Bonding, Structure, and Resonance

By James Ashenhurst

3 Trends That Affect Boiling Points

Last updated: September 16th, 2024 |

Figuring out the order of boiling points is all about understanding trends. The key thing to consider here is that boiling points reflect the strength of forces between molecules. The more they stick together, the more energy it will take to blast them into the atmosphere as gases.

There are 3 important trends to consider.

  • The relative strength of the four intermolecular forces is: Ionic > Hydrogen bonding > dipole dipole > Van der Waals dispersion forces. The influence of each of these attractive forces will depend on the functional groups present.
  • Boiling points increase as the number of carbons is increased.
  • Branching decreases boiling point.

Let’s have a closer look:.

3 important trends that affect boiling points include intermolecular forces molecular weight and branching

Table of Contents

  1. Trend #1: The Relative Strength Of The Four Intermolecular Forces
  2. Trend #2 – For Molecules With A Given Functional Group, Boiling Point Increases With Increasing Molecular Weight
  3. The Role Of Symmetry (or lack thereof) On Melting And Boiling Points
  4. Notes

1. Trend #1: The relative strength of the four intermolecular forces .

Compare the different butane alcohol derivatives shown below. Molecules of diethyl ether, C4H10 O, are held together by dipole-dipole interactions which arise due to the polarized C-O bonds.

Compare the boiling point of diethyl ether (35 °C)with that of Its isomer 1-butanol (117 °C). The greatly increased boiling point is due to the fact that butanol contains a hydroxyl group, which is capable of hydrogen bonding.

Still, the attractive forces in butanol pale in comparison to those of the salt sodium butoxide, which melts at an extremely high temperature (well above 260 °C) and actually decomposes before it can turn into a liquid and boil.

boiling-points-reflect-intermolecular-attractive-forces-for-example-diethyl-ether-versus-n-butanol-versus-sodium-n-butoxide

Then think about butane, C4H10, which contains no polar functional groups. The only attractive forces between individual butane molecules are the relatively weak Van der Waals dispersion forces. The result is that butane boils at the temperature at which water freezes (0° C), far below even that of diethyl ether.

Moral of the story: among molecules with roughly similar molecular weights, the boiling points will be determined by the functional groups present.

You could tell a similar tale for the similar amine and carboxylic acid isomers shown below.

amine-derivatives-of-similar-molecular-weight-with-compare-boiling-points-dipole-dipole-vs-hydrogen-bonding-vs-ionic

For a previous discussion of the 4 intermolecular forces, see the previous article (The Four Intermolecular Forces And How They Affect Boiling Points)

2. Trend #2 – For molecules with a given functional group, boiling point increases with molecular weight.

Look at the dramatic increases in boiling points as you increase molecular weight in all of these series:

alkanes-versus-alcohols-versus-carboxylic-acid-vs-ether-boiling-points

Here’s the question: How, exactly do intermolecular forces increase as molecular weight increases?

Well, the key force that is acting here are Van der Waals dispersion forces, which are proportional to surface area. So as you increase the length of the chain, you also increase the surface area, which means that you increase the ability of individual molecules to attract each other.

On an intuitive level, you could compare these long molecules to strands of spaghetti –  the longer the noodles, the more work it takes to pull them apart. As the chain length increases, there will be regions where they can line up next to each other extremely well.

Individually, each interaction might not be worth very much, but when you add them all up over the length of a chain, Van der Waals dispersion forces can exert tremendous effects.

3. The Role Of Symmetry (or lack thereof) On Melting And Boiling Points

This is another byproduct of the surface-area dependence of Van der Waals dispersion forces – the more rod-like the molecules are, the better able they will be to line up and bond.

