Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 13, Problem 13.1VP

(a)

Interpretation Introduction

To determine: The degree of unsaturation of the given hydrocarbon.

Interpretation: The degree of unsaturation of the given hydrocarbon is to be calculated.

Concept introduction: The degree of unsaturation is defined as the formula which is useful for the determination of chemical structures of organic compounds. The number of rings, double bonds and triple bonds in a compound are drawn by the degree of unsaturation.

The degree of unsaturation (DOU) of the hydrocarbons is calculated by the formula,

DOU=2×C+2+NXH2

(a)

Expert Solution
Check Mark

Answer to Problem 13.1VP

Answer

The degree of unsaturation of the given hydrocarbon is one.

Explanation of Solution

Explanation

The given hydrocarbon is shown in Figure 1.

Chemistry, Chapter 13, Problem 13.1VP , additional homework tip  1

Figure 1

Count the number of elements (C,N,X,H) present in the given hydrocarbon.

The number of carbon present in given hydrocarbon is 7.

The number of nitrogen present in given hydrocarbon is 0.

The number of halogen present in given hydrocarbon is 0.

The number of hydrogen present in given hydrocarbon is 14.

The degree of unsaturation (DOU) of hydrocarbon is calculated by the formula,

DOU=2×C+2+NXH2 (1)

Where,

  • C is number of carbon present in the given hydrocarbon.
  • N is number of nitrogen present in the given hydrocarbon.
  • X is number of halogen present in the given hydrocarbon.
  • H is number of hydrogen present in the given hydrocarbon.

Substitute the values of C, N, X and H in equation (1) to calculate the degree of unsaturation.

DOU=2×7+2+00142DOU=16142DOU=22DOU=1

Hence, the degree of unsaturation of the given hydrocarbon is one.

(b)

Interpretation Introduction

To determine: The degree of unsaturation of the given hydrocarbon.

Interpretation: The degree of unsaturation of the given hydrocarbon is to be calculated.

Concept introduction: The degree of unsaturation is defined as the formula which is useful for the determination of chemical structures of organic compounds. The number of rings, double bonds and triple bonds in a compound are drawn by the degree of unsaturation.

The degree of unsaturation (DOU) of the given hydrocarbons is calculated by the formula,

DOU=2×C+2+NXH2

(b)

Expert Solution
Check Mark

Answer to Problem 13.1VP

Answer

The degree of unsaturation of the given hydrocarbon is two.

Explanation of Solution

Explanation

The given hydrocarbon is shown in Figure 2.

Chemistry, Chapter 13, Problem 13.1VP , additional homework tip  2

Figure 2

Count the number of elements (C,N,X,H) present in given hydrocarbon.

The number of carbon present in the given hydrocarbon is 7.

The number of nitrogen present in the given hydrocarbon is 0.

The number of halogen present in the given hydrocarbon is 0.

The number of hydrogen present in the given hydrocarbon is 12.

Substitute the values of C, N, X and H in equation (1) to calculate the degree of unsaturation.

DOU=2×7+2+00122DOU=16122DOU=42DOU=2

Hence, the degree of unsaturation of the given hydrocarbon is two.

(c)

Interpretation Introduction

To determine: The degree of unsaturation of the given hydrocarbon.

Interpretation: The degree of unsaturation of the given hydrocarbon is to be calculated.

Concept introduction: The degree of unsaturation is defined as the formula which is useful for the determination of chemical structures of organic compounds. The number of rings, double bonds and triple bonds in a compound are drawn by the degree of unsaturation.

The degree of unsaturation (DOU) of the given hydrocarbons is calculated by the formula,

DOU=2×C+2+NXH2

(c)

Expert Solution
Check Mark

Answer to Problem 13.1VP

Answer

The degree of unsaturation of the given hydrocarbon is zero.

Explanation of Solution

Explanation

The given hydrocarbon is shown in Figure 3.

Chemistry, Chapter 13, Problem 13.1VP , additional homework tip  3

Figure 3

Count the number of elements (C,N,X,H) present in given hydrocarbon.

The number of carbon present in the given hydrocarbon is 7.

The number of nitrogen present in the given hydrocarbon is 0.

The number of halogen present in the given hydrocarbon is 0.

The number of hydrogen present in the given hydrocarbon is 16.

Substitute the values of C, N, X and H in equation (1) to calculate the degree of unsaturation.

DOU=2×7+2+00162DOU=16162DOU=02DOU=0

Hence, the degree of unsaturation of the given hydrocarbon is zero.

(d)

Interpretation Introduction

To determine: The degree of unsaturation of the given hydrocarbon.

