I'm asked which one of the alkane has the lowest heat of combustion (note: the question says heat, not enthalpy) and according to the answers, the right one is d.
I'm trying to understand why d is the right answer but with no success. Please help me understand the concept behind it for future problems.
Branched-chain alkanes have lower values of ΔcH⊖ than straight-chain alkanes of the same number of carbon atoms, and so can be seen to be somewhat more stable. https://en.wikipedia.org/wiki/Alkane#Branched_alkanes
The heats of formation are given in https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation and go from -40.0 kcal/mol for a to -41.8 for b to -42.5 for c to -44.5 for d. Compound d has the largest heat of formation, therefore will have the smallest heat of combustion.
The explanation is complicated. Sometimes you have to wonder why teachers spend so much time on effects that are so minor. The data are interesting, in a trivial sort of way. Show the data; it's interesting; move on.
Highly branched alkanes may stretch the bond angle. This can cause steric hindrance, and can substantially increase the reactivity. "Substantially". That's interesting. In addition, the reactivity of highly branched alkanes can be used to increase the octane rating of gasoline, because the stability of highly branched free radicals slows down the combustion and avoids explosions in the cylinder (pinging). Substantial is the key word, whether it it is reactivity or stability. The difference between 40 and 44.5 is hardly substantial. I would call that difference about half of substantial.
Wikipedia goes on: "However, in general and perhaps surprisingly, when branching is not extensive, branched alkanes are actually more thermodynamically stable than their linear (or less branched) isomers. For example, the highly branched 2,2,3,3-tetramethylbutane is about 1.9 kcal/mol more stable than its linear isomer, n-octane. Due to the subtlety of this effect, the exact reasons for this rule have been vigorously debated in the chemical literature and is yet unsettled. Several explanations, including stabilization of branched alkanes by electron correlation, destabilization of linear alkanes by steric repulsion, stabilization by neutral hyperconjugation, and/or electrostatic effects have been advanced as possibilities. The controversy is related to the question of whether the traditional explanation of hyperconjugation is the primary factor governing the stability of alkyl radicals." Surprisingly subtle and unsettled. Not really a great teaching topic.
The question regularly pops up in secondary school exams. Why? The reasoning seems unsettled. How to answer the question? The teacher must have given the answer if the teacher is at all competent and meant to provide the reasoning; so regurgitate the correct answer without reserving a special place in your memory for it. If the teacher did not provide the answer, then you must do your own research, in advance of the test - first, to discover what kind of questions your teacher is likely to ask (can you search out previous year's tests?) and second, to research the possible answers.
If you do well at this, you may be asked to teach the next class.
Although I am conversant with this aspect of thermochemistry, it is not my area of expertise. Nonetheless, your question as to why branching leads to a lower heat of combustion in a set of alkane constitutional isomers is one that has intrigued me for sometime. So much so that, after asking former peers steeped in physical organic chemistry, they were at a loss for a cogent, satisfying explanation. I have not researched this topic recently but in the years 2006-2009 a "dispute" between Gronert1,3 and Schleyer2 et al. was ongoing to account for the phenomenon.
The introduction to Gronert's first paper states the following:
"A new model based on 1,3 repulsive steric interactions (geminal repulsion) is proposed for explaining the variation in the C-H bond strengths of the alkanes. The model builds from the assumption that 1,3 repulsive interactions are the major factor in determining the stability of a C-C or C-H bond in an alkane. From this simple premise, the model successfully reproduces the effect of branching on the stability of alkanes, alkyl radicals, and alkenes. The results suggest that geminal repulsion can provide a simple, unified explanation for these fundamental stability trends..."
On the contrary, the Schleyer paper states:
"The most direct conclusion is that 1,3-alkyl–alkyl interactions stabilize hydrocarbons, not only the branched, but also the linear isomers." and "We define “protobranching” as the net stabilizing 1,3-alkyl-alkyl interactions existing in normal, branched, and most cycloalkanes, but not in methane and ethane."
Subsequently, Gronert's second paper counters with:
"Though expounding upon the potential ramifications of the protobranching
premise, the authors supply no clear explanation of the theoretical basis of the proposed interaction and provide virtually no data to support it despite the fact that the proposed interaction has been controversial for decades..."
If someone out there has a definitive answer, I'd be delighted to hear about it.
1) S. Gronert, An Alternative Interpretation of the C-H Bond Strengths of
Alkanes, J. Org. Chem., 2006, 71, 1209-1219.
2) M. D. Wodrich, C. S. Wannere, Y. Mo, P. D. Jarowski,
K. N. Houk, and P. von R. Schleyer, The Concept of Protobranching and Its Many Paradigm Shifting Implications for Energy Evaluations, Chem. Eur. J., 2007, 13, 7731 – 7744.
3) S. Gronert, The Folly of Protobranching: Turning Repulsive Interactions into Attractive Ones and Rewriting the Strain/Stabilization Energies of Organic Chemistry, Chem. Eur. J., 2009, 15, 5372 – 5382.
It is well known fact that branched-chain alkanes are more stable than the straight-chain alkanes with the same number of carbon atoms. That means branched-chain alkanes have higher values of enthalphy of formation ($\Delta H_f^\circ$) than that of straight-chain alkanes with the same number of carbon atoms.
During the complete combustion reaction with oxygen, all alkanes give off $\ce{CO2 (g)}$ and $\ce{H2O (g)}$ regardless of they are straight-chain or branched-chain:
$$\ce{C_nH_m + (n + m/2) O2 -> n CO2 + m H2O}$$
The heat of combustion, $\Delta H_c^\circ$, is equals to: $n\left(\Delta H_f^\circ \right)_\ce{CO2} + m\left(\Delta H_f^\circ \right)_\ce{H2O} - \left(\Delta H_f^\circ \right)_\ce{C_nH_m}$ (keep in mind that for this question, the difference between heat and entalphy really does not make a difference).
