INSTRUMENTAL ANALYSIS-ACCESS >CUSTOM<
7th Edition
ISBN: 9781337783439
Author: Skoog
Publisher: CENGAGE C
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Chapter A1, Problem A1.2QAP
Interpretation Introduction
(a)
Interpretation:
The absolute error for the mean of each set is to be determined for following data.
| A | B | C | D |
| 61.45 | 3.27 | 12.06 | 2.7 |
| 61.53 | 3.26 | 12.14 | 2.4 |
| 61.32 | 3.24 | 2.6 | |
| 3.24 | 2.9 | ||
| 3.28 | |||
| 3.23 |
The accepted value for the sets are A= 61.71, B = 3.28, C= 12.23, D = 2.75
Concept introduction:
The following formula will be used for the calculation of the absolute error for the mean-
Where,
Also,
Interpretation Introduction
(b)
Interpretation:
The percent relative error of each mean is to be determined.
Concept introduction:
The following formula will be used for the calculation of the percentage of relative error for the mean-
Where,
Also,
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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 A1 Solutions
INSTRUMENTAL ANALYSIS-ACCESS >CUSTOM<
Ch. A1 - Prob. A1.1QAPCh. A1 - Prob. A1.2QAPCh. A1 - Prob. A1.3QAPCh. A1 - Prob. A1.4QAPCh. A1 - Prob. A1.5QAPCh. A1 - Prob. A1.6QAPCh. A1 - Prob. A1.7QAPCh. A1 - Prob. A1.8QAPCh. A1 - Prob. A1.9QAPCh. A1 - Prob. A1.10QAP
Ch. A1 - Prob. A1.11QAPCh. A1 - Prob. A1.12QAPCh. A1 - Prob. A1.13QAPCh. A1 - Prob. A1.14QAPCh. A1 - Prob. A1.15QAPCh. A1 - Prob. A1.16QAPCh. A1 - Prob. A1.17QAPCh. A1 - Prob. A1.18QAPCh. A1 - Prob. A1.19QAPCh. A1 - Prob. A1.20QAPCh. A1 - Prob. A1.21QAPCh. A1 - Prob. A1.22QAPCh. A1 - Prob. A1.23QAPCh. A1 - Prob. A1.24QAP
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