CHM2211L Final Exam UF

temperature at which the solid phase is in equilibrium with the liquid phase

sharp melting point, range is narrow (1-2 degrees)

because the impurity disrupts the crystalline lattice of the sample so less energy is required to convert the solid phase into a liquid phase

no effect

1. improper amount of sample in capillary tube 2) mistaking sagging or shrinking sample as melting 3) presence of solvent cause crystals to "sweat"

122-123°C

120-122°C

100-74.97-8.39 = 16.64% O # of C: 74.97/12 = 6.25/1.04 = 6 # of H: 8.39/1=8.39/1.04= 8 # of O: 16.64/16 = 1.04/1.04 = 1 EmF: C6H8O

No because IR spectroscopy is used to distinguish functional groups presents on molecules. Both 2-p and 2-h contains the same functional group.

analgesic and antipyretic (pain and fever relief)

Williamson ether synthesis

Amide synthesis

solvent that will completely dissolve both desired compound and impurities when the solvent is hot; desire compound must be insoluble in the chosen solved at cold temperature and inorganic remains soluble when cooled

They serve as nucleation sites where more molecules of the desired compound deposit themselves in layers.

- Hot gravity filtration serves to remove any insoluble impurities that do not dissolve when the solution is brought to its boiling point. - it is good to use the stemless funnel for this technique because this would minimize any contact between the hot solution and the funnel. If a funnel with stem is used the droplets of hot solution might get stuck on the stem inner surface and early crystallization of the solution might occur

solvent derived from petroleum consisting of C7 and C8 hydrocarbon, non polar

p-phenetidine + acetic anhydride (solvent: Sodium acetate and hydrochloric acid)

small amount of impurities caused by air oxidation of the amino functional group

decolorizing carbon

this makes the filtration less efficient and does not adequately remove the solvent from the crystal

1) p-acetamidophenol +sodium methoxide, methanol, and ethanol 2) 1st product (addition of sodium) + Bromoethane

Because the hydrogen on the alcohol group of acetamidophenol would react with the bromoethane to create ethane and there would not be any reaction. The sodium acetate turns that alcohol group into a nucleophile that would attack the C of the bromoethane

Allows the reaction to be carried out at the boiling point of the solvent and yet it prevents loss of solvent or reagent due to evaporation. The vapors condense as they rise up the condenser and thus prevent them from escaping.

Water because the solute is completely soluble at boiling point, but the crystals reappear when solution is placed in an ice bath.

helps with gentle boiling

If the sample made through two synthetic routes are identical, one can create a mixture from the same amount of two samples and the MP should be sharp and close to the recorded MP. If MP is lower than theoretical MP and has big range: samples are different from one another.

HCl is a much stronger acid which protonates the amine group of the p-phenetidin, which makes it more soluble in the solution

HCl protonates p-phenetidine and drives the reaction to the right but too much HCl would convert all of the compound to its conjugate acid and no reaction with acetic anhydride would occur. Sodium acetate creates a buffer within the reaction that would control HCl's acidity.

N - 8ppm CH3 (acetyl group) - 2ppm H(2) (closer to N)- 7.4 ppm H(2) (closer to O)- 6.8-7ppm CH2 (closest to O)-4 ppm CH3 (closet to O) - 1.4 ppm

immiscible (Hexane top)

immiscible (Water top)

miscible

immiscible (ether top)

There more stable or the weaker the conjugate base, the more acidic the compound is. When losing a proton, phenol's conjugate base has resonance stability from its double bond.

no since both are very close in acidity; cannot use reagents in this experiment to neutralize one and keep one (stronger acid) as a weak acid; both would be neutralized

Extraction is a technique that chemists use to remove selected components from a mixture. The different types are solid-liquid extraction and liquid-liquid extraction.

It offers a greener alternative to other solvents, it is readily available, nonflammable, nontoxic, and safe for the environment. Ease of removal of solvent at the end of procedure, and it is relatively quick.

The two solvents that are chosen for the extraction must be immiscible. In addition, the compound to be extracted must have different affinities for the two solvents.

NaOH is so strong that it would neutralized both strong acid such as carboxylic acid and weaker ones like phenol. On the other had NaHCO3 is weak enough that it can only react with strong carboxylic acid and remains unreacted with phenol

this protonates carboxylate ions back into carboxylic acid compounds that forms solids when put in ice bath

use ETHER for back extraction

wash the organic layer - removing remaining water in solution

The salt absorbs the water that is dissolved in the organic solvent.

The carboxylate ion is present. In order to regenerate the carboxylic acid it must be added to HCl in an ice bath to induce crystallization. To isolate the crystals, vacuum filtration is performed.

The phenol is present. There may be some water present in the solution so it must be dried with brine and anhydrous magnesium sulfate until it no longer clumps together. The drying agent is then separated from the organic solution through gravity filtration. Once this is done, the ether must be evaporated under reduced pressure so that the pure phenol solid is left in the flask.

