2.10 Organic chemistry
Posted by: intechemistry on: September 23, 2010
32 Responses to "2.10 Organic chemistry"
April 14, 2011 at 6:29 pm
You are right. The OH- acts as a nucleophile in aq conditions, but as a base in ethanolic conditions which would favour elimination. You emphasise aqueous as though that is what the answer scheme is for. Are you sure the answer scheme is specifically for the ‘aqueous’ part of the conditions? because to be frank, I can’t really see the advantage in making the alcohol product dissolve in the ‘aqueous nature’ of the solution rather than having the alcohol solvent dissolve the alcohol product itself.
Which paper is it from?
There’s a point here that because the hydroxide is dissolved in water so when mixed with ethanol, you get aqueous ethanol, so aq. ethanol is kind of unavoidable anyway.
April 15, 2011 at 1:36 am
Thanks for your answer.
The question goes like this, “explain why using aqueous ethanol is more preferable, than say, ethanol itself” (1999 Winter Paper CH6).
Umm.. sorry, my mind draws blank to the statement “There’s a point here that because the hydroxide is dissolved in water so when mixed with ethanol, you get aqueous ethanol, so aq. ethanol is kind of unavoidable anyway.”
Apologies for my ignorance, but can you be a bit specific on that statement ? Sorry sir. Thanks !
April 15, 2011 at 2:56 am
Xavier, I can only find ‘ethanol’ mentioned in Q4 (c) on page 6 and no mention of it anywhere else. I’ve uploaded the papers. They are linked at the top of this page.
I guess your Winter 1999 paper isn’t the same as the one I’m looking at. Can you confirm?
April 17, 2011 at 7:54 pm
In the Summer 2000 CH6 paper I have there is no Q5 d) iii) in the questions yet there appears to be a d) iii) answer.
i.e. the layout of the answers is wrong.
My copy says:
…
Intro: CH3–CH=CH–CH=CH–COOH Sorbic acid
…
Questions:
…
(c) Sorbic acid will react with a solution of bromine in tetrachloromethane.
(i) Write an equation for the complete reaction of sorbic acid with bromine. (1)
(ii) A 7.00g sample of sorbic acid was reacted with a solution of bromine in
tetrachloromethane; 125cm3 of the solution was required. Calculate the
concentration of the bromine solution in mod dm–3. (4)
(d) The bromine-containing product from (c)(i) can be reacted with sodium hydroxide in
aqueous ethanol to give a substitution reaction.
(i) Explain why aqueous ethanol is used as the solvent rather than, say, ethanol alone. (2)
(ii) Write the structural formula of the product.
Answer:
(d) (i) Aqueous ethanol favours substitution whereas pure ethanol
favours elimination (1) 1
(ii) CH CH CH CH CHCOOH
OH OH OH OH(1)
This is consequential on (c)(i) 1
(iii) Large number of OH groups on product will hydrogen bond with water so
soluble (1)
Non–polar chain in bromo– compound inhibits solubility/ cannot hydrogen bond
extensively with solvent /only one –OH group / intermolecular bonds stronger than
bonds formed with water (1) 2
I think it should be:
(d) (i) Aqueous ethanol favours substitution whereas pure ethanol
favours elimination (1) [1 mark]
for 2nd mark, either:
- Large number of OH groups on product will hydrogen bond with water so soluble
OR
- Non–polar chain in bromo– compound inhibits solubility/ cannot hydrogen bond extensively with solvent /only one –OH group / intermolecular bonds stronger than bonds formed with water
May 8, 2011 at 12:56 pm
Hi sir,
According to the specification (2.10.2(c)), it says:
“carry out the preparation of an halogenoalkane from an alcohol and explain why a metal halide and concentrated sulfuric acid should not be used when making a bromoalkane or an iodoalkane.”
Why can’t we use a metal halide and c.H2SO4?
But in Facer [ed: AS book, p264], for bromination of an alcohol is done by heating alcohol and a mixture of KBr and 50% H2SO4 under reflux. The point is just not to use CONCENTRATED but just 50% of the acid, is it?
Thanks!