Interpretation: The degree of unsaturation of the given hydrocarbon is to be calculated.

Concept introduction: The degree of unsaturation is defined as the formula which is useful for the determination of chemical structures of organic compounds. The number of rings, double bonds and triple bonds in a compound are drawn by the degree of unsaturation.

The degree of unsaturation (DOU) of the given hydrocarbons is calculated by the formula,

DOU=2×C+2+NXH2

(d)

Expert Solution
Check Mark

Answer to Problem 13.1VP

Answer

The degree of unsaturation of the given hydrocarbon is three.

Explanation of Solution

Explanation

The given hydrocarbon is shown in Figure 4.

Chemistry, Chapter 13, Problem 13.1VP , additional homework tip  4

Figure 4

Count the number of elements (C,N,X,H) present in given hydrocarbon.

The number of carbon present in given hydrocarbon is 7.

The number of nitrogen present in given hydrocarbon is 0.

The number of halogen present in given hydrocarbon is 0.

The number of hydrogen present in given hydrocarbon is 10.

The degree of unsaturation (DOU) of hydrocarbon is calculated by the formula,

DOU=2×C+2+NXH2 (1)

Where,

  • C is number of carbon present in the given hydrocarbon.
  • N is number of nitrogen present in the given hydrocarbon.
  • X is number of halogen present in the given hydrocarbon.
  • H is number of hydrogen present in the given hydrocarbon.

Substitute the values of C, N, X and H in equation (1) to calculate the degree of unsaturation.

DOU=2×7+2+00102DOU=16102DOU=62DOU=3

Hence, the degree of unsaturation of the given hydrocarbon is three.

Conclusion

Conclusion

  1. a. The degree of unsaturation of the given hydrocarbon is one.
  2. b. The degree of unsaturation of the given hydrocarbon is two.
  3. c. The degree of unsaturation of the given hydrocarbon is zero.
  4. d. The degree of unsaturation of the given hydrocarbon is three.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Please correct answer and don't use hand rating
Pick the aromatic compound: A. 1,2,3 B. 1,2,4 C. 2,3,4 D. 1,3,4
None

