Organic Synthesis - Science topic

Dear Van-Sieu Luc, I have performed many McMurry Reactions however we used actvated Zn dust to reduce TiCl₄ to Ti(0). Prior to reaction, we have purified commercial Zn dust as follows: Zn dust was quickly washed with distilled water containing few drops of conc. HCl (0.5-1% HCl can be used/ I think it's in Vogel's Practical Organic Chemistry book) to remove the oxide coatings in any. Then thoroughly washed with distilled water to make acid free and finally with acetone and dried under vaccum. It can be stored. This Zn dust can be used for reduction of TiCl₄ to Ti (0) in anhydrous dimethoxymethane solvent under reflux under inert atmosphere for 4-5 hours. We used 0.025 gm atom of Zinc for reduction of 0.0126 moles of TiCl₄ and this insitu generated Ti(0) was used for McMurry coupling of 0.003 mole of beta chloro-alpha,beta-unsaturated aldehydes. Yields varied in between 50-92 %. You are welcome to go through the paper "STEREOSELECTIVE SYNTHESIS OF 1,6-DICHLORO-1,3,5-HEXATRIENE DERIVATIVES BY MCMURRY COUPLING

OF beta-CHLOROACRYLALDEHYDE DERIVATIVES" Susmita Gupta, Gandhi K. Kar and Jayanta K. Ray, SYNTHETIC COMMUNICATIONS, 30( 13), 2393-2399 (2000).

Dear Siddhant Singh, in order to get an answer compare their 'standard reduction potential'. Please have a look at the following video. My Regards

Aniketh, This is not true. Chlorine is strongly accepted to the benzene ring.

I have the same problem as you

Thanks a lot @Vladimir Khlebnikov

The question is if product is insoluble in DMF/ DMSO, how to remove Succinimide from the reaction mass.

Also check https://www.organic-chemistry.org/chemicals/oxidations/n-methylmorpholine-n-oxide-nmo.shtm

https://pubs.acs.org/doi/10.1021/acs.chemrev.6b00512

The difference between malonyl and 2,2-dimethylmalonyl dichlorides is that the former possesses hydrogens in the alfa position to the carbonyl group. What often happens with acyl chlorides of this sort in the presence of bases is the formation of ketenes by abstraction of HCl by the base. The stronger the base, the greater the extent. ketenes are capable of acylation, but prone to the formation of other products. Phenylenediamine, even with Boc-protection is not an easy compound either. You judgement was quite correct when you were adding the dichloride dropwise at -20oC. Pyridine as a base is a good choice, too. It is likely that you generated a good yield of the desired malonyl diamide, but some colored impurities are hardly avoidable when you have an acyl chloride as reactive as the one you use. If these impurities are critical or the yield is too low, you may wish to react malonyl dichloride with N-hydroxysuccinimide. Once bis(N-oxysuccinimidyl) malonate, a much milder acylation agent, is formed in situ, it can be reacted with mono-Boc phenylenediamine. Good luck with your chemistry!

Nalidixic acid - A NSAID grs of drugs or pain killer .REF : en.wikipedia.org ( Image Source ) .

Dear Malar vizhi J.M functional groups in 2-bromo-4,5-dimethoxycinnamic acid are principally all substituents other than hydrogen atoms. First you have the bromo substituents, which can be substituted by other moieties. Other functional groups are the two methoxy (–OMe) groups and the carboxylic acid (–COOH) functional group. Finally, the -CH=CH- double bond can also be seen as a functional group as you can do derivative chemistry with it (such as adding Br₂). I hope this answers your question. Good luck with your work and best wishes, Frank Edelmann

1. You must use high hydrogen pressure on a large scale of starting material (2g or more) for rapid hydrogenolysis.

2. Try working with a wet 50% Pd/C, dry Pd/C often fails.

3. An alternative method is to use lewis acids (FeCl3, TiCl4)

Dear Florian and Chen Jingwen unfortunately Alfa Chemistry does not sell quinalizarin either. When you check their website for the term "quinalizarin" you receive a message saying that the product is currently not available (please see attached screenshot).

With best wishes, Frank Edelmann

Dear Sarah Rabbitt the boiling points of propiolic acid and pyruvic acid are 21 deg. apart. Thus it should be possible to separate the two acids by careful distillation though a Vigreux column. In case that this issue is still of interest to you, I think you should not hesitate to ask this as a separate technical question on RG. Good luck with your work!

