INTO TO CHEMISTRY (EBOOK ACCESS CODE)
5th Edition
ISBN: 9781307892864
Author: BAUER
Publisher: MCG
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Chapter 12, Problem 5QC
Interpretation Introduction
Interpretation:
The state of equilibrium for a reaction is to be provided.
Concept Introduction:
Expert Solution & Answer
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(Part A) Provide structures of the FGI products and missing reagents (dashed box)
1 eq Na* H*
H
-H
B1
B4
R1
H2 (gas)
Lindlar's
catalyst
A1
Br2
MeOH
H2 (gas)
Lindlar's
catalyst
MeO.
OMe
C6H1402
B2
B3
A1
Product carbons' origins
Draw a box around product
C's that came from A1.
Draw a dashed box around
product C's that came from B1.
Classify each of the amino acids below.
Note for advanced students: none of these amino acids are found in normal proteins.
X
CH2
H3N-CH-COOH3N-CH-COO-
H3N-CH-COO
CH2
CH3-C-CH3
CH2
NH3
N
NH
(Choose one) ▼
(Choose one)
S
CH2
OH
(Choose one) ▼
+
H3N-CH-COO¯
CH2
H3N CH COO H3N-CH-COO
CH2
오오
CH
CH3
CH2
+
O
C
CH3
O=
O_
(Choose one)
(Choose one) ▼
(Choose one)
G
Another standard reference electrode is the standard calomel electrode:
Hg2Cl2(s) (calomel) + 2e2 Hg() +2 Cl(aq)
This electrode is usually constructed with saturated KCI to keep the Cl- concentration
constant (similar to what we discussed with the Ag-AgCl electrode). Under these
conditions the potential of this half-cell is 0.241 V.
A measurement was taken by dipping a Cu wire and a saturated calomel electrode into a
CuSO4 solution:
saturated
calomel
electrode
potentiometer
copper wire
CuSO4
a) Write the half reaction for the Cu electrode.
b) Write the Nernst equation for the Cu electrode, which will include [Cu2+]
c) If the voltage on the potentiometer reads 0.068 V, solve for [Cu²+].
Chapter 12 Solutions
INTO TO CHEMISTRY (EBOOK ACCESS CODE)
Ch. 12 - Prob. 1QCCh. 12 - Prob. 2QCCh. 12 - Prob. 3QCCh. 12 - Prob. 4QCCh. 12 - Prob. 5QCCh. 12 - Prob. 6QCCh. 12 - Prob. 1PPCh. 12 - Prob. 2PPCh. 12 - Prob. 3PPCh. 12 - Prob. 4PP
Ch. 12 - Prob. 5PPCh. 12 - Prob. 6PPCh. 12 - Prob. 7PPCh. 12 - Prob. 8PPCh. 12 - Prob. 9PPCh. 12 - Prob. 10PPCh. 12 - Consider the following equilibrium:...Ch. 12 - Prob. 12PPCh. 12 - Prob. 1QPCh. 12 - Match the key terms with the descriptions...Ch. 12 - Prob. 3QPCh. 12 - Prob. 4QPCh. 12 - Prob. 5QPCh. 12 - Prob. 6QPCh. 12 - Prob. 7QPCh. 12 - Prob. 8QPCh. 12 - Prob. 9QPCh. 12 - Prob. 10QPCh. 12 - Prob. 11QPCh. 12 - Prob. 12QPCh. 12 - Prob. 13QPCh. 12 - Prob. 14QPCh. 12 - Prob. 15QPCh. 12 - Prob. 16QPCh. 12 - Prob. 17QPCh. 12 - Prob. 18QPCh. 12 - Prob. 19QPCh. 12 - Prob. 20QPCh. 12 - Prob. 21QPCh. 12 - Prob. 22QPCh. 12 - Prob. 23QPCh. 12 - Prob. 24QPCh. 12 - Prob. 25QPCh. 12 - Prob. 26QPCh. 12 - Prob. 27QPCh. 12 - Prob. 28QPCh. 12 - Prob. 29QPCh. 12 - Prob. 30QPCh. 12 - Prob. 31QPCh. 12 - Prob. 32QPCh. 12 - Prob. 33QPCh. 12 - Prob. 34QPCh. 12 - Prob. 35QPCh. 12 - Prob. 36QPCh. 12 - Prob. 37QPCh. 12 - Prob. 38QPCh. 12 - Prob. 39QPCh. 12 - Prob. 40QPCh. 12 - Prob. 41QPCh. 12 - Prob. 42QPCh. 12 - Prob. 43QPCh. 12 - Prob. 44QPCh. 12 - Prob. 45QPCh. 12 - Prob. 46QPCh. 12 - Prob. 47QPCh. 12 - Prob. 48QPCh. 12 - Prob. 49QPCh. 12 - Prob. 50QPCh. 12 - Prob. 51QPCh. 12 - Prob. 52QPCh. 12 - Prob. 53QPCh. 12 - Prob. 54QPCh. 12 - Prob. 55QPCh. 12 - Prob. 56QPCh. 12 - Prob. 57QPCh. 12 - Prob. 58QPCh. 12 - Prob. 59QPCh. 12 - Prob. 60QPCh. 12 - Prob. 61QPCh. 12 - Prob. 62QPCh. 12 - Prob. 63QPCh. 12 - Prob. 64QPCh. 12 - Prob. 65QPCh. 12 - Prob. 66QPCh. 12 - Prob. 67QPCh. 12 - Prob. 68QPCh. 12 - Prob. 69QPCh. 12 - Prob. 70QPCh. 12 - Prob. 71QPCh. 12 - Prob. 72QPCh. 12 - Prob. 73QPCh. 12 - Prob. 74QPCh. 12 - Prob. 75QPCh. 12 - Prob. 76QPCh. 12 - Prob. 77QPCh. 12 - Prob. 78QPCh. 12 - Prob. 79QPCh. 12 - Prob. 80QPCh. 12 - Prob. 81QPCh. 12 - Prob. 82QPCh. 12 - Prob. 83QPCh. 12 - Prob. 84QPCh. 12 - Prob. 85QPCh. 12 - Prob. 86QPCh. 12 - Prob. 87QPCh. 12 - Prob. 88QPCh. 12 - Prob. 89QPCh. 12 - Prob. 90QPCh. 12 - Prob. 91QPCh. 12 - Prob. 92QPCh. 12 - Prob. 93QPCh. 12 - Prob. 94QPCh. 12 - Prob. 95QPCh. 12 - Prob. 96QPCh. 12 - Prob. 97QPCh. 12 - Prob. 