Creating fuel from algae
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Creating fuel from algae
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[ID] => 560285
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[post_date] => 2025-01-12 09:56:16
[post_date_gmt] => 2025-01-12 14:56:16
[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.
Green tides are overgrowths of algae that occur in response to many factors, including excess nutrients in marine water. Up to 106 tons of a particular algae, E. prolifera, can be collected from a single green tide event. Rather than waste this biomass, researchers attempt to convert the algae into a renewable source of fuel called bio-oil using a process known as hydrothermal liquefaction (HTL). In HTL, water is added to dried algae; the mixture brought to sufficiently elevated temperature and pressure conditions to allow the water to decompose the algal biomass. After the reaction mixture is cooled, it can be separated according to the scheme shown in Figure 1.
Figure 1: Schematic diagram for bio-oil production and separation
Figure 2: Constituents of Bio-Oil. Analysis of the bio-oil indicate that it is composed of over 180 compounds. The top five most common constituent molecules are shown below.
Adapted from: Zhou, D., Zhang, L., Zhang, S., Fu, H., & Chen, J. (2010). Hydrothermal liquefaction of macroalgae Enteromorpha prolifera to bio-oil. Energy & Fuels, 24(7), 4054-4061; Zhou, D., Zhang, S., Fu, H., & Chen, J. (2012). Liquefaction of macroalgae Enteromorpha prolifera in sub-/supercritical alcohols: direct production of ester compounds. Energy & Fuels, 26(4), 2342-2351.
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[question] => The researchers could have used many solvents in extracting the oil from the aqueous reaction mixture. What characteristic of CH₂Cl₂ makes it a better solvent than hexane for this procedure?
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[answer] => 2
[description] => Reason for the Correct Answer:
Density can be important in a separation using because it determines which layer will be on top or on bottom (the denser layer will be on bottom). However, density does not make one solvent better than another for a particular separation.
The stated goal of the procedure is to separate out the bio-oil. The main components of bio-oil all have some sort of polar group, such as a carboxylic acid or ketone (even though a large portion of each fatty acid is nonpolar). This means they will probably dissolve better in a solvent that has some degree of polarity.
CH₂Cl₂ is polar (draw out the molecular geometry to see this), while hexane is completely nonpolar. Therefore CH₂Cl₂ will be a better choice for solvent in this experiment.
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[each_answer] => A. CH₂Cl₂ is more dense than hexane.
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[each_answer] => B. CH₂Cl₂ is more polar than hexane.
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[each_answer] => C. CH₂Cl₂ is less polar than hexane.
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[each_answer] => D. CH₂Cl₂ is less dense than hexane.
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => Researchers carry out a similar procedure for producing bio-oil in which methanol was added to the dried algae instead of water in the first step. After liquefaction, cooling, and separation, what type of compound will likely be formed as the major product(s)?
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[answer] => 4
[description] => Reason for the Correct Answer:
This question asks you about the major product that would be formed in a new experiment. The top two major products of the previous experiment were both carboxylic acids (specifically fatty acids).
In the previous experiment, none of the major products contained nitrogen, so it is unlikely that changing from water to methanol (which doesn’t contain nitrogen) will produce a nitrogen-containing functional group. You can eliminate amide and amine.
Methanol is similar in many ways to water as a solvent – both are polar protic solvents with similar pKa values. You can think about them as H-OH (water) and H-OCH3 (methanol). It is likely that it will facilitate a similar reaction as a solvent, while perhaps exchanging the -OH group when water is the solvent with -OCH3 when methanol is the solvent.
Fatty acids (and other carboxylic acids) are known to react with alcohols to form esters (RCOOH + R’OH → RCOOR’ + HOH). Therefore, it is likely that the bio-oil will contain esters when it is produced in the presence of methanol. There is no evidence in the passage to suggest that ethers will be formed.
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[answers] => Array
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[0] => Array
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[each_answer] => A. amide
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[1] => Array
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[each_answer] => B. ether
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[each_answer] => C. amine
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[each_answer] => D. ester
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => What is the hybridization of carbon 3 in compound 5?
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[answer] => 2
[description] => Reason for the Correct Answer:
You can draw out compound 5 by considering its name: 2,3-dimethyl-2-cyclopenten-1-one.
Compound 5 looks like this (with the third carbon circled):
That carbon is trigonal planar with a double bond, meaning that it is sp²-hybridized.
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[0] => Array
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[each_answer] => A. sp
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[each_answer] => B. sp²
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[each_answer] => C. sp³
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[each_answer] => D. dsp³
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[quiz_unique_key] => 2377279144
[question] => How many distinct stereoisomers exist for the hexadecanoic acid and oleic acid, respectively?
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[answer] => 2
[description] => Reason for the Correct Answer:
You can draw out the compounds based on their molecular formulas listed above. Hexadecanoic acid only has one stereoisomer:
Oleic acid has a double bond that allows the molecule to exist as a cis- or trans-stereoisomer. The cis- isomer is shown below:
In summary, hexadecanoic acid exists as one stereoisomer, while oleic acid exists as two stereoisomers. Therefore, the answer is 1,2.
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[each_answer] => A. 1,1
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[each_answer] => B. 1,2
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[each_answer] => C. 2,2
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[each_answer] => D. 2,4
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[question] => Which procedures could the researchers have used to separate and identify hexadecanoic acid and oleic acid as the two largest constituents of bio-oil?
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[answer] => 1
[description] => Reason for the Correct Answer:
Bio-oil is a mixture of products. To figure out what is in it, you would need a separation method and an identification method.
Gas chromatography and thin-layer chromotography are both methods of separating compounds, so it would be difficult to identify compounds with just these two methods.
Although ¹³C-NMR can provide information on the electronic environment of carbon atoms and can help figure out a molecule’s structure, it cannot usually identify specific, complex molecules such as oleic acid.
Mass spectrometry can provide the molecular weight of a compound, which can be combined with the retention time of each molecule in the gas chromatography to identify the molecule. (It is extremely unlikely that two different molecules will have the same retention time in gas chromatography AND the same molecular weight).
Therefore, the correct answer is gas chromatography and mass spectrometry.
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[each_answer] => A. Gas chromatography and mass spectrometry
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[each_answer] => B. ¹³C-NMR and mass spectrometry
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[each_answer] => C. ¹³C-NMR and thin-layer chromotography
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[each_answer] => D. Gas chromatography and thin-layer chromotography
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Figure 2: Constituents of Bio-Oil
