ANY THING RELATED TO 7 MAY 2014 PAPER 22 CHEMISTRY AS LEVEL

[THAIKA96]

DOES ANY ONE HAVE SOME SORT OF SOLVED/ MARKING SCHEME TYPE OF DOCUMENT REGARDING THE PAPER 22 OF AS CHEMISTRY, WE ARE FREE TO DISCUSS ABOUT THE PAPER BECAUSE 24 HOURS HAVE PASSED SINCE EXAMINATION

 

[basimnazir]

1.
a. Nucleon number is the number of neutrons and number of protons in an atom.
b. Relative Isotopic Mass is the weighed average mass of an isotope of an atom compared to C-12 which has exactly 12 units.
c. Abundance = (RAM x [value of bromine])/100 for both the bromine isotope.
d. RAM = (Mr of BR x Mr)+A/100 [I don't remember the exact what I wrote].
(can't recall rest parts of question 1, let me know what they are and I might be able to remember)

2.
a. Oxidation is the loss of Electrons and Gain of Oxidation (emphasis on "loss of electrons" because question wanted us to answer in terms of electron transfer)
[The rest were numericals with balancing and applying moles formulas throughout]

3.
This was an easy question so it might not need any discussion. Though what candidates found tough was between Sulphur and Phosphorus and Argon. This question has been repeated so here's the answer:
Sulphur exists as S8 in its natural state and Phosphorus occurs as P4. There are strong vander waal's forces of attraction in Sulphur and Argon occurs as a monoatomic molecule with the lowest ionisation energy of all.

4.

a. Why are alkanes unreactive?
Because Alkanes have a pi-bond and a sigma-bond and have a stronger attraction whereas alkenes have only pi-bond which is less stronger than pi-sigma bond thus alkanes are unreactive. [NOT JUST: "BECAUSE ALKANES ARE NON-POLAR]

[Don't remember the next parts]

They asked about Ethane and Chlorine to form ChloroEthane (something like that) so you just have to write all the Free-Radical Substitution mechanism.

Later they asked how Bromine could be produced. Simple:
C2H6 + Br2 ---> C2H5Br + HBr (I'm not sure if this is correct but no other possibility could be made)

5.

The decolorising of bromine suggests that Compound Q is unsaturated hydrocarbon and contains the carbonyl group with double bond. Aldehyde and Ketone give a positive test with 2,4-DNPH but only Aldehyde gives a positive reaction with Tollens reagent. So, Ketone is present (C=O)

And then blah blah easy stuff.

That's pretty much it. Let me know if you gave Practical 33.

 

[THAIKA96]

Paper 33 was the easiest thing I have come across, I wanted something like the answer on the question paper like a PDF file of something so I can know the marks for each question

 

[omarjaved619]

I screwed up the Q2 ionic equation.

Also, for the organic question, they said that on oxidation, ONLY ONE ORGANIC product is formed.

So I put a double bond at one end and a methyl group on the second last carbon. Like that we'll get a ketone, and CO2 and H2O on the other side.

BUT, some of my classmates put the double bond in the middle, saying breaking that would give TWO SAME molecules.

Not sure which one is correct.

 

[Uki Malla]

a. Why are alkanes unreactive?
Because Alkanes have a pi-bond and a sigma-bond and have a stronger attraction whereas alkenes have only pi-bond which is less stronger than pi-sigma bond thus alkanes are unreactive. [NOT JUST: "BECAUSE ALKANES ARE NON-POLAR]

Wrong! Alkanes dont have a pi-bond. That is one of the main reasons why they are unreactive than alkenes! Yes, the alkene pi-bonds do have a shorter length, but also have a high electron density, where as alkanes have no such regions (high electron density). This causes the alkenes to be frequently attacked by electrophiles making them more reactive.

 

[sheetalgera]

1.

