H2 Chemistry: Conversion from Nitrobenzene to Phenylamine
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Originally posted by limywv:
Hi guys, I actually have a rather interesting question(well at least for me).
Here is what I understand: Hydrogenation (Ni catalyst, H2 gas, 200degC, 30atm) of Benzene gives a cyclic-hexane.
But then If lets say we were to convert Nitrobenzene into Phenylamine, would it be just another simple case of hydrogenation?
If it is just hydrogenation, would it also affect the aromaticity of the original benzene ring of phenylamine?
Another way to put it: would the process of hydrogenating Nitrobenzene into Phenylamine cause the benzene ring to become a cyclic hexane? Or would it just happen, without the latter occuring?
Hi limywv, are you in JC2 this year?Under extreme conditions (specifically, high pressures required eg. 30atm) that would complete reduce the benzene ring to a cycloalkane (destroying the aromaticity in the process), would also reduce the NO2 substituent to a NH2 substituent.
To answer your question "would the process of hydrogenating Nitrobenzene into Phenylamine cause the benzene ring to become a cyclic hexane? Or would it just happen, without the latter occuring?"
Yes, reduction (ie. addition of hydrogen) of nitrobenzene to phenylamine is selective, the conditions (specifically pressure) are milder and precise reagents required (eg. reflux with Sn/Zn/Fe and conc HCl in excess, or H2(g) and Ni catalyst at low pressure, eg. 1atm), and this reduction only affects the substituent (reducing NO2 to NH2), preserving the aromaticity of the benzene ring.
Any further question?
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Originally posted by limywv:
So what if I just want to convert nitrobenzene into phenylamine, without causing complete breakdown of aromaticity in the ring? Or is it too complicated, thus not in the H2 syllabus?
As I said earlier, you simply use milder (ie. lower pressure) and more precise conditions/reagents (eg. reflux with Sn/Zn/Fe and conc HCl in excess*, or H2(g) and Ni catalyst at low pressure, eg. 1atm), and this reduction only affects the substituent (reducing NO2 to NH2), preserving the aromaticity of the benzene ring.(*subsequently, use excess NaOH(aq) to convert the protonated phenylamine chloride into the desired phenylamine and sodium chloride byproduct. Separate the phenylamine using steam distillation.)
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Originally posted by limywv:
this fact can also be used to highlight the remarkable stability of the benzene ring?
The stability of the benzene ring is due to high resonance energy or stabilization energy (calculated as the difference between the actual energy of the real molecule (ie. the resonance hybrid) versus that of the singlemost stable resonance contributer; for benzene this is 36 kcal/mol).If you're asked "What observation highlights the exceptional stability of benzene and aromatic compounds?", the required answer (for the H2 syllabus) is "Benzene and aromatic compounds, although unsaturated, undergo electrophilic aromatic substitution instead of electrophilic addition reaction; this is due to the high resonance or stabilization energy due to the delocalized pi electrons within the benzene ring or aromatic compounds."
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Originally posted by limywv:
Alright, many thanks for your help!
Welcome!
PS.
Don't forget to attempt this absolutely delightful O.C. (Organic Chem) deductive / elucidation question!
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Originally posted by UltimaOnline:
The stability of the benzene ring is due to high resonance energy or stabilization energy (calculated as the difference between the actual energy of the real molecule (ie. the resonance hybrid) versus that of the singlemost stable resonance contributer; for benzene this is 36 kcal/mol).If you're asked "What observation highlights the exceptional stability of benzene and aromatic compounds?", the required answer (for the H2 syllabus) is "Benzene and aromatic compounds, although unsaturated, undergo electrophilic aromatic substitution instead of electrophilic addition reaction; this is due to the high resonance or stabilization energy due to the delocalized pi electrons within the benzene ring or aromatic compounds."
Aromacity increases the stability of the compound because the delocalised electrons means that there are no (less?) free electrons, not shared in the bonds, in the individuals atoms in the compound to repel each other, thus making the bonds between the atoms stronger and shorter, making the compound more stable. -
me no understand.
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Originally posted by yourmotherrr:
me no understand.
That's understandable.
