Give the (R, S) designation for each compound given as an...

Give the $(R, S)$ designation for each compound given as an answer to Practice Problem 5.29 (a) $\mathrm{C}_{8} \mathrm{H}_{14} \mathrm{O}_{4}$ $^{1} \mathrm{H} \mathrm{NMR}$ Spectrum Triplet $\delta 1.2(6 \mathrm{H})$ Singlet $\delta 2.5(4 \mathrm{H})$ Quartet $\quad \delta 4.1(4 \mathrm{H})$ IR Spectrum $1740 \mathrm{cm}^{-1}$ (b) $\mathrm{C}_{11} \mathrm{H}_{14} \mathrm{O}_{2}$ $^{1} \mathrm{H} \mathrm{NMR}$ Spectrum Doublet $\quad \delta 1.0(6 \mathrm{H})$ Multiplet $\quad \delta 2.1(1 \mathrm{H})$ Doublet $\delta 4.1(2 \mathrm{H})$ Multiplet $\quad \delta 7.8(5 \mathrm{H})$ IR Spectrum $1720 \mathrm{cm}^{-1}$ (c) $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ $^{1} \mathrm{H} \mathrm{NMR}$ Spectrum Triplet $\delta 1.2(3 \mathrm{H})$ Singlet $\delta 3.5(2 \mathrm{H})$ Quartet $\delta 4.1(2 \mathrm{H})$ Multiplet $\quad \delta 7.3(5 \mathrm{H})$ IR Spectrum $1740 \mathrm{cm}^{-1}$ (d) $\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O}_{2}$ 'H NMR Spectrum singlet $\delta 6.0$ Singlet $\delta 11.70$ IR Spectrum Broad peak $2500-2700 \mathrm{cm}^{-1}$ $1705 \mathrm{cm}^{-1}$ (e) $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{ClO}_{2}$ 'H NMR Spectrum Triplet $\delta 1.3$ singlet $\delta 4.0$ Quartet $\delta 4.2$ IR Spectrum $1745 \mathrm{cm}^{-1}$

Give the $(R, S)$ designation for each compound given as an answer to Practice Problem 5.29
(a) $\mathrm{C}_{8} \mathrm{H}_{14} \mathrm{O}_{4}$ $^{1} \mathrm{H} \mathrm{NMR}$ Spectrum
Triplet $\delta 1.2(6 \mathrm{H})$
Singlet $\delta 2.5(4 \mathrm{H})$
Quartet $\quad \delta 4.1(4 \mathrm{H})$ IR Spectrum $1740 \mathrm{cm}^{-1}$
(b) $\mathrm{C}_{11} \mathrm{H}_{14} \mathrm{O}_{2}$
$^{1} \mathrm{H} \mathrm{NMR}$ Spectrum
Doublet $\quad \delta 1.0(6 \mathrm{H})$
Multiplet $\quad \delta 2.1(1 \mathrm{H})$
Doublet $\delta 4.1(2 \mathrm{H})$
Multiplet $\quad \delta 7.8(5 \mathrm{H})$ IR Spectrum $1720 \mathrm{cm}^{-1}$
(c) $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$
$^{1} \mathrm{H} \mathrm{NMR}$ Spectrum
Triplet $\delta 1.2(3 \mathrm{H})$
Singlet $\delta 3.5(2 \mathrm{H})$
Quartet $\delta 4.1(2 \mathrm{H})$
Multiplet $\quad \delta 7.3(5 \mathrm{H})$
IR Spectrum $1740 \mathrm{cm}^{-1}$
(d) $\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2} \mathrm{O}_{2}$
'H NMR Spectrum
singlet $\delta 6.0$
Singlet $\delta 11.70$
IR Spectrum
Broad peak $2500-2700 \mathrm{cm}^{-1}$
$1705 \mathrm{cm}^{-1}$
(e) $\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{ClO}_{2}$
'H NMR Spectrum
Triplet $\delta 1.3$
singlet $\delta 4.0$
Quartet $\delta 4.2$
IR Spectrum $1745 \mathrm{cm}^{-1}$

Key Concepts

Spectroscopic Techniques in Structure Elucidation

Spectroscopic methods such as 1H NMR and IR spectroscopy are vital tools in organic chemistry for determining molecular structure and identifying functional groups. While these techniques do not directly assign R,S configurations, they provide critical structural information that can help locate chiral centers for further stereochemical analysis.