Dear Alexander Sinko, well done ! 95% !!!, as dear Dr Erdal mentioned , I advise you to be careful to remained 5% ( 20 components), you can use Column Chromatography or Solvent extraction........

Consider two small molecule anti-cancer drugs, Enasidenib and Ivosidenib, which are IDH1/2 inhibitors approved for AML treatment. The fluoride moieties are a prominent feature. If you remove them from the molecules, their PK will be completely messed up; in fact, without the fluorines, they will most likely vanish like the clappers once they reach the liver, due to rapid metabolic degradation. Another distinguishing feature is the predominance of six-member aromatic rings and amide/amine linkages. A drug must be reasonably soluble in water and capable of being transported through the bloodstream to be effective when taken orally. Because amines are weak bases, they are frequently converted to salts with some acid, and thus some oral drugs contain amine salts. One reason for their presence is that they provide the drug with some water solubility. Scientists have taken a cue from nature by utilising the versatility of the amide group. The amide chemical group reigns supreme in medicinal chemistry. The group features in almost all of the bestselling drugs. Because amides have the right amount of stability and polarity, they can interact with biological receptors and enzymes. Because drug design is so nuanced and cancers are so variable, determining which functional groups should be inserted into a drug to boost anticancer activity is best left to experience, modeling software, and repeated tests.

Dear Jialiang,

many thanks for sharing this very interesting technical question with other RG members. I did a SciFinder search for this compound and found that it is not known from the previous chemical literature. However, the analogous compound is known in which the ethyl group in para-position of the phenyl ring on the left side is replaced by a methyl group. This compound has been reported in the following article:

Fortunately this paper has been posted by the authors a public full text on RG. Thus you can freely download it as pdf file and print it out if required. The synthetic route is outlined in the attached reaction equation. You just need to replace the starting material 4-iodotoluene by 4-iodoethylbenzene.

Good luck with your reserach work and best wishes, Frank Edelmann

Hey. Its been so long, I can't remember if I got the result that I needed or not.

You could call the company that manufactures the Dowex columns and ask for assistance. They might be able to guide you in the right direction. Good luck.

Dear Bryan,

thank you for posting this interesting question on RG. As an inorganic chemist I'm not a specialist in polymer chemistry enough to give you a qualified answer. In this context I keep wondering about two things: Why are you expecting a soluble polymer when you use aromatic units as hard segments? And why do you want to work in nasty acetic acid when "normal" organic solvents will also work? To the best of my knowledge the formation of low-molecular weight Schiff bases from aldehydes and amines is often catalyzed by small amounts of acetic acid. Thus I assume that a few drops / ml of acetic acid will do in your case too. For some potentially useful information please have a look at the following relevant articles in which the formation of polyimines in various organic solvents has been studied:

and

The first article is freely available as public full text on RG, while the second one is not. However, five of the authors have RG profiles, so that you can easily request the full text from one of them directly via RG (provided that it is of interest to you).

In any case I wish you good luck with your research work!

Try with triethylamune.

You can check the amine value while adding DMS at different times. Once amine value becomes zero or less than 5, you can stop the reaction or addition of DMS.

Barbituric acid slighltly soluble in water. So, you can wash the reaction mixture with NaHCO3 solution after completion. it may remove the unreacted barbituric acid.

Dear @Mrutyunjaya Panda All biochemical reactions are chemical reactions. I agree with Dr @Frank T. Edelmann in that in principle, there is no major difference. The same reaction can occur 'in vivo' and 'ex vivo'. You can also access a similar discussion at the following link:

Best wishes, AKC

Dear Mohamed H. Aboutaleb many thanks for your interesting question. Unfortunately the question is a bit too general in nature to give you a qualified answer. In this point I fully agree with Vladimir Khlebnikov in that it would be helpful if you could specify you aromatic amine / hydrazine precursor in some more detail. If I knew a structural formula I could easily do a SciFinder search for you to find out if reactions with isocyanates or isothiocyanate have already been reported.

The palladium(II) complex is an effective catalyst for the coupling reactions of aryl Chloroaromatics.

The best method of removing solvents like chloroform, THF, dichloromethane without using a rotary evaporator is simple distillation on water bath.

