Organic Chemistry - Standalone book
10th Edition
ISBN: 9780073511214
Author: Francis A Carey Dr., Robert M. Giuliano
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 5, Problem 43P
The reaction of
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1. Answer the following questions about molecule 1.
a) circle the most basic
atom in molecule
b) Provide the structure of the conjugate acid and a resonance structure of
the conjugate acid that explain its bascity.
HO
H+
molecule 1
c) The pKa of the conjugate acid of the most basic atom in molecule is approximately -1. Which of the
following acids can completely protonate the most basic atom? (Keq>10³).
HI
H₂O
IZ
Cl3C
OH
(+)
NH3
Question 3
Some photophysical parameters (lifetime, t, and quantum yield, $) for fluorescence (n = 370 nm)
and phosphorence (p = 580 nm) of pyrene, 1-chloropyrene and 1-bromopyrene are given in the
table below, as measured at room temperature (RT) and at 77 K in a frozen ethanol glass.
фе
τη
Фл
Фр
Tp
(RT)
(RT)
(77 K)
(77 K)
(77 K)
(ns)
(s)
X = H
0.72
530
0.9
< 0.001
0.39
X
X = CI
0.22
75
0.59
0.058
0.10
X = Br 0.032 2
0.17
0.085
0.004
fl = fluorescence; p = phosphorescence
(a) Construct a Jablonski diagram for pyrene (X = H) at 77K.
(b) Pyrene (X = H) has an absorbance maximum, Amax, at 330 nm and a fluorescence
maximum, 11, at 370 nm. Why does this difference in wavelengths occur?
(c) Explain why the lifetime for phosphorescence is longer than that for fluorescence.
(d) Why does the fluorescence quantum yield increase with decreasing temperature?
(e) Explain the trend in phosphorescence quantum yield as X is varied.
Question 4
The photoisomerization of alkenes is a photochemical transformation between the E- and
Z-stereoisomers. The irradiation of the E-isomer (shown below) with radiation at 340 nm gives an
E:Z ratio of 5:95. Some relevant information for each compound is shown in the table below.
Amax
340nm
E290
E340
(L mol¹ cm¹) (L mol¹ cm-1)
E
340 nm
8000
20000
PE-Z = 0.60
Z
290 nm 16000
2000
Oz-E = 0.30
(a) The reaction proceeds through an excited state. Explain the nature of this excited state, and
explain how it allows formation of the E- and Z-isomers. Explain why this isomerisation is
unlikely to occur thermally.
(b) The product of the equilibrium shown above gives a final ratio with substantially more of one
isomer. Explain why this occurs.
(c) Explain why the starting concentration of isomers does not affect the final ratio after
irradiation.
(d) If the irradiating wavelength used was changed to 290 nm and you started with Z- rather
than E-isomer, use the data in the table above to…
Chapter 5 Solutions
Organic Chemistry - Standalone book
Ch. 5.1 - Prob. 1PCh. 5.1 - Prob. 2PCh. 5.1 - Many compounds contain more than one functional...Ch. 5.2 - Prob. 4PCh. 5.3 - Prob. 5PCh. 5.4 - Classify the isomeric C4H10O alcohols as being...Ch. 5.5 - Bromine is less electronegative than chlorine, yet...Ch. 5.6 - Prob. 8PCh. 5.7 - Prob. 9PCh. 5.8 - Prob. 10P
Ch. 5.8 - Prob. 11PCh. 5.9 - Carbocations are key intermediates in petroleum...Ch. 5.9 - Prob. 13PCh. 5.9 - Prob. 14PCh. 5.11 - Prob. 15PCh. 5.13 - Prob. 16PCh. 5.14 - For the reaction of a primary alcohol RCH2OH with...Ch. 5.15 - Prob. 18PCh. 5 - Write structural formulas for each of the...Ch. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Write structural formulas for all the...Ch. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Epichlorohydrin is the common name of an...Ch. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Some of the most important organic compounds in...Ch. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37PCh. 5 - Prob. 38PCh. 5 - Prob. 39PCh. 5 - Prob. 40PCh. 5 - The reaction of 2,2-dimethyl-1-propanol...Ch. 5 - (a) Assuming that the rate-determining elementary...Ch. 5 - The reaction of 3-tert-butyl-3-pentanol with...Ch. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46DSPCh. 5 - Prob. 47DSPCh. 5 - Prob. 48DSPCh. 5 - Prob. 49DSPCh. 5 - Prob. 50DSPCh. 5 - Prob. 51DSPCh. 5 - Prob. 52DSP
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