To take another intuitive pasta example, what sticks together more: spaghetti or macaroni? The more spherelike the molecule, the lower its surface area will be and the fewer intermolecular Van der Waals interactions will operate. Compare the boiling points of pentane (36°C) and 2,2-dimethyl propane (9 °C).

branching-decreases-boiling-point-and-polar-functional-groups-most-exposed-will-elevate-boiling-points-to-greatest-extent

It can also apply to hydrogen bonding molecules like alcohols – compare the boiling points of 1-pentanol to 2-pentanol and 3-pentanol, for instance. The hydroxyl group of 1-pentanol is more “exposed” than it is in 3-pentanol (which is flanked by two bulky alkyl groups), so it will be better able to hydrogen bond with its fellows.

In summary, there are three main factors you need to think about when confronted with a question about boiling points. 1) what intermolecular forces will be present in the molecules? 2) how do the molecular weights compare? 3) how do the symmetries compare?

One last quick question for the road (see comments for answer).

three-amine-isomers-will-have-highest-boiling-point-butylamine-vs-diethylamine-vs-n-n-dimethyl-ethylamine


Notes

P.S. New! Check out this free 3-page handout on solving common boiling point exam problems! 

MOC_Boiling_Point_Handout (PDF)

 

Comments

Comment section

114 thoughts on “3 Trends That Affect Boiling Points

  1. As the number of carbon Atoms increases their molecular mass increases and their vanderwalls force of attraction increases and this leads to increases in boiling points of higher hydrocarbon chains

  2. I am trying to figure out if acetic acid will evaporate or not in the drying step of my compound after treating the solution with NaOH to a pH around 8

    1. Is your compound water soluble or organic soluble? Because if it’s soluble in organic solvent, you can avoid worrying about this by just extracting with something like ethyl acetate.

      If your compound is water soluble and you are drying it through rotary evaporation / distillation, the acetic acid and water will evaporate at similar rates. At the end you will have NaOH / NaOAc salts however.

  3. Thank you James.
    Remaining with the same example (in the extremes as you said), if I make the solution of diluted acetic acid completely boil in a beaker as to remove water, what will remain in the beaker? Acetate salts?

  4. Take for example a diluted solution of acetic acid in water. Will the bp of acetic acid change at acidic pH or basic pH?

    1. It’s hard to answer this question without more specific detail.

      In the extremes, it’s easy to answer. For instance if H2O is protonated *completely* with a strong acid, it becomes H3O+ . Being a charged compound it therefore has a very low volatility (this is why H2SO4 is often used to remove water). The same is true when H2O is completely deprotonated to NaOH.

      The same would hold for acetic acid if it was either protonated completely or deprotonated completely. Salts are not volatile.

    1. Is column chromatography an option? That’s usually the main method unless you have a lot of material to separate. There’s a reason refineries are so large, they’re essentially huge distillation operations!

    1. Good question. I’m not surprised 2-butyne has a higher melting point – it’s more rigid (fewer degrees of freedom for a methyl group compared to an ethyl group) and therefore should stack together much more easily than 1-butyne. I suppose the same factors are at play in the boiling point as well – rigid structure leads to greater overall surface contact between molecules, meaning stronger VDW dispersion factors, meaning higher boiling point.

  5. this is very helpful, boiling point started to make sense already by that list at the beginning. before finding this i was looking through my class notes and textbook trying to find out what made a difference in a molecules boiling point and i couldnt find a good answer.using this page i was able to answer a question on my assignment. i’m going to study the printable sheet so that it’s easier next time.

  6. A little confused on this, isn’t boiling point a colligative property? So the boiling point elevation depends only on the number of particles present in a solution, not the nature of those particles? Or does that colligative property factor only apply for ionic bonds – which is why they’re so much higher boiling points compared to the organic molecules.

    1. We’re not dealing with solutes & solutions here, just the boiling points of the liquids by themselves. Each pure compound has a distinct boiling point which depends on the intermolecular attractive forces.

  7. Can you please explain why ethoxy ethane have a lower boiling point than n-pentane?
    What about the higher boiling point of methoxy ethane when compared to n-butane?
    Isn’t it contradictory?

