Woodward-Fieser rule for calculating absorption maximum in alpha, beta unsaturated carbonyl compound

[@brianlorume8401]

This is extremely helpful. My lecturer made it look like rocket science.

[@labileeducation927]

Thank you

[@JanamatosS2]

Thank you for this class. Helped me a lot!

[@labileeducation927]

You are welcome.

[@theeaglewing7600]

Bro your utube is very useful and way of teaching fantastic

[@labileeducation927]

Thank you so much bro...

[@KhanSb-p8g]

Thank you so much....this lec. Helps me alot....294 is answer i guess

[@labileeducation927]

You are welcome. Answer is 324 nm.

[@Lamar.Lamar239]

Merci beaucoup ❤

[@labileeducation927]

je vous en prie

[@sharikachoudhary]

6 memberrd ring = 215nm
Ring residue at alpha= 10nm
Ring residue at beta = 12nm
Ring residue at delta= 18nm
1 double bond extended= 30nm
Homo annular diene= 39nm
Lamda max = 324nm

[@lyricalmatrix]

there is no homoannular

[@sharikachoudhary]

​@@lyricalmatrixRing B has a conjugated double bond, so it is Homoannular diene!!

[@lyricalmatrix]

Pls check again. The two double bonds in ring B aren't conjugated​@@sharikachoudhary

[@adwoabentuma]

@@lyricalmatrix The double bond is endocyclic for both A and B making it conjugated for B

[@Manar-h8i]

Exocyclic dobule dound 5=329

[@GarimaMall12]

Thankyou so much

[@labileeducation927]

you are welcome

[@raymondmacharia8005]

Thanks big time

[@labileeducation927]

Welcome

[@phyoemaungmaungthaik7588]

Some of reference tell a,b Unsaturated aldehydes base value is 210 nm.

[@AvnishBhatti]

Sir 290 nm answer. Is am right?

[@labileeducation927]

Good try, but answer is 324 nm.

[@BHCAshishkumar]

329 or 324 which is correct

[@labileeducation927]

Very good. 324 is the correct answer.

[@ramireddyvempada9960]

Nice

[@labileeducation927]

Thank you

[@brightoluwafemi5410]

a, b unsaturated (6-membered) =215nm
1-a ring residue = 10nm
1-b ring residue = 12nm
1-∆ ring residue = 18nm
2- exocyclic double bond = 10nm
2- double bond extending conjugation 60nm
Calculated wavelength= 325nm
Am I correct?

[@BHCAshishkumar]

329 nm

[@labileeducation927]

It is too close. it is 324 nm.

[@petermallick1677]

Last one is 290 is the answer

[@labileeducation927]

Good try. Answer is 324 nm.

[@kannanvalli3026]

265nm

[@labileeducation927]

Good try. Answer is 324 nm

[@abioyejohnsonopeyemi4102]

329nm

[@labileeducation927]

You are very close. it was 324 nm.

[@ShakeelAhmad-vm7xb]

324nm

[@labileeducation927]

wow...you are absolutely right.

[@imanbukhariib0514]

Kase Aya 324 kindly bata dein

[@petermallick1677]

329

[@life_hacks803]

How?

[@labileeducation927]

Very close. it is 324 nm.

[@kainatbajwa2957]

320

[@labileeducation927]

It is quite near. answer is 324 nm.

[@majekodunmilateefat1104]

305

[@labileeducation927]

it is close. Answer is 324 nm

[@wajidhussain4542]

215+10+12+18+30+39 = 324

[@wajidhussain4542]

5 exocyclic

[@amolgundpatil2613]

299

[@labileeducation927]

Good try. it was 324.

[@__sana123]

324 nm

[@labileeducation927]

wow... you are absolutely correct.

[@brightoluwafemi5410]

How do you arrived at your answer?

[@shreyanaik768]

329

[@labileeducation927]

Very close. it is 324 nm.

[@manahilzafar6796]

319

[@labileeducation927]

It is very close. answer is 324 nm.