Return to Question of Equivalent hydrogens. Stereotopicity – Equivalent or Not?

Are these two hydrogens truly equivalent?

Seemingly equivalent hydrogens may be homotopic, enantiotopic, diastereotopic.

How to tell: replace one of the hydrogens with a D.

If produce an achiral molecule then hydrogens are homotopic,

if enantiomers then hydrogens are enantiotopic,

if diastereomers then diastereotopic.

We look at each of these cases.

HH

CH3

H3C H

C(CH3)3

Seem to be equivalent until we look at most stable conformation, the most utilized conformation.

H3C

H2C C(CH3)3

CH3

H

H3C

C(CH3)3

H

H

CH3

H

Homotopic

HH replace one H with D DH

Achiral Achiral

The central hydrogens of propane are homotopic and have identical chemical shifts under all conditions.

Enantiotopic

HH replace one H with D DH

Achiral Chiral, havetwo enantiomers.

The hydrogens are enantiotopic and equivalent in the NMR unless the molecule is placed in a chiral environment such as a chiral solvent..

The hydrogens are designated as Pro R or Pro S

DH

This structure would be S

Pro S hydrogen.Pro R hydrogen

DH

H3C

HCl

Diastereotopic

If diastereormers are produced from the substitution then the hydrogens are not equivalent in the NMR. Diastereotopic hydrogens.

The hydrogens are designated as Pro R or Pro S

This structure would be S

Pro S hydrogen. (Making this a D causes the structure to be S.)

Pro R hydrogen

DH

H3C

HCl

HH

H3C

HCl

replace H with D

HD

H3C

HCl

produced diastereomers

H

H3C

OH

H3C

CH3

H

H

a

a'

cd

b

Diastereotopic methyl groups (not equivalent), each split into a doublet by Hc

a and a’

Example of diastereotopic methyl groups.

13C NMR

• 13C has spin states similar to H. • Natural occurrence is 1.1% making 13C-13C spin spin

splitting very rare.• H atoms can spin-spin split a 13C peak. (13CH4 would yield

a quintet). This would yield complicated spectra.• H splitting eliminated by irradiating with an additional

frequency chosen to rapidly flip (decouple) the H’s averaging their magnetic field to zero.

• A decoupled spectrum consists of a single peak for each kind of carbon present.

• The magnitude of the peak is not important.

13C NMR spectrum

4 peaks 4 types of carbons.

13C chemical shift table

Hydrogen NMR: Analysis: Example 1

1. Molecular formula given. Conclude: One pi bond or ring.

2. Number of hydrogens given for each peak, integration curve not needed. Verify that they add to 14!

3. Three kinds of hydrogens. No spin-spin splitting. Conclude: Do not have non-equivalent H on adjacent carbons.

4. The 9 equivalent hydrogens likely to be tert butyl group (no spin-spin splitting). The 3 equivalent hydrogens likely to be methyl group. The two hydrogens a CH2.

Fragments: (CH3)3C-, -CH2-, CH3-

5. Have accounted for all atoms but one C and one O. Conclude: Carbonyl group!

-(C=O)-

6. Absence of splitting between CH2 and CH3. Conclude: they are not adjacent.

O

Example 2, C3H6O

1. Molecular formula One pi bond or ring

2. Four different kinds of hydrogen: 1,1,1,3 (probably have a methyl group).

3. Components of the 1H signals are about equal height, not triplets or quartets

4. Consider possible structures.

Possible structures

OCH3

HO

O

OH

O

O

Figure 13.8, p.505

Chemical shift table… Observed peaks were 2.5 – 3.1

vinylic

ethers

Observed peaks were 2.5 – 3.1. Ether!

Possible structures

OCH3

HO

O

OH

O

O

NMR example

Formula tells us two pi bonds/rings

Three kinds of hydrogens with no spin/spin splitting.

What can we tell by preliminary inspection….

Now look at chemical shifts

2. From chemical shift conclude geminal CH2=CR2. Thus one pi/ring left.

3. Conclude there are no single

C=CH- vinyl hydrogens. Have CH2=C-R2.