Extraction and washing can share very similar procedure but ultimately have different purposes. Extraction removes desire compounds/solutions from a mixture whereas washing filters out the impurities from the desired solution. Keep extraction and discard washed impurities.

Since the heated vapor from the distillation flask will travel from high to low, the water flows in the opposite direction in order to effectively cool the vapor which would liquify and flow down into the collection tube

Since cloves are dried up, soaking them would allow water to penetrate into the cloves and more eugenol would be isolated.

To extract the eugenol from the water. This process is liquid-liquid extraction and the lower layer will be the dichloromethane along with the desired essential oil.

brine and anhydrous magnesium sulfate

-OH group : 3517.6 (3600-3300)--10ppm -OCH3 group: 2934 (3100-2700)--

with distillation, cloves are allowed to sit in water for full penetration which would be more advantageous because it would yield higher % of product. Extraction using CO2 reduces the chance of water remnant and impurities. CO2 extraction would produce better quality result; during the heating process of distillation, a lot of quality from the oil is evaporated

Chlorophyll b should be more polar than chlorophyll a because the former has a CHO group where the latter only has a CH3 group.

xanthophyll would be more polar than B-carotene because of the -OH group at the ends of the molecules. B-carotene contains no oxygen and is quite non polar.

polar solvent (-CO interaction)

the ink in the pen can tamper with the process; the ink would travel up the TLC plate with the solvent

techniques for separating, analyzing, and identifying compounds (column, thin layer, gas, high performance liquid)

the eluent, the has or liquid phase in which the sample mixture is placed in

a solid support that contains an adsorbent

travels more slowly

travels thought the stationary phase more quickly

has many sites for hydrogen bonding; polar compounds would interact with the silva and adsorb readily to the stationary phase

distance from the origin to the center of the spot DIVIDED BY the distance form the origin to the solvent front; Polar compounds interacts more strongly with stationary phase and travel a shorter distance --> the more polar the compound, the smaller the Rf values

UV lights activates a fluorescent indicator on the plates

Acetone breaks down chlorophyll's lipid bonds to a plant's thylakoid structure and suspends the pigment in solution.

Acetone evaporates extremely quickly when interacting with air and all the solution would evaporate

50:50 hexane/ethyacetate

Ethyl acetate is a polar compound while hexane is non polar; one compound interacts with the polar compounds of spinach while the other with the non polar compounds. The 50:50 ratio was enough to separate the polar and non polar compounds of the spinach, at the same time, the polar compounds also get to move up from the origin due to its interaction with the ethyl acetate. If only hexane was use, the chlorophyll a and b would remains at the origin and would not be separated form one another.

because the compounds being separated will oxidize rapidly on exposure to air so it would fade away

B-carotene is the least polar and chlorophyll b is the most polar

an analgesic and antipyretic drug

-OH is an ELECTRON DONATING GROUP, the substituent would be directed to its para and ortho position. -NH2 is an ELECTRON WITHDRAWING GROUP and would direct the substituent to its meta position. The major projects has iodine at the ortho position to -OH and meta position to -NH2. (compare to the iodine group being para to the -OH)I think this would produce a higher % yield because the iodine group is being paced farther away from both functional groups which reduce the strain.

common reaction of benzene and its derivatives; two steps: 1) aromatic ring acts as a nucleophile and attack electrophile 2) aromatic ring is regenerated through the loss of proton

stabilized by resonance

EAS with Cl2 and Br2 can occur in the presence of Lewis acid such as AlCl3, BUT iodine is the least reactive of the halogens

iodination reactions using sodium iodine as the source of iodine and household bleach as an oxidizer (instead of nitric acid and I2)

Dissolve the solid - use warmth of your hand to speed up dissolution

is an oxidizing agent that reacts with sodium iodine and turns it from a stable compound to a strong electrophile that would now react with the salicylamide in the electrophilic aromatic substitution

removes any residual iodine from the solution

acidify the aqueous solution which drives the products to precipitates

815.3cm-1

1095.3

1621.6

3200-3300 (double peaks)

3450.1

Major product gives the less strained position of Iodine that followings the directing effect of both substituent: meta to NH2 and ortho to OH

H between I -NH2: 8.15 ppm (singlet) H closer to I (between I and O): 7.7 ppm (doublet) H closer to O (between I and O): 6.75ppm (doublet)

When a more polar solvent is used instead of non polar one like ligroin, it will interact more with the polar molecules and less with the non polar molecules causing a less dramatic separation. Rf values would be higher.