  8. Great resource, even for Organic Instructors! Thank you for making the content concise and easy to understand!

  9. Here, I summarized some boiling point of halobutane alkyl chain by increasing boiling point order. With that in hands, it’s eazier to see the trend of adding different halogen atom even mixing them on a butane alkyl chain.
    Perfluorobutane bp: -7`C
    Butane bp: -0.5`C
    1-Fluorobutane bp: 32`C
    1-Chlorobutane bp: 77-78`C
    1-Bromobutane bp: 100-104`C
    1-Iodo-tert-butyl bp: 118-120`C
    2-bromo-2-chlorobutane bp: 120`C
    2-bromo-1-chloro-2-methylpropane bp: 127`C
    1-Iodobutane bp: 130-131`C
    1-Bromo-4-fluorobutane bp: 134-135`C
    1-bromo-3-chloro-2-methylpropane bp: 153`C
    1-Bromo-3-chlorobutane bp: 154`C
    1,4-Dichlorobutane bp: 161-163`C
    1-Bromo-4-chlorobutane bp: 177-180`C (80-82`C/30 mmHg)
    1,4-Dibromobutane bp: 195-198`C (63-65`C/6 mmHg)
    1-Chloro-4-iodobutane bp: 201-201`C (88-89`C mmHg)
    1,4-Diodobutane bp: 261-267`C (147-152`C/26 mmHg)

  10. How chlorine will affect the boiling point of a alkyl chain for exemple 1-chloro-octane compare to octane?
    And for 1-bromo-octane?
    And for 1-bromo-8-chloro-octane?

  11. Nice, really appreciate your broadness without sticking to a job. Society might wish in a bigger way than what you had opted earlier…One thing Dear professor James , when we are discussing about boiling point , I have to say that, does the factors like sterically hinderedness of a molecule come to scenario or we shall ignore about the effects.

    Let you know that I am working as an assistant professor in the University, India. Welcome you and your page with open arms.

  12. I want to know, how water content in crystallized substance affects its melting point. The main reason behind this question is that after synthesizing benzoic acid, its melting point was found to be high, i had to give certain errors that increase the % yield. So, to verify that water was present in the solid crystals, If i can justify that melting point was higher than expected due to presence of water, the error of increased % yield will be justified

  13. Why does trans but-2-ene has higher melting point but lower boiling point than cis but-2-ene? Is this because of dipole moment?

  14. Hey there, just wanted to stop by to say how much I appreciate this website. You have an exceptional ability to present material in a concise, understandable fashion. I wish you were the author of my organic chemistry book or my professor, because this year I have been teaching myself organic chemistry using your website and my textbook (lecturer is awful) and received an A+ in ochem 1. Thank you!

  15. Why does NO2 have a greater boiling point than SO2? Both are polar, dipole-dipole interactions, but SO2 has a greater dipole moment, greater molar mass, and greater surface area. I know NO2 is a radical, does that affect boiling point?

  16. Thank you! I’m studying chemistry at my college and this was very helpful! I normally hate and can barely understand chemistry but this made it very clear and easy to understand so thank you :)

  17. I am having a little problem in understanding the melting point trends……why the following melting point increase is uneven……….formic acid(m.p 8.4) then ethanoic acid(m.p 16)…but then propanoic acid (m.p -21)…..then pentanoic acid (m.p -35)…………………It seems appropriate to understand boiling point trends which increase with molecular size as van der waals forces incrase but how does melting point trend works?

    Thanks in advance!

  18. I am also taking Orgo I and came across a problem I couldn’t figure out. Given 3 alcohols : tert-butyl, sec-butyl, and n-butyl with their respective boiling point measurements at : 82C, 100C, and 117C. 4 carbons, same molecular weight for all them, same intermolecular forces? What creates the differences in their boiling point values? Is it the location of the hydroxyl groups?

    Thanks for your help!

    1. The key factor will be surface area. The linear chains have a higher surface area than the branched alkyl groups (which are more “spherical”) and since van der waals forces are proportional to boiling point, the linear alkyl groups should have higher boiling point.

