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(a) Would you expect the viscosity of isopropanol, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH},\) to be larger or smaller than the viscosity of ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) ? (b) Would you expect the viscosity of isopropanol to be smaller or larger than the viscosity of 1-propanol, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{OH}\) ?

Short Answer

Expert verified
(a) Isopropanol would have a larger viscosity than ethanol due to its larger, more branched molecular structure. (b) Isopropanol would have a smaller viscosity than 1-propanol, because the stronger London dispersion forces in 1-propanol's longer carbon chain have a more significant impact on viscosity compared to the branching effect in isopropanol.

Step by step solution

01

Analyze Molecular Structures

Let's first examine the molecular structures of the three compounds. 1. Isopropanol: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}\) 2. Ethanol: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) 3. 1-Propanol: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{OH}\) As mentioned earlier, all three compounds have similar structures. They all have an alcohol group (-OH) and carbon atoms bonded to hydrogen atoms. However, the arrangement of these carbon and hydrogen atoms affects the overall shape and size of the molecules, which in turn impacts their viscosities.
02

Compare Isopropanol and Ethanol

Now let's compare the molecular structures of isopropanol and ethanol to determine which one might have a larger viscosity. Isopropanol has two methyl groups \(\mathrm{(CH}_3)\) directly attached to the carbon with the alcohol group, whereas ethanol has only one methyl group and one hydrogen attached. This makes the isopropanol molecule larger and more branched than the ethanol molecule. Larger and more branched molecules generally have greater viscosities, because they have more complex shapes that can hinder the movement of the molecules relative to one another. Additionally, the increased molecular size might lead to stronger intermolecular forces (e.g., London dispersion forces) between the isopropanol molecules, adding to the resistance against flow. Thus, we would expect isopropanol to have a greater viscosity than ethanol. So the answer to part (a) is that isopropanol would have a larger viscosity than ethanol.
03

Compare Isopropanol and 1-Propanol

Next, let's compare the molecular structures of isopropanol and 1-propanol to determine which one might have a larger viscosity. Isopropanol has its alcohol group attached to a central carbon, which is connected to two methyl groups. On the other hand, 1-propanol has its alcohol group attached to a terminal carbon, which is connected to two other carbon atoms along a straight chain. Although the overall number of carbon atoms is the same in both molecules, their arrangements differ. Isopropanol is more branched, while 1-propanol has a more linear arrangement. While larger and more branched molecules, like isopropanol, generally have greater viscosities, 1-propanol also has a longer carbon chain. Longer carbon chains have more extensive electron clouds that lead to stronger London dispersion forces between the molecules, which can increase viscosity. In this specific comparison, these forces tend to have a stronger impact on viscosity than the branching effect observed in isopropanol. So the answer to part (b) is that isopropanol would have a smaller viscosity than 1-propanol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Structure
Molecular structures are crucial when understanding the properties of compounds, including viscosity. Viscosity, which is a measure of a fluid's resistance to flow, is influenced by how molecules are arranged and bonded. In this problem, we look at three alcohols: isopropanol, ethanol, and 1-propanol.
  • Isopropanol: egin{align*} ( ext{CH}_3)_2 ext{CHOH} ext{ has a more branched structure with two methyl groups (CH}_3 ext{) attached to the central carbon.} ext{This branching affects how the molecules interact with each other.} ext{The shape of the molecule can make it more bulky, hindering its flow.}
  • Ethanol: egin{align*} ext{CH}_3 ext{CH}_2 ext{OH} ext{ is simpler with only one methyl group and a hydrogen attached to the carbon along with the alcohol.} ext{This linear structure promotes smoother flow compared to more branched structures.} ext{As a result, ethanol is usually less viscous than its branched counterparts.}
  • 1-Propanol: egin{align*} ext{CH}_3( ext{CH}_2)_2 ext{OH} ext{ features a longer straight chain compared to ethanol.} ext{The structure leads to different interactions that affect viscosity.} ext{It's essential to understand this branching and chain length relationship to predict viscosity.}
Intermolecular Forces
Intermolecular forces are the interactions that hold molecules together, impacting properties like viscosity. In alcohols such as isopropanol, ethanol, and 1-propanol, these forces play a significant role.
For the alcohols in question:
  • Hydrogen bonding: Present due to the -OH (hydroxyl) group, increasing viscosity because of the strong attraction between these groups across different molecules.
  • London dispersion forces: All molecules exhibit these, but they occur more prominently in larger molecules with greater electron clouds. These forces contribute more significantly in longer or more massive molecules.
  • In isopropanol and 1-propanol, both hydrogen bonding and London dispersion forces are present. However, 1-propanol, with its longer carbon chain, generally exhibits stronger dispersion forces than isopropanol. This often compensates for its more linear structure regarding overall viscosity.
Understanding these intermolecular interactions helps explain why the viscosity differs among different alcohols, even with similar molecular compositions.
Alcohols
Alcohols are organic compounds where a hydroxyl group (-OH) is bonded to a carbon atom. They are characterized by this -OH group, which contributes to distinct chemical and physical properties, such as the ability to form hydrogen bonds. In this exercise, we explore several alcohols to understand how their structure influences viscosity.
  • Functionality: The -OH group in alcohols leads to hydrogen bonding, which not only influences solubility but also impacts viscosity. The more hydrogen bonds, the greater the tendency for higher viscosity.
  • Structure Impact: The particular shape and size of alcohol molecules, which includes factors such as branching and carbon chain length, further affects the viscosity.
This hydroxyl group's interactions make alcohols a perfect example of how molecular structure and intermolecular forces interplay to impact physical properties like viscosity.
Carbon Chain Length
The length of the carbon chain in a molecule can significantly impact its viscosity. For alcohols, the number of carbon atoms in the chain contributes to the overall bulkiness and electronic properties, affecting how molecules move past each other and thus their resistance to flow.
  • Shorter chains, such as in ethanol, lead to more compact molecules that show less resistance to flow and thus have lower viscosity.
  • Longer chains, as seen in 1-propanol, increase the surface area available for intermolecular interactions such as London dispersion forces. These forces are more substantial due to the larger electron clouds, making the molecules experience greater resistance when trying to move past each other, leading to increased viscosity.
In this way, while the overall molecular size and structure play crucial roles, the simple extension of the carbon chain can tip the balance of viscosities between different alcohols.

