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Pentacyclo[4.2.0.02,5.03,8.04,7]octane
I'm pretty sure this is impossible. A tertiary carbon is bonded to three other carbons. Counting each of these 3 as one-half of a carbon-carbon bond (the other half coming from the other carbon), that means that if there are 9 carbons, there must be 27 half bonds in the molecule, which works out to thirteen carbon-carbon bonds plus half a bond left over. Since you can't have half a bond in a real compound, it's not possible to arrange nine carbons so that each one is bonded to three others. You can do it with 8, in which case you get cubane. ---------------------------------------------------------------------------------------------------------- Uh you shouldn't label this as impossible. I reckon its COMPLETELY POSSIBLE! For example: look at this: H H H H H H H H H | | | | | | | | | H-C-C-C-C-C-C-C-C-C-H | | | | | | | | | H H H H H H H H H This is Nonane (C9H20). I'm not too sure if this is right, but so far it looks pretty good-and makes sense. So that puts a whole in your answer up there =P *OMGG! see those vertical hydrogen-carbon bonds (or lines in between the Hydrogens and Carbons) ther actually supposed to connect the 9 carbons to the the 2 hydrogen atoms above and under them. Sorry, i had it perfect before, but when i save it screws up. Hope this helps... **Ink and Paper** ---Yo i think my answer is wrong, but i ain't gonna delete it coz it took me time---
Pentacyclo[4.2.0.02,5.03,8.04,7]octane
There are several. Barrelene, Benzocyclobutene, Cubane, Cuneane, Cyclooctatetraene, Semibullvalene, and Styrene all possess that structure. This is so because C8H8 has a lot of different ways to combine. Further info: http://en.wikipedia.org/wiki/C8H8
I'm pretty sure this is impossible. A tertiary carbon is bonded to three other carbons. Counting each of these 3 as one-half of a carbon-carbon bond (the other half coming from the other carbon), that means that if there are 9 carbons, there must be 27 half bonds in the molecule, which works out to thirteen carbon-carbon bonds plus half a bond left over. Since you can't have half a bond in a real compound, it's not possible to arrange nine carbons so that each one is bonded to three others. You can do it with 8, in which case you get cubane. ---------------------------------------------------------------------------------------------------------- Uh you shouldn't label this as impossible. I reckon its COMPLETELY POSSIBLE! For example: look at this: H H H H H H H H H | | | | | | | | | H-C-C-C-C-C-C-C-C-C-H | | | | | | | | | H H H H H H H H H This is Nonane (C9H20). I'm not too sure if this is right, but so far it looks pretty good-and makes sense. So that puts a whole in your answer up there =P *OMGG! see those vertical hydrogen-carbon bonds (or lines in between the Hydrogens and Carbons) ther actually supposed to connect the 9 carbons to the the 2 hydrogen atoms above and under them. Sorry, i had it perfect before, but when i save it screws up. Hope this helps... **Ink and Paper** ---Yo i think my answer is wrong, but i ain't gonna delete it coz it took me time---
According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 2 words with the pattern --BANE. That is, six letter words with 3rd letter B and 4th letter A and 5th letter N and 6th letter E. In alphabetical order, they are: cubane urbane
According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 6 words with the pattern --B-NE. That is, six letter words with 3rd letter B and 5th letter N and 6th letter E. In alphabetical order, they are: cubane debone rubine sabine unbone urbane
According to SOWPODS (the combination of Scrabble dictionaries used around the world) there are 14 words with the pattern -U--NE. That is, six letter words with 2nd letter U and 5th letter N and 6th letter E. In alphabetical order, they are: butane butene cubane furane humane lumine lupine murine mutine puisne purine rubine rusine supine