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  •  » Deduce 2,3-dimethylbut-1-ene (C6H12) from IR and NMR

#1 2011-07-05 01:39:33


Deduce 2,3-dimethylbut-1-ene (C6H12) from IR and NMR

The following problem was posted on Yahoo!Answers

This seems a rather easy problem. This is how I solve them. Although the integrals are cryptic, the structure is so simple we can still calculate a ratio without resorting to a calculator. From left to right, 2:1:3:6 = 12H.

Let's start with the peak at 1.7 ppm, with 3H's. It is a singlet, therefore we have a CH3-C(no hydrogens).

C6H12 - CH3C = C4H9, the other four carbons have the 9H's.
If we tried to put a double bond with one hydrogen on each carbon, then we would need to have 7H's on the remaining two carbons. Impossible. Therefore, the peaks at 4.6 ppm are a =CH2.

C4H9 - CH2 = C3H7

There is a doublet with 6H's. This must be two CH3 groups. The one CH must have the two CH3 groups attached.

Now, if this were tinkertoys, then we just have to assemble them. A =CH2, =CCH3, and a CH(CH3)2.

That would give the following structure.


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