One side is looking at density - whether it sinks or floats in liquids of different densities - water, alcohol, etc. Geologists do this with minerals - but they use bromoform (2.6 g/mL)!

the other side is using chemistry and solubility - flame tests to show presence of chlorine in PVC, what solvents different polymers dissolve in.

1.You can change solvent methanol Water

2.also use buffer.

Dear Reyna Martha Gallegos Alvarado here is another interesting heterocycle synthesis experiment designed for undergraduate students.

"This experiment aims provide the students with the knowledge about the One-Pot synthesis of dihydropyrans."

Dear Daniyal Ahmed just in case that this question is still of interest to you: The article cited below contains a detailed description of how to make oleum in the laboratory:

Personally I would suggest to buy oleum from commercial sources as it is an industrial product manufactured on a large scale.

Dear Talal Ashraf there are two potentially useful article on this topic available as public full texts right here on RG:

1. Predicting Feasible Organic Reaction Pathways Using Heuristically Aided Quantum Chemistry

2. Neural networks for the prediction organic chemistry reactions

Hi i agree with Saraniya Bharathi Based individual amino acid combination - the peptide validated as "Easy" "Hard" to synthesis.

You may synthesized many peptides but this was failed due to sequence difficult - if you provide sequence means i will validate it and give you clear reply.

Some Points :

1. You used Rink amide resin - so here there was no problem found

2. After first amino acid coupling - amino acids solvation effect and coupling deficiency issues synthesis goes to wrong direction .

3. Each amino acids have different internal properties - Neutral and Hydrophobic amino acids create trouble - so before planning to synthesis - you write protocol for each "Hard" amino acid.

If sequence easy means no issues - if "difficult sequences" means should be careful and modify your procedure in difficult red-zone area of the sequence.

Hope some of my publications have details about this - may helpful for your research.

All The Best For Your Research and Success

- Dr.R.Selvam - Custom Peptide Synthesis Expert

Grey Matter Research Foundation.

Unlike other halides of boron BF3 doesn't undergo hydrolysis. BF3-2H2O is a very stable compound. So BF3 etherate is also hygroscopic and you need to dry it before using it with any sensitive reagent. I dried it using CaH2 and distilled under reduced pressure. It's better to filter off CaH2 and the solid residue if you don't want to distill under vacuum. I also came across someone using P2O5.

I think you can use DCC for better result 📷

For other approachable method you can go to those link where you can find various method of successful bioconjugation

Madhukar Baburao Deshmukh Thankyou Sir.

and who told you that LS are not soluble in organic solvents. Try it first.

You can prepare a mono adduct of amide from the reaction between lactone and ethylenediamine by heating one mole of lactone with half mole of ethylene diamine in ethanolic solution.

There no possibility of nucleophilic aromatic substitution of hydroxyl group without the presence of sulfonic group as OH is not a better leaving group.

agree with Kurt Wachholder

Dear all, what is the role of sonication in your case? Is it for mixing/dispersion or for reaction assisting. If it is for the later option you can use microwaves. My Regards

To remove the KOH from ethanolic etherification without using extraction method , neutralize the reaction mixture obtained after reflection by treating with 2N HCl to PH=7 , organic layer separates out.Extract organic layer with chloroform or dichloromethane. Removal of Organic layer under vacuum will give ether.

1,2,4-triazoles is stable compd. for action of acids and alkalies.At high temperature above 200 deg. it may loose nitrogen in the form of N2.

You may try 0.1 N HCl in hexafluoroisopropanol or trifluoroethanol.

Dear Banchhanidhi Prusti

In addition to the answers afforded by Frank T. Edelmann and Abdelkader BOUAZIZ , I suggest another synthetic way. This includes the formylation of anisole using Vilsmeier–Haack reaction and the product is then oxidized to form the target product.

All the best

Dear Alan

Can you please attached the proposed synthetic scheme to clarify your question and point my answer?

Regards

To protect thiol group convert it into thioether by reacting it with alkyl halide in acetone in presence of anhydrous base as potasiummcarbonate

… Read more

You might have cleavage of the ester depending on the base you use, but most bases traditionally employed in Sonogashira couplings should be safe.

Crystallization of a zwitterion at the isoelectric point is tricky and often inefficient. If you only need a small amount of product and high yield is not the goal, AND you have a good estimate of pI, this can be the quickest way to get what you need. If you are doing this preparatively and/or if the compound does not crystallize well from your original mixture, ion exchange resins will probably be the most effective tools. (If you are trying to recover your product from a complex mixture, you will need to do IEX chromatography; please provide more details.)

