14.1 Introduction to IR Spectroscopy | Organic Chemistry

[@AnnaDesMarais]

Chad, I just wanted to let you know, that I love you. You're my saving grace.

[@ChadsPrep]

I'm happy to hear that I'm helping you!

[@yashtrivedi2671]

Incredible explanation, thank you

[@ChadsPrep]

You're very welcome Yash! I'm actually working on re-filming the replacement for this lesson today. Any suggestions?

[@yashtrivedi2671]

@@ChadsPrep I think this is a great video so I do not believe much more needs to be added. But maybe another video (in addition to the one you already have) on practice problems could be helpful!

[@ChadsPrep]

My premium course on chadsprep.com currently includes 800 practice questions (I'm steadily adding more) including 15 IR questions fyi!

[@mup1537]

Man it's just kind of amazing how different your ranges are from my text book. They're obviously quite similar but sometimes different by 100 cm-1

[@26d8]

thank you so much, why the cis alkene has changed in dipole moment but trans alkene doesn't? arent both symmetric?

[@ChadsPrep]

Dipole moment = (partial charge)(distance of separation) The trans isomer is nonpolar and has a zero dipole moment (sp2 C-H bonds oriented 180 degrees apart). This means there are no partial charges. So whether we stretch the bond or not there is no dipole moment when you plug in zero for the partial charge in the above expression. To absorb IR radiation there has to be a change in the dipole moment which isn't the case here. But the cis isomer while symmetrical in one plane still has a small dipole moment as the sp2 C-H bonds are only oriented 60 degrees apart. With a non-zero dipole moment it will have a partial charge in the equation for the dipole moment and so stretching the bond will change the dipole moment and thus we can expect an IR absorption. This would have been far easier to explain in person but hope this helps!

[@26d8]

@@ChadsPrep thanks alot for your response, that sure helped me alot.