AS Chemistry- helping each other out!

Never learnt why...Tertiary carbocations are more stable than primary carbocation so more chance of formation?
Google told me C-O bond becomes more polar due to release of electrons from alkyl groups but I don't think that's the case..?

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As far as I'm aware the key reason is that the carbocation formed in most mechanisms involving alcohols is more stable if its a tertiary carbocation formed since methyl groups are slightly electron donating which helps to stabilise the +ve charge on the carbon

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Yeah that's all I know
But then this can be taken in two ways; if it's stable it can stay in that state longer so slower reaction, or because it's stable it is more likely to form so faster reaction. Given that tertiary is more reactive the latter is the case I suppose but I haven't found a satisfactory answer to it.


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hello everyone. apologies if this question sounds pretty stupid but if you have an equation and it gives X + aq ----> Y
does the + aq simply mean +water???

Never learnt why...Tertiary carbocations are more stable than primary carbocation so more chance of formation?
Google told me C-O bond becomes more polar due to release of electrons from alkyl groups but I don't think that's the case..?

Posted from TSR Mobile

It does, yes

Hey guys, do you have any predictions to what maybe could come up in these papers lol I've been looking at past papers now from like 2004 to 2014 and I'm trying to figure out what maybe to expect and areas that could be key.

SN1 mechanisms are more suitable because of steric hinderance and induction. ( was testing you) however less frequent SN2 mechanisms will occur with tertiary alcohols.

Yes I know that
But that doesn't explain why tertiary is more reactive precisely

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what exam board?

Yes I know that
But that doesn't explain why tertiary is more reactive precisely

Posted from TSR Mobile

You don't need the formation data for ethene. Learn how to manipulate Hess cycles. I find it easier and less problematic to set up a Hess cycle. You should get -1558.9 kj mol-1

Here's my working using the longer way, my writing or explanation isn't great but that's how I did it.



To prove it: Using the products - reactants equation. I get the same answer.

-1642.5 -(-83.6) = 1558.9 kj mol-1

I should mention also, the enthalpy of combustion is burning 1 mol of a substance...that's why we can't double up the equation. We are burning 1 mol of C2H6.

Thanks!!

I used to know but why don't we count the oxygen atoms?

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Oxygen is in excess and there isn't actually any enthalpy change happening with the 31/2 O2(g). It's zero. How much heat will be evolved by making 31/2 O2 (g) from 31/2 O2 (g). None because It's already in it's standard state O2 (g). So you could draw another arrow with 0 kj mol-1 if you wanted but it doesn't affect the final outcome of the equation in any way.

Thanks for explaining

Exactly. He's answering a different question

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Hi I've learnt carbocations with more alkyl groups are most stable I don't know why though