Principles of Instrumental Analysis
7th Edition
ISBN: 9781337468039
Author: Skoog
Publisher: Cengage
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Chapter 10, Problem 10.8QAP
Interpretation Introduction
Interpretation:
The reason for the ionization interferences to be less severe in ICP than in flame emission spectroscopy needs to be explained.
Concept introduction:
ICP (Inductively coupled plasma) spectroscopy is a kind of mass spectrometry that detects metals as well as non-metals of low concentrations as one part in 1015 (part per quadrillion, ppq) on the low-background isotopes. Flame emission spectroscopy is a traditional method of spectroscopy. In this method, quantitative measurement of the optical emission obtained from the stimulated atoms are used to determine the concentration of an element that is to be analyzed. Ionization interference can be defined as a phenomenon which exhibits a change in emission intensity resulting in the ionization equilibrium to shift in presence of other coexisting elements.
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1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.
Give the product of the bimolecular elimination from each of the isomeric halogenated compounds.
Reaction A
Reaction B.
КОВ
CH₂
HotBu
+B+
ко
HOIBU
+Br+
Templates More
QQQ
Select Cv Templates More
Cras
QQQ
One of these compounds undergoes elimination 50x faster than the other. Which one and why?
Reaction A because the conformation needed for elimination places the phenyl groups and to each other
Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other.
◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other.
Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.
Five isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane
The IR spectra of these five alkenes have the key absorptions (in cm
Compound
Compound A
–912. (§), 994 (5), 1643 (%), 3077 (1)
Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption)
Compound C
Compound D)
–714 (5), 1665 (w), 3010 (m)
885 (3), 1650 (m), 3086 (m)
967 (5), no aharption 1600 to 1700, 3040 (m)
Compound K
Match each compound to the data presented.
Compound A
Compound B
Compound C
Compound D
Compound
Chapter 10 Solutions
Principles of Instrumental Analysis
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