Concept explainers
(a)
Interpretation:
IHD for the compound having the molecular formula
Concept introduction:
In order to determine the IHD of a given molecular formula, first draw any saturated molecule that has the same number of each non-hydrogen atom as in the given formula. The general formula of a saturated hydrocarbon is
![Check Mark](/static/check-mark.png)
Answer to Problem 4.49P
IHD for the compound having molecular formula
Explanation of Solution
The given molecular formula is
Thus, this saturated molecule has 8 additional hydrogen atoms as compared to the given molecular formula. IHD for the given molecular formula is calculated by dividing that number of additional hydrogen atoms by 2. Thus, IHD is
The IHD for the compound with a given molecular formula is calculated by applying the steps above.
(b)
Interpretation:
IHD for the compound having the molecular formula
Concept introduction:
In order to determine the IHD of a given molecular formula, first draw any saturated molecule that has the same number of each non-hydrogen atom as in the given formula. The general formula of a saturated hydrocarbon is
![Check Mark](/static/check-mark.png)
Answer to Problem 4.49P
IHD for the compound having molecular formula
Explanation of Solution
The given molecular formula is
It takes a total of 15 hydrogen atoms to saturate each carbon, nitrogen, and oxygen in this compound. Thus, this saturated molecule has 10 additional hydrogen atoms as compared to the given molecular formula. IHD for the given molecular formula is calculated by dividing that number of additional hydrogen atoms by 2. Thus IHD is
The IHD for the compound with a given molecular formula is calculated by applying the steps above.
(c)
Interpretation:
IHD for the compound having the molecular formula
Concept introduction:
In order to determine the IHD of a given molecular formula, first draw any saturated molecule that has the same number of each non-hydrogen atom as in the given formula. The general formula of a saturated hydrocarbon is
![Check Mark](/static/check-mark.png)
Answer to Problem 4.49P
IHD for the compound having molecular formula
Explanation of Solution
The given molecular formula is
It takes a total of
The IHD for the compound with a given molecular formula is calculated by applying the steps above.
(d)
Interpretation:
IHD for the compound having the molecular formula
Concept introduction:
In order to determine the IHD of a given molecular formula, first draw any saturated molecule that has the same number of each non-hydrogen atom as in the given formula. The general formula of a saturated hydrocarbon is
![Check Mark](/static/check-mark.png)
Answer to Problem 4.49P
IHD for the compound having molecular formula
Explanation of Solution
The given molecular formula is
It takes a total of
The IHD for the compound with a given molecular formula is calculated by applying the steps above.
(f)
Interpretation:
IHD for the compound having the molecular formula
Concept introduction:
In order to determine the IHD of a given molecular formula, first draw any saturated molecule that has the same number of each non-hydrogen atom as in the given formula. The general formula of a saturated hydrocarbon is
![Check Mark](/static/check-mark.png)
Answer to Problem 4.49P
IHD for the compound having molecular formula
Explanation of Solution
The given molecular formula is
It takes a total of
The IHD for the compound with a given molecular formula is calculated by applying the steps above.
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Chapter 4 Solutions
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
- Don't used hand raiting and don't used Ai solutionarrow_forwardHighlight in red each acidic location on the organic molecule at left. Highlight in blue each basic location on the organic molecule at right. Note for advanced students: we mean acidic or basic in the Brønsted-Lowry sense only. Cl N شیخ x Garrow_forwardQ4: Draw the mirror image of the following molecules. Are the molecules chiral? C/ F LL CI CH3 CI CH3 0 CI CH3 CI CH3 CH3arrow_forward
- Complete combustion of a 0.6250 g sample of the unknown crystal with excess O2 produced 1.8546 g of CO2 and 0.5243 g of H2O. A separate analysis of a 0.8500 g sample of the blue crystal was found to produce 0.0465 g NH3. The molar mass of the substance was found to be about 310 g/mol. What is the molecular formula of the unknown crystal?arrow_forward4. C6H100 5 I peak 3 2 PPM Integration values: 1.79ppm (2), 4.43ppm (1.33) Ipeakarrow_forwardNonearrow_forward
- 3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage LearningGeneral, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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