)
O + O3 à 2O2

Q6.
a) i) 256(+) (1)
ii) S8 (1)
iii) S4(+) (1)

b) (33/100 x 194) + (34/100 x 195) + (25/100 x 196) + (8/100 x 198) = 195.2 (2)
c) Breathalyser (1)

d) C-O peak at “” Carbonyl (5)
C=O peak at “” Aldehyde, ketone, ester

C 66.7/12 H 11.1/1 O 22.2/16
5.558 11.1 1.3875
/smallest (1.3875)
41
C4H8O Mr = 72

c-c-c-c=o c-c-(=o)c-c Aldehyde (Butanal) or Ketone (Butanone)



Q7.
a) N (1)
b) i) S (1)
ii) Nucleophilic Substitution
OH- arrow to back of last carbon bonded to Br. Arrow from C-Br to Br. C(delta +) and Br(delta -) OH replace Br, Br- left. (4)

iii) Increase in temp, increase Ek, increase rate of reaction. (1)

c) But-2-ene + HCL à R (2)
d) HCFs (1)
Q8.
a) –C-C-C-C— with CN on 2^nd and 4^th Carbon, H’s on rest. (1)
b) Addition reaction, no waste products, 1 product. (1)

c) Le Chatelier’s Principle – When a dynamic equilibrium is subject to change, the position of the equilibrium will shift to minimise the change.

Increase temp à
Shifts to left, endothermic side of reaction, reduce temp.

Increase pressure à
Shifts to left, less moles of gas, reduce pressure.

Absense of catalyst à
Will not shift equilibriums position. (5)



d) Theoretical = 220 mol
Actual = 11.13x10^3/(12x3)+3+14 = 210
210/220 x 100 = 95.5% (2)

e) Boltzmann for temp increase à shift to the right, Ea same position, but more area thus more molecules reach activation energy. Increase in temp increase Ek, more frequent and successful collision thus increases rate of reaction.

Boltmann for catalyst à shift Ea to left, lowers Ea so more molecules can reach it, more successful collision, provide alternate route, increasing rate of reaction.

Graphs on y-axise: number of molecules, x-axis: energy
(7)

nooo i put decane instead of c10h22 on the first question. What is wrong with me.

With the question where it was why does the reaction have 100% atom economy, even though it didnt as Hydrogen was being produced what did people say? I said because hydrogen is also a desired product as it can be sold on or used in process such as the hydrogenation of an Alkene, an example is the manufacture of margarine... Anyone?