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Topic: Boiling point of cyclic esters  (Read 805 times)

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Offline Cormzy

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Boiling point of cyclic esters
« on: April 13, 2019, 12:24:35 PM »
Hi I have been struggling on this question for quite a while, I was wondering if somebody to explain why cyclic esters (dihydrofuranone) have a higher boiling point than normal esters (ethyl ethanoate) when both are of a similar rmm.

My only guess would be that cyclic esters are less branched and therefore the dipole attractions between the carbonyl groups are stronger.
Thanks.

Offline Enthalpy

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Re: Boiling point of cyclic esters
« Reply #1 on: April 14, 2019, 07:21:55 PM »
One carbon more.

Offline pgk

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Re: Boiling point of cyclic esters
« Reply #2 on: April 15, 2019, 10:55:46 AM »
The ring strain of cylic esters increases their vaporization enthalpy and consequently, it increases their boiling point, too.

Trouton's rule
https://en.wikipedia.org/wiki/Trouton%27s_rule
Ring strain (Baeyer strain)
https://en.wikipedia.org/wiki/Ring_strain

Offline rolnor

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Re: Boiling point of cyclic esters
« Reply #3 on: April 15, 2019, 11:06:48 AM »
does not the ringcompounds have stronger dipole-moment?

Offline pgk

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Re: Boiling point of cyclic esters
« Reply #4 on: April 15, 2019, 11:17:43 AM »
Other factors such as dipole moment and conformers/epimers energy also contribute to the increasing of vaporization enthalpy but the ring strain is the most significant one.
PS: You can check it by applying the ring strain of a five member ring in the Trouton’s rule for γ-butyrolactone in comparison with ethyl acetate.
A simplified form of the Trouton's rule is:
Hv = 21Tb, where Hv is measured in kcal/mol and Tb in oK.
« Last Edit: April 15, 2019, 01:17:43 PM by pgk »

Offline rolnor

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Re: Boiling point of cyclic esters
« Reply #5 on: April 15, 2019, 02:39:12 PM »
I am thinking about THF, it has much higher boilingpoint then diethyl ether, also its water miscible, is not the main reason for this its higher dipole moment?

Offline pgk

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Re: Boiling point of cyclic esters
« Reply #6 on: April 15, 2019, 03:27:11 PM »
Of course and in addition to the ring strain, dipole moment and other factors play an important role in the vaporization enthalpy, as mentioned above.
Anyway, the comparison between γ-butyrolactone and ethyl acetate is an easily comprehensible example of the influence of vaporization enthalpy to the boiling point of cyclic compounds. However, this is more complex because the Trouton’s law does not work well in case of inter/intramolecular interactions, dimers formation, hydrogen bonding, etc. because additional energy is needed for vaporization.
Please, note that the boiling point is the temperature that vapor pressure is equal to 1 atm. But the higher the vaporization enthalpy the lower the vapor pressure at a given temperature is. On the other hand, vapor pressure dependence within temperature is not proportional by a proportion coefficient but it follows the Clausius-Clapeyron equation:
ln(P2/P1) = Hv/R(1/T2-1/T1)
Nevertheless, there is no need to dive in such deep waters of physical chemistry, in order to clarify the issue.
« Last Edit: April 15, 2019, 04:13:53 PM by pgk »

Offline rolnor

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Re: Boiling point of cyclic esters
« Reply #7 on: April 15, 2019, 04:56:12 PM »
Pentane has boilingpoint 36°C, cyclopentane 50°C so here there is no dipole difference.
Diethyl ether has 34°C and THF 66°C so here there is partialy an dipole effect, partialy ringstrain effect?

Offline clarkstill

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Re: Boiling point of cyclic esters
« Reply #8 on: April 16, 2019, 03:10:49 AM »
Is entropy not likely to be a large factor here? The cyclic molecules have minimal rotational and translational degrees of freedom so a smaller amount of entropy is lost when they are condensed from the vapour phase, whereas the acyclic molecules have a much larger decrease in disorder? That said, I'm an organic chemist so have forgotten most of the physical chemistry I ever knew, so this could be utterly misguided.

