
Concept explainers
a.
To convert:
The given condensed structure to skeletal structure.
Introduction:
In the condensed structure, the arrangements of atoms are shown in a single line. The position of branching is not shown in the condensed structure.
The skeletal structure is used to show the arrangement of atoms bonded together in a molecule that forms the basic structure of the compound.
The skeletal structure is shown in two dimensions where carbon atoms are shown by vertices of line segments. Solid lines are used to show bonds between molecules.
Double bonds are shown by two lines and triple bonds are shown by triple lines between the atoms in a structure.
b.
To convert:
The given condensed structure to skeletal structure.
Introduction:
In the condensed structure, the arrangements of atoms are shown in a single line. The position of branching is not shown in the condensed structure.
The skeletal structure is used to show the arrangement of atoms bonded together in a molecule that forms the basic structure of the compound.
The skeletal structure is shown in two dimensions where carbon atoms are shown by vertices of line segments. Solid lines are used to show bonds between molecules.
Double bonds are shown by two lines and triple bonds are shown by triple lines between the atoms in a structure.
c.
To convert:
The given condensed structure to skeletal structure.
Introduction:
In the condensed structure, the arrangements of atoms are shown in a single line. The position of branching is not shown in the condensed structure.
The skeletal structure is used to show the arrangement of atoms bonded together in a molecule that forms the basic structure of the compound.
The skeletal structure is shown in two dimensions where carbon atoms are shown by vertices of line segments. Solid lines are used to show bonds between molecules.
Double bonds are shown by two lines and triple bonds are shown by triple lines between the atoms in a structure.

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Chapter 4 Solutions
GENERAL ORGANIC+BIO...(LL)-W/MOD.ACCESS
- AG/F-2° V 3. Before proceeding with this problem you may want to glance at p. 466 of your textbook where various oxo-phosphorus derivatives and their oxidation states are summarized. Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14: -0.93 +0.38 -0.50 -0.51 -0.06 H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3 Acidic solution Basic solution -0.28 -0.50 3--1.12 -1.57 -2.05 -0.89 PO HPO H₂PO₂ →P → PH3 -1.73 a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH P 0.0 -0.5 -1.0- -1.5- -2.0 H.PO, -2.3+ -3 -2 -1 1 2 3 2 H,PO, b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) H,PO 4 S Oxidation stale, Narrow_forward4. For the following complexes, draw the structures and give a d-electron count of the metal: a) Tris(acetylacetonato)iron(III) b) Hexabromoplatinate(2-) c) Potassium diamminetetrabromocobaltate(III) (6 points)arrow_forward2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation constant for [Fe(CN)6] 4 is ~1032, and that: Fe3+ (aq) + e = Fe²+ (aq) E° = +0.77 V [Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V (4 points)arrow_forward
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- What is the missing reactant R in this organic reaction? N N H3O+ +R + • Draw the structure of R in the drawing area below. • Be sure to use wedge and dash bonds if it's necessary to draw one particular enantiomer. Click and drag to start drawing a structure. fmarrow_forwardThe product on the right-hand side of this reaction can be prepared from two organic reactants, under the conditions shown above and below the arrow. Draw 1 and 2 below, in any arrangement you like. 1+2 NaBH3CN H+ N Click and drag to start drawing a structure. 5arrow_forwardAssign this HSQC Spectrum ( please editing clearly on the image)arrow_forward
- (a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³arrow_forwardfcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transferarrow_forward34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10arrow_forward
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