Thus, the heat of combustion is larger if the heat of formation of initial alkane is lower (less stable).
Estimation of heats of formation of organic compounds by additivity methods introduced by Benson and coworkers (Ref.1 & 2) is well suitable for OP's level of education to estimate heats of formation of given compounds. This method allows the rough calculation and prediction of heat of formation of organic compounds based on additivity. According to article in Wikipedia:
Starting with simple linear and branched alkanes and alkenes the method works by collecting a large number of experimental heat of formation data (see: Heat of Formation table) and then divide each molecule up into distinct groups each consisting of a central atom with multiple ligands. To each group is then assigned an empirical incremental value which is independent on its position inside the molecule and independent of the nature of its neighbors (in $\pu{kcal/mol}$ at $\pu{298 K}$):
- P primary $\ce{C-(C)(H)3}: -10.00$
- S secondary $\ce{C-(CH2)C}: -5.00$
- T tertiary $\ce{C-(CH)C2}: -2.40$
- Q quaternary $\ce{C-(C)4}: -0.10$
- gauche correction $: +0.80$
- 1,5 pentane interference correction $: +1.60$
Let's apply these rules into following examples:
$$ \begin{array}{c|cc} \text{Alkane} &\ce{C-CH3} &\ce{C-CH2-C} &\ce{C-CH-C2} &\ce{C-C4} & gauche\text{ corr.} & \text{Tot.}& \text{expl.}\\\hline n\text{-pentane} & -10 \times 2 & -5 \times 3 & \text{-} & \text{-} & \text{-} & -35 & -35.1\\ \text{2-methylbutane} & -10 \times 3 & -5 \times 1 & -2.4 \times 1 & - & +0.8 \times 1 & -36.4 & -36.9\\ \text{2,2-dimethylpropane} & -10 \times 4 & \text{-} & \text{-} & -0.1 \times 1 & \text{-} & -40.1 & -40.1\\ n\text{-hexane} & -10 \times 2 & -5 \times 4 & \text{-} & \text{-} & \text{-} & -40.0 & -40.0\\ \text{2-methylpentane} & -10 \times 3 & -5 \times 2 & -2.4 \times 1 & \text{-} & +0.8 \times 1 & -41.6 & -41.8\\ \text{3-methylpentane} & -10 \times 3 & -5 \times 2 & -2.4 \times 1 & \text{-} & +0.8 \times 1 & -41.6 & -41.1\\ \text{2,3-dimethylbutane} & -10 \times 4 & \text{-} & -2.4 \times 2 & \text{-} & +0.8 \times 2 & -43.2 & -42.5\\ \text{2,2-dimethylbutane} & -10 \times 4 & -5 \times 1 & \text{-} & -0.1 \times 1 & +0.8 \times 2 & -43.5 & -44.5\\\hline \end{array} $$
By these calculations, it is evident that 2,2-dimethylbutane (d) has the largest heat of formation among hexane isomers, and hence least heat of combustion, which agrees with the experimental values (last column). Yet, keep in mind that this method is an empirical calculation, but suitable for OP's level of education.
References:
This question has already had some pretty good answers by user55119, James Gaidis and so on. However, since the bounty setter was looking for a definitive answer in pertinence with some tests, and I am quite well-versed with the particular exam he is interested in(i.e. the JEE), I would like to provide my final insights along with a summary of the concerned parts so that this thread may be successfully closed.
As it has been pointed out above repeatedly, it's common knowledge that branched alkanes are somewhat more "stable" than their straight-chained counterparts and hence have lower heats of combustion(and it is the knowledge of this fact that this question is trying to judge). However, this variation is quite less, as James has nicely illustrated in his answer by quoting the relevant enthalpy values for an isomeric series.
The reason for this subtle variation in stability has been a matter of debate, and there is no unanimous answer for this. However, a few plausible theories have been discovered, and user55119 has done a remarkable job in providing excerpts and links to the research papers on this topic.
As far as I know, the people who design such exam questions usually do so by keeping any one of the possible explanations in mind. Most commonly, they will keep the idea of the greater steric crowding in case of branched alkenes and the somewhat greater bond energy of 1° C-H vs the 2° C-H.
Another theory which is slightly lesser common in such exam-setters opinion is a possibility of a hyperconjugative stabilisation in alkanes, the idea of "protobranching" like was stated above. But again, the lack of a collective agreement on one of these conjectures makes it tough to look for a logical rationale behind such questions.
Again, my personal take on this situation would be that, as far as these questions are considered, one can keep the fact I stated above as an absolute certainty for solving them. As for the underlying logic, one can marvel at the ingenuity behind these various explanations, because all of them present a good chemical argument, and present fascinating cases in themselves.
Hope this helps :)
For this question, you might ignore the difference between heat and entalphy, it really does not make a difference.
The question basically asks you to find the most stable alkane of the ones listed. The reason for this is that they all burn to produce the same $\ce{CO_2 + H_2O}$, so the heat of combustion is larger if the initial molecular energy is lower.
Now, for the answer: branching increases the stability of alkanes. The reason for this is not completely clear to me, I've heard several explanations over time. It might be helpful to think about the electron cloud being more 'packed', more 'spherical' in branched alkanes, and you may think that a 'spherical' cloud is not so favorable to break than an elongated one. If anyone got a more intuitive explanation, I'd be happy to hear it.
Regardless of the explanation, it is a known fact that the more branches one alkane has, the more stable it is. That is why the answer is d).
