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Anilinium hydrochloride intermediate formation

  1. Apr 11, 2008 #1


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    Acetanilide synthesis
    HCl, aniline, sodium acetate trihydrate=> Acetanilide

    I'm being told that there is an intermediate formed (anilinium hydrochloride) but I can't figure out how it is formed exactly. (Is the reaction, for the anilinium hydrochloride just the aniline attacking a H-Cl thus forming like this ? C6H5NH2 + HCl → C6H5N+H3Cl- but I still can't figure out how this is an intermediate to form acetanilide)

    I know that: supposively electron pair of NH2 group attacks the O^- when the molecule reacts to form bromoacetanilide but not sure as to the formation of aniliinium hydrochloride I' confused.
  2. jcsd
  3. Apr 11, 2008 #2


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    Yes. It is an acid-base reaction but you should represent it as:

    [tex]C_6H_5NH_2 + HCl \rightarrow C_6H_5NH_3^+Cl^-[/tex]
  4. Apr 11, 2008 #3


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    so it's a side reaction then from the main one that I want, not an intermediate ?

    If it is an intermediate I don't see how when I add acetic anhydride, it would react since the way I thought it reacted was that the electrons of the aniline would attack the possitive carbon of the acetic anhydride (after the electrons of one of the double bonds go to the oxygen giving it a negative charge)

    Thanks chemisttree
  5. Apr 11, 2008 #4


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    But if this is an intermediate I don't understand what would happen. (how the aniline would attack the acetic anhydride or if it's the other way around.
  6. Apr 12, 2008 #5
    Christina , (C6H5NH3+Cl-) would be a mutually stable complex compound i don't think actully it would need another step to goes on, unless you run this reaction under a freezing degree of T= 0 C, with aid of HNO2 you will get then a completely spearated diazonium chloride which is ( C6H5N2+Cl), so this compund becomes absolutely stabilizing cpd., but if you put special conditions as i mentioned before it would go for the diazonium chloride formation..
  7. Apr 12, 2008 #6


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    but the point of the reaction is to produce acetanilde and after I added the acetic anyhydride to the anilinium hydrochloride, to the reaction, stirred, and then cooled to [tex]5^oC[/tex] with stirring, acetanilide was formed so I'm still not sure how the reaction would have this compund

    http://img402.imageshack.us/img402/5952/200pxacetanilidemu4.png [Broken]
    Last edited by a moderator: May 3, 2017
  8. Apr 13, 2008 #7
    yeah..thats really weird..hmmmm,i will ask my organic chemistry professor about this i have a lecture after couple hours let see what she is going to say about that....
  9. Apr 13, 2008 #8


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    great, I want to see what you find out.:cool:

    and if anyone else can answer this puzzle I'd appreciate it.:biggrin:
  10. Apr 13, 2008 #9
    hi..well ahhhh i asked her.. there was a BIG misunderstanding in this simple equation..and for coincidence by the way we took this reaction today in the lecture :D

    well Acetanilide that you wish to obtain needs 'Acyl Halide'(CH3COCl) not Hydrocloride(HCl) like you mentioned before..and the reaction should goes like this in order to obtain Acetanilide;

    C6H5-NH2 + CH3COCl ------> C6H5-NHCOCH3 which is Acetanilide,

    so this should bring you to understand that the reaction you mentiond above about the rx. of Aniline with HCl would not absolutely results Acetanilide, it should give as chemisttree and i mentioned above the Anilinium Hydrochloride or 'Aniline Hydrochloride'.
    thats if you want to run this reaction by 'one step formation reaction',..i mean if you want to obtain Acetanilide from Aniline this is the only step you should take over..but if you mean anything else the "chemistry converting puzzle"(obtaining Acetanilide from Anillinium Hydrochloride) it would certainly differ..we should first eliminate the (-NH3+Cl-) group from the benzene ring..by a very strong agent and under extreme conditions..then from benzene we will obtain the Aniline easily from sand-mayer rx, then finally we would reach the final normal step of Acetanilide formation that i told you about it (rx with Acetyl Chloride).
    that were all i could understand and this is what i think it should go like..any inquiries i'd be so glad to listen to them..
  11. Apr 13, 2008 #10
    what is the EXACT procedure you are using? It would be much easier to figure out what you are doing if you list the exact procedure that you are using to make acetanilide.
    Last edited: Apr 13, 2008
  12. Apr 13, 2008 #11


