Can someone explain to me the coupling pattern by proton $5$? Why is it triplet of quintet? Is it long range coupling?
I'm assuming that the proton $5$ couples to two other protons on carbon $4$. But what's next?
$\begingroup$
$\endgroup$
-
$\begingroup$ Sure, it couples with two other protons on C4, hence the triplet. $\endgroup$ – Ivan Neretin Dec 2 '17 at 23:56
-
1$\begingroup$ It's a triplet of septuplets. It's hard to see, but you can see the seventh peak sticking out. $\endgroup$ – orthocresol♦ Dec 2 '17 at 23:56
$\begingroup$
$\endgroup$
You are close. The triplet part is the coupling of the enantiotopic hydrogens at C4. The width of the C5 blowup is ~20 Hz in width which means J = ~7 Hz. The smaller coupling is due to allylic coupling with the methyls (C7 and C8). Assuming J5,7 and J5,8 are the same, then the six hydrogens produce a septet. You just can't see the outer peaks because they are too small. If coupling to the two methyls is different then the septet will appear broadened because of lack of resolution. A larger blowup of C5 would reveal the outer peaks (one and seven).