May 8, 2011 at 2:29 pm
The special words in that are concentrated sulfuric acid”, then bromoalkane and iodoalkane Note chloroalkanes are not mentioned.
So if trying to use a bromoalkane from an alcohol and conc. H2SO4, a metal bromide or metal iodide would be necessary.
You will doubtless recall that when (for example) NaBr and NaI are mixed with conc H2SO4 that not only does HBr or HI form, but then they go on to do a redox reaction with the remaining conc H2SO4. A whole host of compounds are formed as well as a lot of heat. Trying to make R-Br or R-I by this method could give a terrible mess.
A metal chloride however is better as no redox reaction between the HCl and remaining c.H2SO4 happens.
There is another issue here about the concentration of H2SO4: The term used above was “concentrated H2SO4″ It is ambiguous as to exactly how strong it is. Because of that, the possibility of undergoing a redox reaction exists. Specifying the concentration of the conc.acid removes this ambiguity and can be made to work.
Alternatively red phosphorous and Br2 (or I2) which forms PBr3 and PI3 respectively which then reacts with the alcohol may be used instead to halogenate the alcohol.
May 8, 2011 at 11:19 pm
salam
sir, do we use conc. H2SO4 instead of dil. H2SO4 for the oxidation of alcohol to carboxylic acid?
May 9, 2011 at 11:04 am
In paragraph 3, i dont understand why you bold the ‘H’ of HBr and HI.
So, stating that we use 50% of H2SO4 will remove the ambiguity and this can be used in exam?
For bromination of alcohol using:
2P + 3Br2 —> 2PBr3
Then PBr3 goes on to halogenate the alcohol R-OH, apart from the corresponding halogenoalkane R-X, what else would the product be? HBr?
Thank you!
May 9, 2011 at 1:19 pm
@ 8, hidayah, May 8, 2011 at 11:19 pm
Waalaikumassalam.
Do not use concentrated H2SO4, but use dil H2SO4 instead with the oxidising agent (e.g. K2Cr2O7) in the oxidation of alcohols to carbox acids.
If conc H2SO4 is used, I think there is a risk of some amount of the alcohol being lost by an elimination reaction forming an alkene while we are refluxing, and also, the carbox acid product may undergo esterification.
May 9, 2011 at 1:29 pm
@ 9, Xin Ling 10M2, May 9, 2011 at 11:04 am
I bolded the ‘H’ to point out we had gone from the metal halide to the hydrogen halide.
I will side with facer and say if you say 50% H2SO4 then you will get the marks.
For the reaction of 3 R-OH + PX3, you get
3 R-X + H3PO3
where X = Br or I
PCl5 is the one that gives Hydrogen halide.
May 11, 2011 at 6:18 pm
[chem unit 2 june 2010, section A, Q14]
Best method of converting ethanol into C2H5I:
B heat ethanol and KI in the presence of dilute acid, or
D heat red phosphorus, ethanol and iodine under reflux
i chose B but the answer is D.
May 12, 2011 at 11:30 am
salam alayk
sir, referring to GF page 149 there are to reason why -OH in carboxylic acid is more acidic than -OH in alcohol. i can’t really understand the second reason….. need for explanation
thank you
May 12, 2011 at 12:36 pm
Salam
I’ve done a small sketch of what Facer is talking about. The three blue and white orbitals are the 2pz orbitals of each atom in COO-.
http://intechemistry.files.wordpress.com/2010/09/facer-p149-delocalisation-in-carboxylate-anion.jpg
On losing the H+ ion, the 2pz of each atom is in the correct orientation to allow overlap hence the O with the -’ve (which is in the 2pz orbital) can delocalise with the C 2pz which in turn overlaps with the 2pz of the O in the C=O.
The grey angled shapes (one darker than the other just to help visualise things) shows overlap between these orbitals.
As we know delocalising is a stabilizing factor. spread the charge out and you lower the repulsion 
It’s very similar to what happens in benzne.
Here’s some more diagrams which will help you visualise it: http://intechemistry.files.wordpress.com/2010/09/more-carboxlate-diags.jpg
Hope that helps.
