Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group. Concept Introduction: Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses. The number of carbon atoms present in the chain characterize the monosaccharide. They are given below. Carbon chain with three carbon atoms is triose. Carbon chain with four carbon atoms is tetrose. Carbon chain with five carbon atoms is as pentose. Carbon chain with six carbon atoms is as hexose.

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Answer 1

Question

Chapter 14, Problem 14.32AP

a.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

c.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

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Done 11:14 ⚫ worksheets.beyondlabz.com 5 (a). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 6 5 (b). Using the peak information you listed in the tables for both structures, assign each peak to that portion of the structure that produces the peak in the NMR spectrum. Draw this diagram on your own sheet of paper and attach the sketch of your drawing to this question. Question 7 6. Are there any differences between the spectra you obtained in Beyond Labz and the predicted spectra? If so, what were the differences? <

2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled** Peak Chemical Shift (d) 5.7 1 Multiplicity multiplate .......... 5.04 double of doublet 2 4.98 double of doublet 3 4.05 doublet of quartet 4 5 LO 3.80 quartet 1.3 doublet 6 Peak Chemical Shift (d) Multiplicity

Interpreting NMR spectra is a skill that often requires some amount of practice, which, in turn, necessitates access to a collection of NMR spectra. Beyond Labz Organic Synthesis and Organic Qualitative Analysis have spectral libraries containing over 700 1H NMR spectra. In this assignment, you will take advantage of this by first predicting the NMR spectra for two closely related compounds and then checking your predictions by looking up the actual spectra in the spectra library. After completing this assignment, you may wish to select other compounds for additional practice. 1. Write the IUPAC names for the following two structures: Question 2 Question 3 2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled**

Answer 2

Question

Chapter 14, Problem 14.32AP

a.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

c.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

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Answer 3

Question

Chapter 14, Problem 14.32AP

a.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

c.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

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Answer 4

Question

Chapter 14, Problem 14.32AP

a.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

c.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

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Answer 5

Chapter 14, Problem 14.32AP

a.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

b.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

c.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

Answer 6

Chapter 14, Problem 14.32AP

a.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

Answer 7

Chapter 14, Problem 14.32AP

a.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Simplest carbohydrates are known as monosaccharides. They contain three to six carbons generally in a chain form with a carbonyl group present in the terminal or the adjacent carbon atom from the terminal. Monosaccharides that have the carbonyl group at the terminal carbon atom C1 are known as aldoses and the monosaccharides that have the carbonyl group on the adjacent carbon atom C2 are known as ketoses.

The number of carbon atoms present in the chain characterize the monosaccharide. They are given below.

Carbon chain with three carbon atoms is triose.

Carbon chain with four carbon atoms is tetrose.

Carbon chain with five carbon atoms is as pentose.

Carbon chain with six carbon atoms is as hexose.

Answer 8

b.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

Answer 9

b.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

Answer 10

c.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

Answer 11

c.

Interpretation Introduction

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

Answer 12

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Answer 13

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Answer 14

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Answer 15

Ch. 14.1 - Draw a Lewis structure for glucose that clearly...Ch. 14.2 - Prob. 14.2P Ch. 14.2 - Prob. 14.3P Ch. 14.2 - Prob. 14.4P Ch. 14.2 - Rank the following compounds in order of...Ch. 14.2 - Prob. 14.6P Ch. 14.2 - Prob. 14.7P Ch. 14.2 - Prob. 14.8P Ch. 14.2 - Prob. 14.9P Ch. 14.3 - Prob. 14.10P

Solution Summary: The author explains how monosaccharides are classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept explainers

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

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Chapter 14 Solutions

Solution Summary: The author explains how monosaccharides are classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept explainers

Interpretation: Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group. Concept Introduction: Refer part “a.”.

Interpretation: Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group. Concept Introduction: Refer part “a.”.

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Chapter 14 Solutions

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.

Interpretation:

Given monosaccharide has to be classified based on the number of carbon atoms present in the chain and also the carbonyl group.

Concept Introduction:

Refer part “a.”.