A-level Organic Chemistry

[The Ruler]

Yes, I know there probably is a thread on just A-level Chemistry. But I wanted to dedicate a thread just to this d*amned topic anyway.

My question: How do you turn nitrobenzene into 1,3 - dinitrobenzene? And then from that to 1,3,5 trinitrobenzene (?)? I'm not sure if I'm naming the compounds right. Hence, the '?' in brackets.

EDIT: I know you need to add NO2+ ions... But what's the reagent? Conditions?

[~Raynn~]

Hmmm, I wondered that too...

The conditions you need for the nitration of benzene to simply (1-)nitrobenzene is concentrated nitric acid & sulphuric acid, and 50-60 degrees C.... and according to my textbook:
"Careful temperature control is needed to minimize the formation of dinitrobenzene".
What I understood from this is that to add more nitro groups to the benzene ring, you just increase the temperature further...all the other reagents required probably stay the same.

I'm pretty sure you're right about all the naming. :thumbs_up

[Tara x]

Salam
I have done this but dont need to do again but use a chemistry revision guide, should help :D x

[The Ruler]

Hmmm, I wondered that too...

The conditions you need for the nitration of benzene to simply (1-)nitrobenzene is concentrated nitric acid & sulphuric acid, and 50-60 degrees C.... and according to my textbook:
"Careful temperature control is needed to minimize the formation of dinitrobenzene".
What I understood from this is that to add more nitro groups to the benzene ring, you just increase the temperature further...all the other reagents required probably stay the same.

I understand, but what temperature exactly? What range of temperatures?

I have done this but dont need to do again but use a chemistry revision guide, should help

Since you've already done this, perhaps you could help? (^_^) I don't own an A2 chemistry revision guide. Poverty exists even in a Ruler's life.

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[~Raynn~]

Achh, hopefully you won't need to know any exact values, but maybe this'll give you some idea...

http://www.patentstorm.us/patents/54...scription.html

"When a mononitrobenzene product is desired, the reaction is suitably conducted at a reaction temperature not exceeding 80° C, preferably between 0° C. and 60° C, more preferably between 10° C. and 60° C.....
The process for producing dinitrobenzene is suitably conducted at a reaction temperature of between about 30° C. and about 70° C, preferably between about 50° C. and about 70° C, more preferably about 65° C."

I couldn't find anything for trinitrobenzene...it all seems pretty vague... :hmm:

[The Ruler]

^Thank you. Did you ever have to do organic plannings? They're assessments that count towards your final grade. I have one on Tuesday, but am slightly clueless as to what exactly it's about.

[~Raynn~]

Lol, I kinda remember having to do some really confusing thing about identifying which functional group was present in an organic sample...what exactly do you have to plan??

[The Ruler]

Identifying functional groups isn't so bad. But planning an organic synthesis, such as esterification is what we're supposed to do.

Could you tell me about the separating funnel? About the densities of the soluted and things like that? I've been reading my notes... But I don't fully understand.

[The Ruler]

Could someone take me through the steps of saponification?

[~Raynn~]

All I can remember is that it involves the base hydrolysis of an ester, under reflux, to give a carboxylate salt (the soap) and an alcohol. The soap is present in solution, and you precipitate it out by adding excess sodium chloride.

What syllabus do you follow for chemistry, by the way?? Because I can't remember ever even carrying out an esterification, letting alone having to plan it! (so I don't have a clue about the coursework...sorry!)

[The Ruler]

It's not coursework. It's an assessment. We follow the Edexcel syllabus.

I got the summary from my summary sheets. But what I need to know are the steps to it... You know? Like heat under reflux; separating funnel (I don't know if we use it); Buchner funnel + flask (yet again, unsure whether we have to use it) and other apparatus that one might need to carry out stupid saponification.

[The Ruler]

OK... Scrap saponification. Could you help me with the stages of recrystallisation?

