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disproportionation ....

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  • hoay's Avatar
    189 posts since Jul '11
    • In C6H6 the oxidatio number of carbon is -1. In chlorobenzene the oxidation state of Cl is +1 as calculated by the rules of oxidation numbers. 

      The electrophilic substiution of benzene with chorine to produce chlorobenzene and hydrogen chloride can also be described as Dispropportionation reaction. Is it correct to think like that? 

      Like wise in case of nitrobenzene the oxidation number of N is +5. When it is reduced N changes to -1.    

      Why chlorine is +1 in chlorobeznene. Please explain.

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,009 posts since May '05
    • Originally posted by hoay:

      In C6H6 the oxidatio number of carbon is -1. In chlorobenzene the oxidation state of Cl is +1 as calculated by the rules of oxidation numbers. 

      The electrophilic substiution of benzene with chorine to produce chlorobenzene and hydrogen chloride can also be described as Dispropportionation reaction. Is it correct to think like that? 

      Like wise in case of nitrobenzene the oxidation number of N is +5. When it is reduced N changes to -1.    

      Why chlorine is +1 in chlorobeznene. Please explain.


      No, halogenation via EAS is not disproportionation, as Cl is only reduced (OS of the Cl atom decreases from 0 to -1) and C is only oxidized (OS of the C atom bonded to Cl increases from -1 to +1, the OS of the other C atoms on the benzene ring remain -1).

      The clearest instance of disproportionation in Org Chem would be the Cannizzaro reaction.

      When nitrobenzene is reduced to phenylamine, the OS of N decreases from +3 to -3.

      "...as calculated by the rules of oxidation numbers..."

      Those 'rules' are merely simplified guidelines that are of limited use, and should only be applied at O levels.

      The quirkiness about A levels, is that it's caught awkwardly in-between overly-simplified O levels, and the real / correct Uni level Chemistry.

      The overly-simplified OS / ON 'rules' will fail when applied to Org Chem and complicated species in Inorg Chem.

      To correctly determine the correct OS / ON of an atom, you'll have to consider the structure of the species containing the atom, formal charges and electronegativities, as well as resonance contributors and resonance hybrids (where applicable).

      For Org Chem, Cambridge requires the A level student to be familiar with the use of Functional Group Level.

      Edited by UltimaOnline 17 Jan `18, 4:06AM
  • hoay's Avatar
    189 posts since Jul '11
    • the ionic equation for the reduction of nitrobenzene to give phenylamine shows 6e being gained. But you just told that the OS of N changes from +5 to -3 mans it gains 8e. 

      The ionic equation is given in CIE MS June 2016. Refer to Q.6(a)(i). The link is below:

      http://pastpapers.papacambridge.com/Cambridge%20International%20Examinations%20(CIE)/AS%20and%20A%20Level/Chemistry%20(9701)/2016%20Jun/9701_s16_ms_41.pdf

      Based on your explanation that the N's OS in phenylamine is -3 make me to deduce that the OS of the carbon bonded to Nitrogen is +6. carbon loses 2e AND hydrogen loses 6e so a total of 8e BUt the MS shows only 6e that are coming from Hydrogens. 

  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,009 posts since May '05
    • You caught my error, my bad, and thanks.

      Indeed, it's always good to check the balancing of redox equations using both charges, and OSes, they will always match if done correctly.

      When nitrobenzene is reduced to phenylamine, the OS of N decreases from +3 to -3, hence 6 e- is involved in the reduction half-equation.

      Originally posted by hoay :

      the ionic equation for the reduction of nitrobenzene to give phenylamine shows 6e being gained. But you just told that the OS of N changes from +5 to -3 mans it gains 8e. 

