clutch21

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Hey I am having trouble determining the relative boiling points between these six compounds. If someone could help me out with an explanation that would be great, Thanks!

1. C-C-C-C-Cl

2. C-C-C-C-F

3. Cl-C-C-C-C-F

4. C-C-C-C

5. C-C-C-N-C

6. C-C-C-C-N

So far, I think 4 has the lowest BP. Whereas 6 then 5 are the highest BP's.
 

amberkre

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So I think the order goes (lowest bp to highest bp) -- 4, 2, 5,1,6,3

this is my reasoning:

#4 is an alkane which is a gas so it's definitely lowest
#2 also has a low bp--> for the haloalkanes- larger atoms means that the electrons are held further away from the nucleus and better able to form bonds with other molecules. Thus, the london forces are greater between molecules of C-C-C-C-Cl than molecules of C-C-C-C-F, that is why C-C-C-C-F has a lower boiling point (the bonds between the molecules are broken at lower temperatures) also Cl is more electronegative and F
#5- forms weak hydrogen bonds, because this molecule only has one protic hydrogen
#1- relatively high boiling point because this molecule strong dipole-dipole interactions between molecules since the molecule is polar
#6- even higher boiling point than #1 because this molecule will form strong hydrogen bonds (it has 2 protic hydrogens that are capable of hydrogen bonding with the other molecules)
#3- honestly i'm not sure why this molecules, boiling point is the highest but I looked it up and it's 114.699 °C i'm guessing it has to do with the fact that it's very polar?

After I reasoned out the order, I went and checked the boiling points and got:
1. 78 deg C
2. 31.9 °C
3. 114.699 °C
4. -1 °C
5. 63 °C
6. 79 °C

Hope that helps, I'm not sure if its totally correct though!
 
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clutch21

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Thanks! This was a helpful explanation!
 

Ibn Alnafis MD

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So I think the order goes (lowest bp to highest bp) -- 4, 2, 5,1,6,3

this is my reasoning:

#4 is an alkane which is a gas so it's definitely lowest
#2 also has a low bp--> for the haloalkanes- larger atoms means that the electrons are held further away from the nucleus and better able to form bonds with other molecules. Thus, the london forces are greater between molecules of C-C-C-C-Cl than molecules of C-C-C-C-F, that is why C-C-C-C-F has a lower boiling point (the bonds between the molecules are broken at lower temperatures) also Cl is more electronegative and F
#5- forms weak hydrogen bonds, because this molecule only has one protic hydrogen
#1- relatively high boiling point because this molecule strong dipole-dipole interactions between molecules since the molecule is polar
#6- even higher boiling point than #1 because this molecule will form strong hydrogen bonds (it has 2 protic hydrogens that are capable of hydrogen bonding with the other molecules)
#3- honestly i'm not sure why this molecules, boiling point is the highest but I looked it up and it's 114.699 °C i'm guessing it has to do with the fact that it's very polar?

After I reasoned out the order, I went and checked the boiling points and got:
1. 78 deg C
2. 31.9 °C
3. 114.699 °C
4. -1 °C
5. 63 °C
6. 79 °C

Hope that helps, I'm not sure if its totally correct though!
Fluorine forms H-bond whereas Chlorine does not, shouldn't compounds containing fluorine have higher boiling point than the ones containing chlorine?
 

gettheleadout

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So I think the order goes (lowest bp to highest bp) -- 4, 2, 5,1,6,3

this is my reasoning:

#4 is an alkane which is a gas so it's definitely lowest
#2 also has a low bp--> for the haloalkanes- larger atoms means that the electrons are held further away from the nucleus and better able to form bonds with other molecules. Thus, the london forces are greater between molecules of C-C-C-C-Cl than molecules of C-C-C-C-F, that is why C-C-C-C-F has a lower boiling point (the bonds between the molecules are broken at lower temperatures) also Cl is more electronegative and F
#5- forms weak hydrogen bonds, because this molecule only has one protic hydrogen
#1- relatively high boiling point because this molecule strong dipole-dipole interactions between molecules since the molecule is polar
#6- even higher boiling point than #1 because this molecule will form strong hydrogen bonds (it has 2 protic hydrogens that are capable of hydrogen bonding with the other molecules)
#3- honestly i'm not sure why this molecules, boiling point is the highest but I looked it up and it's 114.699 °C i'm guessing it has to do with the fact that it's very polar?

After I reasoned out the order, I went and checked the boiling points and got:
1. 78 deg C
2. 31.9 °C
3. 114.699 °C
4. -1 °C
5. 63 °C
6. 79 °C

Hope that helps, I'm not sure if its totally correct though!
I would have said, in decreasing order of b.p., 6-5-3-2-1-4

As for your explanations, Cl is actually less electronegative than fluorine, London dispersion forces are secondary to dipole-dipole in compounds 1 and 2, 6 and 5 have hydrogen bonding so I would presume they would have the strongest IMF.
 

Ibn Alnafis MD

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I would have said, in decreasing order of b.p., 6-5-3-2-1-4

As for your explanations, Cl is actually less electronegative than fluorine, London dispersion forces are secondary to dipole-dipole in compounds 1 and 2, 6 and 5 have hydrogen bonding so I would presume they would have the strongest IMF.
That was my guess too. 6 and 5 are obvious. 3 and 2 contain fluorine, which is more electronegative than chlorine and it can form hydrogen bond.
 

gettheleadout

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I don't like this question.
 

slz1900

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Why would chlorine or flourine hydrogen bond? I was under the impression that a single electronegative atom attached to an alkane isn't able to pull enough charge away from the carbon to make the hydrogens bound to the carbon capable of participating in H bonds.
 

gettheleadout

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Why would chlorine or flourine hydrogen bond? I was under the impression that a single electronegative atom attached to an alkane isn't able to pull enough charge away from the carbon to make the hydrogens bound to the carbon capable of participating in H bonds.
I didn't even notice that in Alnafis' posts. You're right, fluorine will not exhibit hydrogen bonding in these examples because no hydrogens are covalently bound to fluorine. There is no inductive carryover to C-H hydrogens as far as hydrogen bonding is concerned.

The C-F dipole should be stronger than the C-Cl dipole, however, due to the greater electronegativity of fluorine compared to chlorine.

Edit: I should have remembered this, it's a polarizability issue for compounds 1 and 2. A problem like this tripped me up at one point in organic and I had to look it up to figure this out. The polarizability (basically the max potential strength of an induced dipole) increases with atomic size. Even though the C-X dipole is stronger for C-F than C-Cl, chlorine is much larger than fluorine and thus the ability for it to interact with other molecules through London dispersion / van der Waal's forces is increased. This is more significant than the strength of the dipole in dipole-dipole interactions, and this is just something you have to know.
 
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