I've got my chemistry exams (I'm sitting Edexcel's new spec exams on Wednesday) in a couple of days, and I'll be revising my arse over the next 30 odd hours (don't worry I'll find some point to get some sleep ), and I thought, rather than making lots of small threads with my various enquiries and problems, I decided to make one little thread for them all, and then I won't clutter up the chemistry forum Thanks for everyone's help in advance!
First problem I'm looking at the reactions of the alkenes, and quickly going over the oxidation of alkenes to diols, and it says somewhere in my notes that it requires the following conditions:
Potassium Manganate (VII)/Sodium Hydroxide + Heat; first query: sodium hydroxide? Perhaps I took that down wrong, but if it's sodium hydroxide, then there's no oxidising agent. That's right isn't it? So perhaps I meant sodium (or potassium) dichromate?
I also read elsewhere that dilute acid, like dilute sulphuric acid is required; well my notes don't have it (not that they've been trusty so far ) but is it? I mean, we could oxidise an alkene, and partially reduce, say manganate, to MnO2, rather than fully reduce the manganate to Mn2+? Or is the dilute acid necessary? edit: I think I'm being stupid here: the acidic conditions are probably necessary, when we come to writing ionic equations we'll need some H+s to balance it all out, so acidic conditions are required. Is that right?
If it isn't necessary (sorry for taking this situation to the extreme ), then we require it to be aqueous right? As otherwise there's no source of OH- ions to form a diol; or is the aqueous condition implied by writing state symbols on the manganate?
Thanks again for any help in advance! And sorry for the essay of a question!
Mmm, new spec chemistry! Should be a fun day on wednesday, what with both of those exams and 2 economics exams as well!
Anyway, onto your question - not quite sure about where that sodium hydroxide's coming from - I have acidified potassium manganate (VII) in my notes which also answers your second question - yes, acidic conditions are required.
Also note the colour change - purple to colourless. This can be used to distinguish between alkanes and alkenes in need be, as alkanes will not react ^^ (however bromine water is obviously a far quicker and easier test)
EDIT: After minor research, apparently I lied :p well, sort of. It can also react under non-acidic conditions, but in this instance the colour change will not be purple - colourless, it will be dark green + brown precipitate, as the manganate ions are reduced first to manganate (VI) ions, and then to manganese (IV) oxide. However, since the only reaction in the book includes the purple - colourless colour change, I tihnk it's safe to assume that for our specification, acidic conditions are needed.
Originally Posted by GHOSH-5
Hello!
I've got my chemistry exams (I'm sitting Edexcel's new spec exams on Wednesday) in a couple of days, and I'll be revising my arse over the next 30 odd hours (don't worry I'll find some point to get some sleep ), and I thought, rather than making lots of small threads with my various enquiries and problems, I decided to make one little thread for them all, and then I won't clutter up the chemistry forum Thanks for everyone's help in advance!
First problem: I'm looking at the reactions of the alkenes, and quickly going over the oxidation of alkenes to diols, and it says somewhere in my notes that it requires the following conditions:
Potassium Manganate (VII)/Sodium Hydroxide + Heat; first query: sodium hydroxide? Perhaps I took that down wrong, but if it's sodium hydroxide, then there's no oxidising agent. That's right isn't it? So perhaps I meant sodium (or potassium) dichromate?
I also read elsewhere that dilute acid, like dilute sulphuric acid is required; well my notes don't have it (not that they've been trusty so far ) but is it? I mean, we could oxidise an alkene, and partially reduce, say manganate, to MnO2, rather than fully reduce the manganate to Mn2+? Or is the dilute acid necessary? edit: I think I'm being stupid here: the acidic conditions are probably necessary, when we come to writing ionic equations we'll need some H+s to balance it all out, so acidic conditions are required. Is that right?
If it isn't necessary (sorry for taking this situation to the extreme ), then we require it to be aqueous right? As otherwise there's no source of OH- ions to form a diol; or is the aqueous condition implied by writing state symbols on the manganate?
Thanks again for any help in advance! And sorry for the essay of a question!
Mmm, new spec chemistry! Should be a fun day on wednesday, what with both of those exams and 2 economics exams as well!
Anyway, onto your question - not quite sure about where that sodium hydroxide's coming from - I have acidified potassium manganate (VII) in my notes which also answers your second question - yes, acidic conditions are required.
Also note the colour change - purple to colourless. This can be used to distinguish between alkanes and alkenes in need be, as alkanes will not react ^^
Ah nice. My friend has chemistry, economics and latin on Wednesday Good luck with 4 though! I'm worried about just 2
Cheers for that! Yep purple to colourless, with the MnO2 brown intermediate, although I don't think we need to know about that. It could also work with acidified dichromate then, couldn't it? With an orange to green change, as the Cr2O7 2- is reduced to Cr3+?
Next Problem Do we need to know about the hydration of alkenes? I have some slightly complicated mechanisms of the reaction of the addition of water across the double bond, with the formation of a sulphonate and then the nucleophilic substitution of the water; but I can't seem to find much on that about the textbook. I imagine my teacher was mistaken and taught us that anyway?
EDIT: After minor research, apparently I lied :p well, sort of. It can also react under non-acidic conditions, but in this instance the colour change will not be purple - colourless, it will be dark green + brown precipitate, as the manganate ions are reduced first to manganate (VI) ions, and then to manganese (IV) oxide. However, since the only reaction in the book includes the purple - colourless colour change, I tihnk it's safe to assume that for our specification, acidic conditions are needed.
Ah I see. Yep, as I mentioned the brown MnO2 precipitate; I don't think these are ions though, with zero charge? Hmm. But yeah, I think we can play safe in the exam with acidic conditions, and the full reduction of the permanganate ions through to Mn2+. Cheers again!