      The ionic equation is given in CIE MS June 2016. Refer to Q.6(a)(i). The link is below:

      http://pastpapers.papacambridge.com/Cambridge%20International%20Examinations%20(CIE)/AS%20and%20A%20Level/Chemistry%20(9701)/2016%20Jun/9701_s16_ms_41.pdf

      Based on your explanation that the N's OS in phenylamine is -3 make me to deduce that the OS of the carbon bonded to Nitrogen is +6. carbon loses 2e AND hydrogen loses 6e so a total of 8e BUt the MS shows only 6e that are coming from Hydrogens. 


  • hoay's Avatar
    189 posts since Jul '11
    • the final picture is now:

       

      the carbon bonded to N in both nitrobeznene and phenylamime is +1.

      the carbon bonded to N in both these neither lose or gain electrons.

      All Other Carbons in both these have OS of -1.

       

      Consequently the OS of C bonded to C of CH3 is zero as C-C bond in non-polar but the C of CH3 is bonded to H which makes it -3.

      Edited by hoay 17 Jan `18, 4:34AM
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,009 posts since May '05
    • Originally posted by hoay:

      the final picture is now:

       

      the carbon bonded to N in both nitrobeznene and phenylamime is +1.

      the carbon bonded to N in both these neither lose or gain electrons.

      All Other Carbons in both these have OS of -1.

       


      Yes, exactly correct.

      And the 6 electrons come from the reducing agent involved, for this reaction, can be H2(g)/catalyst or Fe(s)/H+ or Sn2+(aq) or Zn(s), etc. LiAlH4 is not recommended as azo compounds are the major products.

  • hoay's Avatar
    189 posts since Jul '11
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,009 posts since May '05
    • Originally posted by hoay:

      what about the side-chain carbon in themethylbenzene?? it has -3 charge.


      For the reaction of KMnO4 on methylbenzene, the OS / ON (and not formal / ionic charge) of the C atom in the methyl side-chain is -3, and KMnO4 will oxidize methylbenzene to benzoic acid, with the COOH group's C atom now having an OS of +3 (in the major resonance contributor, there are resonance contributors in which the OS of the COOH group's C atom is +2, which does not contradict the 6e- released in the oxidation half-equation, because for those contributors, a ortho/para C atom has OS of 0 instead of -1, ie. total electrons released is still 6e- ; the resonance hybrid follows accordingly with average OSes).

      Edited by UltimaOnline 17 Jan `18, 4:53AM
  • hoay's Avatar
    189 posts since Jul '11
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,009 posts since May '05
    • Originally posted by hoay:

      Thank you for your prompt answers.

      God bless you. 


      No prob, Hoay! :)

      Ok, I'm gonna sleep now... it's 5am here in Singapore...

  • hoay's Avatar
    189 posts since Jul '11
    • me too...It's 2.05 am in Karachi, Pakistan.

       

      Kindly can you explain why disproportionation occurs? with refernce to the metla ions such as Cu+ and Cl2 with NaOH.

      Edited by hoay 17 Jan `18, 5:11AM
  • Moderator
    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
    UltimaOnline's Avatar
    15,009 posts since May '05
    • Originally posted by hoay:

      me too...It's 2.05 am in Karachi, Pakistan.

       

      Kindly can you explain why disproportionation occurs? with refernce to the metla ions such as Cu+ and Cl2 with NaOH.


      Ok, brief answer to this last question before I log off...

      For copper, both Cu(s) and Cu2+(aq) are more thermodynamically stable under standard conditions, hence the position of equilibrium and the Gibbs free energy change favors the disproportionation reaction.

      For chlorine, under alkaline conditions disproportionation is favored, while under acidic conditions comproportionation is favored. This is due to the nucleophilic attack of the OH- Lewis base unto the Cl2(aq) molecule under alkaline conditions. Under acidic conditions, the thermodynamic driving factors are Le Chatelier's principle together with the positive entropy change of Cl2(aq) to Cl2(g), made possible by the double protonation of the ClO- Bronsted-Lowry base followed by nucleophilic attack from the Cl- Lewis base.

      Edited by UltimaOnline 17 Jan `18, 5:48AM
  • Ninipi's Avatar
    2 posts since Feb '18
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