hybridization of halobetasol propionate

halobetasol propionate | orbital hybridization

First draw the structure diagram for halobetasol propionate, and for every non-hydrogen atom, count the σ-bonds. Note that double and triple bonds consist of one σ-bond together with one or two π-bonds: Identify those atoms with lone pairs: Find the steric number by adding the lone pair count to the number of σ-bonds: Consult the following chart to determine the hybridization from the steric number: steric number | hybridization 2 | sp 3 | sp^2 4 | sp^3 5 | dsp^3 6 | d^2sp^3 7 | d^3sp^3 Assign the provisional hybridization based on the table: Next identify any sp^3 atoms with lone pair electrons which can participate in a conjugated π-bond system. These atoms can lower their energy by placing a lone pair in a unhybridized p orbital to maximize overlap with the neighboring π-bonds. Note that halogens and elements from the third period and below do not engage in bond conjugation, except in the case of aromaticity: Adjust the provisional hybridizations to arrive at the result: Answer: | |

formula | C_25H_31ClF_2O_5 name | halobetasol propionate IUPAC name | [(6S, 9S, 16S, 17R)-17-(2-chloroacetyl)-6, 9-difluoro-11-hydroxy-10, 13, 16-trimethyl-3-oxo-6, 7, 8, 11, 12, 14, 15, 16-octahydrocyclopenta[a]phenanthren-17-yl] propanoate alternate names | halobetasol | ulobetasol propionate | ultravate mass fractions | C (carbon) 61.9% | Cl (chlorine) 7.31% | F (fluorine) 7.83% | H (hydrogen) 6.44% | O (oxygen) 16.5%