
Concept explainers
(a)
Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.
Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In

Answer to Problem 14.17P
The labeled proton splits into septet in NMR spectrum of the given compound.
Explanation of Solution
The given compound is
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled proton gives seven peaks in NMR spectrum.
The labeled proton splits into septet in NMR spectrum of the given compound.
(b)
Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.
Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In

Answer to Problem 14.17P
The labeled proton (a), (b), and (c) splits into triplet, multiplet and quintet in NMR spectrum of the given compound.
Explanation of Solution
The given compound is
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled proton (a) gives three peaks (triplet) in NMR spectrum.
Proton (b) is bonded to one methylene group that has two hydrogen atoms and one methyl group that has three hydrogen atoms. The number of peaks is calculated by the formula,
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled proton (b) splits into a multiplet in NMR spectrum.
Proton (c) is bonded to two methylene groups that have four hydrogen atoms. The number of peaks is calculated by the formula,
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled proton (c) gives quintet in NMR spectrum.
Therefore, protons (a), (b), and (c) give
The labeled proton (a), (b), and (c) splits into triplet, multiplet and quintet in NMR spectrum of the given compound.
(c)
Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.
Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In

Answer to Problem 14.17P
The labeled proton (a) and (b) splits into doublet and sextet in NMR spectrum of the given compound.
Explanation of Solution
The given compound is shown below.
Figure 1
There are two labeled protons, (a) and (b). Proton (b) is bonded to a carbon atom that has one hydrogen and one methylene group. The number of peaks is calculated by the formula,
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled proton (b) splits into a sextet in NMR spectrum.
Proton (a) is bonded to a carbon atom that has only one hydrogen atom. The number of peaks is calculated by the formula,
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled proton (a) gives two peaks (doublet) in NMR spectrum.
Therefore, protons (a) and (b) give
The labeled proton (a) and (b) splits into doublet and sextet in NMR spectrum of the given compound.
(d)
Interpretation: The number of peaks present in the given NMR signal of labeled proton is to be calculated.
Concept introduction: In NMR spectrum, peaks are known as resonances, lines or absorptions. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. In

Answer to Problem 14.17P
The labeled proton (a) and (b) splits into triplet, doublet and doublet in NMR spectrum of the given compound.
Explanation of Solution
The given compound is shown below.
Figure 2
There are two labeled protons, (a) and (b). Proton (a) is bonded to a carbon atom that has two hydrogen atoms. The number of peaks is calculated by the formula,
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled proton (a) gives three peaks (triplet) in NMR spectrum.
Both protons (b) are bonded to a carbon atom that has only one hydrogen atom. The number of peaks is calculated by the formula,
Where,
➢
➢
Here,
Substitute the value of
Hence, the labeled protons (b) give two peaks (doublet) in NMR spectrum.
Therefore, protons (a) and (b) give
The labeled proton (a) and (b) splits into triplet, doublet and doublet in NMR spectrum of the given compound.
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Chapter 14 Solutions
PKG ORGANIC CHEMISTRY
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- 3. Use Kapustinskii's equation and data from Table 4.10 in your textbook to calculate lattice energies of Cu(OH)2 and CuCO3 (4 points)arrow_forward2. Copper (II) oxide crystalizes in monoclinic unit cell (included below; blue spheres 2+ represent Cu²+, red - O²-). Use Kapustinski's equation (4.5) to calculate lattice energy for CuO. You will need some data from Resource section of your textbook (p.901). (4 points) CuOarrow_forwardWhat is the IUPAC name of the following compound? OH (2S, 4R)-4-chloropentan-2-ol O (2R, 4R)-4-chloropentan-2-ol O (2R, 4S)-4-chloropentan-2-ol O(2S, 4S)-4-chloropentan-2-olarrow_forward
- Use the reaction coordinate diagram to answer the below questions. Type your answers into the answer box for each question. (Watch your spelling) Energy A B C D Reaction coordinate E A) Is the reaction step going from D to F endothermic or exothermic? A F G B) Does point D represent a reactant, product, intermediate or transition state? A/ C) Which step (step 1 or step 2) is the rate determining step? Aarrow_forward1. Using radii from Resource section 1 (p.901) and Born-Lande equation, calculate the lattice energy for PbS, which crystallizes in the NaCl structure. Then, use the Born-Haber cycle to obtain the value of lattice energy for PbS. You will need the following data following data: AH Pb(g) = 196 kJ/mol; AHƒ PbS = −98 kJ/mol; electron affinities for S(g)→S¯(g) is -201 kJ/mol; S¯(g) (g) is 640kJ/mol. Ionization energies for Pb are listed in Resource section 2, p.903. Remember that enthalpies of formation are calculated beginning with the elements in their standard states (S8 for sulfur). The formation of S2, AHF: S2 (g) = 535 kJ/mol. Compare the two values, and explain the difference. (8 points)arrow_forwardIn the answer box, type the number of maximum stereoisomers possible for the following compound. A H H COH OH = H C Br H.C OH CHarrow_forward
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