orgo uW: Lab techniques & biotechnology

Question 1

What type of molecules absorb UV light?

• Which technique sepertes compounds based on polarity? & can be used to moitor reactions?*
• How are the components visualized?*

Answer 1

• conjugated molecules
  
  • ​Molecules with UV chromophores (double and triple bonds, carbonyls, conjugated systems) can be visualized with UV light.
• Thin-layer chromatography
• After separation, the mixture components are visualized, usually by UV light.

Question 2

Conjugated systems

Answer 2

Question 3

mass spectroscopy

Answer 3

Question 4

Amines

Heterocyclic

Answer 4

• can be classified as primary, secondary, tertiary, or quaternary (ie, one, two, three, or four non-hydrogen substituents, respectively).
• heterocyclic: denoting a compound whose molecule contains a ring of atoms of at least two elements (one of which is generally carbon).

Question 5

Infrared Spectrum (IR)

Answer 5

• displays absorption signals characteristic of the functional groups in a particular molecule based on the IR frequencies absorbed by those functional groups.
• Functional group bands appear in the same region of the IR spectrum regardless of the overall structure of the molecule.

Question 6

Distillation

Answer 6

• Distillation separates compounds based on their boiling point.
• Constitutional isomers can experience different intermolecular forces, contributing to the difference in their boiling points.
• The isomer that experiences increased intermolecular hydrogen bonding has a higher boiling point compared to the isomer that experiences increased intramolecular hydrogen bonding.

Question 7

Ortho Subsituent

Answer 7

• Because the hydroxyl and nitro groups in 2-nitrophenol are ortho and therefore close in proximity to each other, the hydrogen from the hydroxyl group can hydrogen bond intramolecularly with the lone pair of electrons on the nitro group.
• This intramolecular bonding decreases the number of intermolecular bonds that can form, thereby decreasing the boiling point of the compound.

Question 8

Para Subsituent

Answer 8

• Because the hydroxyl and nitro groups on 4-nitrophenol are para, they are able to hydrogen bond intermolecularly but not intramolecularly.
• Intermolecular bonds hold the molecules of 4-nitrophenol together, thereby increasing the boiling point and causing it to stay in the flask while 2-nitrophenol distills.

Question 9

Superheating, and what prevents it?

Answer 9

• Superheating happens when a liquid is heated above its boiling point, but it does not boil.
• Surface tension causes the vapor pressure inside bubbles to increase as they form, causing them to explode at the surface.
• Addition of boiling chips gives the bubbles a surface to form on as the liquid is heated, and allows for even boiling.

Question 10

Thin-Layer Chromatography (TLC) uses what to visualize the results from TLC

Answer 10

Absorption of ultraviolet (UV) light induces electron excitation (with high-energy photons), and a compound must contain a UV chromophore to absorb UV light.

Question 11

Mass Spectroscopy:

Question 12

¹H NMR spectrscopy

Answer 12

• An external magnetic field is applied to a sample in NMR spectroscopy.
• Radio waves, which are low in energy, are used to detect hydrogen atoms and excite them from the α spin to the β spin state

Question 13

Thin Layer Chromatography details

Answer 13

• Thin layer chromatography is used to separate compounds based on polarity.
• The Rf value of a compound is a ratio of the distance up the plate a compound travels to the distance the solvent travels (distance traveled/sovent font).
• A polar compound will have a smaller Rf value than a nonpolar compound
• Rf value is always less than 1, smaller Rf=more polar=less mobile

Question 14

IR stretches

Answer 14

• Characteristic functional group absorptions include
  
  • 3650–3200 cm−1 (alcohol and phenol O−H stretch);
  • 3550–3060 cm−1 (amide N–H stretch);
  • 3100 cm−1 (sp2 C–H stretch);
  • 3000–2875 cm−1 (sp3 C–H stretch);
  • 2260–2,100 cm−1 (triple bonds);
  • 1850–1650 cm−1 (C=O stretch).

Question 15

Extraction

Answer 15

• Extraction requires immiscible solvents of different polarities.
• The position of the layers in the funnel depends on the densities of the solvents. Denser solvents settle at the bottom of the funnel and less dense solvents float on top.
• example image: passage gave that ethyl acetate & chloroform are both nonpolar

Question 16

Acylation reactions

Answer 16

• Acylation reactions between anhydrides and amines generate amides and carboxylic acids.
• amides and carbonxylic acids are polar, water-soluble molecules because they form hydrogen bonds in an aqeous solution
  
  • However, long alkyl side chains (R groups) of amides and carboxylic acids increase the hydrophobic (nonpolar) nature of the molecules and make them insoluble in aqueous solvents.
• The solubility of carboxylic acids in water increases when they are converted into carboxylate anions by a base (such as LiOH, pka=~.36)
• Long-chain amides and carboxylic acids are the products of the reaction between long-chain anhydrides and the amine group of phenylethylamine.
  
  • To separate long-chain amides from carboxylic acids in the organic layer, a base must be added.
  • Bases deprotonate the carboxylic acid and generate carboxylate anions that are more soluble in the aqueous layer than in the organic layer.
  • Dilute LiOH is strong enough to deprotonate a carboxylic acid but not to deprotonate an amide N–H. Therefore, addition of the base LiOH will produce carboxylate anions that enter the aqueous layer and allow long-chain amides to remain in the organic layer.

