Use Woodward-Fieser rules to estimate the λmax values for...

Use Woodward-Fieser rules to estimate the $\lambda_{max}$ values for the following compounds: (1) Calculated $\lambda_{max}$ ____ nm; Observed $\lambda_{max}$ = 234 nm (2) Calculated $\lambda_{max}$ ____ nm; Observed $\lambda_{max}$ ca. 270 nm (3) Calculated $\lambda_{max}$ ____ nm; Observed $\lambda_{max}$ = 283 nm (4) Calculated $\lambda_{max}$ ____ nm; Observed $\lambda_{max}$ =249 nm (5) Calculated $\lambda_{max}$ ____ nm; Observed $\lambda_{max}$ =226 nm (given that cyclopentenone has a wavelength of 202 nm)

Transcript

00:01 Hello student, we have given these, there two structures and we have to find out the value of lambda max for this structure.

00:09 So in the first structure, we have this six membered cyclic enone.

00:14 So we have basic value for this cyclic enone having six membered ring is 215 nanometer.

00:25 Secondly, we have these double bond and these two double bonds are in conjugation.

00:32 So we can write three extended conjugation is present here.

00:41 For one extended conjugation, we add 30 nanometer value and here three are present.

00:49 So total 90 nanometer value is added here.

00:52 And this double bond act as an exocyclic double bond.

01:00 So for exocyclic double bond, we add five nanometer and here this homoannular diene is present.

01:19 So we have to add 39 nanometer for this component.

01:25 And let us count the number of substituent present here.

01:30 So this alpha, beta, gamma and delta positions.

01:35 So at higher position, we have one substituent at gamma and one substituent at this position.

01:43 So we have two substituent at gamma and higher position.

01:57 So for one, we have to add 18 and for two, we have to add 36 nanometer.

02:04 So the sum of all these value will give us the lambda max value.

02:10 So after adding all these value, we will get a value of 385 nanometer.

02:17 So for this first molecule, we have lambda max value is equal to 385 nanometer...

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