Offline pgk

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Re: Boiling point of cyclic esters
« Reply #9 on: April 16, 2019, 11:26:18 AM »
By the chance, let’s try to put everything in order!
Vaporization energy Ev follows the Gibbs free energy equation:
Ev = Hv-TS → Hv = Ev + TS
Up to phase transition and if other parameters do not change (e.g. thermal rearrangement, degradation, polymerization, etc.), entropy numerically coincides with the specific heat though they have completely different physical meanings.
On the other hand, vaporization energy is the overall energy needed, in order to destroy all kind of forces and intermolecular interactions in the liquid state, such as:
- Gravitational forces, which are proportional to the molecular mass.
- Energy changes due to changes of heat distribution within atoms and functional groups, depending on the molecular size, shape, as well as on the flexibility/ rigidity of the molecule and thus, indirectly related with entropy.
- Surface energy per active surface area, which coincides with the gas-liquid surface tension and which also changes within temperature (Eötvös rule).
- Temporary van der Waals forces (Keesom forces) = 0.1-0.5 kcal/mole depending on the size and shape of the molecules.
- Permanent Van der Waals forces (London Forces) = 0.3-3.0 kcal/mole depending on the dipole moment.
- Hydrogen bond ≈ 5 kcal/mole but may be higher if fluorine atoms participate.
- Ionization energy (e.g. carboxylic acids), which also follows a Gibbs equation.
Ei,o = 2.303RTpka
- Activation energy of the thermal reactions during evaporation (e.g. peroxides formation during THF distillation, which also explains why never distilling ethers to dryness).
- Ring strain of cyclic molecules, which is negligible for a six member ring but may reach up to 28 kcal/mole for a three member ring (e.g. epoxides).
- Energy difference between conformers and epimers.
- Etc...
But the most important is that neither vaporization enthalpy nor specific heat, are constant within the overall range of temperature changes. They can be considered as real constants within a narrow range of temperature, only.
As a conclusion, the boiling point macroscopically depends on the nature, size and the particular structure of the molecule. Thus, general rules for comparison of boiling points between molecules with the same atom number, may be unsuccessful; regardless if the corresponding literature and software applications already exist.
« Last Edit: April 16, 2019, 03:02:45 PM by pgk »

Offline rolnor

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Re: Boiling point of cyclic esters
« Reply #10 on: April 16, 2019, 11:41:49 AM »
But the molecules in my example are simple, you dont have hydrogen bonding etc.?

Offline pgk

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Re: Boiling point of cyclic esters
« Reply #11 on: April 16, 2019, 11:47:50 AM »
ROCH2R’ + O2 + light and/or heat  →  ROCH(R’)OOH 

ROCH(R’)OOH + ROCH2R’  →  ROCH(R’)OOH---O(R)CH2R'
« Last Edit: April 16, 2019, 02:12:46 PM by pgk »

Offline rolnor

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Re: Boiling point of cyclic esters
« Reply #12 on: April 16, 2019, 01:30:53 PM »
I dont understand this, is it reply to some other topic?

Offline pgk

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Re: Boiling point of cyclic esters
« Reply #13 on: April 16, 2019, 02:07:00 PM »
This is the reaction scheme of ethers autoxidation in air (THF included), leading the corρesponding α-O-hydroperoxides and which form H-bond dimers with their maternal ethers, in situ.
In a similar way, ethers α,α'-O-dihydroperoxides are also formed that lead to H-bond trimers with their maternal ethers, too.
Diethyl ether hydroperoxide
https://en.wikipedia.org/wiki/Diethyl_ether_peroxide
Hydroperoxide
https://en.wikipedia.org/wiki/Hydroperoxide
« Last Edit: April 16, 2019, 02:39:01 PM by pgk »

Offline rolnor

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Re: Boiling point of cyclic esters
« Reply #14 on: April 16, 2019, 04:17:52 PM »
Yes but what has this to do with the boilingpoint?

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