Question 17

Anhydride to amide (acylation)

Answer 17

Question 18

Relative strength of acids & bases

Answer 18

Question 19

Acid-Base Extraction Example

Which additional extraction steps would cause phosphatidylethanolamine and 2,6-dimethoxyphenol to enter the aqueous layer consecutively?

Answer 19

• Answer: Add 0.05 M H2SO4 (aq) to the organic layer followed by 0.01 M NaOH (aq).
• Explained:*
• Organic compounds with acidic or basic functional groups enter the aqueous layer as ionic salts (which are formed by deprotonation or protonation) when acids and bases are added to an extraction.
• Amines are weak bases that require strong acids (H₂SO₄ or HCl) to be protonated.
• carboxylic acids are strong acids, thus can be deprotonated by strong & weak bases
  
  • Phenols are much weaker acids than carboxylic that are only deprotonated by strong bases (NaOH or KOH)

Question 20

Dilutions

Answer 20

• dilutions reduce the concentration of a solute by transferring a small volume of solute V_T into a larger volume of solvent V_s.
• The dilution factor is calculated by dividing V_T from the stock solution or the previous diluted solution by the final volume of the solution V_F

Question 21

Chemically Equivalent Hydrogens

Answer 21

• Hydrogen atoms or groups on a molecule that display rotational or planar symmetry are in identical magnetic environments. These atoms, called chemically equivalent hydrogens, have the same types of atoms surrounding them.
• Equivalent hydrogens have identical chemical shifts and are represented as a single signal on the NMR spectrum.

Question 22

Rotational Symmetry

Answer 22

Question 23

Planar Symmetry

Answer 23

Question 24

Common NMR splitting patterns

Answer 24

• Spin-spin splitting of peaks in an NMR spectrum results from interactions between nonequivalent hydrogen atoms within three bonds of each other.
• The splitting pattern can be determined by the n + 1 rule, where n is the number of neighboring hydrogen atoms when J is the same for all nonequivalent hydrogen atoms.

Question 25

¹H NMR chemical shift (ppm) scale

Deshielding & sheilding

Answer 25

Question 26

Complex splitting patterns when n+1 rule is not applicable

Answer 26

Question 27

Narrow/weak/strong IR spectrum

Answer 27

Question 28

Intensity of IR Signals

Answer 28

Question 29

Strecths of IR spectrum

Answer 29

3650–3200 cm−1 (O–H stretch)

3300 cm−1 (sp C–H stretch) –>WEAK, narrow

3100 cm−1 (sp2 C–H stretch)

3000 cm−1 (sp3 C–H stretch) –> STRONG

1810–1650 cm−1 (C=O stretch)

Question 30

What is the number of signals corresponding to aromatic hydrogen atoms present in the 1H NMR spectrum of Compound 3?

Answer 30

• Hydrogen atoms in a molecule that have the same types of atoms surrounding them and that exhibit rotational or planar symmetry are chemically equivalent and experience identical magnetic environments.
• Therefore, equivalent hydrogen atoms have the same chemical shift and appear as a single signal in the 1H NMR spectrum.

Question 31

Effect of ester addition to ¹H NMR shift

Answer 31

• Proton nuclear magnetic resonance (¹H NMR)
• The location of signals in the 1H NMR spectrum is known as the chemical shift, measured in ppm.
• The chemical shift depends on the electronic environment around protons.
• Protons with an electron cloud surrounding them are shielded from the magnetic field and exhibit an upfield signal, whereas protons adjacent to electronegative substituents are deshielded from the magnetic field and have a downfield signal.

In Compound 1, He and Hf are on a carbon adjacent to a hydroxyl group. Because oxygen is electronegative, it deshields the neighboring protons, causing a downfield shift (3.80 and 3.71 ppm, respectively) from the chemical shift of alkyl protons (0–2 ppm). Transesterification of Compound 1 transforms the hydroxyl group into an ester (Compound 3), which contains more electronegative atoms than the hydroxyl group. Therefore, the ester withdraws electrons and deshields He and Hf more so than the hydroxyl group and causes the signal to shift farther downfield (4.36 and 4.25 ppm, respectively).​

Question 32

Transesterficiation

Answer 32

• produces an ester from a carboxylic acid and an alcohol
• the ester can be hydrolyzed back to its alcohol & carboxylic acid constituents in acidic or basic condiitons

Question 33

1. ____________________is a purification technique that separates compounds based on polarity and consists of a mobile phase, stationary phase, detector, and computer for data acquisition. The mobile phase is a solvent or mixture of solvents that are pumped through the system under pressure, and the stationary phase is a column made of either a nonpolar or polar material.
2. _________________requires small sample sizes but separates compounds based on boiling point. This method contains a stationary and mobile phase; however, the mobile phase is a gas rather than a solvent.
3. _______________separates compounds based on solubility in a particular solvent; this method uses two immiscible solvents that make up the aqueous and organic layers. No stationary phase, mobile phase, or detector used
4. _______________ separates small sample sizes and includes a stationary phase, solvent mobile phase, and detector (UV).

Answer 33

1. HPLC
2. Gas chromatogrpahy
3. Extraction
4. TLC

Question 34

Which technique is used to evaluate a compound’s purity?

Answer 34

• TLC
• The components of a mixture are separated based on polarity, and a single spot on a TLC plate is indicative of a compound’s purity.