Structure Elucidation and Spectroscopy

Question 1

Describe what an inversion point is. Comment on molecules with inversion points.

Answer 1

In a structure with point symmetry, the IP at the center of the molecule such that if two lines are drawn in opp. directions along the same axis, then both line segments intercept identical atoms at the same distance from the IP. Molecules with IPs are NONPOLAR, Dipole vectors cancel each other out.

Question 2

As symmetry (increases/decreases), the number of signals in the spectroscopic study (increases/decreases).

Answer 2

As symmetry increases, the number of signals in a spectroscopic study decreases.

Question 3

What information can degrees or units of unstaturation give us?

Answer 3

They give us information about number of rings and/or pi bonds present within a molecule. For every extra pair of bonding e- in a molecule, there is a unit of unsaturation.

Question 4

Define alipathic.

Answer 4

Alipathic refers to a structure that has no rings or pi bonds. Alkanes are alipathic.

Question 5

What’s the formula for alipathic alkanes?

Answer 5

CnH2n+2

Question 6

Equation for degrees of unsaturation?

Answer 6

(2(#C) + (#N) + (#H) - (#X) + 2)/2

Question 7

Describe IR spectroscopy in detail.

Answer 7

IR is directed at molecule, which absorbs IR, causing transitions b/t vibrational energy levels within it, so that the molecule vibrates more frequently as E is absorbed. This change in vibrating (stretching) b/t atoms causes a change in the dipole moment, which could be monitored. We record the change in the intensity of IR from when it enters the molecule and when it exits.

Question 8

Describe IR spectroscopy in layman’s terms.

Answer 8

When you bombard a molecule with IR, it causes different bonds to vibrate at distinct frequencies. This is measured in cm^-1.

Question 9

Relate wave number, wavelength, and energy.

Answer 9

1/λ = Wn α Energy that is absorbed by bond

Question 10

When is IR most useful?

Answer 10

Functional groups or which functional groups aren’t present.

Question 11

What are limitations of IR?

Answer 11

Doesn’t tell us where functional group is located or how many of them are in a molecule. Also useful for distinguishing constitutional isomers, but not stereoisomers.

Question 12

What is the wave number for: O-H; C=O; C=C? CN? CC triple?

Answer 12

OH = 3200 - 3600 cm-1; CO = ~1700; CC = 1650; CN or triple CC = 2100 - 2260.

Question 13

What occurs with H-bonding and IR spec?

Answer 13

OH groups in IR spec have a broadened peak. Molecules are in continuous random motion in liquid. Groups have strong H-bonds, weak, or none. This results in a random distribution of H-bonds and thus covalent bonds. H-bond weakens the covalent bonds, lowers the E of the bond and thus lowers the E of absorption for the bond.

Question 14

Which infrared photons affect the vibration energies or a molecule, UV and visible photons affect the ____ energy levels.

Answer 14

Electronic. When UV or visible photons are absorbed by a molecule, an e- is said to be excited from the ground state to an excited state.

Question 15

When is UV-visible spec most useful?

Answer 15

We typically use only UV-visible spec to analyze molecules with pi-bonds, especially in conjugated systems. UV-visible spec in Ochem focuses on transitions between pi and pi* energy levels. Sigma bonds aren’t analyzed bc it requires A LOT of E. If a UV-visible active compound is involved, it can be used to determine a yield of a reaction.

Question 16

In UV-visible spec, as the conjugation increases, the wavelength (decreases/increases)?

Answer 16

As the conjugation increases, so does the wavelength of the λmax ( λ of highest absorbance).

Question 17

In UV-visible spec, how do carbonyls compare with alkenes in terms of absorbance?

Answer 17

Aldehydes and ketones have more intense absorbances that are longer in wavelength than their hydrocarbon counterparts. This is attributed to the n–> pi* transition, possible bc of the lone pairs on the carbonyl O.

Question 18

Explain NMR.

Answer 18

E in the form of EM radiation is added to the sys. and analyzed in terms of what is absorbed. The E levels that are affected are for the spin of a nucleus in the presence of an external magnetic field. Within a molecule, two identical nuclei may be in diff. electronic environments. So, the difference in their local B fields, caused by the moving e-, manes they won’t require the exact same amount of E to excite the nucleus to a higher E spin state.

Question 19

Explain NMR in layman’s terms.

Answer 19

If we give H+ some energy (by radio wave absorption), then the H+ can be promoted (flipped) to the higher energy spin, which is opp to the direction of the external B field. This absorption is called resonance. The resonance frequency is the frequency of the radio wave that’s needed to cause a flip in spin. The resonance frequency energy or field strength) of absorption is called the chemical shift.

Question 20

What is “spin” in terms of NMR?

