Methanol and its metabolites
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- Methanol and its metabolites
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Methanol and its metabolites
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[post_date] => 2025-01-14 06:31:20
[post_date_gmt] => 2025-01-14 11:31:20
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Methanol poisoning is a medical emergency. Methanol is converted to formic acid via formaldehyde, with formic acid being the toxic metabolite that causes damage to the optic nerves leading to permanent blindness.
Oral ingestion of methanol-containing mixtures require immediate medical attention. Once ingested, peak serum levels occur within 30-90 minutes. Methanol is metabolized in the liver in sequential steps to formaldehyde first by hepatic alcohol dehydrogenase, then to formic acid by formaldehyde dehydrogenase and then finally detoxified as carbon dioxide. The slow conversion of formic acid to carbon dioxide forms the basis of methanol toxicity.
In the laboratory, methanol can be converted to formic acid via a series of chemical transformations. The chemical process is summarized in the schematic presented in Figure 1. A student proposes to replicate this chemical pathway and performs two trials but obtains varied results.
Figure 1. Conversion of methanol to formic acid
Trial 1: At the beginning of the experiment, the student mistakenly uses isopropyl alcohol ((CH3)2CHOH) instead of methanol. He succeeds in converting isopropyl alcohol to an unknown substance using pyridinum cholochromate (PCC, C5H6NCrO3Cl) in dichloromethane (CH2Cl2) solvent (this is the same procedure for formaldehyde formation from methanol). However, this unknown substance demonstrates different properties than what he expects from formaldehyde.
Trial 2: The student follows the same protocol with PCC and dichloromethane (as in Trial 1), but with the proper starting reagent of methanol and successfully forms formaldehyde. He proceeds to add Tollen's reagent (Ag2O dissolved in aqueous NH3) to the formaldehyde, successfully yielding formic acid.
Citation: International Programme on Chemical Safety. (2002, May). Methanol. Retrieved from: https://www.inchem.org/documents/pims/chemical/pim335.htm.
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[quiz_unique_key] => 578908434
[question] => What are the chemical processes denoted in A and B in Figure 1?
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[answer] => 1
[description] => Reason for the Correct Answer:
Oxidation is defined as the process in which the central atom of a functional group is transformed into a more highly oxidized form (higher oxidation state). This happens when carbon atom gains bonds to more electronegative elements (for instance: O, N, and Cl).
In the opposite fashion, reduction is defined as the process in which the central atom of a functional group is transformed into a more highly reduced form (lower oxidation state). This happens when carbon atom gains bonds to less electronegative elements (H).
In our example, the oxidation state of the central carbon in methanol is -2 (-1 for each of the three hydrogens, and +1 for the oxygen). For formaldehyde, the oxidation state is 0 (-1 for each of the two hydrogens, and +12 for the oxygen double bond). For formic acid, the oxidation state is +2 (-1 for the hydrogen, and then +12 for the oxygen double bond, then +1 for the other oxygen).
Given the progressive increase in oxidation state of the central carbon, A demonstrates an oxidation process and B similarly demonstrates an oxidation process.
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[each_answer] => A.A – Oxidation, B – Oxidation
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[each_answer] => B.A – Oxidation, B – Reduction
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[each_answer] => C.A – Reduction, B – Oxidation
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[each_answer] => D.A – Reduction, B – Reduction
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => Which statement is true regarding some physical properties of methanol and isopropyl alcohol?
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[answer] => 1
[description] => Reason for the Correct Answer:
Methanol is a primary alcohol, and isopropyl is a secondary alcohol. Primary alcohols have one bond to an organic group from the CH2 group attached to the -OH group. A good example is ethanol (CH3-CH2-OH) or propanol (CH3-CH2-CH2-OH). Methanol technically doesn’t fit the “one linkage” rule, but it is considered a primary alcohol since it is simpler than ethanol containing zero linkages to an organic group from the CH2 group attached to the -OH. Secondary alcohols have two linkages to organic groups from the CH group attached to the -OH. This is evident when the chemical structure of isopropyl alcohol ((CH3)2CHOH) is drawn out. Primary and secondary alcohols have differences in boiling point and solubility.
Solubility for alcohols in water is primarily determined by the hydrogen bonds that form between the alcohol molecules and the water molecules. Smaller alcohols form bonds more easily with the surrounding water, while bulkier alcohols form H-bonds less readily due to the long, nonpolar hydrocarbon tails (hydrophobic effect). This decreases the solubility for large alcohols in water.
