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Hey can anyone help me give a good explanation why phenol, methly etc. groups act as ortho para directing groups. I really want to know the explanation and haven't found a good one as of yet...Thanks.

Have a look at this picture.



Basically, the oxygen atom has two lone pairs. One of them can go towards forming a O=Arene double bond. This produces several resonance forms.

Do you see the location of the negative charges in the resonance forms ? They are ortho and para to the phenol group.

Hence, the electrophile (whatever it maybe) will attack at these locations of high negative charge.

The case with a methyl group is a bit more complicated. With a methyl group we have no obvious source of electrons but what is in fact happening is that the electrons in one of the sigma C-H bonds in the methyl are overlapping with a p orbital on the C atom in the CH3 group producing a sort of weak pi bond.... From here you can see how we get the above resonance forms once more.

Hyperconjugation is the name of this effect :

http://en.wikipedia.org/wiki/Hyperconjugation

Is hyperconjugtion the same as sigma bond conjugation

Ah ok i see, i never actually got to learn that. Thanks a lot for tr explanation! Hopefully may see you at imperial next year if i do well in the interciew!

Is hyperconjugtion the same as sigma bond conjugation


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Help me with this question please.When solid iron(III) oxide reacts with hydrochloric acid, iron(III) chloride solution is formed with water.
Balanced symbol eq: 6HCL + Fe2O3 = 2FeCl3 + 3H2O

I know iron oxide is a base so its obv a neutralisation reaction. But where in this reaction is the H+ + the OH- to give it the name of a neutralisation reaction.

I took a guess saying the Fe initially reacts with the acid to form iron chloride solution and then it gives out hydrogen and the oxygen from the iron oxide reacts with the hydrogen in the acid to form hydroxide which then creates the neutralisation. Am I right or am I just completely wrong?

Well, the Fe ³⁺ and Cl- ions are essentially spectator ions.

Hence, the entire equation could be reduced to the general ionic equation for an acid base reaction :

2H⁺ + O^2- ----> H₂O

Hence, neutralisation.

EDIT : The point I'm trying to make is that you dont necessarily have to have a H+ and OH- ion interacting to have a neutralisation reaction. A simple example would be the reaction of NH3 gas with HCl gas. You get NH4Cl.

This is still a neutralisation reaction even though there are no OH- ions.

I have a question in terms of Isotopes: Is it possible that Isotopes have a higher attractive force? then the centrifugal force of electrons must be weaker. That is to say that the electrons are closer to the nucleus than atoms. In other words: the atomic radius is lesser in an Isotope than in an atom. Am I right with my considerations?

Think about it carefully. What is the source of the electron-nucleus attraction? Does this change between isotopes?

Let me see... the attraction force must be higher, if the nucleus has more protons than electrons on electron shell. Then the centrifugal force must be weaker. But that is not possible. Isotopes have more neutrons than protons and not vice versa. So the attraction force in an Isotope must be the same compare to an atom. That is to say the atomic radius is not lesser in an Isotope than in an atom. That's why I don't think so.

Perhaps there are atoms which have more protons than neutrons in an nucleus?

Let me see... the attraction force must be higher, if the nucleus has more protons than electrons on electron shell. Then the centrifugal force must be weaker. But that is not possible. Isotopes have more neutrons than protons and not vice versa. So the attraction force in an Isotope must be the same compare to an atom. That is to say the atomic radius is not lesser in an Isotope than in an atom. That's why I don't think so.

Perhaps there are atoms which have more protons than neutrons in an nucleus?

Hey Guys, I really need some help with infra red spectra
Can anyone help me ?

Thanks very much guys, and happy new year and i wish everyone has a successful 2013

And I love chemistry, because its more complicated than women haha

x

What is the charge on a neutron ?

As far as I know the charge of a neutron is neutral, whereas the charge of protons is positive and the one of electrons is negative. In terms of attraction force and centrifugal force neutrons have no influence on electrons due to the charge. So I think that the changing of number has no effects on electrostatic attraction, as neutrons neither positive nor negative. That is to say that neutrons are not able to repel or to attract electrons, even if the number of neutrons is greater in an isotope. That's why electrons will not come closer to the nucleus in an isotope. From this perspective and my considerations before I'm sure that isotopes are not able to reduce the atomic radius.

Well, in the first one, you have the nice shoulder peaks just next to the C-H stretches at 2900 and then there's the sharp absorption at 1600. I'm guessing this is a very long chain aldehyde ?

IR on its own is not really that diagnostic. You can't tell all that much from only an IR spectrum.



I don't think the first one has a carbonyl. The strech at ~1450 is too low for that. I reckon it's just a hydrocarbon of some sort, note the similarity of the spectrum to that of nujol. You also can't really tell the length of the hydrocarbon chain with any reliability from IR.



Yeah presence of C-H single bonds and at least one C-H double bond. Could be aromatic though, but I would expect an alkene. Would need NMR or something else to be sure.



Yeh an alcohol.



Be careful here. These streches can be caused by alkenes as well, and the fingerprint region is not very diagnostic. While you may be correct in this case, I think you need to be careful about trying to take too much information from an IR spectrum. However, the spectrum is very similar to that of benzaldehyde so you may well be correct.



You missed the presence of a carbonyl as well, though I don't think that it is part of an amide as the C=O strech is too high.



Agreed.

Overall, you're generally correct, but just be careful as sometimes I think you're coming to conclusions more detailed than you should really be based solely on these spectra.