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Concept explainers
(a)
Interpretation:
Lewis structure for the
Concept introduction:
Lewis structure is a convenient way to convey information such as which atoms are bonded to each other by which type of bond. Only valence electrons take part to draw Lewis structure.
Valence electrons that participate in bond formation are called bonding electron pairs whereas electrons that do not take part in bonding and that remain as non-bonding electrons are termed as lone pair of electrons.
Lewis structure is drawn from the total valence electron count of each atom in the molecule. Skeleton structure for the given molecule is drawn with atoms bonded with a single bond, the central atom is always less electronegative atom. Electrons that participate in forming the bond are bonding electrons. Remaining electrons are distributed first to outer atoms such that each atom completes its octet other than hydrogen.
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Answer to Problem 1.44P
Lewis structure for
Explanation of Solution
The given molecule
Thus the skeleton structure for
Remaining two valence electrons are contributed as lone pair of electrons to the outer atom which is nitrogen.
Hence the complete Lewis structure for
Lewis structure for
(b)
Interpretation:
Lewis structure for
Concept introduction:
Lewis structure is drawn from the total valence electron count of each atom in the molecule. Skeleton structure for the given molecule is drawn with atoms bonded with a single bond, the central atom is always less electronegative atom. Electrons that participate in forming the bond are bonding electrons. Remaining electrons are distributed first to outer atoms such that each atom complete its octet other than hydrogen.
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Answer to Problem 1.44P
Lewis structure for
Explanation of Solution
In
Remaining
Lewis structure for
(c)
Interpretation:
Lewis structure for
Concept introduction:
Lewis structure is drawn from the total valence electron count of each atom in the molecule. Valence electrons participate in bond formation are called bonding electron pairs whereas electrons that do not take part in bonding and that remain as non-bonding electrons are termed as lone pair of electrons. Skeleton structure for the given molecule is drawn with atoms bonded with a single bond, the central atom is always less electronegative atom.
Lone pair of electrons contributed to outer atoms to complete octet except hydrogen.
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Answer to Problem 1.44P
The Lewis structure for
Explanation of Solution
Carbon is the central atom in
Remaining six electrons are contributed as lone pairs to the outer atom oxygen but one electron is involved in bonding with complete octet of oxygen atom.
Lewis structure is drawn from the total valence electron count.
(d)
Interpretation:
Lewis structure for
Concept introduction:
Lewis structure is drawn from the total valence electron count of each atom in the molecule. Skeleton structure for the given molecule is drawn with atoms bonded with a single bond, the central atom is always less electronegative atom. Electrons that participate in forming the bond are bonding electrons. Remaining electrons are distributed first to outer atoms such that each atom completes its octet other than hydrogen.
![Check Mark](/static/check-mark.png)
Answer to Problem 1.44P
Lewis structure for
Explanation of Solution
In
Eight electrons from total
Therefore the Lewis structure for
Lewis structure is drawn from total valence electron count.
(e)
Interpretation:
Lewis structure for
Concept introduction:
Lewis structure is drawn from the total valence electron count of each atom in the molecule. Valence electrons that participate in bond formation are called bonding electron pairs whereas electrons that do not take part in bonding and that remain as non-bonding electrons are termed as lone pair of electrons. Skeleton structure for the given molecule is drawn with atoms bonded with a single bond, the central atom is always less electronegative atom.
Lone pair of electrons contributed to outer atoms to complete octet except hydrogen.
![Check Mark](/static/check-mark.png)
Answer to Problem 1.44P
Lewis structure for
Explanation of Solution
The given molecule
Remaining
Hence the Lewis structure for
Lewis structure is drawn from total valence electron count.
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Chapter 1 Solutions
Get Ready for Organic Chemistry
- 5) Confidence interval. Berglund and Wichardt investigated the quantitative determination of Cr in high-alloy steels using a potentiometric titration of Cr(VI). Before the titration, samples of the steel were dissolved in acid and the chromium oxidized to Cr(VI) using peroxydisulfate. Shown here are the results (as %w/w Cr) for the analysis of a reference steel. 16.968, 16.922, 16.840, 16.883, 16.887, 16.977, 16.857, 16.728 Calculate the mean, the standard deviation, and the 95% confidence interval about the mean. What does this confidence interval mean?arrow_forwardIn the Nitrous Acid Test for Amines, what is the observable result for primary amines? Group of answer choices nitrogen gas bubbles form a soluble nitrite salt yellow oily layer of nitrosoaminearrow_forward3. a. Use the MS to propose at least two possible molecular formulas. For an unknown compound: 101. 27.0 29.0 41.0 50.0 52.0 55.0 57.0 100 57.5 58.0 58.5 62.0 63.0 64.0 65.0 74.0 40 75.0 76.0 20 20 40 60 80 100 120 140 160 180 200 220 m/z 99.5 68564810898409581251883040 115.0 116.0 77404799 17417M 117.0 12.9 118.0 33.5 119.0 36 133 0 1.2 157.0 2.1 159.0 16 169.0 219 170.0 17 171.0 21.6 172.0 17 181.0 1.3 183.0 197.0 100.0 198.0 200. 784 Relative Intensity 2 2 8 ō (ppm) 6 2arrow_forward
- Solve the structure and assign each of the following spectra (IR and C-NMR)arrow_forward1. For an unknown compound with a molecular formula of C8H100: a. What is the DU? (show your work) b. Solve the structure and assign each of the following spectra. 8 6 2 ō (ppm) 4 2 0 200 150 100 50 ō (ppm) LOD D 4000 3000 2000 1500 1000 500 HAVENUMBERI -11arrow_forward16. The proton NMR spectral information shown in this problem is for a compound with formula CioH,N. Expansions are shown for the region from 8.7 to 7.0 ppm. The normal carbon-13 spec- tral results, including DEPT-135 and DEPT-90 results, are tabulated: 7 J Normal Carbon DEPT-135 DEPT-90 19 ppm Positive No peak 122 Positive Positive cus и 124 Positive Positive 126 Positive Positive 128 No peak No peak 4° 129 Positive Positive 130 Positive Positive (144 No peak No peak 148 No peak No peak 150 Positive Positive してしarrow_forward
- 3. Propose a synthesis for the following transformation. Do not draw an arrow-pushing mechanism below, but make sure to draw the product of each proposed step (3 points). + En CN CNarrow_forwardShow work..don't give Ai generated solution...arrow_forwardLabel the spectrum with spectroscopyarrow_forward
- Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
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