The more concentrated sample would produce a more significant or finer separation between polar and non polar compounds. Diluting the sample would cause a blurry separation between complex compounds instead of a fine separation

Non-polar or least polar compound elute first and therefore unreacted ferrocene elute first because it has no polar group. Acetylferrocene elute seconds because it only has 1 polar C=O group compared to diacetylferrocene which would elute last with 2 polar C=O groups.

acetic anhydride reacts with phosphoric acid to form asylum ion which is an electrophile that can be used to add into an aromatic ring

it is about 10^5 times more reactive than benzene in electrophilic aromatic substitution because it consists of an iron atom sandwiched between two cyclopentadienyl anion

neutralize the solution

methylene chloride

Water is very polar and should not be in the mix of the chromatography column. If water is present, it would interact with the acetylferrocene and diacetylferrocene and the whole solution would travel down together with no separation

because the drop would dry out and channels can form in the packing material which leads to inappropriate, poor separation.

To load the sample onto the top of the column

use ligroin

70:30 ligroin:ethyl acetate

pure ethyl acetate

the column chromatography served its purpose of separating and purifying samples. with the crude sample there are ferrocene, acetylferrocene and diacetylferrocene all present in the product. After purifying through the chromatography column, only acetylferrocene was collect.

1,1'-diacetylferrocene would be the isomer present. If 1,2 and 1,3-diacetylferrocene were to be formed, both C2 and C3 would have to be able to accept the acetyl group which means those position has to be nucleophilic. However if looked at the resonance structure of those positions, those carbon has the tendency to be positively charged which means they are electrophilic and would not be able to accept acetyl group. Therefore acetyl group at C1 an C1' is the best option

Helps expose fresh metal surface by removing some magnesium oxide on the exterior.

Cleaning the surface of the magnesium metal.

First bromobenzene reacts with magnesium to form phenyl magnesium bromide. Then by adding the electrophile, the PMB would nucleophilically attack the compound.

Grignard reagents are sensitive to water. If water is introduced into the mix, the benzene will become unreactive because it would take an H from H2O.

Grignard reagents are sensitive to water so if diethyl either containing more water is used it will interfere with the formation of the Grignard reagent.

- ice prior to hydrosulfuric acid to complete the precipitates before it is dissolved by the acid to. - The double bond with Oxygen (benzophene) reacts with the hydosulfuric acid and it is protonated by an addition of hydrogen creating an alcohol

Recrystallization using ethanol. The desired compound is soluble in ethanol at boiling point and is insoluble in ethanol when it cools.

An aqueous layer forms (bottom) and an organic layer forms (top). The desired compound is dissolved in the organic layer (like dissolves like).

While for aldehydes and ketones, after the first Grignard reaction, there is nothing left to react, for esters, after the first reaction the compound becomes a keno or aldehyde and can drive further Grignard reaction

Since the compound is very non polar due to the lack of O group, it can only be remove through a workup step using a non polar substance. By adding the cold ligroin, the biphenyl would be removed which followed the ligroin down the Buchner funnel.

3374

600-800 (ring torsion)

1500

3000

1200

small stretches from 1500-2000

Grignard reagent is extremely sensitive and reactive with water. Since it is a strong nucleophile it would react with water first and will not be able to produce the desirable reaction and product

Phenyl magnesium bromide + h2o --> benzene + hydroxyl magnesium bromide

PARA is difficult to isolate because it has so many resonance that the compound can have many charges on it at different pH. Isoelectric point is where the compound carries not net charges, containing equal amount of (+) and (-) that neutralize each other. It is where it is easiest to gain precipitates and therefore its is very unstable and sensitive to pH.

increase % yield

most widely used herbal remedies, treating various ailments including asthma, macular degeneration, cardiovascular disease, vertigo, and memory loss and dementia associated with Alzheimer's disease

decline with charcoal before proceed

P-toluidine + acetic anhydride (NaOAc, HCl aq) --> N-acetyl-p-toluidine

via halo test: dip the stirring rod into the mixture then touch the tip to a piece of filter paper --> purple halo = incomplete; light pink/clear halo = complete

needs to boil gentle: too little heat will keep the reaction from occurring at a good time, too much heat will cause excessive frothing and possible boo over

this will cause cracks in the filter bed and result in insufficient filtration

this would minimize crystallization of product in the filter --> precipitate should form when being acidified

the precipitates needs a nucleation site that can be stimulated by glass particles.

At the isoelectric pt. the desirable precipitate will form, more acidic = salt precipitates (NH3+), more basic = no precipitate 1) add 3M NH4OH solution --> decrease pH to isoelectric pt 2) Adjust pH of the filtrate until it is just basic 3) adjust pH again with glacial acetic acid

1663

double peaks at ~3000+

we cant oxidize p-touluidine with KMnO4 because the NH2 must first be protected or else it would be oxidized also. the side product would have been a R-NOO-

if esterification occur prior to deprotection then both functional would would be deprotected and we would have to report esterification.