Organic Chemistry / Cumulative General Concepts
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Solution: With current spectroscopic techniques (discussed in Chapters...

Question

With current spectroscopic techniques (discussed in Chapters 15–17), chemists are generally able to determine the structure of an unknown organic compound in just one day. These techniques have only been available for the last several decades. In the first half of the twentieth century, structure determination was a very slow and painful process in which the compound under investigation would be subjected to a variety of chemical reactions. The results of those reactions would provide chemists with clues about the structure of the compound. With enough clues, it was sometimes (but not always) possible to determine the structure. As an example, try to determine the structure of an unknown compound, using the following clues:

  • The molecular formula is C4H10N2.

  • There are no π bonds in the structure.

  • The compound has no net dipole moment.

  • The compound exhibits very strong hydrogen bonding.
    You should note that there are at least two constitutional isomers that are consistent with the information above. (Hint: Consider incorporating a ring in your structure.) 

Video Transcript

Hey guys. So, in this problem we need to go ahead and find a constitutional isomer or two, at least, that have, for the formula C4H10N2 where we don't have a net dipole, we have strong hydrogen bonding and we don't have a PI bond, okay? but guys if we go ahead and run the IHD calculation, right? We're going to see that we're going to need some kind of PI bond your ring, okay? So let's go ahead and figure this out, we've got our IHD formula we know is 2 n plus 2 minus H all divided by 2, right? Guys here, what do we know? Well, we know that X or H is equal to 1, right? We also know that oxygen is equal to 0, right? Let me fix that zero a little bit, and our nitrogen is equal to negative 1, negative 1, okay? So guys, let's go ahead and factor that in, let's go plug it in, we've got 2 times 4 plus 2, okay? Minus 10 plus negative 1, right? But is it negative 1? it's actually negative 2 because we've got n2, right? plus negative 2, okay? So what is this all factoring to, simplifying to 2 plus 2 times 4 is 8 plus 2 that's 10, I don't know why I wrote 2, that's 10, right? And then guys minus 10 minus 2, right? So we're going to have 8, right? So now what we get is 10 minus 8 over 2, so that's we're going to get 2 over 2, which is going to factor, which is going to simplify into 1 and guys if we don't have a double bond, right? We know it needs to be a ring, okay? So we've got 4 carbons and two nitrogens, so what if we did this, we have our two carbons on one side, 2 carbons on the other and our nitrogens here, does that work? Well, now we know that our nitrogens like to have three bonds and here they we have two, right? So let's go ahead and add our hydrogens, okay? And let's see, we've got 2, 4, 6, 8, 9, 10, so guys this one works just fine, so what if we did something else? what if, see how we've got this four member carbon ring inside, right? What if we did something that looks like this, we've got four carbons in a ring and now we need to add our nitrogens, right? So guys, if we add our nitrogens we know that they need to be on opposite sides, so that we don't get a net dipole, right? So what we can do is our nitrogen over here and the nitrogen over here and now we need to add our hydrogens, right? So we can get H2 and H2, right? Do these work? absolutely, right? We've got strong hydrogen bonding because we have both electro, I'm sorry, hydrogen bond acceptors and donors, right? because we've got the hydrogen directly attached to this electronegative n, right? We've got no dipole because we've got a di, we've got no net dipole because we've got a dipole pulling in this direction, cancelling this one, right? Same thing on this side, right? And guys, do we have any PI bonds? no, right? So guys these two are perfect constitutional isomers that follow these criteria. Alright, so let's move on.

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