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11 Nov 2019

In our lab, we will be oxidizing an unknown secondary alcohol and reacting the ketone with phenylhydrazine to create a phenylhydrazone.

NQ1: This experiment is the oxidation of an aliphatic secondary alcohol to an aliphatic secondary ketone (aliphatic: no double bonds or extra functional groups). Think about the starting material and the product, and predict what signals you expect to see in the IR spectra of these compounds. How will you know prior to leaving lab if your experiment is successful?

Reaction Type(s): oxidation secondary alcohol + oxidizing reagent â ketone derivatization ketone phenylhydrazine â phenylhydrazone

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Sixta KovacekLv2

25 Aug 2019

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Â¥Write a brief (less than 150 words, 3 or 4 sentences) of the aim of the experiment

Â¥Address question of what was done, how it was done, and what compounds were used.

Â¥Include Chemical Reaction where relevant.

See text below. Please provide Purpose following the steps above. Thank you.

Alcohols R-O-H are characterized by the presence of the hydroxyl group

Because of intermolecular co-association, alcohols show higher boiling points

than aldehydes, ethers, or ketones of similar mass. Among alcohols, boiling

points increase with increased surface area. Increased surface area occurs as

more carbons are added to the alkyl portion or with decreased branching of th

alkyl group. Thus, 1-propanol has a higher boiling point than ethanol

(increased C#). 1-propanol also has a higher boiling point than 2-propanol

(decreased branching)

The solubility of alcohols in water decreases as the surface area of the

alkyl (nonpolar) portion increases. Thus, alcohols become less soluble in water

as the number of carbons in the alkyl group increases. When comparing

alcohols that contain the same number of carbons, branching of the alkyl group

decreases its surface area and thus increases the water solubility of the alcohol.

1-Propanol is slightly soluble in water while 2-propanol is quite soluble

The limit of water solubility is reached at about 4 or 5 carbons in the

alkyl group. However, it is possible to "salt out" or separate a water-soluble

alcohol from the water it is mixed with by the addition of certain salts such as

K2CO3. If an alcohol-water mixture is saturated with such a salt, the anion

selectively forms strong ion-dipole bonds to water, leaving the alcohol as a

separate layer. Thus, water is separated from the alcohol because the ionic salt

bonds more strongly with water than with the alcohol. There are twice as many

ion-dipole bonds possible with water than with the alcohol.

Alcohols are relatively acidic organic molecules, being weaker acids than

water, yet stronger than acetylene.

Acidity: water > alcohol > acetylene > ammonia > alkanes

Alcohols may be converted to sodium alkoxides by the action of metallic sodium

under anhydrous conditions. Primary alcohols react more readily tharn

secondary or tertiary alcohols due to the relative stability of the alkoxide ions in

the solvent alcohol.

R-o:-

Alcohols undergo nucleophilic substitution reactions with ease under

acidic conditions. The relative ease of such substitution for primary, secondary

and tertiary alcohols serves as a means of classification. Upon treatment with

concentrated HC1, tertiary alcohols are rapidly converted to alkyl chlorides

Secondary alcohols require warming in a boiling water bath for 15 minutes

Primary alcohols do not react at all under these conditions. The alkyl chlorides

formed by this reaction are insoluble in the concentrated HCl and form an

easily detectable second layer

R-O-H HC1 >

Ð-Ð-Ð

Complications arise from the fact that benzylic and allylic alcohols also

react immediately, but yield products that are soluble in the concentrated HCI

Alcohols are easily oxidized by any of these strong oxidizing agents:

dilute KMnO4, NajCr.07 in acid, or CrO3 (chromic anhydride)

acidified sodium dichromate is somewhat helpful in the classification of

alcohols. Prim

ary and secondary alcohols are oxidized rapidly within several

minutes, while tertiary alcohols do not react under these conditions

Oxidation occurs more easily when the carbon bearing the -OH group

bonded to at least one atom of hydrogen. The complex mechanism of

involves the transfer of a hydride ion H:- as a critical step.

