Sarahisme Posted June 13, 2005 Share Posted June 13, 2005 i don't agree with any of these answers: (but i am sure that i am wrong of course ).... so little help please if anyone has time Link to comment Share on other sites More sharing options...
Yggdrasil Posted June 13, 2005 Share Posted June 13, 2005 The answer is v. A carbon has four valence electrons, so when carbon has four electrons, it has a formal charge of 0. Carbon 1 has three bonds and no lone pairs giving it three electrons and thus a formal charge of +1 (less electrons, means more positive charge). Carbons 2 and 3 have four bonds, giving it 4 electrons and no formal charge. Oxygen atoms have six valence electrons, and oxygen 4 has one bond and three lone pairs giving it a total of 7 electrons (1 from the bond, and 6 from the lone pairs). Thus, the oxygen has a formal charge of -1. Link to comment Share on other sites More sharing options...
Sarahisme Posted June 13, 2005 Author Share Posted June 13, 2005 yeah thats what the answer booklet said, and i can see the correct results for 2,3 and 4 carbons but for carbon 1, shouldnt there be 4 bonds to it? Link to comment Share on other sites More sharing options...
Yggdrasil Posted June 13, 2005 Share Posted June 13, 2005 There are only three bonds to the carbon in the diagram, not four. This may seem weird, since for carbon to have a full octet it needs four bonds, but it is possible to have a carbon atom without a full octet. Carbon 1 is what is known as a carbocation, an electron-defficient carbon. If you take organic chemistry, you'll see a lot of reactions involving carbocation intermediates. Link to comment Share on other sites More sharing options...
BenSon Posted June 13, 2005 Share Posted June 13, 2005 yeah thats what the answer booklet said, and i can see the correct results for 2,3 and 4 carbons but for carbon 1, shouldnt there be 4 bonds to it? Yggdrasil is right, also if it was attached to another bond then none of those answers would be right because then carbon one would have a 0 charge and the only answer with 0 for the first carbon is answer (i) and that is clearly wrong. ~Scott Link to comment Share on other sites More sharing options...
Sarahisme Posted June 13, 2005 Author Share Posted June 13, 2005 yeah ok, its just i thought if it did have only 3 bonds, then it would have at least one free electron... ?? Link to comment Share on other sites More sharing options...
BenSon Posted June 13, 2005 Share Posted June 13, 2005 Nope, valance electrons (4) minus lone pairs (0) - Half the total number of electronsin reactions (3) equals formal charge of 1. ~Scott Link to comment Share on other sites More sharing options...
Sarahisme Posted June 13, 2005 Author Share Posted June 13, 2005 yeah its the lone pairs bit, i don't see how there can be no lone pairs (or lone electrons even, that are not in pair (i.e.. just one electron)) ???? Link to comment Share on other sites More sharing options...
BenSon Posted June 13, 2005 Share Posted June 13, 2005 Well there is no electrons paired or otherwise on the diagram around the carbon so i don't see where it would go. I understand where your coming from though it sems like it has four valence electrons and only three are being used so it should have a spare floating around. Aparently thats just not the way it sets itself up structually though, I don't know why it is like this but it is. ~Scott Link to comment Share on other sites More sharing options...
Yggdrasil Posted June 13, 2005 Share Posted June 13, 2005 yeah its the lone pairs bit' date=' i don't see how there can be no lone pairs (or lone electrons even, that are not in pair (i.e.. just one electron))????[/quote'] Basically, a carbocation has an sp2 hybridization and an empty p orbital. The empty p orbital makes carbocations very reactive, so they're highly unstable and you wouldn't be able to isolate one in a laboratory. However, they do exists, if briefly, as an intermediate. Radicals (compounds having lone electrons not in pairs) similarly have an sp2 or sp hybridization, carrying the lone electron in a p orbital. Link to comment Share on other sites More sharing options...
budullewraagh Posted June 13, 2005 Share Posted June 13, 2005 carbocations can be prepared in a laboratory, but in order to remain stable over a span of time, require the presence of superacids (HSbF6, H2FSO3, etc) Link to comment Share on other sites More sharing options...
Sarahisme Posted June 14, 2005 Author Share Posted June 14, 2005 allright, thanks for that Link to comment Share on other sites More sharing options...
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