ORGANIC CHEMISTRY (LOOSELEAF)
6th Edition
ISBN: 9781260475630
Author: SMITH
Publisher: MCG
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Chapter 29.8, Problem 17P
Interpretation Introduction
Interpretation: An explanation corresponding to the given statement that the treatment of cholesterol with mCPBA results in formation of a single
Concept introduction: Steroids, a group of tetracyclic lipids composed of one five-membered ring and three six-membered rings. These rings are lettered by A, B, C, and D. Most of the steroids contain two angular methyl groups. These methyl groups are numbered as
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Students have asked these similar questions
Write the complete common (not IUPAC) name of each molecule below.
Note: if a molecule is one of a pair of enantiomers, be sure you start its name with D- or L- so we know which enantiomer it is.
molecule
Ο
C=O
common name
(not the IUPAC
name)
H
☐
H3N
CH₂OH
0-
C=O
H
NH3
CH₂SH
H3N
☐
☐
X
G
(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
Chapter 29 Solutions
ORGANIC CHEMISTRY (LOOSELEAF)
Ch. 29.2 - Problem 31.1
One component of jojoba oil is a wax...Ch. 29.3 - Problem 31.2
How would you expect the melting...Ch. 29.3 - Problem 31.3
Draw the products formed when...Ch. 29.3 - Problem 31.4
The main fatty acid component of the...Ch. 29.3 - Prob. 5PCh. 29.7 - Problem 31.10
Locate the isoprene units in each...Ch. 29.7 - Problem 31.11
Locate the isoprene units in...Ch. 29.7 - Problem 31.12
Write a stepwise mechanism for the...Ch. 29.7 - Prob. 14PCh. 29.8 - Prob. 15P
Ch. 29.8 - Prob. 16PCh. 29.8 - Prob. 17PCh. 29 - 31.17 Locate the isoprene units in each...Ch. 29 - Prob. 25PCh. 29 - Locate the isoprene units in each compound. a. e....Ch. 29 - 31.27 Classify each terpene and terpenoid in...Ch. 29 - Prob. 33PCh. 29 - Draw three-dimensional structures f or each...Ch. 29 - Prob. 37PCh. 29 - Prob. 38PCh. 29 - Prob. 39PCh. 29 - Prob. 40PCh. 29 - 31.38 Draw the products formed when cholesterol is...Ch. 29 - 31.39 Draw a stepwise mechanism for the following...Ch. 29 - 31.40 Draw a stepwise mechanism for the following...Ch. 29 - Prob. 44P
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- 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+].arrow_forward2. (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_forward
- draw skeletal structures for the minor products of the reaction.arrow_forward1. 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_forward
- Check 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_forwardWhat 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_forward
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