Chemistry Paper 4-theory- doubts =D

[iKhaled]

JazakAllah bro got it btw i guess u made a mistake in the ksp equation it is actually ksp=[Pb^2+][Cl-]^2

thats my Ksp where is the mistake?

 

[Zari]

thats my Ksp where is the mistake?

ohhh sorry i thought u didnt did Cl- to the power of 2 Sorry once again

 

[iKhaled]

ohhh sorry i thought u didnt did Cl- to the power of 2 Sorry once again

hehe its okii

 

[knowitall10]

I looked through it but it doesnt clearly differentiate between teh AS and the A2 components

You need to read the instructions clearly Have a copy of the syllabus next to you as you read.

 

[Xtremite]

There isn't much to this question other than trying to figure out how you could use atoms/elements in the original compound (ammonium nitrate) to make fragments that would add up to each of the peaks in the question. My initial hunch was that A and B would be NH3 and NO2, but those don't balance (I mean even if only one mole of these gases is made, the Mr of NH4NO3 - 62 - is not equal to the combined Mr of both product gases - 17 + 46 = 63, so this must be dismissed.) Instead, chose two gases that would separately make up the maximum m/e values, 18 and 44 using onlu the elements provided. These would be H2O (18) and N2O (44). They even balance! 18 + 44 = 62. Then you can break these up to make each of the smaller values: O+ = 16, OH+ = 17. And N+ = 14, O+ = 16, N2+ = 28, NO+ = 30. (They'll all have positive signs 'cause the spectrometer only deals with positive ions.)

Thank you so much!!

 

[KurayamiKimmi]

Asalamo Alaikum everyone!
Can anyone please help me with this A2 question:-
Q 3 part d(ii) of this paper?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_42.pdf

 

[iKhaled]

Asalamo Alaikum everyone!
Can anyone please help me with this A2 question:-
Q 3 part d(ii) of this paper?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_42.pdf

questions of this type is really annoying because it takes a lot of time to figure out the fragment by trying to multiply the atoms with each other till u get the given Mr so its a big waste of time i see. i would suggest that if a question of this type came in the exam, leave it to the last. its better to waste 5-6 marks instead of leaving 3 full questions or somehting because a question of this type will just waste ur whole time to figure it out..anyone else agrees?

 

[Soldier313]

Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot

 

[knowitall10]

Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot

umm...xhizors i think we need you here

 

[knowitall10]

and Ariel Robert ...u said u like chem...

 

[Zari]

Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot

ohhh srsly i too hav the same dbts :O plzz help

 

[knowitall10]

Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot

Soldier313 how long can u wait? I'll study it and then tell you...

 

[Soldier313]

Soldier313 how long can u wait? I'll study it and then tell you...

I can wait a while Thank you

 

[iKhaled]

Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot

i will talk about the first question u have posted

b) first of all u try to identify the type of proton on each peak.. the numbers on the top of the peak tells u how many hydrogen proton is present in the type of proton..and i believe u do have a data booklet so u can check how the type of protons look like at the chemical shift given..

the first peak at approx 1.2-1.3 shows that it has 3 hydrogen protons and it is split into triplet which means the adjacent carbon must have 2 hydrogen protons on it because of the rule we need to learn (n+1) where n is the number of hydrogen protons on the adjacent carbon. so we have a peak of RCH3 because thats the only one available in the data booklet which has 3 hydrogen protons.

now the peak at approx 2.1 shows that it has 2 hydrogen proton on it so it can be CH2 and it is CH2 because the peak at 1.2-1.3 shows that the adjacent carbon is split into 3 which means the adjacent carbon has 2 hydrogen so yeah t must be CH2 now this speak is split into quartet which means there is 3 hydrogen protons on the adjacent carbon and this is true because see RCH3 at peak 1.2-1.3 has 3 hydrogen carbons..so we can guess we have RCH2R and RCH3 now obviously the peak at 5.5 is OH because this speak always shows as single and doesnt split if u look at the last one (check in the data booklet) it is a benzene ring with a phenol group attached to it but beware that this peak shows that it has 4 hydrogen protons so there are 2 hydrogen which has been removed from the benzene ring. one substituted with a phenol group and the other one will be CH2CH3 so our molecule will be

HO-C6H4-CH2CH3 see the benzene ring peak is split into 4 quartet also and u can see that it is next to a molecule which has one proton and another molecule which has 2 protons so 3+1 using the n+1 rule will split this peak into 4!! so yeah our identification of the aromatic compound is correct

i really hope u understood it because it is very confusing to explain it on the internet instead of face to face

 

[Soldier313]

i will talk about the first question u have posted

b) first of all u try to identify the type of proton on each peak.. the numbers on the top of the peak tells u how many hydrogen proton is present in the type of proton..and i believe u do have a data booklet so u can check how the type of protons look like at the chemical shift given..

the first peak at approx 1.2-1.3 shows that it has 3 hydrogen protons and it is split into triplet which means the adjacent carbon must have 2 hydrogen protons on it because of the rule we need to learn (n+1) where n is the number of hydrogen protons on the adjacent carbon. so we have a peak of RCH3 because thats the only one available in the data booklet which has 3 hydrogen protons.