[~Raynn~]

Recrystallization...okay, basically, you have the sample of (impure) crystals, in a beaker or whatever...you add to it the minimum amount of boiling water needed to completely dissolve the crystals, then allow the mixture to cool (you can scratch the sides of the beaker to promote the growth of crystals from the glass particles you dislodge, but make sure you don't destroy the crystals that do form, lol). The slower you cool it, the larger the crystals'll be.

After it's completely cooled, you need to filter it (that's where you'd use a Buchner funnel...suction filtration) and dry it, to get rid of the excess water. The whole process can be repeated if you want to purify the sample further. That's pretty much it, I think...

[The Ruler]

^Yah, thanks. Because everything seemed hopeless, I ended up memorising the steps. But your explanation makes sense.

[~Raynn~]

No prob...the assessment's tomorrow? Good luck!!

[The Ruler]

Thanks... It's laughable really. I just finished 3 hours of studying... And I don't feel any more wiser than I was. *sigh*

[~Raynn~]

Lol, I so know the feeling...so how'd it go??

[The Ruler]

I forgot all about 2,4 DNP. And thus, probably lost a few marks that way. It was totally different to what I'd expected... So I was a little surprised. I finished the 2 hour thing in 30 minutes and then did a second draft (more so that I'm not the first one out than anything else really). I think that pretty much sums it all up.

It went bad.

[~Raynn~]

Aww...don't worry about it. I'm certain it couldn't have gone too badly...do they give you any kind of second chance, at a similar assessment or something, btw?

And hmmm... 2,4-dnp?? That doesn't seem to be linked to esterification or ester hydrolysis in any way, lol...

[The Ruler]

No... Not esterification, but recrystallisation.

Nope, I don't get a second chance. Wouldn't it be wonderful if I did? *Sigh*

[Musaafirah]

Edexcel sounds so incredibly crap!
I did Chemistry with OCR and I can't remember having to do such work!
Mind you my grades weren't all that, but ah well.

[The Ruler]

But you had coursework, did you not? My friend is doing OCR ans has coursework. Cswrk has its own ups and downs... Really, I'd rather do these practicles than coursework.

[~Raynn~]

Second chances...*sigh* indeed, lol. If only...

And yup, I do ocr too...and am kinda almost wishing I didn't, lol, since I've got a piece of coursework I should really be working on at this very moment...:exhausted

[The Ruler]

That reminds me of a C3 paper I should be doing now.

Just curious... You're an A-level student now, right? What subjects do you do?

[~Raynn~]

I do maths, further maths, physics and chemistry (yup, a total science nerd, if it wasn't already obvious, lol...)

What about you??

[Banu_Hashim]

OCR is by far the worst exam board for every subject!

and... I too have Chem coursework... which I haven't the foggiest idea how to start.

[Whatsthepoint]

I do maths, further maths, physics and chemistry (yup, a total science nerd, if it wasn't already obvious, lol...)

What about you??

Physics and maths are ok, but there's to much information in chemistry that you can't simply calculate.

[The Ruler]

Biology, Chemistry and Maths for A-level. Did English Literature for AS then dropped it. I couldn't deal with memorising quotes anymore.

Well, you're not a total science nerd unless you get excited when you hear people talk about LiAlH4 or vectors... Or integration. I know of one that literally jumps in the air and claps her hands at the mention of those.

[Banu_Hashim]

I do maths, further maths, physics and chemistry (yup, a total science nerd, if it wasn't already obvious, lol...)

I remember my Maths teacher saying that some people debate whether Maths is a science or an art. :skeleton:

Who in their right mind considers maths an art???

[Banu_Hashim]

Biology, Chemistry and Maths for A-level. Did English Literature for AS then dropped it. I couldn't deal with memorising quotes anymore.

Well, you're not a total science nerd unless you get excited when you hear people talk about LiAlH4 or vectors... Or integration. I know of one that literally jumps in the air and claps her hands at the mention of those.

Um.... I have a feeling that Rayn does get excited at the mention of integration. Am I right Raynn ? :p

[The Ruler]

Who in their right mind considers maths an art???