98QPCh. 12 - Prob. 99QPCh. 12 - Prob. 100QPCh. 12 - Prob. 101QPCh. 12 - Prob. 102QPCh. 12 - Prob. 103QPCh. 12 - Prob. 104QPCh. 12 - Prob. 105QPCh. 12 - Prob. 106QPCh. 12 - Prob. 107QPCh. 12 - Prob. 108QPCh. 12 - Prob. 109QPCh. 12 - Prob. 110QPCh. 12 - Prob. 111QPCh. 12 - Prob. 112QPCh. 12 - Prob. 113QPCh. 12 - Prob. 114QPCh. 12 - Prob. 115QPCh. 12 - Prob. 116QPCh. 12 - Prob. 117QPCh. 12 - Prob. 118QPCh. 12 - Prob. 119QPCh. 12 - Prob. 120QPCh. 12 - Prob. 121QPCh. 12 - Prob. 122QPCh. 12 - Prob. 123QPCh. 12 - Prob. 124QPCh. 12 - Prob. 125QPCh. 12 - Prob. 126QPCh. 12 - Prob. 127QPCh. 12 - Prob. 128QPCh. 12 - Prob. 129QPCh. 12 - Prob. 130QPCh. 12 - Prob. 131QPCh. 12 - Prob. 132QPCh. 12 - Prob. 133QPCh. 12 - Prob. 134QPCh. 12 - Prob. 135QPCh. 12 - Prob. 136QPCh. 12 - Prob. 137QPCh. 12 - Prob. 138QPCh. 12 - Prob. 139QPCh. 12 - Prob. 140QPCh. 12 - Prob. 141QPCh. 12 - Prob. 142QPCh. 12 - Prob. 143QP
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- 2. (Part B). Identify a sequence of FGI that prepares the Synthesis Target 2,4-dimethoxy- pentane. All carbons in the Synthesis Target must start as carbons in either ethyne, propyne or methanol. Hint: use your analysis of Product carbons' origins (Part A) to identify possible structure(s) of a precursor that can be converted to the Synthesis Target using one FGI. All carbons in the Synthesis Target must start as carbons in one of the three compounds below. H = -H H = -Me ethyne propyne Synthesis Target 2,4-dimethoxypentane MeOH methanol OMe OMe MeO. OMe C₂H₁₂O₂ Product carbons' origins Draw a box around product C's that came from A1. Draw a dashed box around product C's that came from B1.arrow_forwardDraw the skeletal ("line") structure of the smallest organic molecule that produces potassium 3-hydroxypropanoate when reacted with KOH. Click and drag to start drawing a structure. Sarrow_forwardDraw the skeleatal strucarrow_forward
- ← Problem 14 of 31 Submit Draw the major product for this reaction. Ignore inorganic byproducts. 1. BH3-THF 2. H2O2, NaOHarrow_forwardClassify each amino acid below as nonpolar, polar neutral, polar acidic, or polar basic.arrow_forwarddraw skeletal structures for the minor products of the reaction.arrow_forward
- 1. Provide missing starting materials, reagents, products. If a product cannot be made, write NP (not possible) in the starting material box. C7H12O Ph HO H 1) 03-78 C 2) Me₂S + Ph .H OH + 2nd stereoisomer OH Ph D + enantiomer cat OsO 4 NMO H2O acetonearrow_forwardPlease note that it is correct and explains it rightly:Indicate the correct option. The proportion of O, C and H in the graphite oxide is:a) Constant, for the quantities of functional groups of acids, phenols, epoxy, etc. its constants.b) Depending on the preparation method, as much oxidant as the graphite is destroyed and it has less oxygen.c) Depends on the structure of the graphic being processed, whether it can be more tridimensional or with larger crystals, or with smaller crystals and with more edges.arrow_forwardCheck the box under each a amino acid. If there are no a amino acids at all, check the "none of them" box under the table. Note for advanced students: don't assume every amino acid shown must be found in nature. ནང་་་ OH HO HO NH2 + NH3 O OIL H-C-CO CH3-CH O C=O COOH COOH + H2N C-H O H2N C H CH3-CH CH2 HO H3N O none of them 口 CH3 CH2 OH Хarrow_forward
- What is the systematic name of the product P of this chemical reaction? 010 HO-CH2-CH2-C-OH ☐ + NaOH P+ H2Oarrow_forward1. Provide missing starting materials, reagents, products. If a product cannot be made, write NP (not possible) in the starting material box. a) C10H12 Ph OMe AcOHg+ + enantiomer Br C6H10O2 + enantiomerarrow_forwardDraw the Fischer projection of the most common naturally-occurring form of cysteine, with the acid group at the top and the side chain at the bottom. Important: be sure your structure shows the molecule as it would exist at physiological pH. Click and drag to start drawing a structure. :☐ ©arrow_forward
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