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    1. add 33ml of 0.4N HCl then add 1.2ml of aniline to an erlemeyer flask with stirbar. Stir to effect solution. Warm to 50C
    2. prepare solution containing 2g of sodium acetate trihydrate dissolve in 7ml of water
    3. measure out 1.5ml of acetic anhydride
    4. add acetic anhydride in one portion to the warm solution of anilinium hydrochloride. Stir vigourously and add the sodium acetate immediately and in one portion. Cool reaction to 5C in ice bat anhd continue to stir until crystaline product completely precipiates. Collect crystals of acetanilide by vacuum filtration and air dry
  13. Apr 13, 2008 #12

    To me it seems that the only reason you are making the HCl salt of aniline is so it will be soluble in H20. Your final product is not soluble in water so it will be easy to filter out. Once you make the HCl salt you are basifying the solution with sodium acetate. Once basified, the aniline will react with the acetic anhydride. The HCl salt is just so you can get aniline to dissolve in water.
    Last edited: Apr 13, 2008
  14. Apr 13, 2008 #13


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    hm...okay so is the mechanism different than I thought it was ?

    [tex]C_6H_5NH_2 + HCl \rightarrow C_6H_5NH_3^+Cl^-[/tex]

    and then

    how does the hydrogen leave? does it just leave as HCl again?
    and thus [tex]C_6H_5NH_2[/tex] is reformed?

    because I originally thought that the mechanism of the aniline attacking the acetic anhydride would be the double bond electrons on the acetic anhydride (one side) would go to the oxygen [tex]O^-[/tex] so then the aniline's electron pair would attack the carbon that was +.

    how would the sodium acetate react by basifying the reaction?
    I assume it would take something as a base like an H so would it be from HCl or the anilinium ion, thus forming the aniline right?

  15. Apr 13, 2008 #14
    Precisely, that is the only mechanism you have to worry about. The HCl is just so you can get the aniline to go into solution since HCl salts are usually very water soluble. Acetate is an anion, so it is a base. You are just free basing your aniline HCl salt so that it will react with the acetic anhydride. Hydrochloride salts don't exist in basic conditions.

    That is why you have to cool the flask down before you add the acetate. When you mix acids and bases you are going to have some heat.

    If you just throw in aniline into water along with acetic anhydride you might not get much of a reaction at all because your aniline will float on top or form an emulsion if you are stirring vigoursly. In order to get a good yield, 90% of the time you want your starting materials to be soluble in the solvents you are working in.
    Last edited: Apr 13, 2008
  16. Apr 13, 2008 #15


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    Sure is but I'd still like to know what happens.

    so everything else I said was incorrect?

    if it is a base it should react by accepting a proton (which I think it does but not sure from what molecule(s)) or electrons. Since you said it was an acid base reaction (it forms acetic acid), then it must get the H from somewhere. (from the HCl I'm assuming)
  17. Apr 13, 2008 #16
    Right. The acetate anion is going to suck up the protons from the HCl of the aniline salt giving you acetic acid and aniline again.

    All you are doing is this:

    aniline-------> aniline HCl salt----------> aniline-----------> final product

    It seems silly, but you have to make the HCl salt first to simply get your material to dissolve in water.
    Last edited: Apr 13, 2008
  18. Apr 14, 2008 #17


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    okay...I was curious about that.

    I don't think it's silly if that's what needs to be done to form the product, because technically if you were just looking at a reaction equation it wouldn't tell you these things which I find is quite necessary when you actually want to synthesize compounds in the lab.

    Thanks gravenewworld
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