Chemistry
Chemistry
4th Edition
ISBN: 9780393250985
Author: Gilbert
Publisher: W.W. Norton & Company
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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.

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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
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