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Chapter 13: Problem 30

Would you expect alanine (an amino acid) to be more soluble in water or in hexane? Explain.

Short Answer

Expert verified
Alanine is expected to be more soluble in water than in hexane due to its amphiphilic nature and the solubility principle "like dissolves like". The polar regions of alanine interact favorably with polar water molecules, while the nonpolar side chain has weaker interactions in water. In hexane, the nonpolar side chain interacts favorably, but the polar groups have weaker interactions, leading to overall lower solubility in hexane.

Step by step solution

01

Determine the polarity of alanine

To determine the solubility of alanine in water or hexane, we need to understand its polarity. Alanine is an amino acid with the molecular formula C3H7NO2. Its structure consists of an amino group (-NH2), a carboxyl group (-COOH), and a side chain methyl group (-CH3) attached to a central carbon atom. The amino group and carboxyl group are polar due to the presence of electronegative oxygen and nitrogen atoms. The side chain methyl group is nonpolar. This makes alanine an amphiphilic molecule, having both polar and nonpolar regions.
02

Understanding the solubility principle

The solubility of a substance is determined by the principle "like dissolves like". This means that polar substances tend to dissolve well in polar solvents, while nonpolar substances dissolve well in nonpolar solvents. Water is a polar solvent, with a partial negative charge on the oxygen atom and partial positive charges on the hydrogen atoms due to the difference in electronegativity. Hexane is a nonpolar solvent, with a structure consisting of carbon and hydrogen atoms with relatively equal electronegativity, resulting in a uniform distribution of electron density and no polar regions.
03

Comparing solubility in water and hexane

Alanine, being an amphiphilic molecule, has both polar and nonpolar regions. The polar regions of alanine are expected to interact well with the polar water molecules through hydrogen bonding and dipole-dipole interactions, while the nonpolar side chain will have weaker interactions in water due to hydrophobic effects. In hexane, the nonpolar side chain of alanine will interact favorably through London dispersion forces with the nonpolar hexane molecules, while the polar amino and carboxyl groups will have weaker interactions in the nonpolar hexane environment.
04

Conclusion

Based on the amphiphilic nature of alanine and the solubility principle "like dissolves like", alanine is expected to be more soluble in water than in hexane. This is because the polar regions of alanine will have stronger interactions with the polar water molecules, while the nonpolar side chain will have weaker interactions in water. In hexane, the nonpolar side chain will interact favorably, but the polar groups will have weaker interactions, leading to overall lower solubility in hexane.

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Most popular questions from this chapter

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) \(0.75 \mathrm{~L}\) of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) \(125 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KBr},\) (c) \(1.85 \mathrm{~L}\) of a solution that is \(12.0 \% \mathrm{KBr}\) by mass (the density of the solution is \(1.10 \mathrm{~g} / \mathrm{mL}\) ), (d) a \(0.150 \mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate \(16.0 \mathrm{~g}\) of AgBr from a solution containing \(0.480 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\)

Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of \(0.64 \mathrm{~g}\) of adrenaline in \(36.0 \mathrm{~g}\) of \(\mathrm{CCl}_{4}\) elevates the boiling point by \(0.49^{\circ} \mathrm{C}\). Is the molar mass of adrenaline calculated from the boiling-point elevation in agreement with the following structural formula?

Consider two solutions, one formed by adding \(10 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to \(1 \mathrm{~L}\) of water and the other formed by adding \(10 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) to \(1 \mathrm{~L}\) of water. Are the vapor pressures over the two solutions the same? Why or why not?

List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m} \mathrm{KBr}\), \(0.030 \mathrm{~m}\) phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\)

Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q),(\mathbf{b})\) \(50.0 \mathrm{mg}\) of an aqueous solution that is \(1.50 \mathrm{~m} \mathrm{NaCl}\), (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.
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