3. A stock solution of an organic compound (molecular weight - 187 gimol is prep

Question

3. A stock solution of an organic compound (molecular weight - 187 gimol is prepared by dissolving 7.2 mg in 25 of solvent. A UV sample prepared from 200 of stock and 2.80 mL solvent placed in a 2.0 cm quartz cuvette provides an absorbance of 1.782 at 263 nm. What is the molar absorptivity of this compound at 263 nm 4. Use the Woodward-Fieser Rules in Chapter 15 and Appendix 3 to calculate the predicited Amax value(s) for each of the following compounds. Be sure to show all of your calculations and rationale. (6 pts, total) b. a. C.

Explanation / Answer

3) As per Beers-lambarts law

A = Cl

= A/Cl

Absorbance at 263 nm is 1.782, length is 2 cm

Concentration C is = m.wt is 187, weight = 7.2 mg and volume is 25 ml

molarity = 7.2/187X1000/25

              = 1.54 mmol.

Then 200 microlit stock was placed into 2.8 ml, hence concentration will dilute

So M1V1 = M2V2

1.54X200 = M2X3000 microlit

M2 = 1.54X200/3000

M2 = 0.1 m mol

Hence concentration is 0.1

= A/Cl

= 1.782/0.1X2

= 8.925

4)

For compound a)

Base value = 214

Ring residue = 2X5

Alkyl Residue = 2X5

Toatal = 234 nm

for b)

Base value = 214

Ring residue = 2X5

Alkyl Residue = 2X5

Extended conjugation = 1X30

total = 264

for c)

Base value = 214

Ring residue = 4X5

Alkyl Residue = 1X5

total = 239

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