I once did. I even wrote a note to whoever marked my exam paper in hope to please the person so they're a little less harsher when marking the paper.

[Banu_Hashim]

I once did. I even wrote a note to whoever marked my exam paper in hope to please the person so they're a little less harsher when marking the paper.

lol, if only examiners took bribes...

[~Raynn~]

Lol, okayy...I must admit I do somewhat like integration...it can be fun...but there's nothing exciting about LiAlH4...and I detest vectors...

Maths, an art... hmmm, pure maths can look pretty cool at times, you know...and it's always fun "proving" 1=2...;D
I expect no-one to read through it, but this is just something I came across ages ago...

http://www.maa.org/devlin/LockhartsLament.pdf

NOT that I'm quite that crazy, just to be clear, lol...

[The Ruler]

Help me answer this Raynn:

Magnesium sulphate is soluble in water whereas barium sulphate is almost insoluble. Draw an enthapy level cycle for a Group 2 sulphate to show how the lattice enthalpy and the enthalpy of hydration can be used to explain the difference on the solubilities of magnesium sulphate and barium sulphate.

It's the synoptics. I hate that paper. I could get up and get to my folder several feet away... But it requires much energy I'm not willing to spend. Anyway... What's an enthalpy level cycle?

[~Raynn~]

Magnesium sulphate is soluble in water whereas barium sulphate is almost insoluble. Draw an enthapy level cycle for a Group 2 sulphate to show how the lattice enthalpy and the enthalpy of hydration can be used to explain the difference on the solubilities of magnesium sulphate and barium sulphate.

It's the synoptics. I hate that paper. I could get up and get to my folder several feet away... But it requires much energy I'm not willing to spend. Anyway... What's an enthalpy level cycle?

My guess is it's what I was taught to be a 'Born-Haber cycle'. Have you done Hess' Law?? The enthalpy cycle is basically an application of that, and is used to find the lattice enthalpy, when all other energy values are known...does this kind of thing look familiar?... [NaCl Cycle]. Try looking up born-haber cycles if none of this rings a bell...

As for the differences in the solubility, I'm not sure about the enthalpies of hydration, but with lattice enthalpy... because the magnesium cation is smaller than the barium one, it polarizes the sulphate ion more strongly, distorting the anion so the lattice breaks up more easily. MgSO4 therefore has a lower (less exothermic) lattice enthalpy; it ionizes more readily in water, so is more soluble. This is the explanation that first comes to mind, but I should probably check...

[The Ruler]

*facepalm moment* How could I possibly forget... I see. Thank you.

[~Raynn~]

Lolll, no prob! :statisfie

[The Ruler]

How would I do this:

Solid magnesium sulphate contains water of crystallisation, MgSO4.xH20. 1.23g of magnesiun sulphate crystals were dissolved in distilled water and excess barium nitrate solution was added. The white precipitate of barium sulphate was filtered off, dried and weighed. 1.16g of anhydrous barium sulphate was formed. Calculate x.

[~Raynn~]

Basicallyyy...

- You know the mass of anhydrous BaSO4 (Mr = 233.1), obtained from the hydrated magnesium sulphate used at the start, 1.16g. Use this the find the number of mols of BaSO4 formed. (mols = mass/Mr; you get 0.004976...mols)

- This is also equal to the number of mols of hydrated crystals you started with...and, you're given the mass of the crystals, 1.23g. Use these values to find the Mr of the MgSO4.xH20 crystals. (Mr = mass/mols = 247.166...)

- The Mr of anhydrous MgSO4 is 120.4...subtract this from 247 to find the Mr of the xH20 part to get 126.76...

- Finally, divide this by the Mr of water, 18, to get x = 7.0

Man, this is a rather uncomfortable reminder of the coursework I'm still supposed to be doing...thus far, I have the title - "Determining the formula of Hydrated Copper Sulphate*: Evaluation".
Hold the applause.

*And this is after about 10 mins of debating whether or not to say 'copper sulphate crystals'

[The Ruler]

I see. Thank you.

