magnitudes of heat of combustion 💚🤍

Okay, so when I first tackled organic chemistry, figuring out alkene stability and its link to heat of combustion felt like trying to solve a Rubik's Cube blindfolded! But once you grasp the core idea, it actually makes so much sense. Think of it this way: more stable compounds are already in a lower energy state. When something burns (combusts), it releases energy to reach a much lower, more stable state (like CO2 and H2O). If your starting alkene is already more stable, it means it has less 'excess' energy to release during combustion. That's why a more stable alkene will always have a lower heat of combustion – it's simply got less energy stored within its bonds to begin with. What makes alkenes stable, though? It's not just about 'more alkyl groups' on the double bond, although that's a huge part of it! It also depends significantly on the position of that double bond. Generally, a double bond in the middle of a carbon chain (internal) is more stable than one at the end (terminal). Why? Well, those alkyl groups (like methyl, ethyl, etc.) are electron-donating. When they're attached directly to the carbons of the double bond, they can stabilize the electron density of the pi bond through a phenomenon called hyperconjugation. More alkyl groups mean more hyperconjugation, and thus, greater stability. Let's look at some examples, similar to a problem I once had. Imagine comparing 2-Methylhex-1-ene, (Z)-hept-2-ene, (E)-hept-2-ene, and (Z)-3-methylhex-3-ene. You'd sketch them out first! 2-Methylhex-1-ene: Here, the double bond is terminal (at position 1) and has only one alkyl substituent (the rest are hydrogens on the double bond carbons). This makes it less stable. Hept-2-ene (Z and E isomers): The double bond is internal (at position 2), meaning it has alkyl groups on both sides. Both (Z)-hept-2-ene and (E)-hept-2-ene have two alkyl substituents directly on the double bond carbons. (Z)-3-methylhex-3-ene: This one has the double bond at position 3, and it has three alkyl substituents directly attached to the double bond carbons (one methyl and two ethyl groups). This extensive substitution makes it significantly more stable than the others. Another crucial factor is cis-trans isomerism. Between (Z)-hept-2-ene (cis) and (E)-hept-2-ene (trans), the (E) or trans isomer is typically more stable. This is because in the (Z) or cis form, the bulkier alkyl groups are on the same side of the double bond, leading to steric hindrance and repulsion, which destabilizes the molecule slightly. In contrast, the (E) or trans isomer has these groups on opposite sides, minimizing repulsion and making it more stable. So, when you consider all these factors – the number of alkyl substituents, the internal versus terminal position of the double bond, and cis/trans isomerism – you can confidently predict which alkene is the most stable. And remember, the most stable alkene will always have the smallest magnitude of heat of combustion. It’s all about understanding those energy levels!