boiling point

[akimhaneul]

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There is a question in organic chem odyssey that asks which halide has the greatest boiling point. The choices are 2-bromobutane, 2-chlorobutane, 2-fluorobutane, 2-iodobutane, and all about equal.

Why is the answer 2-iodobutane. It makes sense that iodobutane has more electrons and it's more polarizable, but doesn't fluorobutane have a bigger dipole?

Thanks!

 

[2y4life]

2-iodobutane has the largest molecular mass and since all these are in the same group (halogens), then BP comes down to molecular weight, to make it simple.

 

[DAT Destroyer]

There is a question in organic chem odyssey that asks which halide has the greatest boiling point. The choices are 2-bromobutane, 2-chlorobutane, 2-fluorobutane, 2-iodobutane, and all about equal.

Why is the answer 2-iodobutane. It makes sense that iodobutane has more electrons and it's more polarizable, but doesn't fluorobutane have a bigger dipole?

Thanks!

Wow....an awesome question indeed !!! Many times in Chemistry two or more factors can compete with each other. Normally, dipole trumps Van Der Waals forces,,,,,BUT not here. Iodine gives the compound a slight dipole,,,,and a high weight with a significant Van Der Waals attraction. This beats out the fluorinated compound. The Literature value for 2-fluorobutane is 25 C.....while 2-iodobutane is 119 C.

Hope this helps.

Dr. Jim Romano

 

[Bersabeh]

There is a question in organic chem odyssey that asks which halide has the greatest boiling point. The choices are 2-bromobutane, 2-chlorobutane, 2-fluorobutane, 2-iodobutane, and all about equal.

Why is the answer 2-iodobutane. It makes sense that iodobutane has more electrons and it's more polarizable, but doesn't fluorobutane have a bigger dipole?

Thanks!

this can also be answered by the fact that since iodine bigger in size than fluorine, its ability to hold tight to other elements is more pronounced than fluorine. Hence requiring more heat to boil.

 

[WheatLom]

There is a question in organic chem odyssey that asks which halide has the greatest boiling point. The choices are 2-bromobutane, 2-chlorobutane, 2-fluorobutane, 2-iodobutane, and all about equal.

Why is the answer 2-iodobutane. It makes sense that iodobutane has more electrons and it's more polarizable, but doesn't fluorobutane have a bigger dipole?

Thanks!

The more massive and less branched the higher the BP.

Think about it in a more day-to-day way, that can extrapolate to the chemistry. If you have a phone book and a piece of paper. It will take more energy to raise the temperature of the phone book and it will be harder to burn all the paper at once since they are all so tightly stuck together. A single paper has less mass and it has less "binding" it to other things therefore it would be easier to burn.

 

[akimhaneul]

Wow....an awesome question indeed !!! Many times in Chemistry two or more factors can compete with each other. Normally, dipole trumps Van Der Waals forces,,,,,BUT not here. Iodine gives the compound a slight dipole,,,,and a high weight with a significant Van Der Waals attraction. This beats out the fluorinated compound. The Literature value for 2-fluorobutane is 25 C.....while 2-iodobutane is 119 C.

Hope this helps.

Dr. Jim Romano

Hello Dr. Romano, thanks for your help. Another question...so do van der waals forces always beat out dipole forces when determining boiling point unless it's hydrogen bonding? Is there a case where dipole forces are considered more important in determining boiling point?

 

[DAT Destroyer]

Hello Dr. Romano, thanks for your help. Another question...so do van der waals forces always beat out dipole forces when determining boiling point unless it's hydrogen bonding? Is there a case where dipole forces are considered more important in determining boiling point?

No. Dipole Forces usually trump Van Der Waals forces when comparing molecules of similar carbons and molecular weights . For example, Acetone, C3H6O vs Propane, C3H8. Both have Van Der Waals but acetone has a dipole, thus if we look at the boiling point of acetone we find 56C while propane is only -42C. Thus, as a general rule.....Dipole wins over Van Der Waals. However.....what if we compare octane with acetone. Octane is C8H18.....it is MUCH larger than acetone, MW 58 vs 114....thus Van Der Waals would dominate. The boiling point of octane is 125C. Thus, as you can see, this is a very broad question with no answer. As a rule, however,,,,,,If you compare molecules with dipoles vs. molecules with only Van Der Waals,, pick the one with the dipole, UNLESS the molecular weight is Significantly HIGHER as I just showed you previously. For the DAT exam, it will be very simple comparisons. Hydrogen bonding rarely loses to either dipole or Van Der Waals as seen in the fact that H2O boils higher than H2Te..... I hope this helps.