This rules out a second pi bond as it would have to be fully substituted, CH2=C(CH3)C(CH3)=C(CH3)2 , to avoid additional vinyl hydrogens which is C8H14.

X

In CH2=CR2 are there allylic hydrogens: CH2=C(CH2-)2?

1. Formula told us that there are two pi bonds/rings in the compound.

Do the R groups have allylic hydrogens, C=C-CH?

1. Four allylic hydrogens. Unsplit. Equivalent!

2. Conclude CH2=C(CH2-)2

3. Subtract known structure from formula of unknown…

C7H12

- CH2=C(CH2-)2

------------------------------------------

C3H6 left to identifyRemaining hydrogens produced the 6H singlet.

Likely structure of this fragment is –C(CH3)2-.

But note text book identified the compound as

Infrared Spectroscopy

Chapter 12

Table 12.1, p.472

Energy

Final Exam Schedule, Thursday, May 22, 10:30 AM

Fang, MD10A

Kunjappu, MD10B

Kunjappu, MD10C

320A

Metlitsky, MD10D 1127N

Zamadar 2143N

Infrared spectroscopy causes molecules to vibrate

Infrared radiation does not cause all possible vibrations to vibrate.

For a vibration to be caused by infrared radiation (infrared active) requires that the vibration causes a change in the dipole moment of the molecule. (Does the moving of the atoms in the vibration causes the dipole to change. Yes: should appear in spectrum. No: should not appear.)

A non-linear molecule having n atoms may have many different vibrations. Each atom can move in three directions: 3n. Need to subtract 3 for translational motion and 3 for rotations

# vibrations = 3 n – 6

(n = number of atoms in non-linear molecule)

H

H

H

H

F

F

H

H

Consider C=C bond stretch…

ethylene 1,1 difluoro ethylene

What about 1,2 difluoro ethylene?

Table 12.4, p.478

Different bonds have different resistances to stretching, different frequencies of vibration

Figure 12.2, p.475

Frequency, measured in “reciprocal centimeters”, the number of waves in 1 cm distance.

Energy.

wavelength

Typical Infra-red spectrum.

Figure 12.2, p.475

C=OC-H“fingerprint region”, complex vibrations of the entire molecule.

Vibrations characteristic of individual groups.

Table 12.5, p.480

BDE of C-H

414

464

556

472

Table 12.5, p.480

BDE and CC stretch

727

966

376

Figure 12.4, p.480

Alkane bands

Recognition of Groups: Alkenes (cyclohexene).

Compare these two C-H stretches Sometimes weak

if symmetric

Recognition of Groups: Alkynes (oct-1-yne)

This is a terminal alkyne and we expect to see1. Alkyne C-H2. Alkyne triple bond stretch (asymmetric)

Recognition of Groups: Arenes. (methylbenzene, toluene)

Out-of-plane bend; strong

Recognition of Groups: Alcohols

The O-H stretch depends on whether there is hydrogen bonding present

Compare –O-H vs -O-H….O Hydrogen bonding makes it easier to move the H with H bonding as it is being pulled in both directions; lower frequency

Recognition of Groups: Alcohols

Recognition of Groups: Ethers

No O-H bond stretch present but have C-O in same area as for alcohol.

C-O stretch in assymetric ethers

O

CH3sp3

sp2

Recognition of Groups: Amines

Easiest to recognize is N-H bond stretch: 3300 – 3500 cm-1. Same area as alcohols. Note tertiary amines, NR3, do not have hydrogen bonding.

Hydrogen bonding can shift to lower frequency

Esters

One C=O stretch and two C-O stretches.

Recognition of Groups: Carbonyl

C=O stretch can be recognized reliably in area of 1630 – 1820 cm-1

•Aldehydes will also have C(O)-H stretch

•Esters will also have C-O stretch

•carboxylic acid will have O-H stretch

•Amide will frequently have N-H stretch

•Ketones have nothing extra

What to check for in an IR spectrum

C-H vibrations about 3000 cm-1 can detect vinyl and terminal alkyne hydrogens.

O-H vibrations about 3500 cm-1

C=O vibrations about 1630 – 1820 cm-1

C-O vibrations about 1000-1250 cm-1