hydrogen atom is absent (as in tertiary alcohols), no oxidation can occur

directly. Tertiary alcohols may be oxidized indirectly through the formation of

Alcohols which contain the partial structure C- CH3 give positive

Ð¾Ð½

results in the iodofora reaction. When the alcohol is dissolved in water and

KOH, the addition of a solution of 12 in KI yields a yellow precipitate of

iodoform, CHI3, along with the oxidized portion of the alcohol

Hypoiodite ion is produced by the reaction between molecular iodine and

hydroxide ion:

I2 + 4(OH) -

The hypoiodite ion in turn oxidizes an alcohol with the proper structure

Once this oxidation has occurred, additional hypoiodite ion continues to

react with the aldehyde or ketone produced:

CH,-C

3 (0I)-IC-c3 (OH)-

resulting in the final substitution reaction to give iodoform and the anion of a

carboxylic acid containing one carbon atom less than the original alcohol:

128

lodoform precipitates as fine yellow crystals. This precipitate, a result of a

specific oxidation followed by a substitution, is an indication of the partial

structure of the original alcohol.

Simple alcohols of up to ten carbon atoms may be detected by the

ammonium hexanitrocerrata(lV) reagent, (NH4Ce(NO,). The hexanitro-

cerrate(IV) ion has a distinctive orange-red color. When an alcohol, ROH is

added, one or more OR- groups replace the nitro groups as ligands in the

complex ion. The new complex ion formed has a distinctive color. The general

equation for this reaction is:

This test is positive for aliphatic alcohols up to approximately ten carbons in

length. Phenols or larger alcohols yield insoluble mixtures

Alcohols react with acids or with acid derivatives to form esters. This

characteristic reaction is useful in the formation of specific esters such as ethyl

acetate, acetylsalicylic acid (aspirin) and methyl salicylate (oil of wintergreen). In

this experiment, methyl salicylate will be prepared by reacting methyl alcohol

with salicylic acid in the presence of a trace of sulfuric acid as the catalyst

Lower molecular weight alcohols such as ethanol, methanol, and

isopropyi alcohol absorb water on exposure to moist air or are prepared as

azeotropic mixtures containing water. The presence of water in alcohols may be

detected in several ways. If a small lump of CaC2 is added to any alcohol that

contains a trace of water, a reaction occurs that produces immediate bubbles of

acetylene gas:

Thus, evolution of acetylene gas upon the addition of calcium carbide indicates

the presence of water in the alcohol. Another way to test for the presence of

water in alcohols involves the use of anhydrous cupric sulfate powder (CuSO4)

Anhydrous cupric sulfate is a very pale, almost white powder. In the presence of

water, deep blue crystals of cupric sulfate pentahydrate,

CuSO4-5H20, are formed. Thus, anhydrous cupric sulfate indicates the

presence of water in an alcohol by a simple color change.

Commercial ethanol is produced as an azeotropic mixture that contains

95% ethanol and 5% water. Absolute alcohol contains no water. Many reagent

grade alcohols are composed of a nonpotable mixture of 95% ethanol and 5%

isopropyl alcohol. Thus, absolute ethanol and reagent alcohols are both water

free and give negative results when tested for the presence of water

I need help for my lab report please. It's only a-e Questions for study that I need help with, so any help or advice will be greatly appreciated! Underneath is more info on the lab if needed. Thank you!

Question for Study

1. A student performed this lab exercise, generating the following data:

Mass of H5IO6 used: 1.2320 g

Volume of thiosulfate used in blank titration: 47.93 mL

Volume of thiosulfate used in the unknown titration: 39.23 mL

a. What is the concentration of the sodium thiosulfate solution?

b. How many mmol of I3 â were present in the blank titration?

c. How many mmol of I3 â were present in the unknown titration?

d. How many mmol of glycol were present in the unknown titration?

e. What mass of ethylene glycol should be reported?

