4 | intechemistry
January 30, 2011 at 1:25 am

#2, amirul arif, January 24, 2011 at 11:59 am

Salam.

The sillybus ( 5.4.1.c) says “describe the mechanism of the electrophilic substitution reactions of benzene in halogenation, nitration and Friedel-Crafts reactions including the formation of the electrophile” so, strictly speaking you might be expected to be able so show the “curly arrows” involved in producing the electrophile, e.g. Br+, and really, it’s not that difficult to do.

But from the mark schemes (Sept 2007 specimen and June 2010 papers) the suggestion is that you don’t have to use the curly arrow type mechanism for the generation of E+. If you did then it would obviously be OK, but if you don’t use the curly arrows depiction, you must show the generation of E+ using a chemical equation instead.

e.g:
a) c.H2SO4 + C.NHO3 –> (NO2)+ + (HSO4)- + H2O
b) 2Fe + 3Br2 –> 2FeBr3
FeBr3 + Br2 –> Br+ + [FeBr4]-

26 | intechemistry
April 12, 2011 at 11:38 pm

Hummm… I’m a but rusty on this so here’s a guess ’till I think otherwise.

I can imagine the HO- nucleophile undergoing Nu subn. with the Cl in the protruding CCl3 group of Ar-CCl3. I can see a gem diol type species forming, Ar-C(OH)2Cl. At this point, I can imagine one of the O-H bonds breaking to form a C=O and kicking off the Cl- for entropy reasons forming Ar-C=O(OH), but under the prevailing basic conditions, any carboxylic acid group formed at this point would certainly lose a second proton to become the carboxylate anion, not a carboxylig acid. I presume an acidic work-up is performed?

The C in Ar-CCl3 is already +III (more likely to be +IV considering the e- withdrawing effect of the Ph gp) so in forming the carboxylic acid (or carboxylate anion) mentioned, it still in the same O.S. as before, hence the reaction isn’t an oxidation on the trichloromethylbenzene.

The Q is a bit odd after 2 Cl’s have been substituted by 2 OH’s and I’m fairly sure it isn’t going to come up in your finals.

28 | intechemistry
April 17, 2011 at 6:32 pm

KMnO4(aq)# oxidises the side chain of phenyl based aromatic rings. It has to with the interaction of orbitals in the C bonded to the ring, ultimately allowing C-C bond cleavage. I believe the process shares some similarity to the famous alpha-H’s used extensively in carbonyl chemistry.

Anyway, none of the above is on the syllabus. Oxidation of the side chain using alkaline KMnO4 used to be on the old syllabus.

# I am more familiar with doing it in basic conditions however (but of course would need to undergo acidic work-up afterwards to make the carboxylic acid).

30 | intechemistry
May 8, 2011 at 2:44 pm

This diagram should explain what it expected of you”

http://intechemistry.files.wordpress.com/2010/09/cholesteryl-benzoate-synthesis-from-cholesterol-and-benzoyl-chloride.jpg

Hopefully you can how trivial it is, ‘cos it’s just the usual reaction between an alcohol and an acid chloride.

People are often scared in chemistry by the big names, Don’t let the ‘big names’ disguise the simple chemistry underneath.

34 | intechemistry
May 9, 2011 at 1:50 pm

I should also say, its the experimental procedures which are what the syllabus is emphasising at that point. The cholesteryl benzoate just happens to be an example.

36 | intechemistry
May 9, 2011 at 5:37 pm

I’m not sure what you will make of this, but here goes…

Definition: A chiral molecule cannot be superimposed upon it’s mirror image.

Definition: A reaction in which stereochemically different molecules react differently are called stereospecific reaction
Adaptation of definition given in Morrison and Boyd 6Edn p370

Huh?

Addition of Br2 to alkenes which are geometrical isomers (of more commonly called stereo isomers) of each other gives unique products. An example of a stereospecific reaction is given below:

e.g. cis-but-2-ene + Br2 -> racemic mixture of 2,3-dibromobutane. One isomer has R,R, configuration {a labelling method for identify the chiral centre produced}, the other isomer made has S,S configuration. i.e. we get 2(R),3(R)-dibromobutane and 2(S),3(S)-dibromobutane.

But

trans-but-2-ene + Br2 -> Only one product forms here and is called a meso-isomer. which is 2(R),3(S)-dibromobutane. This meso molecule does not rotate the plane of polarisation of monochromatic plane polarised light because the two chiral centres in the product from the transreaction cancel eachother out in the same molecule. Note 2(R),3(S)-dibromobutane is exactly the same as 2(S),3(R)-dibromobutane

From the cis- isomer, two products from the cis reaction don’t rotate the pop of mono ppl because two types of molecule are created and one type of molecule cancels out the effect of the other If you were to separate the racemic mixture (which contains two types of molecule) then the individual species, i.e. separate 2(R),3(R)-dibromobutane, and separated 2(S),3(S)-dibromobutane would rotate the pop of mono ppl.