Ah I see. Yep, as I mentioned the brown MnO2 precipitate; I don't think these are ions though, with zero charge? Hmm. But yeah, I think we can play safe in the exam with acidic conditions, and the full reduction of the permanganate ions through to Mn2+. Cheers again!
the manganate ions are reduced first to manganate (VI) ions, and then to manganese (IV) oxide
The manganate (VI) ions are the dark green, and are ions (MnO4^2-)
You are right in saying that the brown precipitate isn't an ion though - that's Manganese (IV) oxide (MnO2 as you mentioned - no charge)
Originally Posted by GHOSH-5
Ah nice. My friend has chemistry, economics and latin on Wednesday Good luck with 4 though! I'm worried about just 2
Cheers for that! Yep purple to colourless, with the MnO2 brown intermediate, although I don't think we need to know about that. It could also work with acidified dichromate then, couldn't it? With an orange to green change, as the Cr2O7 2- is reduced to Cr3+?
Next Problem: Do we need to know about the hydration of alkenes? I have some slightly complicated mechanisms of the reaction of the addition of water across the double bond, with the formation of a sulphonate and then the nucleophilic substitution of the water; but I can't seem to find much on that about the textbook. I imagine my teacher was mistaken and taught us that anyway?
RE: acidified dichromate.
and nope, no hydration of alkenes needed! always interesting to know though, and it'll probably come up next year or something anyway - knowledge is never wasted
Oh, and I forgot to say too, one of my friends is also meant to be having chemistry, economics and latin on the same day! He's sitting economics on Thursday instead now, with an overnight supervision ^^
The manganate (VI) ions are the dark green, and are ions (MnO4^2-)
You are right in saying that the brown precipitate isn't an ion though - that's Manganese (IV) oxide (MnO2 as you mentioned - no charge)
Ah sorry I was getting confused: so if we keep reducing we go through the following stages:
Permanganate ion/Manganate (VII) ion = MnO4 -
Manganate (VI) ion = MnO4 2-
Manganese (IV) oxide = MnO2
Manganese ion = Mn2+
Right?
RE: acidified dichromate.
and nope, no hydration of alkenes needed! always interesting to know though, and it'll probably come up next year or something anyway - knowledge is never wasted
Ah OK, just wanted to check that. But yeah, probably will pop up next year
Thanks again for everything so far
Originally Posted by Sockpirate
Oh, and I forgot to say too, one of my friends is also meant to be having chemistry, economics and latin on the same day! He's sitting economics on Thursday instead now, with an overnight supervision ^^
Ah yeh, I think my friend is having to stay on overnight supervision and will sit something on Thursday too
Next Problem: Do we need to know about the hydration of alkenes? I have some slightly complicated mechanisms of the reaction of the addition of water across the double bond, with the formation of a sulphonate and then the nucleophilic substitution of the water; but I can't seem to find much on that about the textbook. I imagine my teacher was mistaken and taught us that anyway?
One sec, may I just enquire as to how you are managing to get a sulfonate from addition of water across the double bond of an alkene?
One sec, may I just enquire as to how you are managing to get a sulfonate from addition of water across the double bond of an alkene?
Should have mentioned that: electrophilic addition of conc. sulphuric acid to form the sulphonate which can then allow for the substitution of water. edit: pretty much what Jooeee said below
Originally Posted by Tinsley
Oooh I'm doing these two too! Can't help you though I'm afraid, but I will be back on this thread in th emorning...
I might lurk round, I'm doing AQA next thursday, and need to revise like hell because I can't remember those squiggly arrows go and ethanoic potassium hydroxide reacting with haloalkenes and all that ********.
I might lurk round, I'm doing AQA next thursday, and need to revise like hell because I can't remember those squiggly arrows go and ethanoic potassium hydroxide reacting with haloalkenes and all that ********.
I think he meant you add sulphuric acid, hence the sufonates and then you add water to hydrolyse and form an alcohol. Also note that the sulphuric acid is reformed and so is a catalyst.
Query 3 A question asks me to state two observations of the incomplete combustion of an alkane; firstly, would this mean to form carbon monoxide or just carbon? As complete combustion goes straight to carbon dioxide.
i.e. Is it:
Or is it:
Or does it imply both simultaneously? I'm not entirely sure about two observations, but I imagine one would be soot? Would the other be an orange flame?
Query 3 A question asks me to state two observations of the incomplete combustion of an alkane; firstly, would this mean to form carbon monoxide or just carbon? As complete combustion goes straight to carbon dioxide.
i.e. Is it:
Or is it:
Or does it imply both simultaneously? I'm not entirely sure about two observations, but I imagine one would be soot? Would the other be a orange flame?
Chemistry discussion, revision and homework help.
Useful Resources:
Ghosh's Not-So-Thin Chemistry Help Thread :S
), and I thought, rather than making lots of small threads with my various enquiries and problems, I decided to make one little thread for them all, and then I won't clutter up the chemistry forum 
Thanks for everyone's help in advance! 
) but is it? I mean, we could oxidise an alkene, and partially reduce, say manganate, to MnO2, rather than fully reduce the manganate to Mn2+? Or is the dilute acid necessary? edit: I think I'm being stupid here: the acidic conditions are probably necessary, when we come to writing ionic equations we'll need some H+s to balance it all out, so acidic conditions are required. Is that right? 
which also answers your second question - yes, acidic conditions are required.
Good luck with 4 though! I'm worried about just 2 
RE: acidified dichromate.
edit: pretty much what Jooeee said below 
Most pointless part of the syllabus ever.
).
