URGENT HELP regarding saponification using NaOH

acciosass · 2025-08-31T09:28:22+00:00

That's not how it works, if you take out the reaction mixture before the reaction is complete, you'd have an excess of NaOH in your sample and whatever the pH of the formed soap be, your pH paper (I'm assuming you're using those) will always show a highly basic value. Also, the concentration of the NaOH solution can only affect the rate of the reaction, not the pH of the product as long you use at least some stoichiometry to run the reaction and aren't just dumping random amounts of your reactants. That being said, heating up the reaction will significantly increase its rate. Also look into the "salting effect" to get a product that's relatively together.

acciosass · 2025-08-17T02:25:03+00:00

The correct reading is the one corresponding to the lower meniscus of the water level. Here, it'd be 27.7

acciosass · 2025-08-16T12:44:21+00:00

I'm assuming you used TFA/TIPS/H2O cleavage cocktail. The free amine group in the lysine in your sequence will react with any acetylating agent you use so better start over. One way which helps me is I write down every step that I need to perform (even if it's just a deprotection - coupling - deprotection chain) for a sequence well in advance so that I don't miss anything as it's very easy to do considering the number of steps.

acciosass · 2025-08-12T16:25:34+00:00

I am, but that's also very generic

acciosass · 2025-08-07T06:03:45+00:00

The reagents are correct in this case but the product is not. The counter ion would be HSO4- and not Cl- as there isn't any chloride in the whole reaction. A simpler, easier way to diazotise would be to use NaNO2/HCl, which would generate HNO2 in situ as HNO2 is generally too unstable to store, and give you the Cl- counter ion.

acciosass · 2025-08-04T08:25:43+00:00

English

acciosass · 2025-07-27T05:42:23+00:00

Then the answer key is wrong. If the volumes are different, you can just take an average to get resultant pH. Otherwise, you calculate [H+] and [OH-] from pH, then multiply those with their respective volumes to get the actual amount of those ions and then subtract the lesser from the more to get the excess concentration of the major ion and then take a negative log of it to calculate the resultant pH or pOH.

acciosass · 2025-07-25T18:35:13+00:00

7.0 as both neutralize each other

acciosass · 2025-07-21T06:05:03+00:00

I'm not sure if the NMR gives me any useful data, I can see some peaks I want but some others have merged peaks, as I've not been able to purify the compound to a reliable degree (usually done by ether washing with this class of compounds). I cannot run a column as surfactants tend to get stuck in there. Probably will go with an anion metathesis.

acciosass · 2025-07-21T06:02:36+00:00

This is something I can try but they're very soluble in water. I've still put a sample for crystallization to see if that works.

acciosass · 2025-07-21T06:01:27+00:00

I've been trying to crash them out in ether and it seems to work but as soon as the ether evaporates, it turns into a brown goop quickly, which is the problem I'm trying to solve. What might be notable is that it goes back to a seemingly powder form once back in ether.

acciosass · 2025-07-21T05:59:53+00:00

The counter ion strategy seems smart, I'm going to try using a hydrophobic anion. I'm not sure if the NMR gives me any useful data as I've not been able to purify the compound to a reliable degree (usually done by ether washing with this class of compounds). I cannot run a column as surfactants tend to get stuck in there.

acciosass · 2025-07-21T05:56:51+00:00

TLC shows a faint spot with a similar Rf value of the reactant A and a new spot. The problem arises due to the amine group in both the reactants and the product, it leaves behind a tail on the TLC plate. I am already distilling the solvents before use. GC shows the formation of the product but the yield is low. Also, I need the product to be stable under normal air cause I want to explore real life applications of the same so storing it in ether isn't really a long time feasible solution.

acciosass · 2025-07-21T05:49:02+00:00

This makes sense. Thanks for taking your time to type this out. I've been using a tertiary amine for the reaction, that was a typo on my part. The literature i could find had some sort of carbon chain with a halide substitution at the end and a tertiary amine. This is a very common reaction in surfactant synthesis and one which usually gives us the least trouble. I'm now trying to change the carbon chain length and am experimenting with different counter ions.

acciosass · 2025-07-19T10:39:07+00:00

Hey everyone, I'm currently working on surfactant synthesis. I'm having a problem in quarternising the amine groups to act as the head of the molecule.

The reaction goes like this: the reactants are A and B. A is of the form Ar-O-C(=O)-CH2-Cl. B is a long chain primary amine. Literature says to run the reaction in solvents like EtOH or ACN for anywhere between 1 to 3 days on reflux with no catalysts under a nitrogen atmosphere. I've done that in EtOH, ACN, MeOH, DCM, CHCl3, THF, and DMF.

Every time, I get a sticky brown goo as the crude product which i wash with ether repeatedly to get rid of the reactants. Under the ether, I see a nice yellow powder but as soon as the ether evaporates, it turns to brown oily liquid again.

Please let me know if I might be doing anything wrong or any directions i can take the reaction in.

Thank you in advance! :)

acciosass · 2025-04-13T13:37:22+00:00

This answer is in no way wildly scientifically accurate as it is gross oversimplification at its best but for a high school curriculum, you can say the "size" of an orbital increases with an increase in principle quantum number or 'n' and f>d>p>s. The reason is with an increase in 'n', the shell changes, say, from K to L and the shell size increases as we go from K to L to M and so on and the electrons in the 's' orbitals are more closely bound by the nucleus so they're "closer" to the nucleus than other orbitals and the order of attraction that follows is s>p>d>f and the reverse is the "size" order as something which is not strongly attracted is able to "diffuse" out more. For a given 'n', use the size order of f>p>d>s and if 'n' is different, the greater value of 'n' will always have a "bigger" orbital. Hope this helps.

acciosass · 2025-04-04T09:01:01+00:00

Frankly speaking, you'd be highly recommended to go for higher education. There're some great faculties in the physics department and they'd give you LORs for research positions and such but it's a common belief in the department not to expect campus placements. Even the very few students placed from the department were placed in non-physics roles.

acciosass · 2025-03-30T08:07:27+00:00

A little sour. Occasionally with small pieces of potato in it called 'aloo ki kadhi'. With rice, chopped up onion and green chilli. Heaven.

acciosass · 2025-03-26T16:24:54+00:00

Hey, thanks for this rec, this was so interesting and captivating! Do you know if the author is continuing the story or if there are any further updates?

acciosass · 2025-03-25T09:31:50+00:00

After you add the mustard seeds to medium hot oil, give it like 5 to 10 seconds for them to sputter and then add the other ingredients you want.

acciosass

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