Structure of an isomer would cause it to auto-ignite easily?

[The Bob]

Just a quick one guys.

I have been revising and I have come across a question that I thought I knew the answer to.

Q. There are many isomers of C₁₆H₃₄. What features of the structure of an isomer would cause it to auto-ignite easily?

I thought that isomers with 'branches' (e.g. methyl- etc. groups) would have a lower boiling point and so react easier in a combustion engine. However, the answer say that straight chain hydrocarbons are more easily auto-ignited.

I was simply wondering why this was the case or if the answers are incorrect.

Cheers.

The Bob (2004 ©)

 

[Gokul43201]

This is something I've wondered about too. It is well-known that straight chain HCs have a lower ignition temperature, and hence, poorer octane rating (recall 'knocking' and n-heptane)...but it doesn't appear obvious why this is true.

 

[GCT]

The straight chain alkane occupies a higher level on the PE diagram, that is it is relatively unstable, and it takes relatively less activation energy for the process to occur than as would a more stable compound which occupies a lower position on the PE diagram.

I have searched throghout my org. text regarding the issue of stability of alkanes, and it is not explained too throughouly. Some parts of it were especially interesting, such as that of the trend in melting points. Basically I found that the chemistry of alkanes is not very straightforward. The text claims the relative stability of isobutane over butane, however it does not give any reasons as to why that is as it does with other chemical groups (alcohol, alkenes, etc...).

The potential energy in this case would refer to mainly the bonds. Perhaps a more substituted carbon structure would offer some sort of hyperconjugation/delocalization stability.

 

[The Bob]

So aliphatic alkanes are less stable so are more ideal for auto-ignition but have a higher boiling/melting point than branched alkanes?

Ok. If that is what it is then fine.

The Bob (2004 ©)

 

[GCT]

less stable as in that it will react faster.

What makes you think that higher boiling/melting point has anything to do with this reaction process? The former has to do with intermolecular phenomena, while the topic you posed relates more to potential energy of the intramolecular bonds.

 

[Gokul43201]

What makes you think that higher boiling/melting point has anything to do with this reaction process?

But it should (at least in my opinion). Ignition is largely (at least in automobile engines) dominated by the reaction of the vapor. The temperature at which the vapor pressure is sufficient to achieve the required vapor:air ratio (for combustion) is the ignition temperature.

I would imagine that the relative instability of long chain alkanes would lead to easier "cracking" or breaking up into shorter chains. This is known to be true. I do see how this instability could lead to a lower combustion (or ignition) temperature...but I'm not entirely convinced yet (maybe it's just me).

 

[GCT]

But it should (at least in my opinion). Ignition is largely (at least in automobile engines) dominated by the reaction of the vapor. The temperature at which the vapor pressure is sufficient to achieve the required vapor:air ratio (for combustion) is the ignition temperature.

perhaps, I'm not too familiar with the engineering aspects, however, I would still imagine that the straight chain would be preferred over the brached; as far as vapor is concerned both isomers would vaporize readily at room temperatures or higher (propane's boiling point is -42.06 C)...at least for the relatively lower molecular weight alkanes.

I would imagine that the relative instability of long chain alkanes would lead to easier "cracking" or breaking up into shorter chains. This is known to be true. I do see how this instability could lead to a lower combustion (or ignition) temperature...but I'm not entirely convinced yet (maybe it's just me).

I'm don't have a complete understanding of the relative stability of the isomers (as I said before, perhaps due to some sort of hyperconjugation/delocalization). The unstable isomer has more PE associated with it's bonds (which can be explained through QM, although I'm not aware of the exact answer), the net enthalpy of reaction is more exothermic. It seems to me that the activation state (transition state) of both isomers is more or less the same, and thus the isomer which is closest to energy in this transition state (which would be the more unstable isomer with a higher PE) would have the lower activation temperature. I'll need to research to be sure.