Introduction

The chemical analysis of organic compounds often presents major experimental difficulties. An accurate analysis depends upon the sample undergoing a single, well-defined reaction which quickly goes to completion. Often, organic compounds undergo several different reactions at the same time, and reaction rates are sometimes quite slow.

A functional-group analysis is one in which a particular grouping of atoms in a compound is made to react in a specific well-defined way with a particular reagent.1 In this experiment use is made of the reaction between an oxidizing reagent and an organic compound with the functional group COH-COH.

Analysis of Ethylene Glycol

Ethylene glycol reacts with periodate ion to form formaldehyde and iodate:

+ ð¼ð4 â â 2 ð»2ð¶ð +ð¼ð3 â + ð»2ð (8-1)

This oxidation-reduction reaction will proceed to completion with negligible side reactions under the appropriate set of conditions. The reaction is slow, but increases with increasing concentration of periodate. Here a known, fixed amount of periodate in excess of that needed to completely react with the glycol is added to the unknown. After the glycol has reacted, an excess of iodide is added to the solution. The iodide reacts with unreacted periodate according to:

ð¼ð4 â + 11 ð¼ â + 8 ð» + â 4 ð¼3 â +4 ð»2ð (8-2)

The iodide also reacts with the iodate produced in reaction 8-1 according to:

ð¼ð3 â + 8 ð¼ â + 6 ð» + â 3 ð¼3 â + 3 ð»2ð (8-3)

The triiodide produced in these reactions is then determined by titration with thiosulfate:

ð¼3 â +2 ð2ð3 2â â 3 ð¼ â + ð4ð6 2â (8-4)

To find the glycol content, a blank solution containing the same fixed amount of periodate but no glycol is treated similarly. The difference in the number of moles of ð¼3 â produced in the blank solution, and that produced in the sample solution, is equal to the number of moles of glycol in the sample solution.

Procedure

1. Prepare a solution from your unknown: a. Your unknown may have recently been refrigerated â make sure it is at room temperature before use. b. Transfer the vial contents quantitatively to a 250 mL volumetric flask. c. Dilute to volume with water and mix well.

2. Prepare a Periodic Acid standard solution (pronounced âpurr-eye-oh-dic acidâ): a. Weigh out 1.16-1.24 g of H5IO6 on an analytical balance. Periodic acid is very corrosive â to your skin and to metals. Do not spill any on the analytical balances. b. Quantitatively transfer the solid to a 100 mL volumetric flask, and dilute to volume with water.

3. Prepare a sodium thiosulfate solution: a. Weigh ~11.2 g of sodium thiosulfate pentahydrate. b. Transfer the solid to a 1L Pyrex bottle and dilute to 500 mL to produce an approximately 0.09 M solution. c. Add 0.10 g Na2CO3 and swirl to mix well.

4. Prepare a ~0.5 M solution of sulfuric acid: a. In a beaker (250 or 400 mL), combine 25 mL of 3 M sulfuric acid and 125 mL of water. Mix. Remember: âAcid Into Waterâ. What glassware is appropriate to measure these volumes?

5. Prepare samples for titration: a. Pipet 20 mL of your diluted unknown and 10 mL of the periodic acid solution into a 250 mL conical flask and swirl to mix. b. Wait a minimum of 45 minutes to allow complete reaction before titrating. Do the blank titrations while you wait! c. Prepare a total of 3 samples in this way.

6. Perform the sample titrations: a. Add ~1.5 g of KI to your conical flask and swirl to dissolve. b. Add 10 mL of your 0.5 M sulfuric acid (use a graduated cylinder). c. Titrate from a dark red initial colour to the first change to yellow. d. Add 2.0 mL of thiodene indicator solution. The solution will become a dark grey/violet colour. e. Finish the titration (fractional drops) to a sharp disappearance of the violet colour. 7. Perform the blank titrations: a. Pipet 10 mL of periodic acid solution into a 250 mL conical flask. b. Add 20 mL of water with a graduated cylinder. c. Add reagents and perform the titration as in Step 6 above. Calculations Report the total mass (in mg) of ethylene glycol in your unknown.

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