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Originally posted by 16/f/lonely:
Aromacity increases the stability of the compound because the delocalised electrons means that there are no (less?) free electrons, not shared in the bonds, in the individuals atoms in the compound to repel each other, thus making the bonds between the atoms stronger and shorter, making the compound more stable.
limywv (and the H2 syllabus) was referring to stability of benzene in the sense of its lack of reactivity towards addition.What 16/f/lonely said, has some truth to it, but is not the full picture in and of itself, due to his lack of precise elaboration. He (not'16'/not'f'/not'lonely'NSF) said, making the bonds "shorter", shorter as compared to what? Note that the resonance hybrid has one-and-a-half bonds between the 6 C atoms, which are certainly 'shorter' than the single bonds of the alternating-single-and-double-bonds Kekule structure of benzene, but they are longer than the double bonds of said model.
For a proper look at the actual reason for the exceptional stability of benzene and all aromatic compounds, you need to understand Molecular Orbital theory (which is beyond the H2 and even H3 syllabuses).
However, in the context of stability of conjugate bases/acids to explain the relative strengths of various acids/bases (which is within the H2, H1 and H3 syllabuses), the JC student is expected to be able to explain using the concept of resonance (ie. delocalization of electrons throughout parts of the molecule), to explain why the negative/positive charge of the conjugate base/acid is stabilized/destabilized by delocalization (ie. resonance) over several electron donating/withdrawing electronegative/electropositive atoms in the molecule.
For instance, why carboxylic groups are acidic.
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Originally posted by UltimaOnline:
limywv (and the H2 syllabus) was referring to stability of benzene in the sense of its lack of reactivity towards addition.What 16/f/lonely said, has some truth to it, but is not the full picture in and of itself, due to his lack of precise elaboration. He (not'16'/not'f'/not'lonely'NSF) said, making the bonds "shorter", shorter as compared to what? Note that the resonance hybrid has one-and-a-half bonds between the 6 C atoms, which are certainly 'shorter' than the single bonds of the alternating-single-and-double-bonds Kekule structure of benzene, but they are longer than the double bonds of said model.
For a proper look at the actual reason for the exceptional stability of benzene and all aromatic compounds, you need to understand Molecular Orbital theory (which is beyond the H2 and even H3 syllabuses).
However, in the context of stability of conjugate bases/acids to explain the relative strengths of various acids/bases (which is within the H2, H1 and H3 syllabuses), the JC student is expected to be able to explain using the concept of resonance (ie. delocalization of electrons throughout parts of the molecule), to explain why the negative/positive charge of the conjugate base/acid is stabilized/destabilized by delocalization (ie. resonance) over several electron donating/withdrawing electronegative/electropositive atoms in the molecule.
For instance, why carboxylic groups are acidic.
Sorry teacher, for the incomplete answer. I understand why my answer has to have 2 marks deducted from the score.
At JC level there's no requirement beyond knowing the standard answer (I forgot what) if they ask why benzene is more resistent to additive reactions.
But if you truly want to understand better and hence not need to learn by rote learning, you should try to find out why instead of just accepting the answer!
I disliked chem....I admit, because I started out with rote learning. After a while, I tried to understand the certain properties that exist and after a while things became clearer. But I still hated questions that involved calculators.
My answer is incomplete because I had no requirement to learn why the answer to the question is so. But it gave me a better understanding of what goes on in chemistry.
Notice chemistry draws many parallels with physics?
Needless to say, I liked physics very much. Ironically my chemistry ended up scoring better than my physics which, by unfortunate circumstances (stomachache during exam -.-) somehow became my weakest link at the final test.
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>>> But if you truly want to understand better and hence not need to learn by rote learning, you should try to find out why instead of just accepting the answer! <<<
Precisely the reason why 'A' level students would want to attend my tuition. JC teachers do not properly explain the H2 Chemistry concepts (simply because they don't have the luxury of time), they have little choice but to instruct the students to "just memorize the notes we give you, don't ask so much. Priority is for you to pass."
Whereas in my tuition, I will guide you to understand first, and then to enjoy, and when you have these two, the last step, memorization, becomes effortless.
One example of evidence for this, is that all my students are able to draw (and enjoy drawing) all the required mechanisms in the H2 Organic Chemistry syllabus, without any memorization at all. Which is in stark contrast to the standard JC approach of "just memorize the mechanisms, you don't need to understand why."So if in 2009, you're in JC1 or JC2, taking H2 or H1 Chemistry, and you're not satisfied with the limited explanations from your JC teacher on Chemistry concepts, you may consider getting a couple of friends to form a group and come for my tuition.
Happy New Year to (not'16'/not'f'/not'lonely'NSF) and everyone else reading this!
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Wah, someone's got the makings of becoming the CEO of a MNC offering tutoring services.