now the peak at approx 2.1 shows that it has 2 hydrogen proton on it so it can be CH2 and it is CH2 because the peak at 1.2-1.3 shows that the adjacent carbon is split into 3 which means the adjacent carbon has 2 hydrogen so yeah t must be CH2 now this speak is split into quartet which means there is 3 hydrogen protons on the adjacent carbon and this is true because see RCH3 at peak 1.2-1.3 has 3 hydrogen carbons..so we can guess we have RCH2R and RCH3 now obviously the peak at 5.5 is OH because this speak always shows as single and doesnt split if u look at the last one (check in the data booklet) it is a benzene ring with a phenol group attached to it but beware that this peak shows that it has 4 hydrogen protons so there are 2 hydrogen which has been removed from the benzene ring. one substituted with a phenol group and the other one will be CH2CH3 so our molecule will be

HO-C6H4-CH2CH3 see the benzene ring peak is split into 4 quartet also and u can see that it is next to a molecule which has one proton and another molecule which has 2 protons so 3+1 using the n+1 rule will split this peak into 4!! so yeah our identification of the aromatic compound is correct

i really hope u understood it because it is very confusing to explain it on the internet instead of face to face

This is amazing! it's exactly what i wanted!
JazakAllah khair ! May Allah bless you inshaAllah!

 

[Zari]

i will talk about the first question u have posted

b) first of all u try to identify the type of proton on each peak.. the numbers on the top of the peak tells u how many hydrogen proton is present in the type of proton..and i believe u do have a data booklet so u can check how the type of protons look like at the chemical shift given..

the first peak at approx 1.2-1.3 shows that it has 3 hydrogen protons and it is split into triplet which means the adjacent carbon must have 2 hydrogen protons on it because of the rule we need to learn (n+1) where n is the number of hydrogen protons on the adjacent carbon. so we have a peak of RCH3 because thats the only one available in the data booklet which has 3 hydrogen protons.

now the peak at approx 2.1 shows that it has 2 hydrogen proton on it so it can be CH2 and it is CH2 because the peak at 1.2-1.3 shows that the adjacent carbon is split into 3 which means the adjacent carbon has 2 hydrogen so yeah t must be CH2 now this speak is split into quartet which means there is 3 hydrogen protons on the adjacent carbon and this is true because see RCH3 at peak 1.2-1.3 has 3 hydrogen carbons..so we can guess we have RCH2R and RCH3 now obviously the peak at 5.5 is OH because this speak always shows as single and doesnt split if u look at the last one (check in the data booklet) it is a benzene ring with a phenol group attached to it but beware that this peak shows that it has 4 hydrogen protons so there are 2 hydrogen which has been removed from the benzene ring. one substituted with a phenol group and the other one will be CH2CH3 so our molecule will be

HO-C6H4-CH2CH3 see the benzene ring peak is split into 4 quartet also and u can see that it is next to a molecule which has one proton and another molecule which has 2 protons so 3+1 using the n+1 rule will split this peak into 4!! so yeah our identification of the aromatic compound is correct

i really hope u understood it because it is very confusing to explain it on the internet instead of face to face

JazakAllah for this May ALLAH bless you in sha Allah

 

[iKhaled]

This is amazing! it's exactly what i wanted!
JazakAllah khair ! May Allah bless you inshaAllah!

thanksss pray for me pls i am solving the other 2 now !

 

[iKhaled]

JazakAllah for this May ALLAH bless you in sha Allah

merciiiii zari

 

[iKhaled]

Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot

for the second question u have posted..

its the same scenario. first u will see how many hydrogen protons available on each type of proton from the number given at the top of each peak and then u look at the splitting of the peak..from the splitting u will be able to know what is the adjacent type of proton sticking to the one u r trying to find.

now the first peak at 1.2 approx is a triplet which means the adjacent group attached to it must contain 2 hydrogen protons..so it must be CH3 attached to CH2.
the peak at approx 2 is a singlet and has 3 hydrogen protons so it must be attached to attached to to something which doesnt split it like an OH or C=O but it cant be an OH since wdont have any peak that shows OH is present so it will obviously be C=O..from here i started noticing that our organic compound might be an ester since there is a C=O and doesnt have OH and it says 2 oxygen is present in it.. so it can be CH3COOR

now the final peak has 2 hydrogen protons in it and it is split into quartet so it must be next to a group that has 3 hydrogen atoms..from here u can notice our compound will be CH3CH2CO2CH3..got it ?

 

[iKhaled]

Aoa wr wb
Can someone please help me with these doubts?

Qn 9 b and d )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_y07_sm_4.pdf


Qn 9 c ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_4.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_ms_4.pdf


Qn 7 c ii )
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_41.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_ms_41.pdf


I would really appreciate if you could provide a detailed explanation for these NMR questions

JazakAllah khair, thanx a lot

the final question!!

this one u will not need to use the n+1 rule since there is no splitting so u can say "yaay" no confusing shit in the question..anyway lets look at the peaks!

first peak at 3.8 approx. has 3 hydrogen protons present in it looking in the data booklet is can be R-O-CH3
second peak at 5.5 is as we know obviously the OH peak so R-OH
now the final peak is at approx 6.8 looking in the data booklet thats a benzene ring but notice that it contains 4 hydrogen and not 6 as a regular benzene ring so it has lost 2 hydrogens at 1st and 4th position so it is R-C6H4-R.

now combine all what we got together u will have CH3-O-C6H4-OH..i guess all of them is a singlet peak because if u see all the groups we found are attached to oxygen so they wont have any splitting ( i am kinda sure of that)

i will be glad if i made u understand the 3 questions u have posted hehe