When amino acid zwitterions are being synthesized, cation exchange resins such as Dowex 50, Amberlite IRC120 and others offer a very simple and efficient method for recovering products. The following process is frequently used following acid treatment, such as amino acid amide hydrolysis in HCl(aq):

RCH(NH[R'C=O])CO2H + xs HCl(aq) --> RCH(CO2H)NH3+ . Cl- + R'CO2H

The hydrolysis mixture consists of RCH(CO2H)NH3+ . Cl-, R'CO2H, water and excess HCl (e.g. R'=CH3 for an amino acid acetamide, where acetic acid is a volatile byproduct).

1) Evaporate excess volatile acid (rotary evaporation will remove water and HCl gas.) Purging HCl gas in a stream of N2(g) beforehand (fume hood) will reduce bumping during evaporation. Vent the vacuum pump exhaust to fume hood.

2) Dissolve the acid evaporation residue in a small volume of water

3) Calculate the number of millimoles of the zwitterionic substance you intend to isolate and pack a short column with 2-5 times this number of milli-equivalents of appropriately-washed NH4-form ion exchange resin (preparation usually done by washing with strong acid solution followed by water to give neutral eluate, then aqueous ammonia treatment to convert RSO3H groups to RSO3- .NH4+, and finally thorough washing with deionized water to give neutral filtrate.)

4) Slowly apply the solution of evaporated hydrolysis mixture residue onto the column, then wash the column with pure water until the eluate is free of Cl- (AgNO3 test) and the pH is neutral (or not lower than the pH of your DI water). At this point, the acidified zwitterion should be completely bound to the resin [RCH(CO2H)NH3+ . resin-SO3-]. (Check the column application eluate to be sure nothing has leaked through. TLC with ninhydrin detection works well for amino acids). You may also check dried drops of eluate with ninhydrin to be sure that displaced NH4+ ion is no longer eluting from the column.

Elute the washed column with aqueous NH4OH, using approximately twice the number of millimoles as meq of resin (e.g a column containing 50 mL of 0.8 meq/mL resin would be eluted with 2 x 50 x 0.8 = 80 mmol of NH4OH(aq), say 40 mL of 2 M). If the product has not completely eluted from the column by this time, more water should be used before you add more ammonia. Very hydrophobic amino acids may elute better if some methanol or ethanol is added to the water.

5) Evaporate the ammonia eluate. As with the HCl solution in step 1, this solution may behave better during rotary evaporation if you first bubble a stream of N2 gas into the eluate to purge excess NH3 (fume hood), followed by vacuum evaporation, only heating once all free ammonia is gone. Nearly all ammonium carboxylate salts (RCO2- . NH4+) decompose upon evaporation with heating to carboxylic acids, so this step is basically:

[RCH(CO2-)NH2 .NH4+](aq) --> RCH(CO2-)NH3+(s) + NH3(g)

i.e. once ammonia gas leaves, this residue is theoretically pure zwitterion.

6) At this point the purification method of choice is usually recrystallization, commonly from water-alcohol mixtures.

Formylation: Vilsmeier-Hack Reaction -> Formylation at C3

it is conspicious, that the presenec of an electron- withdrawing substituent at nitrogen directs electrophilic substituition to C2

https://www.mdpi.org/ecsoc/ecsoc-7/papers/E013/E013.htm

Synthesis of benzoxazole can be achieved by heating o-aminophenol with carboxylic acid in presence of polyphosphoric acid. In another method o-amino phenol is heated with HC(OCH3)3 in presence of HCl.

interesting answers

Dear Gagandeep Kaur,

Polarity of solvent can also have a big influence on the products obtained in organic photochemistry. For example, in the case of photoinduced hydrogen transfer. Two mechanisms are possible : a one-step mechanism where the electron and the proton are transferred in the same time and a two-steps mechanism where electron in transferred first and proton follows. These two mechanisms cannot be distinguished in terms of energy, they are both exergonic. But, for a one-step process, the electron transfer is endergonic where as in the case of a two-steps process, the electron transfer is exergonic. As you could see in our publications (10.1039/C9NJ03061A, 10.1002/chem.200903045), different products can be obtained with different solvent (cage effect discussion is made in the former). $\Delta G^{0}_{el}$, electron transfer exergonicity, is available from the Rehm-Weller equation, as a function of polarity of solvent. For a less thermodynamic point of view and a more kinetic point of view, I suggest you to see this last publication in Science for concerted proton-electron transfer :

10.1126/science.aaw4675.

