Which type of intermolecular force accounts for each of these differences? (a) CH 3 OH boils at 65 °C; CH 3 SH boils at 6 °C. (d) Acetone boils at 56 °C, whereas 2-methylpropane boils at -12 °C.

Hey everyone, we're told that despite having the same molecular formula and Mueller mass, ethanol boils at 78 degrees Celsius, while dimethyl ether boils at negative 24 degrees Celsius identify the type of inter molecular force responsible for the difference. First, let's go ahead and draw this out, drawing out our ethanol, we have two carbons and one hydroxyl group. Next drawing out our dime ethyl ether. We have an oxygen group in between two methyl groups. As we can see right here, ethanol has a hydroxyl group that is able to hydrogen bond. So we have hydrogen bonding present in our ethanol and that is shown here with that hydrogen attached to our oxygen. Looking at dimethyl ether, we have a non polar compound and because we have a non polar compound, we have London dispersion forces. Since ethanol exhibits hydrogen bonding and dimethyl ether doesn't. This is the reason why ethanol has a higher boiling point than dimethyl ether, which only has London dispersion forces. So I hope that made sense. And let us know if you have any questions

Ch.11 - Liquids and Intermolecular Forces

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Brown 14th Edition

Ch.11 - Liquids and Intermolecular Forces

Problem 18a,d

Chapter 11, Problem 18a,d

Which type of intermolecular force accounts for each of these differences? (a) CH₃OH boils at 65 °C; CH₃SH boils at 6 °C. (d) Acetone boils at 56 °C, whereas 2-methylpropane boils at -12 °C.

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Identify the molecular structures of acetone and 2-methylpropane. Acetone is a ketone with a carbonyl group (C=O), and 2-methylpropane, also known as isobutane, is a branched alkane.

Recognize the types of intermolecular forces present in each compound. Acetone can engage in dipole-dipole interactions due to its polar carbonyl group, and also has London dispersion forces. 2-methylpropane only has London dispersion forces as it is a nonpolar molecule.

Compare the strength of the intermolecular forces in each compound. Dipole-dipole interactions (as seen in acetone) are generally stronger than London dispersion forces alone (as seen in 2-methylpropane).

Relate the strength of the intermolecular forces to the boiling points of the compounds. Stronger intermolecular forces result in higher boiling points.

Conclude that the difference in boiling points between acetone and 2-methylpropane is due to acetone's ability to form dipole-dipole interactions, which are absent in 2-methylpropane.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Intermolecular Forces

Intermolecular forces are the attractive forces between molecules that influence physical properties such as boiling and melting points. These forces include hydrogen bonding, dipole-dipole interactions, and London dispersion forces. The strength and type of these forces determine how closely molecules are held together, affecting their state and behavior at different temperatures.

Hydrogen Bonding

Hydrogen bonding is a specific type of strong dipole-dipole interaction that occurs when hydrogen is bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine. In the case of acetone, the presence of a carbonyl group (C=O) allows for hydrogen bonding with other molecules, leading to a higher boiling point compared to non-polar molecules like 2-methylpropane, which lacks such interactions.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure, allowing it to transition from liquid to gas. Factors influencing boiling points include molecular weight, intermolecular forces, and molecular structure. In this context, acetone's higher boiling point compared to 2-methylpropane can be attributed to stronger intermolecular forces in acetone due to hydrogen bonding.