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You should learn these reactions
Because when dilute ammonia aqueous is used dropwise the NH3 not being a stronger base than the OH- cant take part in the reaction. As dilute ammonia aqueous means.
NH3 + 2 H2O -----> NH4+ + 2 OH-
The OH being the stronger one when dil NH3 is used in little amount reacts with Ni to displace Two aqua groups to form a complex [Ni(H2O)4(OH)2].
And as the Ni previously had a charge of +2 and OH is - and there are 2 OH thus the charge neutralises and the complex produced is neutral.
You should learn these reactions
They're in ligand exchange in transition elements
- example given in book is with Cu while here it is Ni but the overall reactions will be the same
thnx guys yeah i kinda new the reaction but didnt exactly know the reason
You got the reason now, right?
The question says Oxygen is divalent i.e it must have 2 bonds each
I will explain for R and then I guess you can go about the others.
Winter 2014 - P43 - Q8-c
Part i I got the same as ms
Can someone please explain how R has 4 absorptions
And how P and Q have 5
I got S as 2 because of the CH3 and CHO groups
P = CH3 , CH2 , CHO , CH3-CO and which other groups??
Similarly which ones for Q and R??
Thanks a lot! Could you please check theseI will explain for R and then I guess you can go about the others.
Okay, so for R the compound is CH3CH2(CH3)CH2CHO
Now for determining the absorption in nmr we have to check the environment on both sides left and right.
1.One would be R-CHO
2.Second would be R-CH2-CHO
3.Third one would be (CH3)2-CH2-R
4.Fourth one would be CH3-CH2-R(Now in this case you will see both the CH3(by being attached to a CH2 group) have the same environment)
Note:The carbon in bold and italics implies one environment.
Hope it helps.
The question says Oxygen is divalent i.e it must have 2 bonds each
Phosphorus is trivalent i.e it must have 3 bonds each
It's just trial and error to try and form a structure in which each phosphorus has 3 bonds and each oxygen has 2 bonds
Try placing them alternately and the joining the bonds to get what's needed
You're welcome!Thanks a lot! Could you please check these
Q = CH3CH2CH(CH3)CHO
1. R-CHO
2. R-CH-CHO
3. R-CH(CH3)-R'
4. R-CH2-CH3
5. R-CH3
P = CH3CH2CH2CH2CHO
1. R-CHO
2.R-CH2-CHO
3.R-CH2-CH3
4. R-CH3
5. I'm not sure of which one will be the 5th one since all the environments are covered right?? :/ or will it be R-CH2-CH2-R'?
S = (CH3)3C-CHO
1. R-CHO
2. R-CH3
Okay thanks!You're welcome!
Okay so you've mostly done it right , just a few things:
Q = CH3CH2CH(CH3)CHO
1. R-CHO
2. R-CH3-CHO(This one won't be CH carbon, rather CH3)
3. R-CH(CH3)-R'
4. R-CH2-CH(Not CH3)
5. R-CH3
P = CH3CH2CH2CH2CHO
1. R-CHO
2.R-CH2-CHO
3.R-CH2-CH3
4. R-CH3
5. I'm not sure of which one will be the 5th one since all the environments are covered right?? :/ or will it be R-CH2-CH2-R'?[Yeah you are right it will be R-CH2-CH2]
Now for this one, frankly speaking, I am myself perplexed when considering what is give in the marking scheme, since as far as my knowledge goes:
S = (CH3)3C-CHO
1. R-CHO would show one absorption
2. R-CH3 would also show one absorption
Moreover, I also feel that R-C-CHO should also give one absorption since that again reflects a different environment but I am really unsure why ms says 2 absorptions.:/ If anyone can elaborate this it would be great?
Yeah you are right that the CH is attached to CHO, I have just denoted it in that way though it isn't exactly appropriate , I should have shown it simply as R-CH3 since it is only attached to CH but I wrote the CHO part to indicate is the branched CH3 and not the first one, or lol it could just be R'-CH3.Okay thanks!
Btw for this one:
"Q = CH3CH2CH(CH3)CHO
2. R-CH3-CHO(This one won't be CH carbon, rather CH3)"
How would it be CH3 cause technically the CH is attached to the CHO and the CH3 is not directly attached to CHO right?
And for S, i think it's only 2 because it's the hydrogen NMR and since that C has no H's bonded to it it won't show any absorption ...?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_12.pdf
can anyone please explain question 19?
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