Edexcel AS Chemistry Unit 2 Thread Watch

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Phalange
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#161
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#161
(Original post by jonathan3909)
But cyclohexene is polar-Its written in the edexcel book:confused:
If that's true, I'm guessing cyclohexanol is more polar due to O-H group while hexene does not have...
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SK-mar
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#162
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#162
(Original post by jonathan3909)
But cyclohexene is polar-Its written in the edexcel book:confused:
in terms of IR spectra then it is very very very slightly polar as its bonds can abosorb radiation. But in the terms of your original question its very slight polarity is irrelevant.
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jonathan3909
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#163
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(Original post by killfestab)
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 ?
For part (iii) I got 0.025 and for (iv) I got 1.775
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jonathan3909
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#164
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#164
(Original post by SK-mar)
in terms of IR spectra then it is very very very slightly polar as its bonds can abosorb radiation. But in the terms of your original question its very slight polarity is irrelevant.
Open page 158 of the book and it says cyclohexene is deflected:confused:
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killfestab
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#165
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#165
(Original post by jonathan3909)
For part (iii) I got 0.025 and for (iv) I got 1.775
i got the same,
cheers

btw, three significant figures...
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jonathan3909
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#166
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#166
(Original post by Phalange)
If that's true, I'm guessing cyclohexanol is more polar due to O-H group while hexene does not have...
The book says in cyclohexene ,ther is a planar C = C part of the molecule so it is polar:confused:
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SK-mar
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#167
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OK theres one thing thats majorrrrrly annoying me! ..... can someone put a step by process of the formation of a BROMO/IODO halogenoalkane from an alcohol please. I know you use phosphoric acid but howwwww? thank you in advance

is it for example:

KBr + H3PO4 = HBr + KH2PO4

C2H5OH + HBr = C2H5Br + H2O

???
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jonathan3909
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#168
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#168
(Original post by killfestab)
i got the same,
cheers

btw, three significant figures...
Last page,Q 7 (b) and (c):confused:
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killfestab
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#169
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#169
one sec lemme do it
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killfestab
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#170
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#170
*SOLVE it :P
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jonathan3909
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#171
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#171
(Original post by SK-mar)
OK theres one thing thats majorrrrrly annoying me! ..... can someone put a step by process of the formation of a BROMO/IODO halogenoalkane from an alcohol please. I know you use phosphoric acid but howwwww? thank you in advance

is it for example:

KBr + H3PO4 = HBr + KH2PO4

C2H5OH + HBr = C2H5Br + H2O

???
http://www.chemguide.co.uk/organicpr...es/making.html
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killfestab
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#172
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#172
7b)
i) The second spectra is for compound C because it is the only compound with an -OH functional group and thus we see a peak around 3200 cm^-1

ii) The other spectrum has a shark peak at around 1750 cm^-1 and that is indicative of the C=O functional group i.e. an aldehyde or ketone

c)
i) Compound C will react with bromine dissolved in hexane, and we will see a colour change from orange/red to colourless

ii) C6H5CHBrHBrCH2OH
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jonathan3909
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#173
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#173
(Original post by killfestab)
*SOLVE it :P
Last spectrum is of C as (O-H) hydrogen bonding peak is there and it also does not have major C=O peak.

C will react in bromine/hexane solution and brown color will disappear??
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Phalange
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#174
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#174
(Original post by SK-mar)
OK theres one thing thats majorrrrrly annoying me! ..... can someone put a step by process of the formation of a BROMO/IODO halogenoalkane from an alcohol please. I know you use phosphoric acid but howwwww? thank you in advance

is it for example:

KBr + H3PO4 = HBr + KH2PO4

C2H5OH + HBr = C2H5Br + H2O

???
Works with any halogenoalkane to be honest, but the reason why you do this with bromo/iodo is because HBr and HI are not easy to get hold of so you need to make it

I always wrote it but...
(1) Add damp red phosphorous to Br2 (or I2) to produce a hydrogen halide
H3PO4(l) + 3KBr(s) --> K3PO4 + 3HBr(g)

(2) React the hydrogen halide with the alcohol
CH3OH + HBr --> CH3Br + H2O
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Phalange
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#175
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#175
(Original post by jonathan3909)
The book says in cyclohexene ,ther is a planar C = C part of the molecule so it is polar:confused:
Ok this is what I think. It is polar but it is such a small degree it doesn't matter (cyclohexene). OH is much more significant factor. Don't worry too much we don't have to know a lot on cycloalkanes...

