Identification of Acids, Bases and Neutral Compounds

Purpose: Identify the individual components previously separated from a mixture by extraction.

Preparation:
  Prepare a test strategy based on the outcome of your separation. If one component of your mixture is an organic acid, that substance could be a carboxylic acid (as was shown in the examples for the scheme) or it could be a phenol. To make a determination between these you need to read McMurry Sections 17.3 and 20.3. In particular you need to check the differences in the IR spectra of Figs 17.13 and 20.5 and make a listing of the expected absorptions in the NMR spectra. On the other hand, if one component of your mixture is a base, you need to see McMurry Section 24.4 and Figure 24.8. In these sections you will learn how to distinguish among the various amine types (1o, 2o or 3o); in particular, check the differences in the IR spectra of Figure 24.8 and list the PMR absorptions expected for each amine type. Have these data ready in advance as initial entries in a Table of Chemical Tests. Further chemical tests given below will aid in most determinations.

The neutral compound can be an aldehyde, ketone, alcohol, alkyl halide, alkene, ether, etc. Clearly further tests and preliminary IR and PMR data are needed to determine the functional groups in the neutral compound. Prepare a separate Table of Chemical Tests for this component.

Lists of candidates are provided in the table of links below:

Carboxylic Acid Aldehyde Alcohol Alkene Amine
Phenols Ketone Alkyl halide Ether  


Procedure:
The following chemical tests and spectroscopic details should be applied to the appropriate separate component of your unknown.

Examination and reporting of IR Data. As in laboratories in Chem 3012, the IR spectra have to be correctly labeled. Diagnostic peaks must be measured after calibration and noted on the spectrum. The specific feature of the diagnostic peak must be noted e.g. "C=O stretch" and pointed out. These features are then reported in the Table of Chemical Tests.

Examination and Reporting of PMR data.  All signals must be identified.

Phenols - Ferric chloride test. Most phenols, enols and a number of other compounds with moderately acidic OH groups react with ferric chloride to give colored complexes.

Dissolve 30 mg of unknown in 1 ml of water, or a mixture of water and alcohol, and add up to 3 drops of 1% aqueous ferric chloride. A red, blue, purple or green coloration indicates the presence of a phenol. In some cases the color changes with time. Special directions should be consulted for compounds that are believed to be phenols but which give negative tests with aqueous ferric chloride

Phenols - Solubility Test. If the compound is soluble in aqueous NaOH it is an organic acid - either a carboxylic acid (R-CO2H) or a Phenol (Ar-OH):

R-CO2H

+

OH-

---->

R-CO2-

+

H2O

water insoluble

water soluble

Ar-OH

+

OH-

---->

Ar-O-

+

H2O

water insoluble

water soluble

Carboxylic acids are relatively strong and will also dissolve in a 1.1 M NaHCO3. However most phenols are relatively weak acids and will not react with aqueous bicarbonate to form a soluble ion:

R-CO2H

+

HCO3-

---->

R-CO2-

+

H2O

+

CO2

water insoluble

water soluble

bubbles

Ar-OH

+

HCO3-

<----

Ar-O-

+

H2O

+

CO2

water insoluble

(not formed)

Set up a three test tube solubility experiment as before in the Mixture Separation by Extraction with the following test solvents: 2 mL of 2.5 M NaOH, 2 mL of 1.1 M NaHCO3 and 2 mL of deionized water. Make careful drawings of your results. Try this test on known compounds for confirmation.

Phenols decolorize bromine in CCl4 just like alkenes - Refer to "Alkyl Halide/Alcohol" in the Chem 3012 directory.

Test for Aromaticity (includes phenols):

Place 30 mg of anhydrous Aluminum Chloride in a 3 inch test tube and gently heat it over a microflame until the AlCl3 has sublimed on the walls of the test tube. Cool the tube in air for 30 sec and then allow 1-2 drops of a solution of your unknown in chloroform (5 mg in 5 drops of CHCl3) to flow down the wall of the test tube while rotating the tube. A color will develop on the wall of the test tube if you have an aromatic compound. Simple benzene derivatives give a yellow-orange or red color; bicyclic aromatics give a blue of purple color; and more complex aromatic systems give a green color. The reaction involves -complex formation between the aromatic system and the AlCl3.

Spectroscopic Differences Between Carboxylic Acids and Phenols (M&B Sections 19.22 and 24.17).

In the IR, phenols will show very distinctive O-H absorptions in the 3200-3600 cm-1 range as alcohols. Furthermore a carboxylic acid must have a strong carbonyl (C=O) at 1680-1725 cm-1. In order to be a phenol, the IR spectrum must have aromatic absorptions at ~1500 and ~1600 cm-1 (some carboxylic acids are also aromatic so this criterion is not definitive). When observed in the PMR, the carboxylic acid OH proton absorbs in the range 10.5-12 while the phenol OH normally absorbs in the 4-7 range (again the phenols must have aromatic protons   6-8.5 - as well). These peaks are typically broad.

