
(a)
Interpretation:
The major advantage and disadvantages of two dimensional NMR methods over conventional one-dimensional techniques are to be stated.
Concept introduction:
Nuclear magnetic resonance spectroscopy is a method used to determine the physical and chemical nature of atoms in different compounds. In two-dimensional nuclear magnetic resonance spectroscopy information about atoms is plotted in two frequency axes.

Answer to Problem 19.42QAP
The major advantage of two dimensional NMR method is that chemical shift and coupling constants are obtained and disadvantage is the complex analysis of two dimensional NMR method.
Explanation of Solution
The advantages and disadvantages of two dimensional NMR methods over conventional one-dimensional techniques are as mentioned below.
S. No. | Advantages | Disadvantages |
1 | One dimensional NMR gives information about simple organic compounds while two dimensional NMR gives information about complex organic compounds. | The two dimensional NMR consists of different phases of cross and diagonal peak multi-pet. |
2 | Two dimensional NMR gives information about the chemical shift, coupling constant and signals of compounds. | The analysis of two dimensional NMR is complex as compared to one dimensional NMR. |
3 | The spectrum of complex compounds like proteins, heterocyclic compounds can only found by two dimensional NMR. | The intense peak may get obscured by cross-peaks in two dimensional NMR. |
(b)
Interpretation:
The pulse sequence used in COSY and HETCOR experiments and behavior of magnetization vector
Concept introduction:
The structure of heterocyclic complex compounds is determined by two dimensional NMR. The two dimensional NMR involves the techniques COSY and HETCOR. In the HETCOR experiment, the behavior of a single bond between two different nucleoli is observed. The COSY experiment interprets the resonance of correlating protons when there is spin-spin bond coupling in protons.

Answer to Problem 19.42QAP
The pulse sequence used in COSY and HETCOR experiments explains the single bond coupling between two different nucleoli and resonance in the compound. The magnetizing behavior of vector
Explanation of Solution
The COSY and HETCOR are two pulse sequence experiments. There is an evolution period between two pulses. The evolution period is the difference between time periods of pulses. In the chemical shift, coupling constant arises between this evolution period. The COSY and HETCOR experiments give the spectrum of the complex compound. The two pulses are applied with some time duration on a sample of the compound. This process is applied again after the sample of the compound achieves equilibrium. The spectrum of a compound is obtained by repeating this process several times. The chemical shift, coupling constant and signals are obtained by analyzing the spectrum.
The behavior of the magnetization, vector
(c)
Interpretation:
The identification of resonances in
Concept introduction:
The resonances in the compound are found by the spectrum of the compound.

Answer to Problem 19.42QAP
The resonances in atom
Explanation of Solution
The resonance in the compound is found by the environment of
The below figure represents the proton NMR of
The
The double bond is also present between atoms
(d)
Interpretation:
The differences in the resonance of atoms
Concept introduction:
The resonance in different atoms can be distinguished by chemical shift, coupling constant and signal of pulses.

Answer to Problem 19.42QAP
The resonance of atoms
Explanation of Solution
The atom
On the basis of chemical shift resonance of atoms
(e)
Interpretation:
The resonances in
Concept introduction:
The resonance of different atoms in a compound is explained on the basis of chemical shift. The chemical shift is the resonate frequency of the nucleus in the magnetic field applied. On the basis of the position and chemical shift of nucleus resonance in the compound can be explained.

Answer to Problem 19.42QAP
The atoms
Explanation of Solution
The peaks of
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Chapter 19 Solutions
Principles of Instrumental Analysis
- Using the bond energy values, calculate the energy that must be supplied or is released upon the polymerization of 755 monomers. If energy must be supplied, provide a positive number; if energy is released, provide a negative number. Hint: Avogadro’s number is 6.02 × 1023.arrow_forward-AG|F=2E|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: Acidic solution -0.93 +0.38 -0.51 -0.06 H3PO4 →H4P206 H3PO3 H3PO2 → P→ PH3 -0.28 -0.50 → -0.50 Basic solution 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 H4P2O6 (-0.93/+0.38V). Calculate the values of AG's for both processes; comment. (3 points) 0.5 PH, 0.0 -0.5- 2 3 9 3 -1.5 -2.0 Pa H,PO H,PO H,PO -3 -1 0 2 4 Oxidation state, N 2 b) Frost diagram for phosphorus under acidic conditions is shown. Identify possible disproportionation and comproportionation processes; write out chemical equations describing them. (2 points) c) Elemental phosphorus tends to disproportionate under basic conditions. Use data in…arrow_forwardThese two reactions appear to start with the same starting materials but result in different products. How do the chemicals know which product to form? Are both products formed, or is there some information missing that will direct them a particular way?arrow_forward
- What would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 1 2 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure. 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Priva ×arrow_forwardPredict the products of this organic reaction: Explanation Check IN NaBH3CN H+ ? Click and drag to start drawing a structure. D 5 C +arrow_forwardPredict the products of this organic reaction: H3O+ + ? • Draw all the reasonable products in the drawing area below. If there are no products, because no reaction will occur, check the box under the drawing area. • Include both major and minor products, if some of the products will be more common than others. • Be sure to use wedge and dash bonds if you need to distinguish between enantiomers. No reaction. Click and drag to start drawing a structure. dmarrow_forward
- Iarrow_forwardDraw the anti-Markovnikov product of the hydration of this alkene. this problem. Note for advanced students: draw only one product, and don't worry about showing any stereochemistry. Drawing dash and wedge bonds has been disabled for esc esc ☐ Explanation Check F1 1 2 F2 # 3 F3 + $ 14 × 1. BH THE BH3 2. H O NaOH '2 2' Click and drag to start drawing a structure. F4 Q W E R A S D % 905 LL F5 F6 F7 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility < & 6 7 27 8 T Y U G H I F8 F9 F10 F11 F12 9 0 J K L P + // command option Z X C V B N M H H rol option commandarrow_forwardAG/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_forward
- 4. 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_forward5. Consider the compounds shown below as ligands in coordination chemistry and identify their denticity; comment on their ability to form chelate complexes. (6 points) N N A B N N N IN N Carrow_forward
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