Answer 20

As the nucleus precesses ( twist due to ext torque), it generates a weak B-field ( just as spinning e- do). When the atomic nucleus has an odd number of protons, the spins cannot pair up to cancel each one another out. For H spec, protons have a spin up or spin down.

Question 21

Explain broad peaks in NMR.

Answer 21

Protic hydrogens can be distinguished from other signals by its braodness. Broadening results from H-bonding in solution.

Question 22

What three things must you consider in analyzing NMR Spectrum?

Answer 22

Integral (number of H that make up a signal; splitting pattern ( coupling between hydrogen neighbors); shift value ( determined by local B field caused by lone pairs e- in the motion or electronic density associated with EN atoms);

Question 23

In a CH2 group of 3-pentanone, how many spin combos does it have? What are they. How does this split a neighboring CH3?

Answer 23

The two H of CH2 have one of four possible spin combinations: up/up, up/down, down/up, down/down. Every CH3 group next to a up/up CH2 has a slightly higher signal. Every CH3 group next to down/down CH2 has a lower. The CH3 signal would appear in a 1:2:1 signal. A Triplet.

Question 24

The splitting patterns follows the 2n+1 rule. True or false?

Answer 24

It’s the “n+ 1” rule. If there are two neighboring non-equivalent H, then that H gets split into three. It’s a triplet. 2 +1 = 3.

Question 25

How does deshielding affect shift values?

Answer 25

Consider 3-pentanone, where the CH2 group adjacent to the C=O group is deshielded by the magnetic field on the neighboring oxygen and thus requires a stronger external B-field to energize the spin levels than a CH2 group that is next to an alkyl chain. The CH2 group is said to be downfield.

Question 26

Whenever you see a triplet and quartet in a 3:2 ratio, you should conclude there is a _____ group.

Answer 26

An isolated ethyl group ( H3C–CH2 )

Question 27

Whenever you see a doublet and a septet in a 6:1 ratio, you should conclude there is an _____ group.

Answer 27

An isolated isopropyl group. (H3C)2 –CH

Question 28

What does zero units of unsaturation and one oxygen suggest?

Answer 28

The molecule is an alipathic ether or alipathic alcohol.

Question 29

The only way to get two types of hydrogen on a disubstituted benzene is to have two equal substituents on benzene (ortho/meta/para) to one another.

Answer 29

Para.

Question 30

What are the shifts for the following H? Carboxylic acid, aldehyde, aromatic, vinylic (alkene), alkoxy, alpha.

Answer 30

COOH: 10-12ppm; aldehyde: 9-10ppm; aromatic: 7-8 ppm; vinylic = 5 -6; alkoxy: 3.5 - 4ppm; alpha: 2 - 2.5

Question 31

For disubstituted ortho and meta molecules, they have ___ unique hydrogens in the aromatic region of the spectrum.

Answer 31

They have four unique H.

Question 32

Para is the easiest arrangement to distinguish because it has a ______.

Answer 32

Double of doublets in the aromatic region.

Question 33

Because the solvent is higher [ ] than the solute, it is imperative that the solvent not have any hydrogens. If the solvent has H nuclei, then it would produce the largest signal in the spectrum. How do we avoid this?

Answer 33

Solvents are chosen to have deuterium (^2H). The issue that can arise here is D2O, D will gradually replace protic hydrogens capable of undergoing exchange. The signal for the protic H will dissapear gradually.

Question 34

Explain the process of mass spec.

Answer 34

The molecule is converted into a cation following the addition of some input energy (high speed collision or incident EM radiation). The cation is then accelerated through an e- field after which it is deflected along a circular path by a perpendicular B-field. The radius of the circular path corresponds to the particle’s mass to charge ratio. By comparing to standards, masses can be determined.

Question 35

What information does mass spec give us?

Answer 35

It is a quick way to determine the molecular mass of an organic molecule and to identify if the molecule contains either chlorine or bromine.

Question 36

The material under mass spec should be analyzed using (high/low) pressure and (gas/liquid) phase.

Answer 36

Lower pressure - so highly reactive free radicals and catonic particles don’t collide and undergo reactions; gas - bc particles travel independently of one another in the apparatus.

Question 37

If a molecule contains bromine, what would appear on the mass spec data?

Answer 37

Elemental Br is 50.7% 79Br and 49.3% 81Br, so mass spec results for compounds with Br show two peaks roughly equal in size spearated by 2 AMU.

Question 38

If a molecule contains chlorine, what would appear on the mass spec data (AMU)?

Answer 38

Elemental Cl is 75.7% 35Cl and 24.3%37Cl, so mass spec results for compounds with a single chlorine shows two peaks with a 3:1 size ratio separated by 2 atomic mass units.