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|  0 recommendations | Message 1 of 10 in Discussion |
| | (Original Message) | Sent: 4/9/2008 1:56 AM |
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| | From:  ·Steve· | Sent: 4/9/2008 3:26 AM |
Here are some hints! There are lots of possible structures, but to get an even number of hydrogens in Compound B, we need to multiply its empirical formula by 2. Doing that to the formula of Compound A also, you get the following still simple formulas to work from: Compound A: C6H12O2 Compound B: C6H10O2 1) What is the degree of unsaturation of these compounds? That is helpful information. 2) Can you think of an oxidation reaction that lowers the number of hydrogens by two? You have probably seen oxidations of alcohols and aldehydes; what happens to the formulas after the oxidation?  >> provide all possible structures of Compound A and of Compound B <<  In principle, there are an infinite number of possibilities, depending on what factor you multiply the empirical formulas by, for starters! Does the problem give you any additional information, such as a molecular weight? Steve |
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| | 0 recommendations | Message 3 of 10 in Discussion |
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| | | From: ·Steve· | Sent: 4/9/2008 3:54 PM |
Secondary alcohol is correct. For example, the oxidation of 2-propanol to acetone results in the loss of two hydrogens: OH O | oxidation || CH3CHCH3 –––––––––�?gt; CH3CCH3 So, I would start with a secondary alcohol that has the formula C6H12O2 which has has a structure analogous to benzoin. Steve |
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| | | From: what464 | Sent: 4/9/2008 7:45 PM |
Im not sure if this right but im going to take a shot...
Compound A can be:
CH3CH(OH)C(=O)CH2CH2CH3
CH3CH(OH)CH2C(=O)CH2CH3
CH3CH(OH)CH2CH2C(=O)CH3
and Compound B is the same except the alcohol groups are ketones now... is there anything I forgot? |
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| | | From: ·Steve· | Sent: 4/9/2008 11:02 PM |
Those work! Other structures that are higher multiples of the empirical formulas are possible, and there are many variations that could occur in all of them, such as branched carbon chains and cyclic structures. Steve |
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| | | From: what464 | Sent: 4/10/2008 4:24 AM |
| I found a few others that work (6 total), but no compounds after 6 carbons. Do you know any compounds that would fit the empirical formula (more than 6 carbons), such as the branched carbon chains or cyclic structures that you previously stated? |
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| | 0 recommendations | Message 8 of 10 in Discussion |
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| | | From: ·Steve· | Sent: 4/10/2008 6:50 AM |
I'm thinking that in theory you can have any molecular formula for Compound A that is an even multiple of the empirical formula C3H6O. For example, X2 A = C6H12O2 B = C6H10O2 A has 1 OH and 1 C=O X4 A = C12H24O4 B = C12H20O4 A has 2 OH and 2 C=O X6 A = C18H36O6 B = C18H30O6 A has 3 OH and 3 C=O etc.
If you know that only one equivalent of HNO3 reacts with Compound A to form Compound B (assuming the stoichiometry is 1-to-1), then only one OH group can be present which would narrow the possibilities to the C6H12O2 formula. Many cyclic structures with formula C6H12O2 containing an alcohol group are possible. For instance, imagine a 6-membered heterocyclic ring containing one oxygen atom. Put a methyl group and and OH group on the ring somewhere, and that opens up a whole new can of worms!  Steve |
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