That's why assessed practicles are so much better. ^_^

[The Ruler]

I would say that you ought to be off LI and carrying on with your coursework, but I'm in need of help yet again.

- A sample of ammonia was dissolved in water to produce 100 cm3 of a solution, X. 10 cm3 of this solution was made up to a volume of 250 cm3. 25.0 cm3 of this diluted ammonia solution was then titrated with aqueous hydrochloric acid of concentration 0.110 M. 37.10 cm3 of the acid was required to neutralise the ammonia solution.

Write and equation for the reaction between ammonia and HCl, and calculate the number of moles of ammonia in solution X. Calculate the concentration of solution X in moldm^-3.

As for the equation, this is what I said:

NH3 + HCl = NH4+ + Cl-

Then I said that once mole of acid reacts with 1 mole of ammonia.
Thus the calculated moles of acid = the number of moles of acid. (for 25cm3 of the ammonia solution)
Then for 250 cm3, I multiplied it by 10. Now I'm lost as to how to continue.

[~Raynn~]

Wow, that's a scary-looking question...okayy, you're right in saying the moles of acid = moles of ammonia. You can find the concentration of the 25cm3 of ammonia by doing conc = mols/volume. From what I understand, this concentration is equal to the concentration of the 250cm3 (you're just taking a 25cm3 sample of that 250cm3 of solution for the titration.
However, this 250cm3 is obtained by diluting 10cm3 of X, i.e. diluting X by 25 times. The concentration of the original solution is therefore 25 times the concentration calculated for the 250cm3. Hope that makes sense...

(Btw, I've hardly ever used the unit, so I was wondering, is 0.110 M simply equivalent to 0.110 mol/dm3??)

[The Ruler]

^Hmmm. I may need some elaborate explanation... But I'll ask for that later.

And as far as my knowledge goes, yes, M = mol/dm3.

[Whatsthepoint]

I'd go about it like this:
you have X moles in 100mL, in 10mL there's 0.1X moles (because you equally divided th solution so 1 tenth of X to every 10mL), in 250mL still 0.1mL(because you just added water), then you divided it by 10 again, so in 25mL tehre's 0.01X moles.

0.01X =n(Hcl) = c(HCl)×V(HCl)
X = 100×0.110 M×0.0371L = 0.4081 moles

ml = cm3
L = dm3

[~Raynn~]

^ I completely agree. So if there's 0.4081 mols in 100cm3, there's 4.081 mols in 1 dm3 (i.e. the concentration is 4.081mols/dm3). I think I obtained the same.

[The Ruler]

I understand. I just need to put it to paper to make it clear.

[The Ruler]

Ah... I just checked my working out the first time I did it. And it seems I got the same answer. Just wasn't sure if it was right.

[The Ruler]

Is there any reason why ethanolic solutions are used for particular reactions?

[Whatsthepoint]

Is there any reason why ethanolic solutions are used for particular reactions?

they dont mix with water?

[Banu_Hashim]

Is there any reason why ethanolic solutions are used for particular reactions?

I'm not sure but they're used in the lipid emulsion test arn't they? To break up tryglycerides (esters) into fatty acids (carboxylic acids) and gycerol (alcohol).

So basically for hydrolysis for this particular reaction.

[The Ruler]

^Hmmm... Yes, I know. But I don't see any water when a halogenoalkane reacts with KCN or ammonia.

- This was in reply to wtp's post.

[~Raynn~]

I think basically, if there's any water present, different products are formed (the reaction of the halogenoalkane with the water itself becomes one of the steps of the overall reaction). Depending on what you wanna get, you use an ethanolic solution instead of an aqueous solution (the ethanol won't react with anything as easily...)

[The Ruler]

I still don't fully understand.

[~Raynn~]

You need a medium/solvent in which the reaction (between the halogenoalkane and ammonia, for example) can take place. The solvent that'd usually be used is water, but its not suitable in this case, because the water itself would react with the halogenoalkane. So, the alternative is ethanol...