May 12, 2011 at 7:48 pm
Hi again sir:)
Unit 2 Jan 2010 Section B
Q18(f)(ii)* [QWC ques]:
Both ethanol and water contain hydrogen bonds. By considering the hydrogen bonding on these two solvents, suggest why 2-chlorobutane is more soluble in ethanol than in water. (2 marks)
Marking scheme:
Water has 2 hydrogen bonds per molecule (on average) whereas ethanol only has 1 (1)
ALLOW
Water has more hydrogen bonds (per molecule) than ethanol
Needs more energy to break H bonds in water (so less soluble) / H bonding (ALLOW
intermolecular forces) stronger in water (1)
So, just confirming, if a substance is insoluble in water, we always say that this is because the substance ‘does not form H-bond with water’. The more in-depth explanation to this is that the energy needed to break the H-bond between water molecules is too high for the substance?
Therefore, if a substance is soluble in water, this means it can break the H-bond between H2O molecules, then form H-bonds with the H2O molecules?
Thank you!
May 15, 2011 at 10:37 pm
Hi Xin Ling.
I’m afraid it’s very much question and case specific. There are substances like aspirin that are poorly soluble in water, but it has a COOH group, so can H bond with water.
Generally, if something doesn’t dissolve in water, then the energy needed to overcome the H2O-H2O Hydrogen bonds and the compound-compound intermolecular forces, is more endothermic than the energy of the H2O – compound intermolecular forces.
This vague energy can be broken down more accurately for an A2 discussion like this: If delta(S)dissolution – T x delta(H)dissolution is not negative then it’s not going to dissolve, or if you are discussing enthalpy and you have an ionic solid, you can use the dleta(H)hyd – delta(H)LatticeEnthalpy.
At AS level, the general statement will probably suffice.
As the actual Q above, it’s a comparative type Q, you have to mention what’s different between the two, and how that difference affects the issue at hand; Here the solubility of the chloroalkane.
So in this case, you could have said
The energy needed to overcome the H2O-H2O Hydrogen bonds and the chloroalkane-chloroalkane intermolecular forces minus the energy of the H2O – chloroalkane intermolecular forces, is more endothermic than the energy of the ethanol – ethanol bonds and the chloroalkane-chloroalkane intermolecular forces minus the H2O – ethanol intermolecular forces.
It’s a a powerful answer, but wordy.
May 22, 2011 at 12:13 am
salam sir
sir, why is it exactly the rate of reaction of tertiary halogenoalkane is higher than that of secondary which in turn higher than primary?
is it due to the methyl group?
May 22, 2011 at 3:46 pm
Salam.
It’s higher for 3o halogenoalkanes only within the SN1 mechanism: because for 3o halogenoalkanes, the energy of the intermediate 3o carbocation (and hence the Ea for that process) is low compared to the other carbocation that could form if a 2o, 1o or methyl halogenoalkane were to undergo reaction.
For SN2 mechanisms it’s actually slower, mainly because of steric hindrance, but also in general because the charge on the carbon carrying the halogen is less d+ than carbons in 2o, 1o or methyl halogenoalkanes.
May 27, 2011 at 12:05 pm
For reaction of halogenoalkane with NaOH, if the question asks for the conditions required, do we say its ‘aqueous NaOH solution’ or ‘ethanol’ (for mixing the immiscible layers of NaOH(aq) and the halogenoalkane)? How about rx with water containing AgNO3(aq)?
How about rx with NH3? just state that the conditions are ‘excess concentrated NH3′ or ‘heat in a sealed tube’? cos i saw somewhere that NH3 has to be mixed with ethanol, NOT water, to help it mix with R-X.
As for elimination rxs of R-X, the condition IS ‘heat under reflux in solution in ETHANOL’, right?
All in all, i’m mainly worried and confused about the conditions. Cos some books say ‘ethanol’ as a solvent but i think the marking scheme doesnt accept that. Do correct me if i’m wrong thanks!
May 27, 2011 at 3:09 pm
For nucleophilic substitution of C-X to C-OH, say this: aqueous ethanolic NaOH solution & heat under reflux.
For elimination, NaOH dissolved in ethanol [is unambiguous] and heat under reflux is a good answer. High temp assists the elimination.
