OCR A Chemistry F324 Rings, Polymers and Analysis Thu 26 Jan 2012

Original post by haybo

do we need to know friedel crafts reactions (acylation and alkylation) ? its not in the text book yet my teacher taught us it? has it been removed from the syllabus?

No but isn't it when you add Chlorine to Benzene known as friedel crafts reaction? It's not the spec. though

Reply 382

Can anyone help me with this?

Reply 383

RukayahR

Original post by haybo

do we need to know friedel crafts reactions (acylation and alkylation) ? its not in the text book yet my teacher taught us it? has it been removed from the syllabus?

I want to know the answer to this aswell!
Acylation/alkylation isn't in my notes but I remember my teacher going on about it!

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Original post by LifeIsGood

Can anyone help me with this?

Pretty sure it is Benzeneacetaldehyde because u have a huge peak at 2.5 on proton NMR, 2 H environments and a double bond O --> C8H80

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Original post by RukayahR

I want to know the answer to this aswell!
Acylation/alkylation isn't in my notes but I remember my teacher going on about it!

Alkylation is pretty easy;

Start with CH4 + Cl2 in presence of UV
This makes CH3CL
React this and AlCl3 together to form CH3+ + AlCl4-
CH3+ is an electrophile then its just a basic electrophilic substitution onto the benzene ring

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Original post by cheesebase

Pretty sure it is Benzeneacetaldehyde because u have a huge peak at 2.5 on proton NMR, 2 H environments and a double bond O --> C8H80

There's some sort of calculation but I'm not sure this is relevant to our spec specifically cuz it's from an old past paper, it's only 3 marks but I don't understand what they did to get the number of carbons.

Markscheme:

Reply 387

Killjoy-

8

Original post by LifeIsGood

There's some sort of calculation but I'm not sure this is relevant to our spec specifically cuz it's from an old past paper, it's only 3 marks but I don't understand what they did to get the number of carbons.

Markscheme:

It's to do with the fact that carbon-13 makes up 1.11% of all carbon atoms in organic compounds. I'm pretty sure its not part of our spec.

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Original post by LifeIsGood

There's some sort of calculation but I'm not sure this is relevant to our spec specifically cuz it's from an old past paper, it's only 3 marks but I don't understand what they did to get the number of carbons.

Markscheme:

Ok - so to do that;

You start with 120.
You know there is only one oxygen
So 120-16 = 104 split between Hydrogen and Carbon
Now just think about the possible combinations
if you have 7 carbons thats 7X12 which is 84 so that would mean u would need 20 Hydrogens to make the total Mr 104

If you have 8 carbons thats 8X12 which is 96 so then you need 8 hydrogens to make total Mr 104

If you have 9...it goes over 104

No compound would be C7H20O so it must be C8H8O..confirmed by the spectra - that make sense?

Reply 389

VQG

11

If you have an assymetrical anhydride (e.g. ethanoic propanoic anhydride) what ester and carboxylic acid would form?

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Original post by VQG

If you have an assymetrical anhydride (e.g. ethanoic propanoic anhydride) what ester and carboxylic acid would form?

An acid anhydride is made from 2 carboxylic acids....

if you are referring to a reaction with an alcohol e.g. ethanol

then you would get I believe equal mixtures of ethyl propanoate and ethyl ethanoate - however there may be a rule for this but its not sure...

Reply 391

Stevo F

18

pretty sure my teacher said ratio between peak heights of mass spec wasnt in our specification... anyways pretty sure the answer's C8H8O, peak 1640-1750 is a C to O double bond and as this is the only O in the compound, must be either a ketone or aldehyde. mr is 120, therefore 120-16 (for oxygen) is 104. Playing around with carbon and hydrogen atoms implies the only sensible values are for C7H20O C8H8O, more or less carbons than this just doesnt work with the number of H left considering the mr. Peaks around 7.5 to 8 suggest benzene ring, which in turn suggests C6H6 meaning its impossible to have a ratio of 7 carbons to 20 hydrogens.

My answer to this question is a bit of a process of elimination but i dont see why this wouldnt be right.

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ZTSR

Just got in from school from Biology exam :redface:

. Taking a Power nap. And let the Chemistry marathon begin. :colondollar:

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