  19. Thanks James! extremely helpful. Now iam clear about the concepts. Please do keep posting such concepts.

  20. What a great summary, and really reafable. I’ve been struggling to get to grips with Chemistry as part of a course I’m doing. This is the best site I’ve found, thank you for putting things in terms I truly understand!

  21. I cannot not leave a comment. Thank you so much :)) I’m currently taking up Organic Chemistry and this has helped a lot!

  22. Determine the order with lowest boiling point on the left:

    A = (CH3)3CCH2CH2CH2CH2CH2C(CH3)3
    B = CH3CH2C(CH3)2C(CH3)2C(CH3)2CH2CH3
    C = (CH3)2CHCH2C(CH3)2C(CH3)2CH2CH2CH3

    All three have the same number of carbons and the same number of branch points. Help!

    1. Branching reduced the ability of a molecule to ‘stack,’ making its state less solid and its boiling point lower.

    1. In this case we’d be comparing two different variables (chain length *and* polarity of functional groups) and it is difficult to predict which factor is more important. I would look for examples of compounds you are curious about and compare their known melting points/boiling points.

    1. It’s probably way too late..but no, it isn’t the same thing. Dipole-induced dipole is describing the attractive forces between a polar molecule (dipole) and an induced dipole (dispersion, london, van der waals). Instead of looking at one isolated force (such as disperson), dipole-induced dipole is describing two that are attracted to eachother…temporarily in the case of induced dipole.

  23. Thank you so much, i’ve been looking for something clear and easy to understand about trends in boiling points and this has answered so many of my questions! Keep up the fantastic work.

  24. do we only can compare the molecular size and molecular shape for the similar molecules or we can compare the size and shape for molecules with different functional groups ? for example can i compare the size and shape of pentanoic acid with heptane ?

    1. It’s tricky to compare two variables (chain length and functional group) at once. Generally you just want to know the impact of each variable individually.

  25. Hi thanks for this info helps a lot but I have a question. It concerns organic compounds. Say you have a ketone such as heptanone. So the boiling point of this compound should be relatively high because it has a large surface area and it is a polar molecule so there are dipole -dipole forces present. Now take a alcohol that is a shorter chain such as propanol. Alcohol should have a higher boiling point because it has a hygrogen bond. But the propanol molecule is a much shorter chain so it has a smaller surface area than the heptanone. So how can we work out which one has a higher boiling point? (In this case heptanone has a higher boiling point than propanol but I would like an explanation as to why.)

    1. Hi Lil, it’s difficult to compare two variables at once because it’s hard to gauge the relative importance of each variable from first principles alone [this is what experiments are for]. In the example mentioned, comparing heptanone and propanol is difficult because we’re comparing two variables at once: chain length and the polarity of the functional group. The key takeaway from this article is that if all variables except one are the same, you can predict boiling points based on these concepts. You’ll find that exams will generally test one variable too.

  26. If we have a molecule with hydrogen bond and one alkane. If alkane has greater molecular weight than hydrogen bond, which on has highest boiling point?

    1. You’re comparing two different variables there; chain length (8 carbons vs. 5 carbons) and polarity (chlorine is electronegative). It’s hard to predict how boiling points are affected when you change more than one variable.

    1. Higher molecular weight is one thing. C-H does not count as a “hydrogen bond’ because carbon is not electronegative enough. Only O-H, N-H and F-H bonds participate in hydrogen bonding.

  27. this was quite insightful, but i have a problem i was given a question to compare pentane and diethyl ether. Based on the information i have seen it looks like pentane has a higher boiling point than di-ethyl ether, based on actual BP but your information tells me that it should be otherwise as diethyl ether has dipole-dipole interactions and pentane has van der waal forces. Could you clear up for me which should actual have the higher BP and give a reason for the answer?

    1. Excellent point! I don’t have a good answer as to why pentane’s BP (36 deg C) is higher than Et2O(34.5 deg C). Interestingly the boiling point of tetrahydrofuran (C4H8O) is 66 deg C, because the ring gives it a permanent dipole moment.