Later they asked how Bromine could be produced. Simple:
C2H6 + Br2 ---> C2H5Br + HBr (I'm not sure if this is correct but no other possibility could be made)

.

as far as i remember, it was butane which they asked us to make

 

[THAIKA96]

as far as i remember, it was butane which they asked us to make

This is wrong, butane forms in the termination step when 2 free ethyl radicals react , c2h5• + c2h5• -----> c4h10

 

[kingo44]

wait for organic part u simply need to write the correct roducets that u get eg from 24 dnp u dnt need to react with hot conc kmno4 it is just observation for alkene my friend said u break the hot con then u get carboxylic acid nd react it with 2 4 dn but this is wrogn as carboxylic acikd doesnt form additrion reaction with 2 4 dnp i simpl wrote ketone and alkene nd explain 4 marks

 

[sheetalgera]

This is wrong, butane forms in the termination step when 2 free ethyl radicals react , c2h5• + c2h5• -----> c4h10

yeah i know that butane is made in the terminaion step. i was correcting the question only.

 

[Dibbo 10]

okay so i dont know what went wrong during my exam but for some reason i gave it an aldehyde instead of a keoton. my mind went nuts during the exam and as a result i got the rest of the part in that question wrong.
will i lost all 5 marks or will there be an error carried forward?

 

[AbbbbY]

a. Why are alkanes unreactive?
Because Alkanes have a pi-bond and a sigma-bond and have a stronger attraction whereas alkenes have only pi-bond which is less stronger than pi-sigma bond thus alkanes are unreactive. [NOT JUST: "BECAUSE ALKANES ARE NON-POLAR]

Ummm.. Excuse me?
What sort of alkanes have a pi bond and a sigma bond. And what sort of chemistry exists where a compound has JUST a pi-bond.

They asked about Ethane and Chlorine to form ChloroEthane (something like that) so you just have to write all the Free-Radical Substitution mechanism.

Later they asked how Bromine could be produced. Simple:
C2H6 + Br2 ---> C2H5Br + HBr (I'm not sure if this is correct but no other possibility could be made)

This isn't correct. Although, from what I recall, the question was how can butane be produced, but if you were to produce Bromine in an FRSR mechanism, it'll be in termination.

Br* + Br* -> Br2 (where * shows a radical)

CH3CH2* + CH3CH2* -> C4H10 was what I, and everyone else I've asked answered.

The decolorising of bromine suggests that Compound Q is unsaturated hydrocarbon and contains the carbonyl group with double bond. Aldehyde and Ketone give a positive test with 2,4-DNPH but only Aldehyde gives a positive reaction with Tollens reagent. So, Ketone is present (C=O)

What??

The compound did not contain a Carbonyl group!! Carbonyl group was formed after reacting it with hot KMnO4 and it tested NEGATIVE with Tollens reagent so THAT was a Ketone, not the initial, starting compound. Besides, C6H12 can never be a carbonyl anyway -_-

Edit: That's weird! I posted this yesterday but it was posted posted today when I came to this thread
Internet disconnectivity, I guess.

 

[kingo44]

u stupidnoif u cut the double bond when reacted with hot conc kmn04 u wood get co2 and h20 not a ketone as only two hydrogen atoms are presnet near the double bond when u cut it there is no alkyl group to give a ketone so how can aketone be formed c6h12 no but but 2 ene maybe but not c6h12 it is a simplke test to show u a double bond is present they dont expect u to react with hot conc km04 they didnt say that

 

[AbbbbY]

u stupidnoif u cut the double bond when reacted with hot conc kmn04 u wood get co2 and h20 not a ketone as only two hydrogen atoms are presnet near the double bond when u cut it there is no alkyl group to give a ketone so how can aketone be formed c6h12 no but but 2 ene maybe but not c6h12 it is a simplke test to show u a double bond is present they dont expect u to react with hot conc km04 they didnt say that

The hell are you saying mate?!
Try writing all this in English, perhaps?

 

[kingo44]

bottom line i wrote hexene and hexan 3 one

 

[kingo44]

nd for geometrical isomerism i drew c6h12 cis nd trans

 

[AbbbbY]

bottom line i wrote hexene and hexan 3 one

Writing hexene is a common mistake, but hexan-3-one?! Wow. 07/04/14. Chemistry was made.

 

[AbbbbY]

nd for geometrical isomerism i drew c6h12 cis nd trans

The question asked for 3 different STRUCTURES of C6H12 that show cis-trans, not the same structure as cis and trans!

 

[kingo44]

jesus abbu u think thats right i doont think so see they didnt tell us to react with conc kmn04 there was no space they only told us that alkene is present thus they showed us it reacted with kmn04 im goin crazzy here lol

 

[kingo44]

i knpw i mean hexan 3 on lol

 

[kingo44]

abby wow i mea n wow