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(a)
Interpretation:
The electron-pair geometry for each carbon atom in
Concept introduction:
The electron pairs in Lewis diagrams repel each other in real molecule and thus they distribute themselves in positions around the central atoms which are far away from one another. This arrangement of electron pairs is called electron-pair geometry. The electron pairs may be shared in covalent bond, or they may be lone pairs.
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Answer to Problem 24E
The Lewis diagram for the molecule
The electron pair geometry for the carbon atoms
Explanation of Solution
To write the Lewis diagram for a compound, first the number of valence electrons is to be calculated. In the molecule,
The atom which is least electronegative and lesser in number is the central atom. In
Figure 1
The electron-pair geometry depends on the number of electron pairs around the central atoms. In the molecule
The Lewis diagram for the molecule
(b)
Interpretation:
The molecular geometry predicted by the valence shell electron-pair repulsion theory for each carbon atom in the molecule
Concept introduction:
Molecular geometry is the precise term that is used to describe the shape of molecules and arrangement of atoms around the central atom. The molecular geometry of a molecule is predicted by valence shell electron-pair repulsion theory or in short VSEPR theory. VSEPR theory applies to substances in which a second period element is bonded to two, three, four, or other atoms.
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Answer to Problem 24E
The Lewis diagrams for
The molecular geometry is tetrahedral for the carbons
Explanation of Solution
To write the Lewis diagram for a compound first the number of valence electrons is to be calculated. In the molecule
The atom which is least electronegative and lesser in number is the central atom. In
Figure 1
The molecular geometry depends on the number of electron pairs as well as number of unpaired electron on the central atoms. In the molecule
The Lewis diagram for the molecule
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Chapter 13 Solutions
Bundle: Introductory Chemistry: An Active Learning Approach, 6th + LMS Integrated for OWLv2, 4 terms (24 months) Printed Access Card
- Use the References to access important values if needed for this question. What is the IUPAC name of each of the the following? 0 CH3CHCNH₂ CH3 CH3CHCNHCH2CH3 CH3arrow_forwardYou have now performed a liquid-liquid extraction protocol in Experiment 4. In doing so, you manipulated and exploited the acid-base chemistry of one or more of the compounds in your mixture to facilitate their separation into different phases. The key to understanding how liquid- liquid extractions work is by knowing which layer a compound is in, and in what protonation state. The following liquid-liquid extraction is different from the one you performed in Experiment 4, but it uses the same type of logic. Your task is to show how to separate apart Compound A and Compound B. . Complete the following flowchart of a liquid-liquid extraction. Handwritten work is encouraged. • Draw by hand (neatly) only the appropriate organic compound(s) in the boxes. . Specify the reagent(s)/chemicals (name is fine) and concentration as required in Boxes 4 and 5. • Box 7a requires the solvent (name is fine). • Box 7b requires one inorganic compound. • You can neatly complete this assignment by hand and…arrow_forwardb) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;' The compound has the ff electronic transitions: 0%o* and no a* 1H NMR Spectrum (CDCl3, 400 MHz) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 13C{H} NMR Spectrum (CDCl3, 100 MHz) Solvent 80 70 60 50 40 30 20 10 0 ppm ppm ¹H-13C me-HSQC Spectrum ppm (CDCl3, 400 MHz) 5 ¹H-¹H COSY Spectrum (CDCl3, 400 MHz) 0.5 10 3.5 3.0 2.5 2.0 1.5 1.0 10 15 20 20 25 30 30 -35 -1.0 1.5 -2.0 -2.5 3.0 -3.5 0.5 ppm 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppmarrow_forward
- Part I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forward
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