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# A2 chemistry watch

1. "Benzene reacts with reagent A to produce a hydrocarbon, B containing 90.57% carbon by mass. The relative molecular mass of B is between 100 and 150.

(i) Deduce the empirical formula of B
(ii) Suggest a possible structure for B
(iii) Suggest the identity of reagent A

So for empirical formula, mass of Carbon is, 90.57/12 = 7.5
mass of Hydrogen = 9.43/1 = 9.43

1:1.25 ratio for CH-----> not sure what the empirical formula would be though(having a DUH moment here)

and tbh got no idea on the other two.... is it easier than it looks?

2. (Original post by boromir9111)
"Benzene reacts with reagent A to produce a hydrocarbon, B containing 90.57% carbon by mass. The relative molecular mass of B is between 100 and 150.

(i) Deduce the empirical formula of B
(ii) Suggest a possible structure for B
(iii) Suggest the identity of reagent A

So for empirical formula, mass of Carbon is, 90.57/12 = 7.5
mass of Hydrogen = 9.43/1 = 9.43

1:1.25 ratio for CH-----> not sure what the empirical formula would be though(having a DUH moment here)

and tbh got no idea on the other two.... is it easier than it looks?

Multiply the ratio so that you get whole numbers for both

Then, n(empirical formula) = somewhere between 100 to 150.

Multiply the empirical formula by n to get the molecular formula, then draw it

Not sure about reagent A.. from my knowledge, benzene is pretty unreactive ..
3. Multiply the ratio by 4, giving C4H5. Relative molecular mass of this is 53, so multiply this by 2 to get 106. The emperical formula is therefore multiplied by 2, giving the molecular formula C8H10. You have to think what kind of reactions benzene can undergo. Alkylation? Nitration? Acylation? Halogenation? The adding of 2 more carbon atoms suggests alkylation with a reagent with 2 carbon atoms, giving ethylbenzene (structure B) which conveniently has a relative molecular mass of 106. The reagent A must therefore be an alkylating agent like chloroethane (with Friedel Crafts catalyst) for an alkylation.
4. (Original post by Kyri)
Multiply the ratio by 4, giving C4H5. Relative molecular mass of this is 53, so multiply this by 2 to get 106. The emperical formula is therefore multiplied by 2, giving the molecular formula C8H10. You have to think what kind of reactions benzene can undergo. Alkylation? Nitration? Acylation? Halogenation? The adding of 2 more carbon atoms suggests alkylation with a reagent with 2 carbon atoms, giving ethylbenzene (structure B) which conveniently has a relative molecular mass of 106. The reagent A must therefore be an alkylating agent like chloroethane (with Friedel Crafts catalyst) for an alkylation.
I was hoping i wouldn't have to go down that road for this ...... thank you for your reply.

Query 1:
please go a little slower on how you got the empirical formula please edit----> we basically want the CH in the simplest whole number ratio so we are just looking for WHOLE numbers(simplest) that can get into that ratio and not decimals??

Query 2:
Why are you multiplying by 2 here? edit -----> i get this part now(sorry i am bit slow in the morning lol)

Edit - for nitration there is addition of NO2 by removal of a proton. For Alkylation is the addition of alkane group by removal of a proton which is attached to Halogen

Query 3: for alkylation when asked what the reagent would be, how would you do that? because of course the halogen will react with the proton of the benzene and the alkane groups will move on to carbon of benzene. Therefore it requires an Halogen carrier and a catalyst as well? this is known as the Friedel-Crafts reaction.

Query 4 : For halogenation we require an halogen carrier (they are normally Anhydrous Aluminium(III) chloride or Iron(III) chloride) and this is an electrophilic substitution reaction well they all are tbh. This doesn't need a catalyst i believe? ------ ain't come across acylation, so got no idea there.

I think it gives it away in the question that it forms an "alkane B..." and therefore would go with Alkylation right?

5. Sorry for going too fast. I wrote my reply really quickly before I left for work in the morning.

Query 1: Correct. I multipled the ratio by 4 to get the lowest possible whole numbers for each atom.

Query 2: Correct. Multiplying the RMM by 2 is the only way to get it between 100 and 150. You are also correct about the nitration and alkylation parts.

Query 3: You are correct in suggesting the Friedel Crafts reaction. And yes, it does need a halogen carrier / haloalkane as well as a cataylst, usually AlCl3 or FeCl3.

However, I only knew that it was this reaction after looking at the mass/forumla of the product B. Benezene is C6H6 and product B is C8H10. So it has to be something with a mass of 106 which has an extra C2H4 i.e. substitution on a hydrogen by an ethyl group. Confirm it by finding the mass of ethylbenzene and if it doesn't have a mass of 106 then you know that's not product B. You can also do it by finding the difference in mass between benzene and product B and see what you can substitute on to add this mass.

So now that we are sure product B is ethylbenzene, we can be pretty sure that the reagent to form product B is chloroethane (the halogen carrier you mentioned). This is reagent A. To be on the safe side, mention a catalyst as well. Either AlCl3 or FeCl3 will do.

Query 4: For halogenation, you use Cl2 for example and the same Friedel Crafts catalysts as for alkylation. Acylation is the reaction of an acid chloride and a Friedel Crafts catalyst to give phenyl ketones. Look this up if you're unsure. And yes, they are all electrophilic substitutions.

Also, the question didn't say "alkane B" so be careful. It said "hydrocarbon B" which is only confirming that the product is an organic compound, in this case an aromatic ring with an alkyl side-chain.
6. (Original post by Kyri)
Sorry for going too fast. I wrote my reply really quickly before I left for work in the morning.

Query 1: Correct. I multipled the ratio by 4 to get the lowest possible whole numbers for each atom.

Query 2: Correct. Multiplying the RMM by 2 is the only way to get it between 100 and 150. You are also correct about the nitration and alkylation parts.

Query 3: You are correct in suggesting the Friedel Crafts reaction. And yes, it does need a halogen carrier / haloalkane as well as a cataylst, usually AlCl3 or FeCl3.

However, I only knew that it was this reaction after looking at the mass/forumla of the product B. Benezene is C6H6 and product B is C8H10. So it has to be something with a mass of 106 which has an extra C2H4 i.e. substitution on a hydrogen by an ethyl group. Confirm it by finding the mass of ethylbenzene and if it doesn't have a mass of 106 then you know that's not product B. You can also do it by finding the difference in mass between benzene and product B and see what you can substitute on to add this mass.

So now that we are sure product B is ethylbenzene, we can be pretty sure that the reagent to form product B is chloroethane (the halogen carrier you mentioned). This is reagent A. To be on the safe side, mention a catalyst as well. Either AlCl3 or FeCl3 will do.

Query 4: For halogenation, you use Cl2 for example and the same Friedel Crafts catalysts as for alkylation. Acylation is the reaction of an acid chloride and a Friedel Crafts catalyst to give phenyl ketones. Look this up if you're unsure. And yes, they are all electrophilic substitutions.

Also, the question didn't say "alkane B" so be careful. It said "hydrocarbon B" which is only confirming that the product is an organic compound, in this case an aromatic ring with an alkyl side-chain.
Oh, i get it you was in a rush no worries then but thank you for replying back..... i get it now and i will go over this again so it stays in mind and not slip out again. Thank you once again mate for your help!!!! i am in a rush now, i gotta head off to library lol..... take care mate

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