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Hi Tom, yes I do. If you click on "Playlists" and then select the topic you are interested in, it will show you the videos in the exact order that I intended for you to watch. Hope you like the content.

Awesome! Glad the video was helpful to you.

Thank you for watching and I am glad the video was useful to you. I hope the exam went well.

Thank you for the kind words. I am glad you found the video useful.

Thanks for watching the video and glad to hear that I was of help on the other side of the globe.

Really glad to hear that you found the video useful and thanks for watching and commenting.

I’m glad you found the video useful. Thanks for watching.

The only way I can think of doing it is to first convert it to an acyl chloride and then react it with either a mild reducing agent (e. g. NaBH4, LiCuR2) to get to the alcohol or ketone. Alternatively, after conversion to an acyl chloride, you can reduce it to an aldehyde by using Li[HAl(O-tBu)3] at low temp.

That's what Sia and David Guetta would say 😂

@@aluminumchemist2586 try Bert Chandler and Lars Grabow my man 😂🎓

@@aluminumchemist2586 in all honesty thank you for the video. I’m trying to modify this equation for a high impact journal

Thank you. I’m glad you found the video useful.

Yes, that is correct. In essence, you will have to solve for the potential by plugging in the concentrations of H+ and Cu2+ directly into the original Nernst equation. I wouldn't recommend the pH derivation in that situation. Generally speaking, the lower the concentration of reactant, the lower the potential value. This means that the reactant oxidative power is diminished with lower concentration.

I'm very happy the video was of help to you and thanks for watching.

Thanks a lot. For some reason I didn't see your reply until today. Glad you found the video useful.

Thank you for the suggestion. It is a work in progress for sure. In the near future I hope to start uploading the second part of O-Chem.

😋

Glad it was helpful! And thanks for commenting.

Thank you for the kind comments and I am glad you found the video helpful.

You are welcome. I am glad you found it useful.

Thanks. I am glad u found the video useful.

You are very welcome. I'm glad you found the video useful.

Glad to be of help.

But values are different, like for chlorination are 1,3.8,5 And for bromination are 1,82,1600

@@anshahuja07 Thanks for letting me know. Generally, the values are used were obtained from experiments done in the 90's, so it is possible that those values have been updated since then. In any case, I'm glad you found the videos useful.

You are welcome. I'm glad you found the material useful and on a funny note: it was node a problem 😄

@@aluminumchemist2586 😂

Organic Chemistry by David Klein

@@aluminumchemist2586 thanks

I obtained all those values from Gary Wulfsberg's Inorganic Chemistry textbook. He himself got the values from 1) R. T. Myers, J. Chem. Educ., 1990, 67, 307. 2) R. G. Pearson, Inorg. Chem., 1991, 30, 2856. Thanks for commenting.

@@aluminumchemist2586 👍👍👍 The lessons on periodic properties have been productive and interesting. Thanks a lot.

@@marcoroman3245 Thank you so much. I really appreciate the feedback.

You are very welcome. I'm glad the video was helpful to you.

You are very welcomed. I am glad you found the video useful.

Thanks. Glad you liked the video.

If what you are asking is how to get the oxidation state of each carbon in insole, then you do the same procedure as presented in the video. On the other hand, if you are comparing insole to another derivative then you have to determine whether or not the derivative has electron-withdrawing groups (oxidized) or electron-donating groups ( reduced). I hope that helps.

Hi Lance, I wouldn't refer to it as being amphoteric, as that implies both acidic and basic behavior. The fact that Eu changes from 3+ to 2+ simply points out that the metal has redox properties. I hope that helps.

Thank you. Cheers!

Very glad that you found the video useful.

Glad to be of help.

You are very welcome. I'm glad you found the video useful.