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chemistry question boiling point of molecules

Hi, please could I have help on this question comparing which molecule has a higher boiling point: CH3 –CH2 –CH3 and CH3 –O–CH3? I thought it would be CH3 –O–CH3? because it has more electrons so stronger id-id force but the mark scheme is talking about dipoles so probably pd-pd forces? However, I thought id-id forces were more relevant that pd-pd when comparing boiling points?
Thank you!
Original post by anonymous56754
Hi, please could I have help on this question comparing which molecule has a higher boiling point: CH3 –CH2 –CH3 and CH3 –O–CH3? I thought it would be CH3 –O–CH3? because it has more electrons so stronger id-id force but the mark scheme is talking about dipoles so probably pd-pd forces? However, I thought id-id forces were more relevant that pd-pd when comparing boiling points?
Thank you!

when they are saying dipoles its is induced dipole-dipole attractions as one dipole induces a dipole in the other one permanent is different, hope that help you where right!
Original post by anonymous56754
Hi, please could I have help on this question comparing which molecule has a higher boiling point: CH3 –CH2 –CH3 and CH3 –O–CH3? I thought it would be CH3 –O–CH3? because it has more electrons so stronger id-id force but the mark scheme is talking about dipoles so probably pd-pd forces? However, I thought id-id forces were more relevant that pd-pd when comparing boiling points?
Thank you!

You could talk about the numbers electrons as both molecules have temporary dipole-dipole interactions, but permanent dipole-dipole interactions would be worth discussing in this case.

Carbon and hydrogen have similar electronegativities and so hydrocarbons are basically nonpolar and thus only have relatively weak temporary dipole-dipole interactions.

Oxygen is very electronegative and so the C-O bonds in CH3OCH3 are polar. Since the dipoles do not face opposing directions (since the oxygen will have a bent shape like in water), they do not cancel. This gives rise to permanent dipole-dipole interactions, which generally speaking are stronger than temporary dipole-dipole interactions and thus require more energy to overcome them.

Edit: actually both molecules have the same number of electrons - the temporary dipole-dipole interactions should be roughly of the same strength, so the permanent dipole-dipole forces exerted by methoxymethane result in the intermolecular forces overall being stronger.
(edited 2 months ago)
Original post by aditya nair12344
when they are saying dipoles its is induced dipole-dipole attractions as one dipole induces a dipole in the other one permanent is different, hope that help you where right!

Please correct me if I’m wrong, but my interpretation of this was that this is roughly meant to read as “It is referring to induced dipoles and not permanent dipoles, which are different. I hope that helps, you were right.”.

Unfortunately this doesn’t appear to be correct. The numbers of electrons in each molecule are the same, so the temporary/induced dipoles should be of roughly the same strengths (in fact, I think actually propane would have ever so slightly stronger temporary dipole-dipole interactions than methoxymethane as the slightly larger central bond angle would allow the molecules to have more contact with each other). The difference is caused by the fact methoxymethane has permanent dipoles as well and the combined strength of these and the induced/temporary dipoles is greater than that of the temporary/induced dipoles in propane.
(edited 2 months ago)
Original post by TypicalNerd
Please correct me if I’m wrong, but my interpretation of this was that this is roughly meant to read as “It is referring to induced dipoles and not permanent dipoles, which are different. I hope that helps, you were right.”.
Unfortunately this doesn’t appear to be correct. The numbers of electrons in each molecule are the same, so the temporary/induced dipoles should be of roughly the same strengths (in fact, I think actually propane would have ever so slightly stronger temporary dipole-dipole interactions than methoxymethane as the slightly larger central bond angle would allow the molecules to have more contact with each other). The difference is caused by the fact methoxymethane has permanent dipoles as well and the combined strength of these and the induced/temporary dipoles is greater than that of the temporary/induced dipoles in propane.

I see, but CH3 –CH2 –CH3 has 44 electrons and CH3 –O–CH3 has 46 electrons. Nevertheless, are the pd-pd forces more significant because the number of electrons are basically the same?
Original post by anonymous56754
I see, but CH3 –CH2 –CH3 has 44 electrons and CH3 –O–CH3 has 46 electrons. Nevertheless, are the pd-pd forces more significant because the number of electrons are basically the same?

You’ve calculated the relative molecular masses - not the numbers of electrons.

Other than that, the contents of your post are correct.
Original post by TypicalNerd
You’ve calculated the relative molecular masses - not the numbers of electrons.
Other than that, the contents of your post are correct.

