For each pair of compounds, circle the compound you expect to have the higher boiling point. Explain your reasoning. a. (CH3)3C—C(CH 3 ) 3 or (CH 3 ) 2 CH—CH 2 CH 2 —CH(CH 3 ) 2 b. CH 3 (CH 2 ) 6 CH 3 or CH 3 (CH 2 ) 5 CH 2 OH c. CH 3 CH 2 OCH 2 CH 3 or CH 3 CH 2 CH 2 CH 2 OH

Hey everyone, let's solve this problem. It says determine which compound you expect to have the higher boiling point for each of the following pairs. And we have three pairs. Explain your reasoning. So when we're looking at increased boiling point we're looking at how strong the inter molecular forces are within a compound. Right? And the stronger the inter molecular forces the higher the boiling point. Because it will be harder to sort of rip those molecules apart from each other. And we might have to look at increased which one has a higher molecular weight because that would have more electrons. Right? Or we might have to look at um if there's higher surface area because the larger size will create stronger inter molecular forces. Okay, so let's look at the compounds that we have. Well looking at the first compound, we have ch 33 on both ends. Okay, that's three methyl groups coming off the end carbons. And in our second compound there's only two methyl groups coming off of the end carbons and we have a longer carbon chain in the center. Okay, well if we're comparing their inter molecular forces, they're both just hydrocarbons. Right. So there's no hydrogen bonding, no dipole dipole interactions. What about molecular weight? Well, if you calculate the molecular weight of these, they're actually the same. They have the same number of carbons and hydrogen. But how are they arranged? Well we can look at surface area. Like I said, the first compound has more branches because those it has three methyl groups coming off of those n carbons. Whereas in the second group in the second compound, rather there's only two methyl groups on each end carbon and there's more carbons in the center of the chain. So our second compound is going to be more spread out larger because it has less branches. When it has more branches it sort of makes the compound shorter and smaller. Right? So because our second compound will have less branches, it will have more surface area and therefore stronger inter molecular forces. So we can eliminate answer choice A. Right? Because that said, the other compound has a higher boiling point. And let's see it says our second compound has a higher boiling point because it has less branches. Right? That is what we just said. Cool, let's look at our second compound or a second pair rather. So we have a hydrocarbon and then I see an alcohol, right? There's an O. H. Group at the end of that second compound. Well, what type of inter molecular forces will we have in the first compound? It's just a hydrocarbon. So there's no h bonding again, no dipole dipole. Just London dispersion forces. But our second compound because of that O. H bond, it can have hydrogen bonding. Right? So because of that hydrogen bonding, we know that's a strong and a molecular force and it's definitely stronger than the London dispersion forces. So that alcohol will have the higher boiling point. So that can eliminate answer choice B. Right? Because it said the hydrocarbon has a higher boiling point. And these answer choices say that the alcohol has a higher boiling point because it conforms stronger inter molecular interactions hydrogen bonding, right? Okay. In our third compound or a third pair, we have an ether right? In our group and O. And oxygen and another R group. So in either and our second compound is another alcohol. Okay, well they both have oxygen which makes them polar, right? The carbon oxygen bond is polar. So there's some dipole dipole interactions. But again that alcohol has hydrogen bonding which we know is stronger than dipole dipole interactions. And that ether does not have a church bonding because the oxygen is not bonded to a hydrogen. So again, our alcohol will be have the higher boiling point which eliminates answer choice D. Leaving answer choice C. To be our correct answer. I hope this video is helpful and thanks for watching

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

2. Molecular Representations

Intermolecular Forces

Organic Chemistry

2. Molecular Representations

Intermolecular Forces

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4:50 minutes

Problem 6a,b,c

Textbook Question

For each pair of compounds, circle the compound you expect to have the higher boiling point. Explain your reasoning.
a. (CH3)3C—C(CH₃)₃ or (CH₃)₂CH—CH₂CH₂—CH(CH₃)₂
b. CH₃(CH₂)₆CH₃ or CH₃(CH₂)₅CH₂OH
c. CH₃CH₂OCH₂CH₃ or CH₃CH₂CH₂CH₂OH

Verified step by step guidance

Step 1: Understand the factors affecting boiling points. Boiling points are influenced by molecular weight, intermolecular forces (such as hydrogen bonding, dipole-dipole interactions, and London dispersion forces), and molecular structure (branching vs. linear).

Step 2: Analyze compound pair (a): (CH3)3C—C(CH3)3 vs. (CH3)2CH—CH2CH2—CH(CH3)2. Both are hydrocarbons, but the first compound is more branched. Branching generally reduces boiling points due to decreased surface area and weaker London dispersion forces. Therefore, the less branched compound is likely to have a higher boiling point.

Step 3: Analyze compound pair (b): CH3(CH2)6CH3 vs. CH3(CH2)5CH2OH. The second compound contains an alcohol group (-OH), which can form hydrogen bonds, significantly increasing its boiling point compared to the first compound, which is a simple alkane.

Step 4: Analyze compound pair (c): CH3CH2OCH2CH3 vs. CH3CH2CH2CH2OH. The second compound is an alcohol, capable of hydrogen bonding, whereas the first is an ether, which cannot form hydrogen bonds as effectively. Hydrogen bonding in alcohols leads to higher boiling points.

Step 5: Summarize the reasoning: For each pair, consider the presence of functional groups capable of hydrogen bonding and the degree of branching. Compounds with hydrogen bonding capabilities or less branching typically have higher boiling points.

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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, including hydrogen bonding, dipole-dipole interactions, and London dispersion forces, significantly influence boiling points. Compounds with stronger intermolecular forces generally have higher boiling points. For example, hydrogen bonding, present in alcohols, is stronger than the dispersion forces in alkanes, leading to higher boiling points.

Molecular Structure and Branching

The structure and branching of a molecule affect its boiling point. Linear molecules can pack closely together, increasing intermolecular forces, while branched molecules have less surface area contact, reducing these forces. Thus, linear alkanes typically have higher boiling points than their branched counterparts due to increased van der Waals interactions.

Functional Groups

Functional groups in organic compounds determine the type and strength of intermolecular forces. Alcohols, with hydroxyl groups, can form hydrogen bonds, leading to higher boiling points compared to ethers or alkanes, which primarily exhibit weaker dispersion forces. The presence of polar functional groups enhances intermolecular attractions, raising boiling points.

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For each pair of compounds, circle the compound you expect to have the higher boiling point. Explain your reasoning.a. (CH3)3C—C(CH3)3 or (CH3)2CH—CH2CH2—CH(CH3)2 b. CH3(CH2)6CH3 or CH3(CH2)5CH2OH c. CH3CH2OCH2CH3 or CH3CH2CH2CH2OH

Key Concepts

Intermolecular Forces

Molecular Structure and Branching

Functional Groups

Watch next

For each pair of compounds, circle the compound you expect to have the higher boiling point. Explain your reasoning.
a. (CH3)3C—C(CH₃)₃ or (CH₃)₂CH—CH₂CH₂—CH(CH₃)₂
b. CH₃(CH₂)₆CH₃ or CH₃(CH₂)₅CH₂OH
c. CH₃CH₂OCH₂CH₃ or CH₃CH₂CH₂CH₂OH