Chapter 13 Solutions

Chemistry

Ch. 13.7 - Prob. 11PECh. 13.8 - Prob. 12PECh. 13.8 - Prob. 13PECh. 13 - Prob. 13.1VPCh. 13 - Prob. 13.2VPCh. 13 - Prob. 13.3VPCh. 13 - Prob. 13.4VPCh. 13 - Prob. 13.5VPCh. 13 - Prob. 13.6VPCh. 13 - Prob. 13.7VPCh. 13 - Prob. 13.8VPCh. 13 - Prob. 13.9VPCh. 13 - Prob. 13.10VPCh. 13 - Prob. 13.11VPCh. 13 - Prob. 13.12VPCh. 13 - Prob. 13.13VPCh. 13 - Prob. 13.14VPCh. 13 - Prob. 13.15VPCh. 13 - Prob. 13.16VPCh. 13 - Prob. 13.17QPCh. 13 - Prob. 13.18QPCh. 13 - Prob. 13.19QPCh. 13 - Prob. 13.20QPCh. 13 - Prob. 13.21QPCh. 13 - Prob. 13.22QPCh. 13 - Prob. 13.23QPCh. 13 - Prob. 13.24QPCh. 13 - Prob. 13.25QPCh. 13 - Prob. 13.26QPCh. 13 - Prob. 13.27QPCh. 13 - Prob. 13.28QPCh. 13 - Prob. 13.29QPCh. 13 - Prob. 13.30QPCh. 13 - Prob. 13.31QPCh. 13 - Prob. 13.32QPCh. 13 - Prob. 13.33QPCh. 13 - Prob. 13.34QPCh. 13 - Prob. 13.35QPCh. 13 - Prob. 13.36QPCh. 13 - Prob. 13.37QPCh. 13 - Prob. 13.38QPCh. 13 - Prob. 13.39QPCh. 13 - Prob. 13.40QPCh. 13 - Prob. 13.41QPCh. 13 - Prob. 13.42QPCh. 13 - Prob. 13.43QPCh. 13 - Prob. 13.44QPCh. 13 - Prob. 13.45QPCh. 13 - Prob. 13.46QPCh. 13 - Prob. 13.47QPCh. 13 - Prob. 13.48QPCh. 13 - Prob. 13.49QPCh. 13 - Prob. 13.50QPCh. 13 - Prob. 13.51QPCh. 13 - Prob. 13.52QPCh. 13 - Prob. 13.53QPCh. 13 - Prob. 13.54QPCh. 13 - Prob. 13.55QPCh. 13 - Prob. 13.56QPCh. 13 - Prob. 13.57QPCh. 13 - Prob. 13.58QPCh. 13 - Prob. 13.59QPCh. 13 - Prob. 13.60QPCh. 13 - Prob. 13.61QPCh. 13 - Prob. 13.62QPCh. 13 - Prob. 13.63QPCh. 13 - Prob. 13.64QPCh. 13 - Prob. 13.65QPCh. 13 - Prob. 13.66QPCh. 13 - Prob. 13.67QPCh. 13 - Prob. 13.68QPCh. 13 - Prob. 13.69QPCh. 13 - Prob. 13.70QPCh. 13 - Prob. 13.71QPCh. 13 - Prob. 13.72QPCh. 13 - Prob. 13.73QPCh. 13 - Prob. 13.74QPCh. 13 - Prob. 13.75QPCh. 13 - Prob. 13.76QPCh. 13 - Prob. 13.77QPCh. 13 - Prob. 13.78QPCh. 13 - Prob. 13.79QPCh. 13 - Prob. 13.80QPCh. 13 - Prob. 13.81QPCh. 13 - Prob. 13.82QPCh. 13 - Prob. 13.83QPCh. 13 - Prob. 13.84QPCh. 13 - Prob. 13.85QPCh. 13 - Prob. 13.86QPCh. 13 - Prob. 13.87QPCh. 13 - Prob. 13.88QPCh. 13 - Prob. 13.89QPCh. 13 - Prob. 13.90QPCh. 13 - Prob. 13.91QPCh. 13 - Prob. 13.92QPCh. 13 - Prob. 13.93QPCh. 13 - Prob. 13.94QPCh. 13 - Prob. 13.95QPCh. 13 - Prob. 13.96QPCh. 13 - Prob. 13.97QPCh. 13 - Prob. 13.98QPCh. 13 - Prob. 13.99QPCh. 13 - Prob. 13.100QPCh. 13 - Prob. 13.101QPCh. 13 - Prob. 13.102QPCh. 13 - Prob. 13.103QPCh. 13 - Prob. 13.104QPCh. 13 - Prob. 13.105QPCh. 13 - Prob. 13.106QPCh. 13 - Prob. 13.107QPCh. 13 - Prob. 13.108QPCh. 13 - Prob. 13.109QPCh. 13 - Prob. 13.110QPCh. 13 - Prob. 13.111QPCh. 13 - Prob. 13.112QPCh. 13 - Prob. 13.113QPCh. 13 - Prob. 13.114QPCh. 13 - Prob. 13.115QPCh. 13 - Prob. 13.116QPCh. 13 - Prob. 13.117QPCh. 13 - Prob. 13.118QPCh. 13 - Prob. 13.119QPCh. 13 - Prob. 13.120QPCh. 13 - Prob. 13.121QPCh. 13 - Prob. 13.122QPCh. 13 - Prob. 13.123QPCh. 13 - Prob. 13.124QPCh. 13 - Prob. 13.125QPCh. 13 - Prob. 13.126QPCh. 13 - Prob. 13.127QPCh. 13 - Prob. 13.128QPCh. 13 - Prob. 13.129QPCh. 13 - Prob. 13.130QPCh. 13 - Prob. 13.131QPCh. 13 - Prob. 13.132QPCh. 13 - Prob. 13.133QPCh. 13 - Prob. 13.134QPCh. 13 - Prob. 13.135APCh. 13 - Prob. 13.136APCh. 13 - Prob. 13.137APCh. 13 - Prob. 13.138APCh. 13 - Prob. 13.139APCh. 13 - Prob. 13.140APCh. 13 - Prob. 13.141APCh. 13 - Prob. 13.142APCh. 13 - Prob. 13.143APCh. 13 - Prob. 13.144APCh. 13 - Prob. 13.145APCh. 13 - Prob. 13.146APCh. 13 - Prob. 13.147APCh. 13 - Prob. 13.148APCh. 13 - Prob. 13.149APCh. 13 - Prob. 13.150APCh. 13 - Prob. 13.151AP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Enzymes - Effect of cofactors on enzyme; Author: Tutorials Point (India) Ltd;https://www.youtube.com/watch?v=AkAbIwxyUs4;License: Standard YouTube License, CC-BY
Enzyme Catalysis Part-I; Author: NPTEL-NOC IITM;https://www.youtube.com/watch?v=aZE740JWZuQ;License: Standard Youtube License