Timeline for Are bridge bonds in diborane longer than the terminal H bonds?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2018 at 0:53 | vote | accept | M. S. L | ||
| Apr 9, 2018 at 0:53 | |||||
| Apr 5, 2018 at 7:49 | vote | accept | M. S. L | ||
| Apr 5, 2018 at 7:49 | |||||
| Mar 27, 2018 at 23:30 | comment | added | MollyCooL | @orthocresol Oh, yes I get that, only the combined stabilty is more, so we cannot compare the individual stability directly. So is there anyother reason that I missed? | |
| Mar 27, 2018 at 20:57 | comment | added | orthocresol | Your last comment is not 100% rigorous either. The B–C bond does not move in isolation: moving a bridging methyl group to a terminal position means that you have to move a terminal hydrogen into the bridge. If the rearrangement is thermodynamically favoured, it only implies that the bridging B–C–B bond plus the terminal B–H bond, combined, are weaker than a terminal B–C bond plus a bridging B–H–B bond. | |
| Mar 27, 2018 at 12:27 | comment | added | MollyCooL | @M.S.L No, it actually means the opposite I would say. Terminal position is energitically we say, meaning if the B-C bond stays in the bridge, its actually unstable there, which implies the bridge is weaker, and that is why it re-arranges to terminal position. I hope that makes sense! | |
| Mar 27, 2018 at 12:20 | comment | added | M. S. L | Yes i read that..but i am sorry, i didn't quite get it why its more energetically favorable .Correct me if i am wrong but if the bridge is reformed again like you said that means the bridge renders some stability to the molecule and that would mean the bridge bond is more stronger than the terminal ones..but then again u said the bridge bonds are weaker ......I am sorry but what am i missing? | |
| Mar 27, 2018 at 11:02 | comment | added | MollyCooL | @M.S.L Thats what I told you, its apparently not that the terminal bonds are broken. Even if the bridged hydrogen is released, the B-C bond formed because of alkylation, returns to the terminal position because it is energitically favourable. | |
| Mar 27, 2018 at 10:02 | comment | added | M. S. L | So can anyone explain why the terminal bonds rather than the bridge bonds are lost during alkylation? | |
| Mar 26, 2018 at 15:58 | history | edited | MollyCooL | CC BY-SA 3.0 |
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| Mar 26, 2018 at 15:35 | history | edited | MollyCooL | CC BY-SA 3.0 |
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| Mar 26, 2018 at 14:44 | comment | added | MollyCooL | Also, a source on quora claims that on alkylation, whichever H atom is released, the B-C bond will remain/transfer to a terminal position because that would be energetically more favorable. I’m not sure about it though. | |
| Mar 26, 2018 at 14:23 | history | edited | MollyCooL | CC BY-SA 3.0 |
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| Mar 26, 2018 at 14:18 | history | answered | MollyCooL | CC BY-SA 3.0 |