The best reagent for the reduction of NO2 group to NH2 group is heterogeneous catalytic reduction using H2/Pt in acetic acid.

There is a common precautions to consider when it comes to storing conditions and storage time for the chemicals to preserve their initial structure. Such substances can be decomposed with the effect of light and heat. Also if you open them without the right provision ,the material can be subjected to air. Such substances can react with oxygen and water moisture.

You have to keep such materials sealed under the proper storage conditions till their use.

The other thing is that their recovery can be very difficult to you.

So, i would advice you to use fresh chemicals for your experiment.

For sure the cause of the failure of the experiment is the expired Bu3SnCl .

Best wishes

Please check also the following related RG discussions: https://www.researchgate.net/post/I_have_L-histidine_hydrochloride_monohydrate_Help_me_please_how_can_I_eliminate_HCl_and_water_molecule_to_obtain_L-histidine_only https://www.researchgate.net/post/What_is_the_best_method_for_conversion_of_guanidine_hydrochloride_to_free_base_guanidine

In general:

MOFs are "bulk" materials, usually contain more than one ion-pairs and not "isolated" ions/ion pairs unlike in a real solution. That means, on the nanometer scale the particles considerably fewer ions on the surface than the total number. And so, the ligand exchange is restricted to the surface. Additionally, not all the surface ligands can be exchanged.

An example (because I am not an expert of Cu3HHTP2, but I assume that it is general): rutile builds up tetragonal unit cells, where the unit cell dimensions are a = b = 0.458 nm, and c = 0.295 nm. The Ti2+ ions are surrounded by 6 oxygen atoms, forming an octahedron, and the unit cell contains more than one TiO2. Not to say anything about the intercalated other ions.

That means, > 1 nm particle sizes contain fewer replaceable, in the case of rutile the unit cell contains (as average, as Axel wrote) 5 TiO2 while on the surface (of the unit cell) only 8 (of 10) Oxygen is active. In theory on a ~1x1x1 nm particle, while the number of constituting TiO2 is 30, the number of surface O is only 20 (of 60). And usually, the nanoparticles are larger than 1 nm.

Of course, in (theoretical) two-dimensional cases the numbers are different.

Additionally, multidentate ligands further reduce potential substitution sites. So, the mentioned 2:1

Finally, Balakrishna's comment is an inappropriately undervalued comment.

Regarding your example.

There are many missing data, and I assume they are also hidden for you. For the solid synthesis, you mentioned 2:1 ratio. In solid syntheses, there is a size limit, which is generally higher than 1 nm. After that a) I have doubts and the 2D structure might be true only; b) using a 2:1 molar ratio seems to be too low.

Due to solubility issues in solution and lack of yield, isolation method, etc., the 10:1 can be reasonable.

Please, also note that thin films are not necessarily 2D structures, and the reactions are rarely so ideal as they are described in publications - as you are getting more experienced you will see the difference between a publication and real life. Many tricks remain hidden in publications, independently from the journal rank.

So, firstly try to reproduce the published method (=do strictly what is written even if you are not in agreement with the authors!), then, based on your experiments and comparison with the written matters, try to find explanations and understand the authors' synthetic method. This way usually works well. Thinking a lot of unknown things, then to conclude finally to repeat the published method is many times a waste of time.

Good luck,

Laszlo

Thankyou Dear Grant Simpson For Sharing This Valuable Information.

I am sure this will be helpful.

Try using a zeolite resin. It works for removing Sodium in water OR you can try 'reverse osmosis'. Follow was Waters or other water suppliers use.

Could be explained by radical mechanism and allylic resonance. Hydroxyl radical attacks the double bond. Secondary carbon is more stable as a radical so the chlorine attacks carbon no 2.

I don't think you really need such a strong base as n-BuLi to deprotonate the methylene group in this substrate; bases like NaH of t-BuOK would be sufficient (even K₂CO₃ in aprotic solvents is used successfully for alkylation of malonates). In this particular case perhaps the electrophile is not reactive enough.