Chirality

Chirality is the property of a molecule that makes it non-superimposable on its mirror image. This arises typically from a carbon atom bonded to four different substituents, giving rise to stereoisomers that can exhibit distinct physical and biological behaviors.

Cahn-Ingold-Prelog (CIP) Priority Rules

The CIP rules provide a systematic method for assigning priorities to the substituents attached to a chiral center based on atomic number and connectivity. This ranking is essential for determining the absolute configuration (R or S) of the chiral center.

Absolute Configuration (R,S) Designation

Absolute configuration involves assigning a spatial arrangement to the substituents around a chiral center using the R (rectus) or S (sinister) nomenclature. This designation is made by orienting the molecule so that the lowest priority group is directed away and then determining the order of the remaining groups as either clockwise (R) or counterclockwise (S).

Transcript

00:03 This is the answer to chapter 21, problem number 81 from the smith organic chemistry textbook.

00:10 This problem gives us four sets of data and asks us to come up with a structure for each set of data.

00:17 And so i'm going to jump right in because this is quite a large problem.

00:24 Okay, so the starting point for each of these is going to be to calculate an hdi based on the formula that we're given.

00:32 And so remember that's going to be two times the number of carbons, which in this case is five plus two minus the number of hydrogens, which in this case is 10.

00:45 And all of that over two, so we have 12 minus 10 is 2, over 2 is 1.

00:50 So we've one degree of unsaturation in this molecule.

00:53 So one double bond or one ring.

00:56 Looking at the ir data, we have a peak at 1728.

01:01 That is going to be indicative of a carbon oxygen double bond.

01:06 And then the other two peaks that we have between 2 ,700 and 2791, that is going to indicate an aldehyde.

01:16 So that's our one degree of unsaturation.

01:19 So then looking at the nmr signals, our second signal at 9 .48 ppm with an integration of 1 is very far downfield.

01:28 And so that is indicative of an aldehyde.

01:34 Okay, and we only have one other signal, 1 .08 ppm singlet with an integration of nine.

01:42 So this is going to be three methyl groups.

01:46 So three, ch3s.

01:51 And so that's four carbons accounted for.

01:55 So i'm going to say that what we have here is a central carbon.

02:00 With three methyl groups on it, and then an aldehyde.

02:13 Okay, so there's our structure for a.

02:16 So now we need to do this for the other three sets of data that we have.

02:20 So for b, the formula is exactly the same as a, so we don't need to calculate the hdi.

02:26 It's going to be exactly the same as a's was.

02:29 So again, our hdi is going to be one, so one ring or one double bond.

02:34 Our only ir signal is at 1718 wave number.

02:40 So this is, again, indicative of a carbon oxygen double bond.

02:45 I'm going to say probably a ketone this time.

02:49 And then looking at our nmr data, we have a 1 .10 ppm doublet with an integration of 6.

02:59 And so this is going to be two equivalent methyl groups.

03:05 And they're going to be next to a carbon with a single proton on them, which is why they're split into a doublet.

03:14 And so they are, the first and third signal are splitting each other, basically.

03:20 And so we know that our third signal is going to be a piece that looks like this.

03:28 So we have two methyl groups and then a bond to something else.

03:35 And then the signal is actually coming from a hydrogen right here.

03:39 So we have these two pieces figured out.

03:42 And then we have a singlet at 2 .14 ppm that integrates to three.

03:48 And so this is probably another methyl group.

03:53 Okay.

03:54 And so when we put all of this information together, it kind of has to look like this.

04:01 So here's our methyl group.

04:04 Our ketone is going to be right here.

04:06 And then we have the rest of the molecule here.

04:12 Okay, and so that's the structure for b.

04:17 So moving on to c, we need to calculate an hdi here.

04:23 So for this formula, it's going to be 2 times 10.

04:29 It's the number of carbons plus 2 minus 12, which is the number of hydrogens.

04:35 And so 22 minus 12.

04:38 Is 10 divided by 2 is 5.

04:42 So we have an hdi of 5 here.

04:44 So right away, anytime i see four or more, i'm thinking an aromatic six -membered ring, since that accounts for four degrees of unsaturation.

04:58 And let's jump to the nmr.

05:01 So 7 .24 ppm double it that integrates to 2.

05:07 So this is probably two aromatic protons.

05:14 And the last signal at 7 .85 ppm, also a doublet, also an integration of two.

05:20 That's going to be two more aromatic protons.

05:23 So we have an aromatic ring here, and it's dis substituted...

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