    2. pentane has higher bp due to more no. of C atoms.
      but butane has lower bp than diethyl ether due to the reason you’ve mentioned

  28. This is an amazing reference that simplifies properties affecting boiling point, which can be confusing when it comes to the many factors that can come into play!! It does a much better job than my professor, who’s really good, or the book were able to do in demystifying all of the above. Thank you to whoever made it!!

  29. Can somebody help me with boiling points of these substances, (just comparison)
    CH3 NH2, CH2 F2, CH3 OH, CH3 CH2 CH2 CH3

  30. Thanks. This piece is very educative.
    (1) Why does butanol (bp- 117) have a higher boiling point than butanamine (bp-77.8)
    (2) How does intramolecular Hydrogen bond affect the boiling point of an organic molecule

    1. Good question. It is likely due to the greater electronegativity of oxygen (3.4) versus nitrogen (3.0) leading to a larger dipole, which means that the molecules will have a greater force holding them together.

  31. thanks i found answers to my questions.. one thing ,, how does functional group affects boiling point in most understandable way ? thanks

    1. If you had to focus on one thing, I’d look for hydrogen bonding groups such as OH and NH. This will increase the boiling point significantly.

  32. gen chem 1 sin city. this site is really helpful. it’s like playing tennis with a much better player; you can only learn and get better from a pro if ya want too! Thanks! Great stuff

  33. Answer to question in post: the amine with the NH2 will have highest boiling point (most hydrogen bonds) followed by NH and then the molecule on the farthest right.

  34. How does branching affect melting point?
    I get the fact that branching decreases bp because of the decrease in surface area, hence less opportunities for Van de Waals interactions. However, how does it work for melting point? When it is branched it is harder to stack, right? Then, a lower melting point?
    Thanks! :)

    1. Branching actually increases the Melting point of a molecule because it decreases the surface area causing it to become more tightly packed.

      However, it decreases the BP because less Van de Waal interactions are able to occur.

  35. I’m having a problem with the branching rule you pointed out in this article. Wouldn’t branching increase the boiling point as it leads to the molecule’s shape being more spherical and tightly packed. So with a tightly packed molecule, it takes more energy to take it apart thus, branching should increase b.p.

    1. More tightly packed would give it a higher melting point. But the lessened surface area would result in a lower boiling point due to lessened Van der Waals interactions.

    2. i think there is a conceptual problem ,,,, b.p doesn’t means the attraction ,,, or packing of a molecule,,, its a physical property,,, hence depends on the interaction b/w the MOLECULES OF A COMPOUND ,,,now,,, as it is spherical in nature ,, hence the area of contact decreases ,,,,,and therefore,,,, interaction becomes weaker ,, and hence b.p decreases ………………
      for instance —– we have a glass of water,, in which there r many H2O,, molecules,,, here we r not talking abt ,,,, h-o bond,, its abt the interaction b/w 2 H2O molecules ,,,,,,,wat we generally says , intermolecular forces ……..

      hope u r now, clear with ur concept …………

  36. 1.ionic bonding is not an intermolecular force; it is an intramolecular force? network covalent, ionic, metallic… -inorganic compounds!
    2. there is something called optical isomerism. R and S enantiomers’ mp and racemate’s mp are different; important stuff if you’re mastering organic chem.

    1. why can’t it be both? I’m thinking mostly of salts of organic compounds like Me4N+ Cl- which have very high melting points as opposed to network solids like NaCl or other inorganic compounds which are not really “molecules” per se.
      thanks for mentioning the R/S as well, that’s an important point!

    1. Can somebody help me about this? Compare boilnig points of
      a) NH3 b)H2O2 c)H20
      a)CF4 b)CCL4 c)CH4
      a)H2O b)H2S c)SIO2

          1. it depends on the extent of h-bonding not on strength nh3 can form 3 h bonds while h2o can only form 2

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