5.0 equiv. of the aa and the coupling cocktail you use is more than enough. Good luck!

Thank you Sir, for your reply

I am using rink amide resin, loading 0.8-1.0 mmol/g N Loading.

So please can you tell me how much resin is required, if i use 5 equivalent AA

Then your calculations should be based on 0.9 mmol. Thus, 0.9 mmol x 5 x M(aa) will give you the mass of the aa you have to use. While is the same count for the coupling cocktail:

0.9 mmol x 5 x M(coupling cocktail components) / density (of each) will give the mass or volume (if necessary)

Good luck!

Thanks Sir,

According to your answer, I found- mmol of resin, required equivalent of aa. but my main question is-

how much to take resin for swelling in starting time (?mg)

That depends on the scale you want to conduct your experiment. For a 6 mL syringe you should take up 100 umol of resin as much.

So for one syringe you divide:

0.1 mmol/(0.9 mmol/g) = 111 mg

Hope this answer your question!

Thanks so much Sir for your supporting.

1) can you tell me which rink amide resin will be good for good yield, i mean which range of loading capacity and also mesh size of resin will be good?.....

Please read it and suggest me -

here i am using mesh size of rink amide resin is 50-90 and 100 mmol (equivalent of resin 0.1mmol).

then follow the standard protocol

after final coupling of aa, break the peptide resin using cocktail 95% TFA, 2.5% TIPS, 2.5% Water (2 ml ), and wash with TFA, kept for 3hrs . then filter the peptide-resin collect in a rb and add diethyl ethyl then centrifuge, some small quantity is obtained. reaction failed. I tried many times but not seen good result.

so i request to you give me your best opinion what do i do and where i am wrong?

can you please give me your email id....?