The boiling points of alcohols increase with additional carbons present in the compound. When more carbons are present, additional energy is required to overcome the increased intermolecular forces (Van der Waals forces) in the conversion from liquid to vapor phase.
Methanol has a lower boiling point and a higher solubility in water than isopropyl alcohol.
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[each_answer] => A.Methanol has a lower boiling point and a higher solubility in water than isopropyl alcohol.
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[each_answer] => B.Methanol has a lower boiling point and a lower solubility in water than isopropyl alcohol.
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[each_answer] => C.Methanol has a higher boiling point and a lower solubility in water than isopropyl alcohol.
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[each_answer] => D.Methanol has a higher boiling point and a higher solubility in water than isopropyl alcohol.
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => Do you expect the unknown substance synthesized in Trial 1 to form a carboxylic acid compound (such as formic acid in Figure 1) more easily or less easily than formaldehyde? Why?
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[answer] => 2
[description] => Reason for the Correct Answer:
First, it is important to determine the nature of the unknown substance synthesized in Trial 1. As seen in Figure 1, the oxidation of primary alcohols (methanol) forms aldehydes (formaldehyde). However, isopropyl alcohol is a secondary alcohol, and oxidation of secondary alcohols leads to the formation of ketones.
In Figure 1, formaldehyde is converted to formic acid due to reactions involving the lone H attached to the main carbon which is oxidized to form the carboxylic acid group (-COOH). However, ketones do not have this additional -CH group and thus is resistant to oxidation.
The unknown substance synthesized in Trial 1 will form a carboxylic acid compound less easily than formaldehyde due to the lack of a lone H attached to the main carbon in the unknown substance.
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[each_answer] => A.The unknown substance synthesized in Trial 1 will form a carboxylic acid compound more easily than formaldehyde due to the presence of a lone H attached to the main carbon in the unknown substance.
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[each_answer] => B.The unknown substance synthesized in Trial 1 will form a carboxylic acid compound less easily than formaldehyde due to the lack of a lone H attached to the main carbon in the unknown substance.
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[each_answer] => C.The unknown substance synthesized in Trial 1 will form a carboxylic acid compound more easily than formaldehyde due to the lack of a lone H attached to the main carbon in the unknown substance.
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[each_answer] => D.The unknown substance synthesized in Trial 1 will form a carboxylic acid compound less easily than formaldehyde due to the presence of a lone H attached to the main carbon in the unknown substance.
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[quiz_unique_key] => 2377279144
[question] => What is the simplest way that the student can confirm the successful formation of formic acid using Tollen’s Reagent in Trial 2?
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[answer] => 4
[description] => Reason for the Correct Answer:
The oxidation of formaldehyde to formic acid using Tollen’s Reagent is given by the following reaction: formaldehyde + Ag2O -> formic acid + Ag. (the oxygen group in Ag2O gets incorporated as the -COOH in the formic acid molecule, and solid silver is precipitated out of solution)
Since the solid silver precipitates, it forms a covering on the test tube (often called “silver mirror test “) and this can be directly observed by the experimenter. Odor, boiling point, and density are other methods that can be used to confirm formic acid formation, but it is not the simplest way. The key here is to recognize that solid Ag will precipitate out of solution.
The simplest way that the student can confirm the successful formation of formic acid using Tollen’s Reagent in Trial 2 is by observing a silver deposition in the test tube.
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[each_answer] => A.By measuring the density of the liquid and comparing to published values.
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[each_answer] => B.By boiling the liquid and comparing the boiling temperature to published values.
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[each_answer] => C.By noting a characteristic sharp odor emanating from the test tube.
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[each_answer] => D.By observing a silver deposition in the test tube.
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[quiz_unique_key] => 2261298308
[question] => What is the unknown product formed in Trial 1?
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[answer] => 2
[description] => Reason for the Correct Answer:
Draw out the structure of isopropyl alcohol and predict the oxidation reaction product. As above, we know that secondary alcohols form ketones with oxidation, so formaldehyde and acetyaldehyde will not be correct answers
The reaction is given as follows:
Given that the resulting ketone contains three carbons, the product is propanone, or more commonly known as acetone.
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[answers] => Array
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[each_answer] => A.Butanone
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[each_answer] => B.Acetone (propanone)
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[each_answer] => C.Formaldehyde (methanal)
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[each_answer] => D.Acetylaldehyde (ethanal)
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