(a) Interpretation: The position of protons bonded to the s p 2 hybridized carbons (absorbs in the 1 H NMR spectrum) in the 1 H NMR spectrum of given compound is to be estimated. Concept introduction: The protons present on s p 2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5 − 8 ppm . The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5 − 6 ppm .

Question

Imagine an electrochemical cell based on these two half reactions with electrolyte concentrations as given below: Oxidation: Pb(s) → Pb2+(aq, 0.10 M) + 2 e– Reduction: MnO4–(aq, 1.50 M) + 4 H+(aq, 2.0 M) + 3 e– → MnO2(s) + 2 H2O(l) Calculate Ecell (assuming temperature is standard 25 °C).

Answer 1

Question

Chapter 17, Problem 17.9P

Interpretation Introduction

(a)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(b)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(c)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) of given compound are to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

Imagine an electrochemical cell based on these two half reactions with electrolyte concentrations as given below: Oxidation: Pb(s) → Pb2+(aq, 0.10 M) + 2 e– Reduction: MnO4–(aq, 1.50 M) + 4 H+(aq, 2.0 M) + 3 e– → MnO2(s) + 2 H2O(l) Calculate Ecell (assuming temperature is standard 25 °C).

: ☐ + Draw the Fischer projection of the most common naturally-occurring form of aspartate, 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. ✓

For a silver-silver chloride electrode, the following potentials are observed: E°cell = 0.222 V and E(saturated KCl) = 0.197 V Use this information to find the [Cl–] (technically it’s the activity of Cl– that’s relevant here, but we’ll just call it “concentration” for simplicity) in saturated KCl.

Answer 2

Question

Chapter 17, Problem 17.9P

Interpretation Introduction

(a)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(b)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(c)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) of given compound are to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 3

Question

Chapter 17, Problem 17.9P

Interpretation Introduction

(a)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(b)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(c)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) of given compound are to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 4

Question

Chapter 17, Problem 17.9P

Interpretation Introduction

(a)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(b)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(c)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) of given compound are to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 5

Chapter 17, Problem 17.9P

Interpretation Introduction

(a)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(b)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Interpretation Introduction

(c)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) of given compound are to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Answer 6

Chapter 17, Problem 17.9P

Interpretation Introduction

(a)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Answer 7

Chapter 17, Problem 17.9P

Interpretation Introduction

(a)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Answer 8

Interpretation Introduction

(b)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Answer 9

Interpretation Introduction

(b)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) in the 1H NMR spectrum of given compound is to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Answer 10

Interpretation Introduction

(c)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) of given compound are to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Answer 11

Interpretation Introduction

(c)

Interpretation: The position of protons bonded to the sp2 hybridized carbons (absorbs in the 1H NMR spectrum) of given compound are to be estimated.

Concept introduction: The protons present on sp2 hybridized carbons in aromatic hydrocarbon are highly deshielded and show absorption at 6.5−8 ppm. The protons present in non aromatic hydrocarbon are highly shielded and show absorption at 4.5−6 ppm.

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 15

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 16

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 17

Trending nowThis is a popular solution!

Answer 18

Trending nowThis is a popular solution!

Answer 19

Trending nowThis is a popular solution!

Answer 20

Trending now

Answer 21

This is a popular solution!

Answer 22

See solution

Check out sample textbook solution

Answer 23

Ch. 17 - Prob. 17.1P Ch. 17 - Problem 17.2 What orbitals are used to form the...Ch. 17 - Problem-17.3. Give the IUPAC name for each...Ch. 17 - Prob. 17.4P Ch. 17 - Problem-17.5 What is the structure of propofol,...Ch. 17 - Problem 17.6 What is the structure of a compound...Ch. 17 - Problem 17.7 How many NMR signals does each...Ch. 17 - Prob. 17.8P Ch. 17 - Prob. 17.9P Ch. 17 - Prob. 17.10P

Solution Summary: The author explains that the position of protons bonded to the sp2 hybridized carbons is to be estimated.

Concept explainers

Chapter 17 Solutions