i know heat of combustion is inversly proportional to stability. also to solve i did try making chair conformations of the substituted cyclohexane. in same type of say 1,3,5 conformations one can do comparsions by seeing equatorial axial positions, as bulky group tend to be in equatorial postions. but how to compare and arrange when there is 1,2,3 and 1,2,4 and 1,3,5 substitutions
Ashish: Welcome to SE! You are quite correct that heats of combustion and formation are inversely related. Since all of your compounds are trimethylcyclohexanes (C9H18), their sum will be equal to the heat of combustion of 9 moles each of graphite and hydrogen. To approach the solution of this problem, we can start with methylcyclohexane and evaluate the effect of adding ultimately two more methyl groups. In the somewhat busy diagram below, all numbers are in kcals/mol. Known heats of formation (NIST) are located inside the ring. Calculated heats of formation (ΔHfo) are in red and values over the arrows are the incremental changes in heats of formation (ΔΔHfo). The more stable conformation of all compounds is assumed. Equatorial (e) and axial (a) assignments are made.
There is no methyl/methyl interaction in the formation of 1,4-isomers A (cis) and B (trans) or 1,3-isomers E (cis) and F (trans). The addition of an equatorial methyl (B and F) is worth -7.0 and 7.1 kcal/mol, respectively, while addition of an axial methyl (A and E) has a value of -5.2 kcal/mol. Note that the difference between an axial and equatorial methyl is worth 1.9 kcal mol. The axial methyl necessarily makes the ΔHfo value less negative than addition of an equatorial methyl. However, there are gauche butane interactions in 1,2-isomers. Because C is less stable than D, the addition G --> C (cis) is -4.1 kcal/mol and G --> D equals -6.0 kcal/mol.
Using a value of -7.0 kcal/mol for the addition of an equatorial methyl, the conversion G --> R gives a value of ΔHfo = -51.0 kcal/mol [NIST = -51.48 kcal/mol]. Note that 1,3,5-isomer U's value is what is expected given the difference between a non-interacting equatorial vs. axial methyl group.
The reader should be able to corroborate the values assigned to P, Q, S and T. The stability (heats of formation) of P through T are R>S>T>Q>P. Therefore, the heats of combustion are in the order R<S<T<Q<P.
The order of stability is R>T>S>Q>P
Then the heat of combustion will be P has the highest heat of combustion and R the less.
The difference in kj/mol are:
If R is 0 then T has 5 more kj/mol, R 14 kj/mol, Q 20 kj/mol and P 34 kj/mol. This are due to the steric strains.