But as the reaction product from the trans- reaction only contains one type of molecule, you can’t separate and so that molecule will never be optically active.

If you made models of these molecules, you would be able to see easily what I’m going on about.

I guess from your question that there may be some discussion of a chiral molecule that reacts differently from it’s stereoisomer (more specifically it’s optical isomer). This can happen. One optical isomer (say A1) can react in a different way with another optically active species (B1) , but the other optical isomer (A2) won’t react in the same way that with optically active species(B1). A classic example is enzymes. They react with different optical isomers in very different ways. The body’s enzymes react and process L-amino acids, but not D-amino acids.

If none of that makes sense to you don’t worry. I’m curious as to where you came across this.

38 | intechemistry
June 6, 2011 at 3:48 am

Xin Ling.
In general, anything that appeared in AS can be brought up in A2. Especially when combined with a chemicals mentioned in A2. E.g. Halogenoalkane as used in the Friedel-Crafts alkylation, but in my opinion, it’s very unlikely that you will get a ‘pure AS’ question coming up in A2.

HALOGENOALKANE: #1 and #2
Please read Q2) here. (http://intechemistry.wordpress.com/2011/04/20/even-more-qs-to-edexcel/)
Not too sure what you mean by this: RX + R’2NH —> RR’2NH + HX, but from you later Q, it’s perfectly valid to write the HX as reacting with the initial amine or the newly formed amine product if you wish. From what Rod Beavon said, a general equation showing a product based in the nucleophilic attack of N in the amine (or ammonia) and loss of HX will be good enough to gain credit. I would say however, that if the 1o amine was needed, then it’s important to state excess ammonia. If anything ‘special’ is needed in terms of the exact product formed, then that will be indicated in the question.

ACID CHLORIDES:
#1
Similar to what I said above, both equations are fine and would likely get credit as they show the correct understanding of reactants and products which can form. Once more, if anything ‘special’ is required e.g. the question states: “write down ALL species formed”, then you would either write eqn (2) by itself or eqn (1) accompanied with the separate HCl + NH3 —> NH4Cl reactions afterwards.
#2 Yes it will react further (although some of the HX(g) will probably escape also, so not not all of it will react and so you would get some amine salt formed also.

Overall, you can show the amine or ammonium salt being produced. Rod Beavon’s response makes me think Edexcel will not be too fussy here and accept a range of answers. The important thing seems to be that the correct reactants and a correct product are shown and exactly which product you should show is likely to be indicated in the Q.

Does that help?

40 | intechemistry
June 12, 2011 at 12:02 pm

Hi Emrys
Grignard reagents were taken out of the “New Syllabus”

44 | intechemistry
June 22, 2011 at 10:45 pm

@42, hidayah, June 22, 2011 at 8:25 pm

Salam Hidayah. As you have found out, for the benefit of all, I’ll say benzenediazonium chloride (BDC) can be converted easily into phenol by warming up the BDC with water. N2(g) will also be produced and HCl(aq).

Re: phenol from phenylamine [ i.e. from phenylamine,C6H5-NH2, to phenol,C6H5-OH ] oh, yes, you definitely need a reagent. to do it.

If you converted phenylamine to the dizonium salt first (using NaNO3 and c.HCl), and after did steam distillation, then that last stage, BDC to phenol wouldn’t require any reagents,

although it’s not familiar to me to have that BDC→phenol done in one step using steam distillation. I usually see BDC warmed in water, ut I guess it wouldn’t really matter if you did say do steam distillation on BDC.

It never occurred to me before . Perhaps your method is more efficient

46 | intechemistry
June 23, 2011 at 4:34 pm

There are ways to describe what’s called the absolute configuration of a chiral centre by putting the H atom down and at the back, with the three other atoms pointing towards you. From this you can easily tell if which enantiomer (R or S) the molecule is. These considerations don’t feature on the specification.

But it’s often necessary to draw the bonds coming from a chiral i.e. asymmetric carbon using the 3D bonds. Two bonds in the plane of the paper, one coming out and one going into the plane of the paper. Don’t have the bonds in the paper appearing as hough they are 180o apart. So the ‘rules’ at A-level are very simple.

There are other techniques for other chiral molecules too.

