A level Chem P3 organic qn

• contemporarydancer

  8 Nov 08, 13:09

  I know it is over but there's this irritating question that is quite wierd to answer.
  
  It is Qn 5 where they ask how to convert a compound similar to phenol into 3-amine,4-nitrophenol in 2 steps?
  
  It is not possible right unless...??

  My guess was to
  
  1) do an electrophilic substitution such that you get 2 nitro groups substituted
  2) followed by, Sn/conc. HCl/ reflux under some kind of controlled conditions so that one nitro group can reduced but not the other?
  
  My main doubt is how is that possible? one nitro group becomes amine but not the other?

   

  
  

• UltimaOnline

  8 Nov 08, 13:59

  I didn't see the actual question, and you (or someone else who sat for the exam) will have to give me the exact compounds ("a compound similar to phenol"?) and exact question for me to give the solution.

   

  Offhand, I'll point out the following :

   

  1) OH (substituent on a benzene ring) has a lone pair available for delocalization and is hence electron donating by resonance and is therefore ortho and para directing.

   

  2) NO2 (substituent on a benzene ring) has a +ve formal charge and is hence electron withdrawing by induction, and is therefore meta directing.

   

  3) NH2 (substituent on a benzene ring), on its own (assuming it has already been reduced by tin with conc HCl from NO2 to NH2), has a lone pair available for delocalization and is hence electron donating by resonance and is therefore ortho and para directing.

   

  4) Consider steric effects too (para position is favoured over ortho positions due to steric hinderance).

   

   

  As for your line of thought regarding protecting one group while reacting another similar or identical group; this is an organic chemistry synthesis strategy of converting/replacing one group (to be protected) to a temporary inert/non-reactive (protected) group, while you carry out the reaction for the non-protected similar group to become the desired group, afterwhich you re-convert the inert/non-reactive (protected) group back into its original group.

   

  Although such synthesis strategies are not usually taught in any significant depth by the JCs at H2 level and it may not be relevant to this particular question.

   

  Perhaps lets wait for the exact question (and its exact compounds... what exactly is the "compound similar to phenol"?) to be posted by someone, or published in next year's TYS?

   

• contemporarydancer

  8 Nov 08, 17:40

  Sorry forgot to attach the question.
  
  
  

• arigatoast

  8 Nov 08, 20:49

  Originally posted by contemporarydancer:

  Sorry forgot to attach the question.
  
  
  

  hi.

  1st step :the electrophilic addtion of 2 mols of NO2+ electrophile for 1 mol of R like what u have suggested.

  2nd step : For both NO2 to get reduce, u need 2 mols of H2 . so to get T, u need only   1 mol of H2 for 1 mol of S. i think need to state the mol ratio. Thats what i think.

• rs rs

  8 Nov 08, 21:26

  sorry i cant find my paper now. but anyone knows how to do question 2 last last part? the one on isomerism.

• secretliker

  8 Nov 08, 22:28

  2(g) In another separate experiment, a new alkane F, C6H14, was produced. When reacted with bromine under ultraviolet light, F produced only two isomeric monobromo compounds G and H, with the formula C6H13Br. Compound H was chiral. Suggest the structures of F, G and H, explaining your reasoning. [3]

• UltimaOnline

  8 Nov 08, 23:10

  

   

  Ok, now with the question, here we go.

   

  R is an ether (R-O-R).

   

  The OCH2CH2CH3 substituent is electron donating by both resonance and induction, hence is ortho and para directing.

   

  arigatoast suggested :

  >>> 2nd step : For both NO2 to get reduce, u need 2 mols of H2 . so to get T, u need only   1 mol of H2 for 1 mol of S. i think need to state the mol ratio. Thats what i think. <<<

   

  That's smart thinking; although in practice, instead of uniformly obtaining only one product of 5-nitro-2-propoxyaniline, you would end up with a mixture of several different products (eg. in some molecules, both nitro groups would be reduced; in some other molecules, neither of the nitro groups would be reduced; in yet other molecules, the para instead of ortho nitro group would be reduced) in other words, a horribly low yield).

   

   

  Which is why, while in theory it seems very easy : just add aqueous HNO3, then reduce one of the nitro groups to amine group (by using Sn and conc HCl).

   

  But in practice, here are two serious problems :

  1) When using the aqueous HNO3, how can the chemist ensure only di-nitration occurs, instead of mono or tri nitration? (the propoxy group directs the incoming nitro electrophile to the ortho and para positions).

  2) When carrying out reduction, how can the chemist ensure only one of the two nitro groups (and furthemore, to be precise, the nitro group ortho to the ethoxy group)  gets reduced to amine?

  I would think the yield would be too low for such a synthesis pathway to be viable.

   

  However, since this compound is a commercially synthesized artificial sweetener (albeit a toxic one), there should be a better synthesis strategy (which will require a lot more than 2 steps) for this compound, and hence, it is possible that this 2008 H2 SEAB-Cambridge exam question is not a fair one (ie. it probably isn't a fair question if the only 2-step synthesis possible, gives too low a yield to be viable).
   

   

  Ah well.

• Chocoholicxx

  8 Nov 08, 23:21

  nitro benzene doesn't get reduced by h2. u nid con hcl and Sn. i tink it shud b due to stearic hindrance that only one nitro grp gets reduced.

   

  2(g)
  
  
             CH3   H   
              |      |
  F : H3C-C  -  C -  CH3
              |       |  
              CH3  H
  
  den i tin from there u shud know wad G and H is (:  

• Chocoholicxx

  8 Nov 08, 23:27

  nitro benzene doesn't get reduced by h2. u nid con hcl and Sn. i tink it shud b due to stearic hindrance that only one nitro grp gets reduced.**

   

  im not sure bout this :p
  
  but i really regretted doing qns 2.. made me so depress after that .. no mood to study for the maths paper 2 haha..

• UltimaOnline

  8 Nov 08, 23:30
• 2008 'A' Level Qn (posted by secretliker)

>>> 2(g) In another separate experiment, a new alkane F, C6H14, was produced. When reacted with bromine under ultraviolet light, F produced only two isomeric monobromo compounds G and H, with the formula C6H13Br. Compound H was chiral. Suggest the structures of F, G and H, explaining your reasoning. <<<

 

 

The only C6H14 alkane that when monobrominated would yield only 2 isomers one of which is a chiral compound, is 2,3-dimethylbutane.

 

Notice that 2,3-dimethylbutane has a mirror plane of symmetry. Substituting away H on the 4 terminal C atoms (1st, 4th, 2 methyl and 3 methyl carbons) are equivalent. That leaves the 2nd and 3rd C atoms, which are also equivalent thanks to this mirror plane.

 

To obtain the chiral compound, substitute away H on (any of the 4) terminal carbons. The 2nd (or 3rd) C atom would then be the chiral carbon.

 

Drink more chocolate.