AgNO3 thing is to identify which halogen was in the haloalkane. It’s a separate issue.
For NH3 both ‘excess concentrated NH3′ or ‘haloalkene plus NH3 heated in a sealed tube’ should be fine. I’ve seen statements declaring the ‘sealed tube’ way is fine. You don’t HAVE to add ethanol. You can simply do the reaction with a magnetic stiffer to get reasonable reaction mixing. The ethanol may be tricky to remove later.
February 17, 2012 at 9:57 pm
Gud day sir ,
why does the water in the reflux condenser flow from the down to up ?is it to ensure that the product vapour condenses as soon as it is produced ?
why is one of the condition for chlorination is to ensure that dry alcohol is used ?
thanx a million sir
February 18, 2012 at 12:13 am
Good night Draco
There are a number of ‘explanations’. Perhaps the easiest and the best is:
By going from the bottom to the top, air is expelled from the ‘jacket’ surrounding the central tube holding the gas to be condensed. The presence of air would reduce cooling efficiency, hence gas more likely to energy from end of condenser! (=bad!)
Not sure which chlorination reaction you are talking about, but Cl2 dissolves in water and some of it also reacts with water (about 30% of it reacts with water if I’m not mistaken). Hence “anhydrous” alcohol is used.
May 23, 2012 at 2:43 pm
Hello sir. For secondary haloalkane, will it undergo Sn1 or Sn2 mechanism? This question came out in our trial paper recently, and I thought it is faster if it undergo Sn1 mechanism, but the answer scheme follows Sn2 mechanism..Why is this so?
May 23, 2012 at 3:19 pm
2o compounds can undergo both.
According George Facer, P21 of the A2 book, for 2o species, SN1 is favoured which is what you mention above. This implies that steric hindrance is a a very strong preventative force, i.e. when it can be an issue, then it is an issue.
BUT, one can adjust the conditions of a reaction to try and favour one mechanism over the other. That’s actually a little step beyond A-level however. The mark scheme MAY have accepted an SN1 mechanism as an alternative. Unless there is a ‘clue’ as to which mechanism you should have given.
If you were given an open Q as to the nucleophilic substitution in a 2o species, choose the mechanism that you best remember.
Do keep an eye out for ‘clues’ as to what choice you should make. e.g. (actually this is an issue for Unit 4 organic) the info in the Q may say the reaction produces two enantiomers. In that case you would HAVE to do an SN1 mechanism…. so watch out for clues. Edxcel seem to do this ‘clue dropping’ quite a lot so that if you don’t pick up on it, then you fall into a trap and give the wrong answer.
Read the Q’s carefully. If the mark scheme does NOT give the SN1 mechanism as an acceptable alternative then my guess is there was indeed a clue somewhere. I memory serves correct for this U2 paper did indeed accept SN1 and SN2 mechanisms, although took the SN2 mechanism as the exemplar one for the mark scheme.
May 23, 2012 at 4:03 pm
In Unit 4 perhaps – but usually only if instructed. Most often, the two SN1 (chiral !!!) products will be asked for in a separate part of the question, usually not in the mechanism itself. So usually just draw one product molecule, and expand upon this later if necessary
Leave a Reply
Authors
umiazuraRecent Comments
 | Kosher on 2.8 Kinetics |
 | Awesome on Smartie-pants |
| Iintechemistry on Smartie-pants |
| Awesome on Smartie-pants |
| Iintechemistry on Smartie-pants |
| Awesome on Smartie-pants |
| Iintechemistry on May/June 2012, Unit6b |
| Iintechemistry on Smartie-pants |
| Awesome on May/June 2012, Unit6b |
| Awesome on Smartie-pants |
April 13, 2011 at 4:01 pm
Why is the solvent AQUEOUS ethanol preferable when performing Nucleophilic Substitution for halogenoalkanes with OH ?
I understand if it’s ethanol liquid, the competing elimination reaction may become dominant. But answer scheme has something like “large number of OH groups will allow product to form H-bonds with water hence soluble in water”. What is the significance of dissolving the product ? So that it won’t form a secondary elimination reaction with the remaining OH ?