Concept explainers
Interpretation:
The complete MO picture and energy diagram for the allyl radical
Concept introduction:
The overlap of the two atomic orbitals (AOs) results in the formation of two molecular orbitals (MOs). One of these, called the bonding MO, is formed as a result of constructive interaction. It is lower in energy than the original AOs. The other, called the antibonding MO, is formed as a result of destructive interaction. It is higher in energy than the original AOs. Antibonding MOs have a node (a nodal plane) situated between the two atoms. The overall character of the MO is determined by the number of bonding and antibonding interactions between the AOs. If the bonding interactions are more, the overall character is bonding. If the antibonding interactions are more, the overall character is antibonding. If they are equal or there are none, the MO is nonbonding.
Depending on whether the AOs overlap along the bond axis or away from the bond axis (sideways), the MOs are designated as
In a molecule with double bonds, all bonding
For a linear system of conjugated p orbitals, all nodal planes in a resulting
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ORGANIC CHEMISTRY SG/SM PA +SQUARECAP E
- For the species below draw additional resonance structures (where all atoms have access to an octect of electrons) for species A, B, C, and D. Determine the bond order of the bond from the underlined C to underlined O for the lowest energy structure(s). You may need to evaluate the formal charge of the additional species you draw. These are species in an aprotic solvent, so only electrons may be moved, not hydrogen atoms. For each lettered structure below, input the bond order as an integer (1,2,etc) or improper fraction (4/3, 5/4, etc). A Structure Bond Order A HO: (f) H B B С Which structure A-D has the longest underlined C-O bond? :N=C=0: D Darrow_forwardIn the following Lewis structure of [(CH3)2OH]*, every atom, bond and lone pair is positioned. To complete the structure, drag the formal charge tags to the appropriate atom(s). Each marker may be used more than once, or not at all. If an atom has a formal charge of zero, do not drag a tag to it. When you drag the marker in, place the little crosshairs in the upper left corner of the marker directly over the atom(s) in question (not above them). H. H-C Н-С-О-С-Н C-H H HH 2- II 2-arrow_forwardDescribe with words how you would draw the curved arrow(s) to get to the more stable resonance form for the left-hand pair. Be sure to include how many curved arrows are needed and what atom (or bond) the electrons are coming from or going to. Do the same for the resonance pair on the right-hand side. more stable more stablearrow_forward
- a) What would the formula (CxHy) of the compound cyclobutyne be? b) What is the hybridization of the carbon atoms in the triple bond? What bond angle is this normally associated with? c) Cyclobutyne is not a stable compound and has never been observed. Suggest a reason why.arrow_forwardProblem Draw Lewis structures for the following:(a) Ethylene (C2H4), the most important reactant in the manufacture of polymers(b) Nitrogen (N2), the most abundant atmospheric gasPlan We show the structure resulting from steps 1 to 4: placing the atoms, counting the total valence electrons, making single bonds, and distributing the remaining valence electrons in pairs to attain octets. Then we continue with step 5, if needed.arrow_forwardProblem Draw a Lewis structure and identify the octet-rule exception for (a) H3PO4 (draw two resonance forms and select the more important); (b) BFCl2.Plan We draw each Lewis structure and examine it for exceptions to the octet rule.(a) The central atom is in Period 3, so it can have more than an octet.(b) The central atom is B, which can have fewer than an octet of electrons.arrow_forward
- What is the state of hybridization of each of the labeled atoms in the neutral compound below.arrow_forwardHow do I calc the formal chargers of nitrene, oxonium ion, carbene, acetylidearrow_forwardUne or more valid answers. Explain the answers. The localization or resonance energy: a) Justify the increased stability of molecules that have unlocated electrons b) It is a consequence of the mobility of the electrons that form the “sigma” bonds c) It occurs in compounds with conjugated double bonds, but not in aromatic compounds d) It is shown by the experimental values of the enthalpies of formation.arrow_forward
- A student argues that the two nitrogens in the compound below are sp? hybridized, but atom a is still more basic than atom b. Is the student correct? Explain. •• -N-CH2CH3 | b CH2CH3 aarrow_forwardDoes cyclooctene have two distinct configurations about its C=C bond? Hint: Consider using a model kit. Cyclooctenearrow_forwardBelow is the SN2 reaction between iodocyclohexane and cyanide (CN–). Draw the missing curved arrow notation in the first box to reflect electron movements. In both boxes, add lone pairs of electrons and nonzero formal charges.arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning