hybridization of 1,4-butanediol diacrylate

1, 4-butanediol diacrylate | orbital hybridization

First draw the structure diagram for 1, 4-butanediol diacrylate, 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 | (-CH_2CH_2OCOCH=CH_2)_2 Hill formula | C_10H_14O_4 name | 1, 4-butanediol diacrylate IUPAC name | prop-2-enoic acid 4-(1-oxoprop-2-enoxy)butyl ester alternate names | 1, 4-butylene diacrylate | 4-prop-2-enoyloxybutyl prop-2-enoate | acrylic acid 4-acryloyloxybutyl ester | butanediol diacrylate | tetramethylene diacrylate | tetramethylene glycol diacrylate mass fractions | C (carbon) 60.6% | H (hydrogen) 7.12% | O (oxygen) 32.3%