Amines - Hinsberg Test. The Hinsberg test  is used to distinguish 3o, 2o and 1o amines:

CH3CH2NH2

+ C6H5SO2Cl

+ NaOH

--->

C6H5SO2NHCH2CH3

+ H2O + NaOH

ethyl amine

benzenesulfonyl

a benzenesulfonamide

(a 1o amine)

chloride

derivative

C6H5SO2NHCH2CH3

+ NaOH

--->

[C6H5SO2NCH2CH3] - Na+

+ H2O

water soluble

 

(CH3CH2)2NH

+ C6H5SO2Cl

+ NaOH

--->

C6H5SO2N(CH2CH3)2

+ H2O + NaOH

diethyl amine

a benzenesulfonamide

(a 2o amine)

derivative

C6H5SO2N(CH2CH3)2

+ NaOH

--->

no reaction

 

(CH3CH2)3NH

+ C6H5SO2Cl

+ NaOH

--->

no reaction

triethyl amine

(a 3o amine)

The product from the 1o amine is soluble in base whereas the product from a 2o amine is insoluble. Secondary and tertiary amines may be distinguished on the basis of solubility in dilute HCl.  The unreacted 3o amine will be soluble in aqueous HCl whereas the benzenesulfonamide product from the 2o amine will be insoluble.

Place 1 mL of 10% sodium hydroxide in a small test tube and dilute with 4 mL of water. Add five drops (use an equivalent amount if a solid) of amine and 9 drops of benzenesulfonyl chloride. Shake the mixture vigorously for 5 minutes. Warm it slightly and continue shaking, When the odor of benzenesulfonyl chloride has dissipated, cool the mixture. If an oil separates, try to induce crystallization by rubbing the inner wall of the teat tube with a glass rod and chilling in an ice bath, Any solid produced should be separated by filtration; it is the benzenesulfonamide derivative of the unknown. A portion of it should be tested in dilute sodium hydroxide solution.

Spectroscopic Analysis of Amines. Attention is focused on the number of H atoms (if any) bonded to nitrogen. In the IR, the N-H band in the 3200-3500 cm-1 region provides the clearest distinction between 1o, 2o and 3o amines. In the PMR the N-H proton band falls in the 1-5 range but their splitting is not simple.

Spectroscopic Analysis of Aromatic Compounds. In the IR there are two regions that pertain to the kind of substitution on the benzene ring, 2000-1600 cm-1 (Mayo, et al., p 196) and 870-675 cm-1 (Mayo, et al., p.197). In the PMR aromatic protons are in the 6-8.5 range. The amount of substitution is evident from the integration of this signal; for example, 5 H clearly indicates a monosubstituted benzene ring whereas fewer protons indicate greater degrees of substitution.

Data Organization and Results. As stated above, a Table of Chemical Tests (see Common Practices... in the Chem 3102 directory) is required for each component. Spectroscopic data is entered as usual for the IR.  For the PMR, each piece of the data is entered separately, an inference is drawn and a comment is made. For example the Test may be "PMR" the Result may be "signal at 6.9" the Inference may be "probably aromatic" and the Comment may be "agrees with IR peaks at 1500 and 1600 cm-1". The next entry would focus on the same signal's multiplicity: the Test may be "PMR signal at 6.9"; the Result may be "c = complex signal"; the Inference may be "typical splitting among aromatic hydrogens"; and the Comment may be "example spectrum in McMurry ". The next line focuses on the same signal's integration: the Test would be "PMR complex signal at 6.9"; the Result would be "area = 5H after normalization"; the Inference would be "the benzene ring is monosubstituted"; and the Comment may be "agrees with IR pattern 870-675 cm-1". Each set of PMR data needs to be dealt with in detail.

Each component is also reported with a separate Table of Candidates (examples in Common Practices). While a column for derivative melting points will be necessary only if a neutral component is an aldehyde or a ketone, a separate column should be labeled "Agreement of PMR with Structure". In it the predicted PMR of the candidate is compared, at least in part, with the actual spectrum provided. For example, if the component's actual spectrum has a combination quartet + triplet combination with signal areas normalized to 2 and 3 hydrogens respectively, the inference would be that the component must have an ethyl group. For those candidates lacking this feature, the response in the Agreement column would be "doesn't have an ethyl group".

Part of your grade is based on your identification of the unknowns.  The remainder is derived from your thorough and accurate discussion and  interpretation of the results.

Conclusion: "Based on the data summarized in the Tables of Chemical Tests and Tables of Candidates above, I conclude that the unknown number ____ consists of ______ and _____."

Reference:

"Mayo et al.": Mayo, D.W., Pike, R.M., Butcher, S.S. and Trumper, P.K. Microscale Techniques for the Organic Laboratory; Wiley: New York, 1991.

"McMurry": McMurry, John, Organic Chemistry, 5th ed., Brooks/Cole Publishing Co.: Pacific Grove, CA, 2000.

Revised Summer, 2000.