Edexcel AS Chemistry Unit 2 Thread Watch

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killfestab
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#141
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(Original post by Phalange)
D
All others are non-polar.
If you put an electrostatically charged rod or substance near a jet of water or a polar liquid, the liquid moves towards the rod (or away) but it's not important the direction just that it will move

Edit: Think of it like magnets... XD

On a side note..........
Which methods do people have to make haloalkanes? ( WIll use the shortest carbon chain to save time xD)
I have these:

CHLORO ALKANES
1) Add PCl5
CH3OH + PCl5 ---> CH3Cl + POCl3 + HCl

2) Add HCl to a TERTIARY alcohol (normal rxn)

3) Add PCl3
3CH3OH + PCl3 ----> 3CH3Cl + H3PO3

BROMO/IODO ALKANES:
1) Add damp red phosprous to the halogen (e.g. Br2), then add alcohol
2P + 3Br2 ----> 2PBr3 *see how we have PBr3 like PCl3 before??
CH3OH + PBr3 ---> CH3Br + H3PO3

2) Add phosphoric (V) acid to metal halide then add alcohol
3NaBr + H3PO4 ---> Na3PO4 + 3HBr
CH3OH + HBr ----> CH3Br + H2O

Ok here is where I am confused. My CGP book says NOT to use sulphuric acid for bromo/iodo but phosphoric because:
We do not use H2SO4 beacuse HBr/HI are oxidised by it so you end up with a range of products .e.g Br2 and so a reduced yield of the desired haloalkane
BUT my text book ("endorsed by edexcel") says you can but only 50% for a bromoalkane

1) Add 50% H2SO4 + NaBr, then add an alcohol
NaBr + H2SO4 ---> NaHSO4 + HBr
CH3OH + HBr ----> CH3Br + H2O

Any insight??
just adding to your list of making chloroalkanes....

another very useful method is the use of SOCl2

The advantage with SOCl2 is that all the products are gaseous except your halogenoalkane - and so separation is easy


and answering your question about the preparation of Iodoalkanes and Bromoalkanes

The best method in my opinion is the Red Phosphorus Reflux with Bromine/Iodine and Alcohol..

However, The Phosphoric Acid method is also good and is better than the sulfuric acid method, because sulfuric acid is an extremely good oxidising agent... but then agen, phosphoric acid isnt the easiest acid to find off the shelf...and the lab method would prolly find it easier to use 50% sulfuric acid... using 50% tries to minimize the oxidation of Bromide to Bromine
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jpm
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#142
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#142
nope since wen thts polar??!!!
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jonathan3909
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#143
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But the book says its polar??
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jonathan3909
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#144
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Why is the B.P. of water greater than ammonia and hydrogen fluoride?
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Phalange
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(Original post by jonathan3909)
Cyclohexene is not polar:confused: !!!!
Cyclohexanol is slightly polar due to OH group??
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jpm
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#146
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#146
dare more h bonds per water molecule den ammonia n HF
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pervy_sage
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#147
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it has more hydrogen bonds i guess
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jpm
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#148
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#148
ya i think its polar due 2 da polar grp-OH
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Phalange
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#149
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(Original post by killfestab)
just adding to your list of making chloroalkanes....

another very useful method is the use of SOCl2

The advantage with SOCl2 is that all the products are gaseous except your halogenoalkane - and so separation is easy


and answering your question about the preparation of Iodoalkanes and Bromoalkanes

The best method in my opinion is the Red Phosphorus Reflux with Bromine/Iodine and Alcohol..

However, The Phosphoric Acid method is also good and is better than the sulfuric acid method, because sulfuric acid is an extremely good oxidising agent... but then agen, phosphoric acid isnt the easiest acid to find off the shelf...and the lab method would prolly find it easier to use 50% sulfuric acid... using 50% tries to minimize the oxidation of Bromide to Bromine
Thanks much clearer than my teachers
Whats SOCl2 reaction??
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killfestab
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#150
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#150
CH3CH2OH + SOCl2 --> CH3CH2Cl + HCl (g) + SO2(g)
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SkyNinja
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(Original post by SK-mar)
how do you know there will be an acid-base titration question?? pluss could you expand on that chlorine bleach thing pleaseee.. cheers
Theres always an acid base calculation as acid base calculations is on its own topic, if you look at the past 3 exam papers they all have it

Heres a question on the chlorine bleach that could come up.
10.00cm3 of a sample of bleach was diluted into a 250cm3 with water. 25.0cm3 of this solution was pipetted into a flask. An excess of aqueous potassium iodide was then added to liberate iodine:

Its just a normal calcualtion like that its gonna take too long to type everything up lol hope that helps if you still want more ill trype it up..
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jonathan3909
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(Original post by pervy_sage)
it has more hydrogen bonds i guess
but in HF, F is more electronegative than oxygen so stonger hydrogen bond but still:confused: :confused:
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jonathan3909
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(Original post by Phalange)
Cyclohexanol is slightly polar due to OH group??
So what is the answer to the MCQ??
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Phalange
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(Original post by dimi3)
hi guys, umm quick question. do we have rates of reactions for unit 2? as in like calculations fr that stuff...?
Uhhh I don't think so.. no calcs for rates.. like gradient of curse
Altho you need to know what speeds up the rate e.g. temp and equilibra, thats about it
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Phalange
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(Original post by jonathan3909)
So what is the answer to the MCQ??
Q Polar liquids are affected by electric fields. For which of the following liquids would a jet
of the liquid be affected by an electric field?
A hexane
B cyclohexane
C cyclohexene
D cyclohexanol

D will be affected! As it is slightly polar due the OH bond... all the others are not
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killfestab
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can anyone else do QUESTION 4 on the iodine/thiosulfate titration in the EDEXCEL AS CHEMISTRY page 253, and tell me what answer they get ?
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jonathan3909
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(Original post by Phalange)
Q Polar liquids are affected by electric fields. For which of the following liquids would a jet
of the liquid be affected by an electric field?
A hexane
B cyclohexane
C cyclohexene
D cyclohexanol

D will be affected! As it is slightly polar due the OH bond... all the others are not
But cyclohexene is polar-Its written in the edexcel book:confused:
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jonathan3909
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Wait-Let 'me solve it!
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SK-mar
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#159
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(Original post by SkyNinja)
Theres always an acid base calculation as acid base calculations is on its own topic, if you look at the past 3 exam papers they all have it

Heres a question on the chlorine bleach that could come up.
10.00cm3 of a sample of bleach was diluted into a 250cm3 with water. 25.0cm3 of this solution was pipetted into a flask. An excess of aqueous potassium iodide was then added to liberate iodine:

Its just a normal calcualtion like that its gonna take too long to type everything up lol hope that helps if you still want more ill trype it up..
ok sorry to nit pick at that titration but how are we supposed to find out moles of anything in that as we only have volumes given to us?? can you please give me an example of a question for this titration and answer it..
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killfestab
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Cyclohexene is not polar, C and H have similar electronegativities and thus centres of positive and negative charge must cancel. Edexcel must have made a typo
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