Does Mixing 2-Propanone and 2-Ethylhexanol Create C11 H24 O2?

[manhattan1nyc]

If I add the chemicals C3 H6 O (2-propanone) and C8 H18 O (2-ethylhexanol)
will I wind up with the chemical formula C11 H24 O2 ?

thanks

 

[mrjeffy321]

Are you asking if you simply mix the two substances together does the new, combined chemical formula become the chemical formula for one of them, plus the chemical formula for the other?

2-Propanol: C3H8O
2-Ethyl Hexanol: C8H18O

C3H8O + C8H18O = C11H26O2
(Above written in the mathematical sense, not in the chemical [equilibrium] reaction sense.)

No, it does not work like that.
You would simply have a mixture (a solution in this case) of one in the other.
If you re-write the chemical formula like that you are saying that a completely new substance is formed
Unless a chemical reaction occurs between the two substances they will stay in their original form.

 

[manhattan1nyc]

In the example I used it was C3 H6 O (2-propanone), I think you may have thought I meant C3 H8 O (2-Propanol).

So if I mixed the 2 chemicals: C3H8O and C8H18O it would NOT be the same as using the chemical C11H26O2 ? In other words I could NOT substitute the chemical C11H26O2 for the solution of C3H8O and C8H18O ?

 

[mrjeffy321]

In the example I used it was C3 H6 O (2-propanone), I think you may have thought I meant C3 H8 O (2-Propanol).

Oh, that you did, I misread it.

So if I mixed the 2 chemicals: C3H8O and C8H18O it would NOT be the same as using the chemical C11H26O2 ? In other words I could NOT substitute the chemical C11H26O2 for the solution of C3H8O and C8H18O ?

Correct, it would NOT be the same.

 

[chemisttree]

If I add the chemicals C3 H6 O (2-propanone) and C8 H18 O (2-ethylhexanol)
will I wind up with the chemical formula C11 H24 O2 ?

thanks

Yes you will, to some degree. The reaction is known as an alcoholysis reaction of ketones. It will produce some of the hemiketal of acetone and 2-ethylhexanol and will have the empirical formula that you indicated. If two equivalents of the alcohol are added, the ketal will be produced (empirical formula $$C_{19}H_{40}O_2$$)

 

[mrjeffy321]

Yes you will, to some degree.

It will? How much is “to some degree”? Is it some tiny equilibrium amount, or is a significant reaction going on?
This is why I don’t do any organic chemistry.

 

[Cesium]

I believe ketalization requires an acid catalyst so just mixing the two chemicals would, practically speaking, give 0 of the hemiketal/ketal. You also need to distill off the water that is produced as a by-product because this adversely affects the equilibrium.

 

[chemisttree]

I believe ketalization requires an acid catalyst so just mixing the two chemicals would, practically speaking, give 0 of the hemiketal/ketal. You also need to distill off the water that is produced as a by-product because this adversely affects the equilibrium.

There is no water produced in the hemiketal (the OP) example. Hemiketals are produced in some quantity by mixing the alcohol with the ketone without acid catalysis. The equilibrium constant strongly favors the ketone and the alcohol. In the example of 1 M acetone in methanol (no acid, RT), the equilibrium concentration of the hemiketal has been measured at approximately 7 mM.

 

[chemisttree]

It will? How much is “to some degree”? Is it some tiny equilibrium amount, or is a significant reaction going on?
This is why I don’t do any organic chemistry.

Yes, it is some small equilibrium amount. Its a shame that you don't do any organic chemistry. Your answer was essentially correct.

 

[Cesium]

In the example of 1 M acetone in methanol (no acid, RT), the equilibrium concentration of the hemiketal has been measured at approximately 7 mM.

Well 7mM is essentially nothing (it is undetectable for someone like me without special equipment). I bet you that with the bulkier 2-ethylhexanol the equilibrium amount would be even less.

There is no water produced in the hemiketal (the OP) example.

It must be with ketals that water is formed then, because I remember reading procedures where it was key to distill off water to shift equilibrium.

Just by using one equivilant of alcohol, is it possible to easily isolate the hemiketal? I am guessing that you would always have some ketal formation which would then have to be separated in some way. What's the standard procedure for this?

 

[chemisttree]

Well 7mM is essentially nothing (it is undetectable for someone like me without special equipment). I bet you that with the bulkier 2-ethylhexanol the equilibrium amount would be even less.

Absolutely!

It must be with ketals that water is formed then, because I remember reading procedures where it was key to distill off water to shift equilibrium.

Quite correct. The ketal product is usually favored thermodynamically over the ketone.

Just by using one equivilant of alcohol, is it possible to easily isolate the hemiketal? I am guessing that you would always have some ketal formation which would then have to be separated in some way. What's the standard procedure for this?

My point regarding adding two equivalents of alcohol to the ketone producing the ketal was poorly worded. I meant that if two equivalents of alcohol are added to the ketone, the ketal is produced. It isn't quite the same as saying that the ketal will form if two equivalents of alcohol are present with one equivalent of ketone.

To answer your question, no, as you isolate the hemiketal, it will spontaneously decompose to produce an equilibrium mixture of hemiketal, alcohol and ketone. The solution that it was isolated from will spontaneously form additional hemiketal... right up to the equilibrium point. The standard procedure to produce ketals varies according to the materials under consideration. As you indicated in your earlier post, ketals are produced under forcing conditions that can include high pressure/temperature and removal of one or more of the products such as water. Acid catalysis speeds the reaction along and is generally employed.

Can you isolate the hemiketal? I suppose that you could isolate the hemiketal through some derivatization scheme such as conversion into a silyl ether. It would be tricky...

 

[Cesium]

Thanks chemisttree, very informative answer!

 

[manhattan1nyc]

Thank you Chemisttree,

do I have to add equal amount of C3 H6 O (2-propanone)
and C8 H18 O (2-ethylhexanol) to form C11 H24 O2 ? or willa solution for example of 80% C3 H6 O (2-propanone) and 20% C8 H18 O (2-ethylhexanol)
form C11 H24 O2 ?

 

[chemisttree]

Thank you Chemisttree,

do I have to add equal amount of C3 H6 O (2-propanone)
and C8 H18 O (2-ethylhexanol) to form C11 H24 O2 ? or willa solution for example of 80% C3 H6 O (2-propanone) and 20% C8 H18 O (2-ethylhexanol)
form C11 H24 O2 ?

That mix will form some of the hemiketal, but as Cesium and Mrjeffy already noted, it will not form much.