Chemistry: Post your doubts here!

[biba]

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

okay in d ( i ) we found the molecular formula as SiCl3... now in d ii the first Mr is 133 that is SiCl3 cuz (28+35+35+35=133)... now we have 247... let's see 247-133=114, we knw the formula of 133, find the formula of 114 which is SiCl2O (28+35+35+16=114) ,add both formulas of 133 and 114 i.e SiCl3 + SiCl2O = Si2Cl5O(Cl 3Si-O-SiCl 2 )
do the same for 263.. 263 - 247 = 16,meaning only O is added hence add Si2Cl5O + O= Si2Cl5O2 (.Cl 3Si-O-SiCl 2-O)
..

 

[biba]

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 can draw the structures but I dunno how to explain them to you..
I made some notes on NMR u can check it out , may b it helps :

 

[sagar65265]

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 2007 specimen, 9 (b) & (d):

The first clue is the empirical formula calculated for the earlier part question; it is C8H10O, and since the number at the top of each peak, that describe the number of hydrogen atoms per group, add up to 10, it looks like the molecular formula is also C8H10O. This is very useful information!

The first two peaks from the right hand side of the graph are typical of a CH3CH2 group;
i) the first group has 3 hydrogen atoms and it is a triplet, so it is attached to a carbon atom with 2 hydrogen atoms attached to it.
ii) the seconds peak is a carbon atom with 2 hydrogen atoms attached to it and since it is a quadruplet, it is attached to a carbon atom with 3 hydrogen atoms on it; the two of them describe each other!
The last peak describes 4 equivalent hydrogen atoms and is at about 6.8 ppm; this indicates either a phenol group or a benzene group (this can be confirmed by checking the empirical formula; subtracting the 2 carbons and 5 hydrogens from the CH3CH2 group gets us a remainder of C6H5O - this looks an awful lot like a phenol group with one extra substituted side chain!)

From this information, it looks like the molecule is an ethyl group attached to a phenyl group; since the 4 phenyl hydrogens are equivalent, the -OH group and the -CH2CH3 seem to be attached to opposite ends of the molecule - and that should be the answer!

d) If the isomer shows no change after adding D2O, this means that there is no -OH group in this isomer, else the third peak on this NMR spectrum would have disappeared. Since there is no change, there is no -OH OR -COOH group or any other ionising group, and there are only two groups that follow this pattern of containing only 1 oxygen - a carbonyl group or an ethoxy group.

Since a carbon double bonded to an oxygen can never occur on a benzene ring and this occurring on the side chain would remove some hydrogens, voiding the "isomer" statement, the compound is an ethoxy compound, mostly ethoxybenzene (or whatever the actual name is, sorry i'm not being more specific!). The observable changes would have to be the loss of the third peak from the right, the peak caused by a methoxy group (at about 3.3 - 4.0 ppm) and the change in splitting of the CH3 peak, I guess.

I'll try out the others ASAP, sorry this is all i've been able to do so far!

Hope this helped!
Good Luck for all your exams!

 

[Jiyad Ahsan]

sanii94

hey open J/08/04 Q6 and 7.. check these questions out

 

[Soldier313]

biba
sagar65265

Thank you so much for your help guys! It really helped a lot!
I really appreciate it !

 

[biba]

biba
sagar65265

Thank you so much for your help guys! It really helped a lot!
I really appreciate it !

no problem .. just pray for me

 

[Soldier313]

no problem .. just pray for me

i definitely will inshaAllah! May Allah grant you success in all that you do inshaAllah

 

[biba]

i definitely will inshaAllah! May Allah grant you success in all that you do inshaAllah

In sha allah! may Allah bless u with success too
Ameeen !

 

[Soldier313]

In sha allah! may Allah bless u with success too
Ameeen !

Aameen inshaAllah

 

[sagar65265]

Guys, I just have a question - In the electrolysis of aqueous solutions containing halide ions, how do we determine using standard electrode potential values (E nought values) what is produced at the anode? Assuming that conditions are all standard and concentration doesn't have a role to play, how do we determine what is produced at the anode?

Thanks in advance!

 

[littlecloud11]

Guys, I just have a question - In the electrolysis of aqueous solutions containing halide ions, how do we determine using standard electrode potential values (E nought values) what is produced at the anode? Assuming that conditions are all standard and concentration doesn't have a role to play, how do we determine what is produced at the anode?

Thanks in advance!

The anions in the solution are the halide ion and OH-
Identify the equations relating to these two ions in the data booklet. The one with the more negative E value is the one which is oxidized and liberated at the anode.

 

[Jaf]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_4.pdf
http://papers.xtremepapers.com/CIE/.../Chemistry (9701)/9701_s03_ms_1+2+3+4+5+6.pdf

Question 3)b).
Why doesn't SiO2 react with NaOH? The ER bluntly states that it just does not. The coursebook says that it DOES infact react with NaOH when the latter is hot and concentrated. So if the argument is that NaOH (aq) won't react with SiO2, then why is it reacting with SnO2. The coursebook says the NaOH needs to be hot and concentrated to react with SnO2 too.

 

[sagar65265]

The anions in the solution are the halide ion and OH-

Identify the equations relating to these two ions in the data booklet. The one with the more negative E value is the one which is oxidized and liberated at the anode.

Then how are we supposed the do Question 3 (c) for the following paper:

The first two solutions I can understand, Fluorine and Sulfate are not released, but why does the marking scheme say Bromine is released at the anode? is it because of the lack of standard conditions or is it some other factor we have to take into consideration? Because the electrode potential for Bromine is much higher than the electrode potential for OH- ions (the difference is more than that 0.30 V value) so Oxygen should either ways be released!

Thanks a lot in advance!
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_42.pdf
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_ms_42.pdf

 

[littlecloud11]

Then how are we supposed the do Question 3 (c) for the following paper:

The first two solutions I can understand, Fluorine and Sulfate are not released, but why does the marking scheme say Bromine is released at the anode? is it because of the lack of standard conditions or is it some other factor we have to take into consideration? Because the electrode potential for Bromine is much higher than the electrode potential for OH- ions (the difference is more than that 0.30 V value) so Oxygen should either ways be released!

Thanks a lot in advance!
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_qp_42.pdf
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s11_ms_42.pdf

Yes, I noticed that too. The only way Br2 could have been produced was if the conditions were non-standard but they really should have mentioned this in the question. Br2 is not produced at the anode when the solution is aqueous and under standard conditions.

 

[sagar65265]

Yes, I noticed that too. The only way Br2 could have been produced was if the conditions were non-standard but they really should have mentioned this in the question. Br2 is not produced at the anode when the solution is aqueous and under standard conditions.

Neat, thanks a lot!

Good Luck for all your exams!

 

[littlecloud11]

Neat, thanks a lot!

Good Luck for all your exams!

You too!

 

[daniyal007]

In oct nov 2012 p42 Q4 , i have solved the whole question correct by drawing correct structural formulae..But in mark scheme skelatal formulae are written..will i be getting marks????

 

[strangerss]

How can prepare for the practical exam at home ?

 

[strangerss]

allasalamu allaykom ,Can some one please help in paper34 Oct/Nov 2012 question 1c? I have no clue how to calculate the concentrations

 

[Soldier313]

littlecloud11 sagar65265

so for the qn 3 c in mj/11 p42, i gather the answer in the marking scheme is wrong?