Also, let's not forget that we have an amide NH in a molecule, which is also pretty acidic, perhaps more acidic than methylene. Chances are that you may need to use more than one equivalent of the base. And of course for good results the reaction media should be anhydrous.

https://www.ncbi.nlm.nih.gov › articles › PMC6419043

Refer to our publication," synthesis of phthalonitriles using a palladium catalyst".

I deed reaction of benzyl alcohol derivative with 2-fluro toluene using KOH as a base and DMF as a solvent at 80-120°C. Reaction time was 5-12 hr. It was yielding 65-80% desired product. Reaction was almost clean on TLC observed under UV but reaction mass was thickened and darkened with time. What I am assuming is that DMF degradation is happening but with a lower rate than the desired reaction.

Hi, I saw this in Apay's link. The answer to your question is that it depends on the reaction mode. If it's a standard Lewis acid catalysed reaction then the most stable Lewis acid base complex is formed (even although its a kinetic process of course). If you read through George Olah's work you'll find your answer and many others. He was one on my chemistry idols when I was working on acylations!

Hi;

Steric hindrance (the reaction site is blocked by F and B(OH)2 groups forming part of the molecule).

Best regards

Hi;

Substituted rings are divided into two groups based on the type of the substituent that the ring carries:

•Activated rings: the substituents on the ring are groups that donate electrons. •Deactivated rings: the substituents on the ring are groups that withdraw electrons.

Examples of activating groups in the relative order from the most activating group to the least activating: -NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups

with R as alkyl groups (CnH2n+1)

Examples of deactivating groups in the relative order from the most deactivating to the least deactivating: -NO2, -CF3> -COR, -CN, -CO2R, -SO3H > Halogens

The activating group directs the reaction to the ortho or para position, which means the electrophile substitute the hydrogen that is on carbon 2 or carbon 4. The deactivating group directs the reaction to the meta position, which means the electrophile substitute the hydrogen that is on carbon 3 with the exception of the halogens that is a deactivating group but directs the ortho or para substitution. Best regards

Sodium Hydide in DMF at RT works like a charm.

For UV TLC plates the compound fluorescence is not a requirement because the UV plate is fluorescent and the absorption of the UV light shadows the plate fluorescence. If your compound has UV absorption in the range of 250-370 nm you can see black spots under the UV lamp. Because your compound contains aromatic groups you have good chances to see it. Please, also note that there are two commercially available UV active TLC sheets: 254 nm and 366 nm. My experience is that the 254 is better. Please, also note that the quantitation on the TLC by UV is not always obvious (because of the spot-spreading) unlike the charring or other visualization methods.

4-Methoxybenzenecarbothioic acid will be the by-product of the Lawesson's reagent after the functional conversion of the carbonyl group (C=O) to the thiocarbonyl group (C=S). That might be the spot you are seeing at the baseline of your TLC and this by-product is also UV active which is in agreement with your observation.

Dear David Oulton Sir,

Thankyou Once Again For Your Valuable Opinion.

Regards,

Daniyal

Hi Ahed,

All aromatic azo compounds are completely conjugated.For example from aniline prepare diazonium salt then couple it with phenol you will get the required azo compound for procedure use book ( practical organic chemistry by Vogel) in this book you will see many other azo compound preparation procedures. Good luck

Try to use AgClO4. Both AgCl and KClO4 are badly soluble in water and polar solvents

There are a few patents as mentioned below which talks of similar methods. The feasibility part is for you to decide.

EP0031296B1

CN102070627A

US3641044A

Do esterification/acid halide conversion (with equimolar ratio) of carboxylic acid first followed by careful addition to amine. Works well for amide preparation without affecting alcohol functional group.

You can check by NMR, elemental analysis, LC-MS or GC-MS.

I think it is difficult to prepare the monohydrazide as the compound will have a tendency to form phthalylhydrazide (2,3-dihydrophthalazine-1,4-dione) on heating. I am not sure whether it will work or not, one can try the following procedure : Stirring of phthalic anhydride in large excess (say 10 times or more but without solvent) of hydrazine hydrate at room temperature (or at lower temperature) for a long time. It is expected to obtai the monohydrazide carboxylic acid salt of hyazine. Careful neutralization with with dil HCl under cold condition will generate the free acid and hydrazine will be removed as salt. Thus the monohydrazide will be precipitated. Filter it to collect the solid (excess of HCl to be avoided as it would make the hydrazide soluble as a salt). It is to be dried under vacuum without heating (heating can convert it to phthalylhydrazide again).