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Most popular questions from this chapter

Indicate whether each statement is true or false: (a) The liquid crystal state is another phase of matter, just like solid, liquid, and gas. (b) Liquid crystalline molecules are generally spherical in shape. (c) Molecules that exhibit a liquid crystalline phase do so at well-defined temperatures and pressures. (d) Molecules that exhibit a liquid crystalline phase show weaker-than- expected intermolecular forces. (e) Molecules containing only carbon and hydrogen are likely to form liquid crystalline phases. (f) Molecules can exhibit more than one liquid crystalline phase.

You are high up in the mountains and boil water to make some tea. However, when you drink your tea, it is not as hot as it should be. You try again and again, but the water is just not hot enough to make a hot cup of tea. Which is the best explanation for this result? (a) High in the mountains, it is probably very dry, and so the water is rapidly evaporating from your cup and cooling it. (b) High in the mountains, it is probably very windy, and so the water is rapidly evaporating from your cup and cooling it. (c) High in the mountains, the air pressure is significantly less than \(101.3 \mathrm{kPa}\), so the boiling point of water is much lower than at sea level. (d) High in the mountains, the air pressure is significantly less than \(101.3 \mathrm{kPa}\), so the boiling point of water is much higher than at sea level.

Based on the type or types of intermolecular forces, predict the substance in each pair that has the higher boiling point: (a) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) or \(n\) -butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right),(\mathbf{b})\) diethyl ether \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\right)\) or 1 -butanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) (c) sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) or sulfur trioxide \(\left(\mathrm{SO}_{3}\right)\), (d) phosgene \(\left(\mathrm{Cl}_{2} \mathrm{CO}\right)\) or formaldehyde \(\left(\mathrm{H}_{2} \mathrm{CO}\right)\). Look up and compare the normal boiling points and nor-

Which member in each pair has the greater dispersion forces? (a) \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH},\) (b) \(\mathrm{NH}_{3}\) or \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\), (c) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) or \(\mathrm{CH}_{2} \mathrm{Br}_{2}\)

True or false: (a) Molecules containing polar bonds must be polar molecules and have dipole-dipole forces. (b) For the halogen gases, the dispersion forces decrease while the boiling points increase as you go down the column in the periodic table. (c) In terms of the total attractive forces for a given substance, the more polar bonds there are in a molecule, the stronger the dipole-dipole interaction. \(\mathbf{d}\) ) All other factors being the same, total attractive forces between linear molecules are greater than those between molecules whose shapes are nearly spherical. (e) The more electronegative the atom, the more polarizable it is.

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