Cyclohexanol definately is not symmetrical so it has polar OH bond that will be affected by water

Edit: LOL I've been thinking about this too and now I'm ******** myself I can't find the right answer

EDIT2: WAit... is that a past paper question or a question in a book? If it was in a book, there can be more than one answer (I would pick cyclohexanol + cyclohexene) BUT if it was in exam I'd pick cyclohexanol as it is a strong more obvious answer
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killfestab
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#176
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(Original post by Phalange)

I always wrote it but...
(1) Add damp red phosphorous to Br2 (or I2) to produce a hydrogen halide
H3PO4(l) + 3KBr(s) --> K3PO4 + 3HBr(g)
Doesn't Refluxing with red Phosphorus follow this equation:

2P(s) + 3Br2 (l) --> 2PBr3 (l)
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SK-mar
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#177
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#177
(Original post by Phalange)
Works with any halogenoalkane to be honest, but the reason why you do this with bromo/iodo is because HBr and HI are not easy to get hold of so you need to make it

I always wrote it but...
(1) Add damp red phosphorous to Br2 (or I2) to produce a hydrogen halide
H3PO4(l) + 3KBr(s) --> K3PO4 + 3HBr(g)

(2) React the hydrogen halide with the alcohol
CH3OH + HBr --> CH3Br + H2O
thankssss, does my equation work too or is yours better to use. ALsooo, is it ok just to put 'red' phosphorous instead of concentrated phosphoric acid? cheers
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dunnoaboutme
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#178
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can someone please help me with one thing, basically a test for group 7 halides is when you add AgNO3, but then you can add NH3, then you can add an organic solvent such as hexane to confirm the test am i right?

so when you add hexane to bromine i know you get a red solution, but what do you get with iodine and chlorine??
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Phalange
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#179
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#179
(Original post by SK-mar)
thankssss, does my equation work too or is yours better to use. ALsooo, is it ok just to put 'red' phosphorous instead of concentrated phosphoric acid? cheers
Oops wrong equation BUT you can use any...
Red yes you have to say red. Here wait...

BROMO/IODO ALKANES:
1) Add damp red phosprous to the halogen (e.g. Br2), then add alcohol
2P + 3Br2 ----> 2PBr3
CH3OH + PBr3 ---> CH3Br + H3PO3

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

These are the best ways. My Edexcel book says use 50% concentrated sulphuric acid but the problem is that sulphuric acid would oxidise the HBr to produce Br2 (a by-product, reducing yield) it even oxidses HI even more... so I would not use sulphuric acid at all but sometimes in questions they use 50%..
Here:
1) Add 50% H2SO4 + NaBr, then add an alcohol
NaBr + H2SO4 ---> NaHSO4 + HBr
CH3OH + HBr ----> CH3Br + H2O

(Original post by killfestab)
Doesn't Refluxing with red Phosphorus follow this equation:

2P(s) + 3Br2 (l) --> 2PBr3 (l)
Yes sorry I put down phosphoric acid method by accident
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Phalange
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#180
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#180
(Original post by dunnoaboutme)
can someone please help me with one thing, basically a test for group 7 halides is when you add AgNO3, but then you can add NH3, then you can add an organic solvent such as hexane to confirm the test am i right?

so when you add hexane to bromine i know you get a red solution, but what do you get with iodine and chlorine??
Chloride:
AgNO3 - WHITE PPT
then NH3 - soluble in dilute or conc NH3

organic solvent - pale green colour
water - pale green colour
room temp - green gas

Bromide:
AgNO3 - CREAM PPIT
then NH3 - soluble in ONLY conc NH3

organic solvent - orange-red
water - yellow orange colour
room temp - red-brown liquid

Iodide:
AgNO3 - YELLOW PPT
then NH3 - not soluble in any

organic solvent - pink-violet
water - brown
room temp - dark grey solid

You do not really need to add organic solvents because as far as I know AgNO3 + NH3 are okay.
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