My bad, they are the same, sorry for that! Makes a lot more sense now :smile:
Dimethyl ether (CH3–O–CH3) has a higher boiling point than propane (CH3–CH2–CH3) because of the presence of permanent dipole-permanent dipole forces. When comparing boiling points, you should consider both pd-pd and id-id forces, but the dominant type of intermolecular forces in the molecule will typically determine the boiling point.
Bye,
Sandro
This topic piqued my interest, so I looked into it further.
Dimethyl ether (CH3-O-CH3) has a higher boiling point than propane (CH3-CH2-CH3) due to differences in intermolecular forces:
Dipole-dipole (pd-pd) forces:
Dimethyl ether has a polar C-O-C bond, creating a permanent dipole.
Propane is non-polar with only C-C and C-H bonds.
The stronger pd-pd forces in dimethyl ether contribute to a higher boiling point.
Induced dipole-induced dipole (id-id) forces:
Both molecules experience weak id-id (London dispersion) forces.
Propane has slightly stronger id-id forces due to more electrons/larger electron cloud.
However, this effect is outweighed by the pd-pd forces of dimethyl ether.
Molecular shape:
Dimethyl ether's bent shape allows for closer packing of molecules.
Propane's linear shape results in weaker intermolecular interactions.
Molar mass:
Dimethyl ether (46 g/mol) and propane (44 g/mol) have similar molar masses.
The difference in molar mass has a minimal effect on the boiling point in this case.
The presence of stronger pd-pd forces in dimethyl ether overcomes the slightly stronger id-id forces of propane, resulting in the higher boiling point of dimethyl ether (-24.8°C vs. -42.1°C for propane).

Ciao,
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My signature: "Regardless of where you may be, expressing gratitude is a universally cherished gesture."
Original post by Nitrotoluene
This topic piqued my interest, so I looked into it further.
Dimethyl ether (CH3-O-CH3) has a higher boiling point than propane (CH3-CH2-CH3) due to differences in intermolecular forces:
Dipole-dipole (pd-pd) forces:
Dimethyl ether has a polar C-O-C bond, creating a permanent dipole.
Propane is non-polar with only C-C and C-H bonds.
The stronger pd-pd forces in dimethyl ether contribute to a higher boiling point.
Induced dipole-induced dipole (id-id) forces:
Both molecules experience weak id-id (London dispersion) forces.
Propane has slightly stronger id-id forces due to more electrons/larger electron cloud.
However, this effect is outweighed by the pd-pd forces of dimethyl ether.
Molecular shape:
Dimethyl ether's bent shape allows for closer packing of molecules.
Propane's linear shape results in weaker intermolecular interactions.
Molar mass:
Dimethyl ether (46 g/mol) and propane (44 g/mol) have similar molar masses.
The difference in molar mass has a minimal effect on the boiling point in this case.
The presence of stronger pd-pd forces in dimethyl ether overcomes the slightly stronger id-id forces of propane, resulting in the higher boiling point of dimethyl ether (-24.8°C vs. -42.1°C for propane).
Ciao,
The flag of Italy.pngSandro
My signature: "Regardless of where you may be, expressing gratitude is a universally cherished gesture."

Propane and dimethyl ether have equal numbers of electrons - the atomic numbers of H, C and O are 1, 6 and 8, respectively.

e^- in CH3OCH3: 6 + 3(1) + 8 + 3(1) + 6 = 26
e^- in CH3CH2CH3 = 6 + 3(1) + 6 + 2(1) + 6 + 3(1) = 26

The amount of contact between the molecules will play a role in how much the molecules can interact and thus this will affect the extent to which the temporary dipoles interact with one another. I would have guessed that propane molecules have more contact with each other, though I accept my initial attempt to justify this may be wrong now upon closer inspection. Regardless, the difference in bond angle is pretty small and so the effect this has should be pretty negligible.

Other than the implication that each molecule has different numbers of electrons, your post seems fine to me.
Original post by Nitrotoluene
This topic piqued my interest, so I looked into it further.
Dimethyl ether (CH3-O-CH3) has a higher boiling point than propane (CH3-CH2-CH3) due to differences in intermolecular forces:
Dipole-dipole (pd-pd) forces:
Dimethyl ether has a polar C-O-C bond, creating a permanent dipole.
Propane is non-polar with only C-C and C-H bonds.
The stronger pd-pd forces in dimethyl ether contribute to a higher boiling point.
Induced dipole-induced dipole (id-id) forces:
Both molecules experience weak id-id (London dispersion) forces.
Propane has slightly stronger id-id forces due to more electrons/larger electron cloud.
However, this effect is outweighed by the pd-pd forces of dimethyl ether.
Molecular shape:
Dimethyl ether's bent shape allows for closer packing of molecules.
Propane's linear shape results in weaker intermolecular interactions.
Molar mass:
Dimethyl ether (46 g/mol) and propane (44 g/mol) have similar molar masses.
The difference in molar mass has a minimal effect on the boiling point in this case.
The presence of stronger pd-pd forces in dimethyl ether overcomes the slightly stronger id-id forces of propane, resulting in the higher boiling point of dimethyl ether (-24.8°C vs. -42.1°C for propane).
Ciao,
The flag of Italy.pngSandro
My signature: "Regardless of where you may be, expressing gratitude is a universally cherished gesture."
Thank you for the additional information!

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