Write structural formulas for all the constitutional isomers of
So, let's go ahead and come up with all the constitutional isomers and structural formulas that we can proceed through h4. Now, just by looking, right? if we follow a 2 n plus 2 rule we can see that we have an ihd of 2, so this is 2 times 3 plus 2, so that's going to get us a total of 8 minus 4 all divided by 2, 8 minus 4 divided by 2 is going to give us an IHD of 2, which means that we can have a ring and a pi bond, 2 pi bonds or 2 rings but we can't actually have two rings because we've only got three carbons, right? It just doesn't work, so let's go ahead and do this, we can say that we've got our 3 carbons there and a double bond here and a double bond here, right? But this isn't actually the structure that works, you can say that we've got 3 carbons, because it's linear, okay? Because of the way the double bonds have to set up their PI orbitals, their P orbitals make those PI bonds. So, here are our hydrogens, okay? So, to convert this into structural formula we would just say H2C, double bond, C, double bond, CH2, right? Very, very similar. Now, we can also do something like this, right? We can say what if we get our ring, right? there's one IHD out of the way, I can't draw a triangle with you guys, sorry, and then have a double bond on one of them, okay? That actually gives us 2 ihd and for that we would say that we've got CH, CH2, CH and then our an extra bond down there, can we really do anything else? actually there is one more thing we can do, right? What if we did carbon, single bonded to carbon, triple bonded to another carbon. Now, we've got three hydrogens here and we've only got one hydrogen here, right? And we know this is linear because of our triple bond, right? for the same reason as this, we need to organize our P orbitals, so now this would just look like CH3, C triple bond CH. Alright guys, let me know if you have any questions, if not, let's move on.
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True or False: The following 2 compounds are constitutional isomers of each other.
Draw a constitutional isomer of the compound given below
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Draw structural formulas for 5 (and only 5) constitutional isomers that have one (and only one) ring and that one ring must be a three-membered ring and the molecular formula C5H7F. Your structural formulas must have no nonzero formal charges and the most typical arrangements of bonding and nonbonding valence electrons. Note that if you give the same isomer more than once, points will be deducted for that isomer. If you give more than 5 constitutional isomers, points will be deducted. For this problem, you must give complete structural formulas showing all atoms and bonding valence electrons. Do not give condensed structural formulas or bond-line formulas.
Choose the compound pair that are not constitutional isomers:
Draw structural formulas for 5 constitutional isomers that have one (and only one) three-membered ring and the molecular formula C6H12. Yes, each of the constitutional isomers that you give must be a substituted cyclopropane. Your structural formulas must have no nonzero formal charges and the most typical arrangements of bonding and nonbonding valence electrons.
Note: For this problem, you must give complete structural formulas showing all atoms and bonding valence electrons. Do not give condensed structural formulas or bond-line formulas.