
Concept explainers
(a)
Interpretation:
The MO resulting from the given orbital interaction is to be drawn.
Concept introduction:
When atomic orbitals (AOs) of the same phase interact, they result in a bonding molecular orbital (MO) that is lower in energy than the individual AOs.

Answer to Problem 3.1P
The MO resulting from the given orbital interaction can be drawn as follows:
Explanation of Solution
The orbital interaction shows two orbitals that are lightly shaded. Both have the same negative phase. Since the phases of both are the same, there will be constructive interference, resulting in a bonding MO of negative phase.
The MO resulting from the given orbital interaction can be drawn as follows:
The interaction between orbitals of the same phase results in the formation of a bonding MO.
(b)
Interpretation:
Whether the resulting MO is unique compared to the one shown on the right of Figure 3-6a is to be determined.
Concept introduction:
When atomic orbitals (AOs) of the same phase interact, the resulting molecular orbital (MO) has a lower energy than the separate AOs. The phases of the interacting orbitals may be both positive or both negative. The resulting stabilization (lowering of energy) is the same for both.

Answer to Problem 3.1P
The MO resulting from the interaction shown will not be unique compared to the one shown in Figure 3-6a.
Explanation of Solution
The interaction in this case is between AOs of negative phases (light shading). Since the phases are the same, the interaction will result in constructive interference, increasing the electron density between the two nuclei. This will lower the energy of the MO compared to the individual AOs. The extent to which the energy is lowered will be the same as in case of the interaction shown in Figure 3-6a. This is because the interacting orbitals are same except for a different phase.
Therefore, the MO resulting from the interaction shown will not be unique compared to the one in Figure 3-6a.
The interaction between AOs of same phase results in a bonding MO with the same stabilization, whether their phases are both positive or both negative.
Want to see more full solutions like this?
Chapter 3 Solutions
Organic Chemistry: Principles And Mechanisms: Study Guide/solutions Manual (second)
- Draw the product of the E2 reaction shown below. Include the correct stereochemistry. Ignore any inorganic byproducts. + Br CH3 Q Strong Base Drawing Atoms, Bonds and Rings Charges Undo Reset H "Br H N Br. Remove Done .N. Drag To Panarrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the curved arrows to draw the product of this elementary step in an elimination mechanism. Include all lone pairs and charges as appropriate. Ignore stereochemistry. Ignore byproducts. + Br: .. 8 0.01 M NaOH heat Drawing Q Atoms, Bonds and Rings Charges and Lone Pairs Draw or tap a new bond to see suggestions. Undo Reset Remove Done + Drag To Panarrow_forward+ Draw the product of the E2 reaction shown below. Include the correct stereochemistry. Ignore any inorganic byproducts. Ph CH2CH3 H H3C H Br DBN [૪] Drawing Atoms, Bonds and Rings H | OH Charges ―00 H. C | Undo Reset Br I Remove Done Drag To Pan +arrow_forward
- Reaction A Now the production A Œ In the product of reaction i 12 Dear the product of actionarrow_forwardMacmillan Learnin When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M* = 163 m/z) is formed. In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm-1. The 13C NMR and DEPT spectra are provided. Draw the structure of the product as the resonance contributor lacking any formal charges. 13C NMR DEPT 90 200 160 120 80 40 0 200 160 120 80 DEPT 135 200 160 120 80 40 0 Draw the unknown amide. 40 40 0arrow_forwardDraw the major product karmed when I reach with the epoxide. Use walge dah bonds, including hydrogen al alcach genic center, to show the chemistry of the product Beeldraw any hydrogen akams on coxygen where applicablearrow_forward
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. H I Select to Add Arrows + H H 'H Q H2O H2O CI:O .H H H H I Select to Add Arrows I : C H2O H H H Select to Add Arrows 'Harrow_forward+ Draw an alkyl halide that produces ONLY the following alkene in an E2 elimination. Ignore any inorganic byproducts. Drawing Strong Base Q Atoms, Bonds and Rings Charges HO Br H2N Undo Reset Remove Done Drag To Panarrow_forwardFor the dehydrohalogenation (E2) reaction shown, draw the major organic product. Хок Br tert-butanol heat Select Drew Templates More Erase CH QQQarrow_forward
- Macmillan Learning Draw the major, neutral organic product for each substitution reaction. For this question, assume that each substitution reaction goes to completion. Disregard elimination. Reaction A. CI H₂O Select Draw Templates More Erase C Harrow_forwardMacmillan Learning Reaction B: CI HO_ 곳으 / Select Draw Templates More с € H D Erasearrow_forwardWhen 2-bromo-93-dimethylbutane is heated with sodium methoxide, one majors.. në la formed. 4th attempt Part 1 (0.5 point) t Ji See Periodic Table See Hint Draw the major alkene product and all other byproducts. Be sure to include lone-pair electrons and charges. Part 2 (0.5 point) What type of mechanism is occuring? Choose one: AS1 3rd attempt X H 41 See Hint Part 1 (0.5 point) Feedback See Periodic Table See Hintarrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning

