Chemistry: Post your doubts here!

[Mohammed salik]

Thanks again and what about 36th?

Welcome!!

 

[princeali97]

q2(D)....http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_21.pdf

 

[salvatore]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_43.pdf
qn no. 7(iii)
I never understand how to solve the NMR questions. Could anyone please do me a favour by explaining the solution to this question?
I'm desperately in need of your help

 

[princeali97]

q3(c)(iii)...http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_21.pdf

 

[ZaqZainab]

q3(c)(iii)...http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_21.pdf

the marking scheme says it all
something like this
sorry coundnt do an exact drawing but i guess will now understand what the ms says

 

[Nisa Mirza]

Well i guess its cause Cl^- is smaller than I^-
So fitting 7 atoms of F around a Cl atom is difficult due to a lot of repulsion between the electrons.
But cause an I atom is bigger it can attach to 7 F atoms easily

 

[sqqqqqqqqsq]

Sorry I can't.all I can say is that for option a charge on the cation will only determine strength of metallic bonding and lattice energy.ratio of charge will only tell the formula.like Mg has charge +2 and cl-1 so compound formula will be mgcl2.Option c seems appropriate.never heard anything like sum of charges so no simply.

Listen what i can understand from this question without calculating anything is that the second equation shows the enthalpy change of atomization which is an endothermic process (the value is +ve) and the only value which is positive is D

 

[saadgujjar]

Plz explain all parts

 

[Abdel Moniem]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_2.pdf
Question 4e)

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_2.pdf
Question 4e)

You can check the ms (Mark Scheme).
The answer is both oxidation and reduction has occurred in one step.
Allyl alcohol is converted into propanal in two steps with out Catalyst:
The first step is reduction (Reduction is Gain of hydrogen, in this case two hydrogen is gained) from the figure while the second is oxidation (As primary alcohol is converted to aldehyde )
But with catalyst as in one step reaction in c, the oxidation and reduction both occur. So that's it.
Hope this helps!

 

[AbbbbY]

Abby, did you have time to find any physics "factsheets" ?

I've them spread over my hard drive I'm just compiling them. Having a tough time finding the individual documents in data over a terabyte. Their names are weird numbers so cant do a search. Slightly more time please. Hopefully will upload them tonite

 

[Abdel Moniem]

How does oxidation and reduction occur when ally alcohol is converted into propanal using a ruthenium catalyst

You can check the ms (Mark Scheme).
The answer is both oxidation and reduction has occurred in one step.
Allyl alcohol is converted into propanal in two steps with out Catalyst:
The first step is reduction (Reduction is Gain of hydrogen, in this case two hydrogen is gained) from the figure while the second is oxidation (As primary alcohol is converted to aldehyde )
But with catalyst as in one step reaction in c, the oxidation and reduction both occur. So that's it.
Hope this helps!

 

[Abdel Moniem]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_2.pdf
Q2ci

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_2.pdf
Q2ci

Oxidising Agent?

 

[AbbbbY]

Plz explain all parts View attachment 40307

i- The double bond will split. The left side will form a ketone and the right side a Carboxylic acid.

ii- both the side chains will change to COOH. I believe it'll be called 1,4 di-benzoic acid.

iii- The secondary alcohol (left) will change to a ketone, the primary (right) will change to a carboxylic acid [i believe aldehyde will be acceptable too as they haven't mentioned if it's heated under reflux or not].

e) Just add two OH across the double bond.

reaction 1: cold KMnO4
reaction 2: K2Cr2O7 + H+ + heat under reflux

If you don't get the products, let me know and I'll draw them.

 

[saadgujjar]

Plz explain

 

[saadgujjar]

i- The double bond will split. The left side will form a ketone and the right side a Carboxylic acid.

ii- both the side chains will change to COOH. I believe it'll be called 1,4 di-benzoic acid.

iii- The secondary alcohol (left) will change to a ketone, the primary (right) will change to a carboxylic acid [i believe aldehyde will be acceptable too as they haven't mentioned if it's heated under reflux or not].

e) Just add two OH across the double bond.

reaction 1: cold KMnO4
reaction 2: K2Cr2O7 + H+ + heat under reflux

If you don't get the products, let me know and I'll draw them.

thnx.....can u give me any link where I can read about the splitting of double bonds?

 

[AbbbbY]

Plz explainView attachment 40309

lys-val-ser-ala-gly-ala-gly-asp ?

 

[AbbbbY]

thnx.....can u give me any link where I can read about the splitting of double bonds?

http://www.chemguide.co.uk/organicprops/alkenes/kmno4.html

 

[Snowysangel]

Can someone please explain protonation??