Proton geometry

Because adjacent bonds are eclipsed when the H—C—C—X unit is syn coplanar a transition state with this geometry is less stable than one that has an anti coplanar rela tionship between the proton and the leaving group... 

Effects that arise because one spatial arrangement of electrons (or orbitals or bonds) IS more stable than another are called stereoelectronic effects There is a stereoelec tromc preference for the anti coplanar arrangement of proton and leaving group in E2 reactions Although coplanarity of the p orbitals is the best geometry for the E2 process modest deviations from this ideal can be tolerated In such cases the terms used are syn periplanar and anti periplanar... 

Because there are two changes ia material composition near the active region, this represents a double heterojunction. Also shown ia Figure 12 is a stripe geometry that confines the current ia the direction parallel to the length of the junction. This further reduces the power threshold and makes the diffraction-limited spreading of the beam more symmetric. The stripe is often defined by implantation of protons, which reduces the electrical conductivity ia the implanted regions. Many different stmctures for semiconductor diode lasers have been developed. 

Compare atomic charges and electrostatic potential maps for imidazole NH protonated and imidazole Nprotonated. In which ion is the positive charge more delocalized Compare carbon-nitrogen bond distances in each ion to those in imidazole as a standard. Are these distances consistent with the bonding patterns shown above for each ion Draw whatever Lewis structures are needed to describe each ion s geometry and charge distribution. 

The first step in the addition of an electrophile such as HBr to an alkyne involves protonation and subsequent formation of an intermediate vinyl cation. Where does propyne protonate Compare energies of 1-methylvinyl and 2-methylvinyl cations. Which is more stable Why Measure CC bond distance in the more stable cation. Does the cation incorporate a full triple bond (as in propyne) or a double bond (as in propene). Examine atomic charges and electrostatic potential maps to locate the positive charge in the two cations. Is the more stable ion the one in which the charge is better delocalized Use the charges together with information about the ions geometry to draw Lewis structures (or a series of Lewis structures) for 1-methylvinyl and 2-methylvinyl cations. 

Obtain the energies of the different possible carbanions alleyne-H+). Which one is most stable Does it correspond to removal of the most electron-poor proton Examine the geometry and atomic charges of the favored carbanion. Where is the negative charge Draw the Lewis structure of this ion. Predict the structure of the Sn2 product. 

In order to predict the structure of the product, you must identify the factors that will tend to favor selective ketal formation. Consider selective carbonyl protonation first. Obtain energies and atomic charges, and display electrostatic potential maps of the alternative protonated ketones (protonated ketone A, protonated ketone B). Identify the more stable isomer. Compare geometries and draw whatever Lewis structures are needed to account for your data. Why is one isomer more stable than the other Is the more stable isomer also that in which the positive charge is better delocalized Will the more stable isomer undergo nucleophilic attack more or less easily than the other Explain. 

Protonation and subsequent loss of water should generate a earbocation. Examine all of the carbocations derived from protonation of (3-D-glueose. Identify the most stable carboeation (this is the one that will form most readily), and draw whatever resonance eontributors are needed to describe the geometry, energy, and atomie charges in this cation. Can you explain why substitution oeeurs selectively at Ci ... 

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