48 | intechemistry
June 23, 2011 at 8:38 pm

1. Having never tried the reaction myself, I am forced to look for references to it in books. According to A New Introduction to Organic Chemistry by G.I.Brown, ISBN 0 582 35128 6, (Pub: Longman) p205 it does. But no elaboration is made, only a small entry in a table. So it’s probably not a particularly good reaction. So does SOCl2. OH gets substituted and a Cl replaces it.

But the reverse reaction gets a lot more mentions, even though it’s harder for OH- nucleophiles to displace the Cl in C-Cl as the bond is stronger in halogenated aryl groups than in alkyl based (“straight” chain) halogenoalkanes, further indicating that phenol to chlorobenzene is a poor reaction.

for simplicity sake, Looks like it’s better to consider the reaction doesn’t take place.

——————–

2. As for ‘dry conditions’ it’s going to be because the water will react with something if present. H2O is a nucleophile and (very) weak acid. Electrophilic substitution often involves electrophiles. Water (a nucleophile) would react with the electrophile. I guess it’s possible that may be asked.

52 | intechemistry
June 24, 2011 at 11:28 am

#50, emrys ong, June 23, 2011 at 10:57 pm
Hydrogen bond is a special type of dipole, so is counted differently.
So say hydrogen bond and London dispersion forces for phenol.

#51
For lower melting point, impurities present in the lattice from forming properly. hence more energy (in the form of heat) must be taken out of the system to make the lattice form.

For higher bpt, a liquid begins to boil when it’s vapor pressure is the same as atmospheric pressure. Let me quote “A simple explanation of this is that the impurities dilute the concentration of water (the number of water molecules per unit volume decreases), and the number of molecules that can vaporize at any give temperature decreases. The result is that a higher temperature is required to achieve the same vapor pressure.” source

54 | intechemistry
June 24, 2011 at 4:07 pm

56 | intechemistry
January 10, 2012 at 12:00 pm

Hi Max.

#1 One important reaction of alcohols involves nucleophilic substitution of the OH group. A nucleophile attacks the C atom carrying the OH group because that C has become delta+ because the O withdraws the C-O bonds electron dessity onto itself and away from the C.
In phenol the ring withdraws e- density into the ring and hence rhe C carrying the O in is more electron rich, so changing the ability to undergo nucleophilic substitution.
That is one difference between them. There are others (e.g. acidity)

#2 I can’t say for certain what edexcel will penalise, but if you write a formula when a mechanism involving the ring is the focus of the question, then I guess that’s when they would deduct marks. In other cases – unless there are specific directions in the question – then it I guess it should by OK
Actually you can ask edexcel examiners questions on their website. They usually reply too

#3 Ka is the equilibrium constant for (generally speaking) weak acids. The greater the Ka, the greater the values of [H+] and [A-] hence the greater the Ka, the stronger the acid (usually RELATIVE strength of one weak acis to another weak acid, i.e. amongst the weak acids)
Kb has the same consideration, but for bases. So the higher the Kb the greater the base strength of the base (again, usually to establsh rank amongst the weak bases).
Edexcel don’t mention Kb on the specification, so no point in wasting time on it.

#4 Will answer later.

68 | intechemistry
January 28, 2012 at 2:27 am

#66, George, January 27, 2012 at 9:47 pm

The slide headings are the wrong way around.
So please go to “Aspirin (synthesis #2)” first then after that, go to “Aspirin (synthesis #1)”.

The correct sequence is CO2 added to phenoxide anion (basic conditions, high pressure, high temp) , then at the end acidified.
After, react acyl chloride with it and it reacts with the phenol based OH group.

The incorrect order was mentioned in class.

70 | intechemistry
February 16, 2012 at 1:31 am

Salam Shahrizan.

The reactants when initially being added together (in a reaction flask – round bottomed or pear-shaped flask) are mixed under cooling (usually using an ice bath). Once the reactants are fully mixed then the refluxing at 55oC (or 60oC) are done.

Generally, radicals are VERY reactive. So much so that they can force benzene can react with them. Remember hydrocarbons? Generally unreactive and contain strong bonds, but with Cl2 and UV light the Cl radical can force hydrocarbons to react and become functionalised.

Also, H2 can react with benzene under high temp and with a catalyst to form cyclohexane… the point being benzene can be forced to react in other ways other than electrophilic substitution.

76 | intechemistry
February 20, 2012 at 1:56 pm

One answer of A-level standard would be:
To ensure the phenylamine is not protonated which would make it ionic and very difficult to separate for the other dissolved ions present and would prevent steam distillation as the ionic compound would be much less volatile than the non-ionic phenylamine itself.