Lewis structure for crotonaldehyde is to be drawn. Concept introduction: In order to draw a Lewis structure for a molecule, start by counting the total number of valence electrons in a molecule. The number of valence electrons by each atom is the same as its group number. Generally, the condensed formula depicts the connectivity of atoms. For the given skeleton of the molecule, distribute the remaining electrons as lone pairs. In doing so, start with the outer atoms and work inwards. Try to achieve an octet on each atom other than hydrogen. If there is an atom with less than an octet, increase the atom’s share of electrons by converting lone pairs from neighboring atoms into bonding pairs thereby creating double or triple bonds. For an uncharged atom, carbon atoms will have maximum four bonds, Nitrogen will have three bonds and one lone pair, while oxygen will have two bonds and two lone pairs. Hydrogen always contributes to one bond. The number of bonds in case of halogen is one, while there will be three lone pairs of electrons on halide atoms.

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Chapter 1, Problem 1.25P

Interpretation Introduction

Interpretation:

Lewis structure for crotonaldehyde is to be drawn.

Concept introduction:

In order to draw a Lewis structure for a molecule, start by counting the total number of valence electrons in a molecule. The number of valence electrons by each atom is the same as its group number. Generally, the condensed formula depicts the connectivity of atoms. For the given skeleton of the molecule, distribute the remaining electrons as lone pairs. In doing so, start with the outer atoms and work inwards. Try to achieve an octet on each atom other than hydrogen. If there is an atom with less than an octet, increase the atom’s share of electrons by converting lone pairs from neighboring atoms into bonding pairs thereby creating double or triple bonds. For an uncharged atom, carbon atoms will have maximum four bonds, Nitrogen will have three bonds and one lone pair, while oxygen will have two bonds and two lone pairs. Hydrogen always contributes to one bond. The number of bonds in case of halogen is one, while there will be three lone pairs of electrons on halide atoms.

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Part V. Label ad match the carbons in compounds Jane and Diane w/ the corresponding peak no. in the Spectra (Note: use the given peak no. To label the carbons, other peak no are intentionally omitted) 7 4 2 -0.13 -0.12 -0.11 -0.10 -0.08 8 CI Jane 1 -0.09 5 210 200 190 180 170 160 150 140 130 120 110 100 -8 90 f1 (ppm) 11 8 172.4 172.0 f1 (ppr HO CI NH Diane 7 3 11 80 80 -80 -R 70 60 60 2 5 -8 50 40 8. 170 160 150 140 130 120 110 100 90 -0 80 70 20 f1 (ppm) 15 30 -20 20 -60 60 -0.07 -0.06 -0.05 -0.04 -0.03 -0.02 -0.01 -0.00 -0.01 10 -0.17 16 15 56 16 -0.16 -0.15 -0.14 -0.13 -0.12 -0.11 -0.10 -0.09 -0.08 -0.07 -0.06 -0.05 -0.04 17.8 17.6 17.4 17.2 17.0 f1 (ppm) -0.03 -0.02 550 106 40 30 20 20 -0.01 -0.00 F-0.01 10 0