Propose possible structures consistent with each set of...

Propose possible structures consistent with each set of data. Assume each compound has an $sp^3$ hybridized $C-H$ absorption in its IR spectrum, and that other major IR absorptions above 1500 $cm^{-1}$ are listed. a. a compound having a molecular ion at 72 and an absorption in its IR spectrum at 1725 $cm^{-1}$ b. a compound having a molecular ion at 55 and an absorption in its IR spectrum at - 2250 $cm^{-1}$ c. a compound having a molecular ion of 74 and an absorption in its IR spectrum at 3600-3200 $cm^{-1}$

Propose possible structures consistent with each set of data. Assume each compound has an $sp^3$ hybridized $C-H$ absorption in its IR spectrum, and that other major IR absorptions above 1500 $cm^{-1}$ are listed.
a. a compound having a molecular ion at 72 and an absorption in its IR spectrum at 1725 $cm^{-1}$
b. a compound having a molecular ion at 55 and an absorption in its IR spectrum at - 2250 $cm^{-1}$
c. a compound having a molecular ion of 74 and an absorption in its IR spectrum at 3600-3200 $cm^{-1}$

Transcript

00:03 This is the answer to chapter 13, problem number 44 from the smith organic chemistry textbook.

00:10 And so this problem is giving us three sets of mass specs and irs, and asking us to draw a suitable molecule for each set of data.

00:25 And so for a, we're given a molecular ion of 72 and an ir peak at, 1725 wave numbers.

00:37 And so basically we know that that irp should correspond to a ketone.

00:45 And so we, i wrote that down in green.

00:48 So that's the first piece of pertinent information.

00:52 And then we'll look at the mass spec.

00:55 So since we know that there's a ketone, we know that there's an oxygen.

00:58 So oxygen accounts for 16 atomic mass units, 72 -mars.

01:03 Minus 16 gets us 56.

01:06 And then i think the strategy for this is to start by assuming that you have as many carbons as you can to get to that number without going over.

01:23 So carbon accounts for 12 atomic mass units.

01:28 So four carbons would be 48 and then fill in the remaining.

01:34 Mass that you have with hydrogens.

01:36 So eight more hydrogens, 48 plus eight, gets us to 56.

01:41 And then we have the one oxygen, pardon me, the one oxygen that we said was present.

01:47 And so given this formula and the fact that there has to be a ketone in it, this is really the only structure that we can draw.

01:56 And so for b, we're going to follow the same process.

02:00 So our molecular ion is at 55 for b.

02:04 And this stands out right away because it's an odd number...