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Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed)

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Reply 640
Avatar for sounique
sounique
Original post by posthumus
Okay so transitional metal complexes are very slightly magnetic so they lie parallel to a strong magnetic field (caused by the ion having one or more unpaired d-electrons).

I haven't used the edexcel book for over a year! funny enough since I started using it I haven't been doing so well lol :tongue: But I have an obsession for learning every detail and further :smile:

I think it would help if you feel the edexcel book does not explain some things properly to you. I use the edexcel revision guide instead of the actual text book (its brilliant).


I think I might just do loads of questions because I think that's the key thing with chemistry.

Cool thanks for the explanation.
Reply 641
Avatar for sounique
sounique
Original post by jojo1995
yes i am, but we have an easter revision session :redface: chemistry is acc so confusing - my head is spinning right now




yes... but you see in cu(h20)6 - i cant find my square brackets, do excuse me :tongue:
you would always write cu(h20)6 even though 2 h20s are not planar right ?


So lucky! Just because I go to a grammar school, our teachers assume we're all going to understand everything! My chem teacher she may/may not do chem revision sessions just before the exams. And then they want us to all miraculously get A*s in every subject. So much pressure and so little help.
Reply 642
Avatar for jojo1995
jojo1995
2
Original post by sounique
So lucky! Just because I go to a grammar school, our teachers assume we're all going to understand everything! My chem teacher she may/may not do chem revision sessions just before the exams. And then they want us to all miraculously get A*s in every subject. So much pressure and so little help.


Benefits of going to a non selective state school eh :wink:

aww don't feel under pressure - do you do economics? i just found some good notes :smile:

your hard work will pay off dw - youll get those A*S in the end - you'll see :smile:
Reply 643
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jojo1995
2
Original post by posthumus
No you can write [Cu(H2O)6]^2+ as [Cu(H2O)4]^2+ as well :smile:

that's why I was asking :tongue:


omg .... ididnt get taught that :redface:

thank you so much posty :smile:
it all makes sense now - yeah but my teacher did say for the nh3 one you can write in the h20s if you want but you dont needtoo
Reply 644
Avatar for sounique
sounique
Original post by jojo1995
Benefits of going to a non selective state school eh :wink:

aww don't feel under pressure - do you do economics? i just found some good notes :smile:

your hard work will pay off dw - youll get those A*S in the end - you'll see :smile:


I know right! Ahaa
No I don't, but thanks anyway. You're so sweet, thanks its nice to hear that support once in a while. :smile:
Reply 645
Avatar for sounique
sounique
Original post by posthumus
No you can write [Cu(H2O)6]^2+ as [Cu(H2O)4]^2+ as well :smile:

that's why I was asking :tongue:


Sorry, but I thought you can never get [Cu(H2O)4]2+ and that it's always 6 waters..?
Original post by sounique
Sorry, but I thought you can never get [Cu(H2O)4]2+ and that it's always 6 waters..?


My point was that they're both the same thing... when writing it you don't have to include the 2 water molecules that aren't in the same plane as the complex ion.
Reply 647
Avatar for jojo1995
jojo1995
2
Original post by sounique
I know right! Ahaa
No I don't, but thanks anyway. You're so sweet, thanks its nice to hear that support once in a while. :smile:


Aw you're welcome - anytime :smile:
Reply 648
Avatar for sounique
sounique
Has anyone done or has access to the Jan 2013 Mock?
Reply 649
Avatar for sounique
sounique
Do we have to learn the complementary colours?
Hey, me again with more questions haha.

For coloured TM ions is it the light that causes it to split the d orbitals? Or is it the ligands? My books are quite vague lol. Then once the d orbitals split is it the absorption of energy that moves one electron up to a higher energy which equates to the wave length of colour absorbed?

Also, for a complex ion with four ligands how do you know if the shape is planar or tetrahedral?

Thank you. :smile:
Original post by LeaX
Hey, me again with more questions haha.

For coloured TM ions is it the light that causes it to split the d orbitals? Or is it the ligands? My books are quite vague lol. Then once the d orbitals split is it the absorption of energy that moves one electron up to a higher energy which equates to the wave length of colour absorbed?

Also, for a complex ion with four ligands how do you know if the shape is planar or tetrahedral?

Thank you. :smile:


The light causes the splitting of course :tongue: because that is what is absorbed :smile: It is absorbed due to an energy difference between the 5 different types of d orbitals.

Now... ligands can alter the colour - the stronger the ligand is, the greater the splitting of the d orbitals. That means great energies are needed to 'fill the split', therefore the stronger the ligand the higher the frequency of light!

& yup, energy is directly proportional to the frequency, as you can see from this equation:

E=hf
(edited 11 years ago)
How to get from alkene to secondary haloalkalkane? what is an example equation?
Original post by posthumus
The light causes the splitting of course :tongue: because that is what is absorbed :smile: It is absorbed due to an energy difference between the 5 different types of d orbitals.