[The Ruler]

Aha... Wakarimasu (I understand). Thanks.

[The Ruler]

Sodium Oxide is a basic oxide.
Write the equation for the reaction of sodium oxide with water

I said: Na2O + H2O ----> 2NaOH

Then, state the type of bonding in sodium oxide and explain how particles react to form the product in the above equation.

Type of bonding: Ionic. But I don't exactly understand what the second part of the question is looking for.

EDIT: Is this equation correct - Al2O3 + 2NaOH ---> 2NaAlO2 + H2O

[noorahmad]

u shud get JCR A level Briggs, helped me a lot for my A level.
i dont think its corect, both are base, it shud be a displacement, well i think so!!! u shud check it

[~Raynn~]

Hmmm... I haven't really done this topic, but this might be helpful: Chemguide - Period 3 Oxides. It gives the following equation for the reaction of aluminium oxide with (aqueous) sodium hydroxide...
Al2O3 + 2NaOH + 3H2O --> 2NaAl(OH)4
(Btw, aluminium oxide is what is called an amphoteric compound...it reacts as both an acid and as a base)

As for how the first reaction occurs, I'd probably start by saying how the sodium oxide dissociates in solution into the sodium & oxide ions (and of course, water dissociates into H+ and OH- ions).

According to chemguide, lol, oxide ions are strong bases and have a "high tendency to combine with hydrogen ions", and thereby form another hydroxide ion.

At the same time, the sodium ions will combine either with the hydroxide ions dissociated from water, or one of the hydroxide ion formed in the step above, to form the final product, sodium hydroxide...atleast, that's how I imagine it :><:...hope that helps!

Edit: I was just thinking, and wanted to add an equation (I use the term loosely; I doubt it's ever written like this) which may or may not make things clearer...

Na2O + H2O --> 2(Na+) + (O2-) + (H+) + (OH-).......[Ions dissociate]
(O2-) + (H+) --> (OH-)
2(Na+) + (OH-) + (OH-) --> 2(Na+) + 2(OH-) --> 2NaOH

Lol, superscripts would be nice...

[Whatsthepoint]

According to chemguide, lol, oxide ions are strong bases and have a "high tendency to combine with hydrogen ions" and thereby form another hydroxide ion...

You used lol and oxide ions in the same sentence:-[

[~Raynn~]

You used lol and oxide ions in the same sentence:-[

:giggling:

Trust me, I'm capable of lol'ing in any sentence...the people I talk to have probably learnt to simply edit out all the laughter when listening to me...

[The Ruler]

I'm going through my notes on Benzene chemistry... And I realised that something doesn't make sense.

In my notes, at first I wrote that the oxidation of methyl benzene with hot, alkaline KMnO4 was a *special case*

Then I wrote: "Using hot alkalike KMnO4 will oxidise the carbon nearest to the benzene ring, even if it isn't the one with the -OH group attached"... Which I understand.

But then: "Using more gentle oxidising conditions produces the expected product"... Followed by the diagram of (benzene ring)(CH2-CH2-OH) going to (benzene ring)(COOH) with the aid of hot, alkaline KMnO4.

I am asking you to tell me what my notes mean... If it's correct or not. I suppose quick note making isn't always a good idea. =_=

[noorahmad]

as far as i remember its correct, its acidified KMnO4, that oxidized it to -COOH

[The Ruler]

as far as i remember its correct, its acidified KMnO4, that oxidized it to -COOH

Nope... it's conc NaOH i.e. alkaline.

I cleared whatever questions I had on benzene today. v_v

[noorahmad]

yeps, the mixture is heated under reflux with alkaline KMnO4 and then acidified with dil.H2SO4...
didn't touch my chemistry book for kinda7 months:),
should have checked it b4 posting... :$

[The Ruler]

^That's fine.

- What diffference is there when Benzene and a normal alkene react with manganate (VI)?
- In a reaction, why does hydrogen add onto the Carbon atom with the most hydrogen attached to it? Something to do with Markownikov's rule...?