Oxygen molecule bonding

This greater reactivity of the silanes may be due to several factors, for example, the easier approach of an oxygen molecule (which may attach initially to the silane by use of the vacant silicon d orbitals) and the formation of strong Si—O bonds (stronger than C—O). 

Neither sodium borohydride nor lithium aluminum hydride reduces isolated carbon-carbon double bonds This makes possible the selective reduction of a carbonyl group m a molecule that contains both carbon-carbon and carbon-oxygen double bonds... 

Now let us investigate the oxygen molecule, which experiment tells us has the molecular formula O2. We might begin by considering formation of a single bond between two oxygen atoms, as represented by the orbital representation... 

The state of the oxygen molecule, 1.62 v. e. above the normal state, probably corresponds to a double bond between the two atoms. This gives 3.47 v. e. for 0 0. 

The revision leads to a difference of 0.06 A. between the interatomic distance in the normal oxygen molecule and the sum of the double-bond radii. This may be attributed to the presence of an unusual structure, consisting of a single bond plus two three-electron bonds. We assign this structure both to the normal 2 state, with ro = 1.204 A., and to the excited 2 state, with ro = 1.223 A., the two differing in the relative spin orientations of the odd electrons in the two three-electron bonds. We expect for the double-bonded state the separation n 1.14 A. 

Active Substrate. If a new stereogenic center is ereated in a molecule that is already optically active, the two diastereomers are not (except fortuitously) formed in equal amounts. The reason is that the direction of attack by the reagent is determined by the groups already there. For certain additions to the carbon-oxygen double bond of ketones containing an asymmetric a carbon. Cram s rule predicts which diastereomer will predominate (diastereo-selecti vity). ... 

To estimate the amount of energy absorbed or released in this reaction, we must compile an inventoiy of all the bonds that break and all the bonds that form. A ball-and-stick model shows that propane contains 8 C—H bonds and 2 C—C bonds. These bonds break in each propane molecule, and one ODO bond breaks in each oxygen molecule. Two CDO bonds form in each CO2 molecule, and two O—H bonds form in each H2 O molecule. In summary ... 

The basic difference between the two pathways resides in the fate of the 0-0 bond in the original oxygen molecule, which in the first pathway is broken in the initial reaction step. The resulting oxygen atoms are adsorbed on the electrode and then are reduced electrochemically ... 

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