Now... ligands can alter the colour - the stronger the ligand is, the greater the splitting of the d orbitals. That means great energies are needed to 'fill the split', therefore the stronger the ligand the higher the frequency of light!

& yup, energy is directly proportional to the frequency, as you can see from this equation:

E=hf

Thanks. :smile:
Original post by Knoyle quiah
How to get from alkene to secondary haloalkalkane? what is an example equation?


Hmmm not too sure about this one :confused:

I would have thought for example:

CH3CH=CHCH2CH3 + HBr + NaOH ----> CH3CHBrCH(OH)CH2CH3 + NaBr

Can someone kindly confirm if this reaction is feasible :smile: Thanks

I would assume they would give this as a question asking for mechanics, not sure if this is unit 1 or 2 though :smile:
Original post by Knoyle quiah
How to get from alkene to secondary haloalkalkane? what is an example equation?


Well, react an alkene with a hydrogen halide generally, the hydrogen will go to the carbon with the most hydrogens to form the major product, most likely the one on the end (if it's propene or the double bond is between the first two carbons of the alkene)
They often ask you to draw the electrophilic addition mechanism of it.

So an example is

CH2CHCH3 + HBr -----> CH3CHBrCH3

Addition.

Original post by posthumus
Hmmm not too sure about this one :confused:

I would have thought for example:

CH3CH=CHCH2CH3 + HBr + NaOH ----> CH3CHBrCH(OH)CH2CH3 + NaBr

Can someone kindly confirm if this reaction is feasible :smile: Thanks

I would assume they would give this as a question asking for mechanics, not sure if this is unit 1 or 2 though :smile:


Why the NaOH? I dont think you need that :redface:
That looks to me like a Haloalkane reaction to make an alcohol in which case
You'd do the haloalkane + NaOH to make the alcohol and NaBr or NaCl
Mechanism* and it's a Unit 1.7 mechanism :smile:
(edited 11 years ago)
Original post by Mollymod
Well, react an alkene with a hydrogen halide generally, the hydrogen will go to the carbon with the most hydrogens to form the major product, most likely the one on the end (if it's propene or the double bond is between the first two carbons of the alkene)
They often ask you to draw the electrophilic addition mechanism of it.

So an example is

CH2CHCH3 + HBr -----> CH3CHBrCH3

Addition.



Why the NaOH? I dont think you need that :redface:
That looks to me like a Haloalkane reaction to make an alcohol in which case
You'd do the haloalkane + NaOH to make the alcohol and NaBr or NaCl
Mechanism* and it's a Unit 1.7 mechanism :smile:


Oops for some reason I thought it was to make a halogenoalcohol :colondollar: (If there is such thing!). But I assume even then it is wrong :smile:

Can you go from alkene to 'halogenoalcohol'?
Original post by posthumus
Oops for some reason I thought it was to make a halogenoalcohol :colondollar: (If there is such thing!). But I assume even then it is wrong :smile:

Can you go from alkene to 'halogenoalcohol'?


In some steps, and I can only think of it working with a diene.

So octan-1,7-diene with Bromine, for example would make 1,2,7,8-tetrabromooctane. Then, you could react that with or water (not aqueous hydroxide ions in this case) to make 1,8-dibromooctan-2,7-diol which is a halogenoalcohol :P You wouldn't use aqueous hydroxide ions because they'd make all the bromo groups into alcohols. Water won't react with primary halogenoalkanes, so it won't change the bromo groups on the 1st and 8th carbon :P
(edited 11 years ago)
Original post by Mollymod
In some steps, and I can only think of it working with a diene.

So octan-1,7-diene with Bromine, for example would make 1,2,7,8-tetrabromooctane. Then, you could react that with or water (not aqueous hydroxide ions in this case) to make 1,8-dibromooctan-2,7-diol which is a halogenoalcohol :P You wouldn't use aqueous hydroxide ions because they'd make all the bromo groups into alcohols. Water won't react with primary halogenoalkanes, so it won't change the bromo groups on the 1st and 8th carbon :P


Does the water have to be acidic ? (include H+)

And would you care to explain why the water would not attack the primary halogenoalkane groups :redface:

Thanks !
Original post by posthumus
Does the water have to be acidic ? (include H+)

And would you care to explain why the water would not attack the primary halogenoalkane groups :redface:

Thanks !


No it's just water as far as I know.

It's a hydrolysis reaction and the reaction rate is very very very slow.
The polarity of the C-X bond is smaller in a primary halogenoalkane compared with a tertiary one, because the carbon in a tertiary one is stabilised by the 3 other carbons surrounding the Carbon bonded to the halogen. Because of this increased polarity, I think the carbon in the C-X is more delta positive, so it's more likely to be attacked by the OH- in water to break the bond. Water's a weaker nucleophile than say, cyanide ions which have a full on negative charge while the O in water is delta negative.

So a weakish nucleophile + Slow rate + haloalkane that has a less polarity between the C-X bond = probably not a feasible reaction (at least, not at RTP).
This reaction would take loads of energy to catalyse and it'd just be so slow that it just might as well not happen.
(edited 11 years ago)

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