00:01
In this problem, we start with compound a that has c8 h10 as its molecular formula.
00:09
So we're told this undergoes nitration to give two products, and then we're told about our proton nmr spectrum.
00:19
So before we look at our nmr data, let's find degrees of unsaturation.
00:25
So this will tell us the number of rings and or pie bonds.
00:30
So we multiply the number of carbons by two and add two, subtract the number of hydrogens, and divide by two.
00:41
So here we get four.
00:43
Now, four is very common when you have a benzene ring.
00:48
So you have three pi bonds and one ring to get four.
00:51
If we look at our nmr data now, we have five hydrogens around in between seven and eight parts per million.
01:02
That's where we're going to find aromatic hydrogens.
01:05
So this tells us that it's mono substituted because we're going to have five hydrogens on our ring...