[ABA2015]

hi friends
I have synthesized 2-Aminobenzophenone with mentioned procedure below, when I wanted to recrystallize the product with boiling ethanol, there was some black participate which showed explosive manner in boiling ethanol but didn't solved even in boiling ethanol, can you suggest
1- what is that black participate?
and also
2-what is the role of acidic reflux in this procedure?
thanks

To 21.0mmol of the corresponding benzoyl chloride, 8.4mmol of p-chloroaniline
were added. Then the mixture was heated to about 180–200
C, 10.08mmol of anhydrous
ZnCl2 were added slowly, and the temperature was gradually increased to
220–230
C. The reaction mixture was maintained at reflux for 3 h. Then it was cooled
to 120
C, and hot water was added to remove the benzoic acid, giving a green-brown
product. The obtained green-brown product was added to a mixture of H2SO4,
CH3COOH, and H2O (5.0:3.5:3.0mL) and refluxed for 40 min. Then, about 20–25mL
of ice-water mixture was added, and the reaction mixture was subsequently extracted
using dichloromethane. The organic phase was washed three times using water and a
solution of 10% ammonium hydroxide to neutralize it. The organic phase was dried
with anhydrous sodium sulfate. Its vacuum evaporation yielded a brown-yellow solid,

[discodermolide]

The black precipitate may be finely divided zinc metal. That will react rapidly with ethanol giving H₂. I would suggest that the cid reflux at the end is to remove any zinc residues to make them water soluble?
The conditions you used seem very harsh with high temperatures involved, maybe there is a better way of doing this Friedel-Crafts?

[ABA2015]

thanks discodermolide, I have searched and I didn't find anything better but actually as you said, I think there must be some thing better, which I couldn't find anywhere even in reaxys database

[kamiyu]

I am sorry that I cannot agree with the suggestion that acid reflux is for removal of zinc:
First reason: removal of zinc can also be done much more simply by dissolving the organic stuff in DCM, followed by filtering.
Second reason: This is FC acylation reaction and I cannot think of any reason why the zinc chloride has been reduced to become metallic zinc

The black precipitate may be finely divided zinc metal. That will react rapidly with ethanol giving H₂. I would suggest that the cid reflux at the end is to remove any zinc residues to make them water soluble?
The conditions you used seem very harsh with high temperatures involved, maybe there is a better way of doing this Friedel-Crafts?

[kamiyu]

Indeed, I am thinking about why this reaction works. The reaction between NH2 and COCl is obviously much faster than nucleophilic attack from the ring electrons. This is sure, no need to argue.

I am still thinking about the whole reaction. Hope I can come up with an explanation..

[discodermolide]

I said zinc residues, not zinc will be removed.
The OP said some black particulate metal, in my opinion this can only be some form of zinc metal. Not much but enough to be highly reactive.

[kamiyu]

Referring to your "new" post:

hi all
I have tried to synthesis 2- aminobenzophenone via the following procedure
1-heated benzoyl chloride to 120oc
2-added p-chloro aniline slowly (in 4 steps)
3-increased temperature to 180oc
4-added ZnCl2 slowly in 4 steps
5- refluxed in 2000c for 2 h
6- waiting cooled to 120 and added plenty of boiled water to remove acid benzoic
7- in this stage I have obtained a black participate
8-I filtered the participate and refluxed with H2SO4:CH3CO2H:H2O (5:3.5:3mL) for 40 min
9- added ice water
10-washed aqueous layer with ethyl acetate
11- organic layer was washed with 50mL NaOH (10%)
12- organic layer was dried over NaSO4
13- evaporated with heater
14- a black and little amount of participate was obtained
the TLC showed benzoic acid+p-chloroaniline+2 spot in visible region+1 in UV
I couldn't receive the desired product with recrystallization and also I doubt that benzophenone has been synthesized even
I'm really confused and also I don't know what is the problem, pls help

I would like to add several points:

Firstly, check this synthesis is from single source or that many groups report the same synthesis. If this synthesis has been done by a single group, be alert to falsification (not uncommon that many procedures are indeed not reproducible, even those in Science and Nature)

Secondly, you cannot judge benzoic acid by TLC because it does not move at all. For spot like this, it may be salt or the initial decomposition of your product when you first spot the sample of the TLC.

[ABA2015]

Referring to your "new" post:
 
Firstly, check this synthesis is from single source or that many groups report the same synthesis. If this synthesis has been done by a single group, be alert to falsification (not uncommon that many procedures are indeed not reproducible, even those in Science and Nature)

yes I have found that in another paper the process of 10 and 11 are different, and they neutralize the acidic mixture and then they extract organic layer with an organic solvent.

