Question

6. UV: Determine the max for each of the following structures. Show your work. D. 2max E. Lmax HN G. 2max Part B: NMR Spectroscopy (read TOC Chapter 22) Go to: H. max htip://www2.chemistry m suedu/faculty reusch/VmTxt m Spectrpy in r nmr htm#nm 1 Read the background material. At the end of this information, there are some "Practice Problems" Do the practice problems indicated below and put answers on this sheet (You can check your answers at the website. You may also do the rest of the problems for further practice) NMR "Practice Problems" Question 1 A to atoms #Haps | Question 2 A #C atoms 3.Label as i, ii, or ii. 1)1,2,3-trichlorobenzene 1,2,4-trichlorobenzene 1,3,5-trichlorobenzene 2). 1,2,3,4-tetrachlorobenzene 1,2,3,5-tetrachlorobenzene 1,2,4,5-tetrachlorobenzene

Answer 1

6. UV lambda-max

A. Homoannular diene = 253 nm

extended conjugation = 30

alkyl substituent = 3 x 5 = 15

lambda-max = 253 + 30 + 15 = 298 nm

B. heteroannular diene = 214 nm

extended conjugation = 30

exocyclic double bond = 5 nm

alkyl substituent = 4 x 5 = 20

lambda-max = 214 + 30 + 5 + 20 = 269 nm

C. homoannular diene = 253 nm

alkyl substiuent = 4 x 5 = 20 nm

lambda-max = 253 + 20 = 273 nm

D. alpha,beta-unsaturated ketone = 214 nm

homodiene = 39 nm

exocyclic double bond = 5 nm

ring substiuent at beta = 12 nm

ring substiuent at other = 18 nm

double bond extending conjugation = 30 nm

lambda-max = 214 + 5 + 39 + 12 + 18 + 30 = 318 nm

E. alpha,beta-unsaturated ketone = 214 nm

double bond extending conjugation = 30 nm

exocyclic double bond = 5 nm

ring substiuent at beta = 12 nm

ring substiuent at other = 2 x 18 = 36 nm

lambda-max = 214 + 30 + 12 + 36 = 292 nm

F. alpha,beta-unsaturated ketone = 214 nm

homodiene = 39 nm

ring consituent at alpha = 10 nm

ring substiuent at beta = 12 nm

ring substiuent at other = 2 x 18 = 36 nm

double bond extending conjugation = 30 nm

lambda-max = 214 + 39 + 10 + 12 + 36 + 30 = 341 nm

G. Heteroannular diene = 214 nm

OR in alpha position = 35 nm

NHR substiuent = 95 nm

extended consjugation = 30 nm

exocyclic double bond = 5 nm

alkyl substituent at beta = 12 nm

alkyl substiuent at others = 2 x 18 = 36 nm

lambda-max = 214 + 35 + 95 + 30 + 5 + 12 + 36 = 427 nm

H. 5-membered alpha,beta-unsaturated ketone = 202 nm

alkyl substiuent at beta position = 2 x 12 = 24 nm

lambda-max = 202 + 24 = 226 nm

I. alpha,beta-unsaturated ketone = 214 nm

extended conjugation = 30 nm

homodiene = 39 nm

O at alpha-position = 35 nm

lambda-max = 214 + 30 + 39 + 35 = 318 nm