[ABA2015]

also I have found these problems on starting material do you think they prevent the product formation
1-the TLC of acyl halide shows 3 spot which I guess are a- anidride b-benzoic acid c-acyl halide
2-ZnCl₂ is a little wet
also in the synthetic process I guess I should neutralize the acidic mixture and then extracting organic layer because I guess the product in acidic mixture forms salt and it remains in aqueous layer and maybe it decomposes to the starting material (the amine and benzoic acid)
what do you think am I right?

[kamiyu]

After about 10 minutes of deep thought, I think I have come up with the way how this reaction works. Actually this reaction is quite easy to understand. The following is the step-by-step explanation of mine:

1) "To 21.0mmol of the corresponding benzoyl chloride, 8.4mmol of p-chloroaniline
were added."

This is to react the amine with the acid chloride to form an amide first. The reason for this is to protect the amine group. Thus, the reagent acid chloride functions BOTH protecting group (one-pot protection/deprotection) and reactant.

2) "Then the mixture was heated to about 180–200
C, 10.08mmol of anhydrous
ZnCl2 were added slowly"

Zinc chloride is a well-known Lewis acid. In this case, it binds with the oxygen atom of the remaining acid chloride (pay attention to the equvialent of acid chloride used) to increase the reactivity of the acid chloride so the RING ELECTRONS can attack the acyl carbon

3) and the temperature was gradually increased to
220–230
C. The reaction mixture was maintained at reflux for 3 h.

This is to make sure that the nucleophlic attack from the ring is complete.

4) Then it was cooled to 120
C, and hot water was added to remove the benzoic acid, giving a green-brown product.

To remove "benzoic acid" is equivalent to saying remove excess acid chloride
The green product is the N-acylated AND ring acylated product.

5) The obtained green-brown product was added to a mixture of H2SO4,
CH3COOH, and H2O (5.0:3.5:3.0mL) and refluxed for 40 min.

Acid hydrolysis to turn amide back to amine (salt form)

6) Then, about 20–25mL of ice-water mixture was added, and the reaction mixture was subsequently extracted using dichloromethane. The organic phase was washed three times using water and a
solution of 10% ammonium hydroxide to neutralize it.

Note that the procedure use DCM to extract the product (salt form) from water. Probably this action minimizes the impurities extracted and the product, despite in salt form, has a largely organic part which makes it still soluble in DCM. The final neutralization by NH4OH is crucial.

[kamiyu]

also I have found these problems on starting material do you think they prevent the product formation
1-the TLC of acyl halide shows 3 spot which I guess are a- anidride b-benzoic acid c-acyl halide
2-ZnCl₂ is a little wet
also in the synthetic process I guess I should neutralize the acidic mixture and then extracting organic layer because I guess the product in acidic mixture forms salt and it remains in aqueous layer and maybe it decomposes to the starting material (the amine and benzoic acid)
what do you think am I right?

To your second question, I think a little wet ZnCl2 does not matter, considering the temperature when adding the ZnCl2 (>100oC) and a little bit water does not affect the reaction much.

I think the reaction is quite easy to understand and I think the procedures make sense. I suggest you follow the steps carefully and I wish you success.

[ABA2015]

thanks for all replies, I think the key of my problem is in considering that benzoyl chloride is a protecting group too (which kamiyu mentioned, and I did not pay attention to before) and also the acidic solution (H2SO4:CH3CO2H:H2O) seems is not efficient enough, because when I changed it to HCl (3N) and H2SO4 (70%) (mentioned in a patent and a paper in elsevier) I saw the green-brown solid which I should receive it after addition of hot water to the mixture of the reaction in the previous procedure.

but something interesting to me is after addition of HCl 3N (3 times for removing benzoic acid) and H2SO4 (70%) and refluxing for 2h, a yellow-brown solid was appeared  which was not soluble in such an acidic medium, but after adding NaOH to the solution the color of the participate was slowly changed to brown and it was dissolved slowly in aques medium, I guess this participate is my desired product of course in amine form (not salt or amide), am I right?

[kamiyu]

Have you managed to get your product??

[ABA2015]

No,  also I have borrowed the blank of the desired product and  added benzoyl chloride to prepare the blank of intermediate (the protected 2-aminobenzophenone with acyl) and compared to the reaction mixture after addition of ZnCl2 despite the spots of the reaction mixture are very near to each other but they seems different, the spot of reaction mixture is a little more polar than the blank of protected 2-aminobenzophenone and also the spot pervades (like benzaldehyde spot or the same compounds) but the blank's spot doesn't