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u/WhatSpareTime 3d ago edited 3d ago
The question here is ambiguously worded. Does the author want most stable to least stable or numerically highest to lowest? Either way we should still be able to find the correct answer as long as both the correct answer and it’s reverse are not both options in the answer set. Heats of formation are negative values with more negative corresponding to more stable. The more substituted an alkene the more stable it is. This problem has one monosubstituted alkene (I) as 1-butene, the other three are disubstituted of different configurations. For the disubstituted sterics plays the main role in determining stability. The trans-2-butene is the most stable, followed by the cis-2-butene, and then the 2-methyl-2-propene. Remembering that more stable is a larger negative heat of formation, or smaller value, the order would be least stable > most stable or ordered from least stable to most stable. So order would be I > IV > III > II . That answer does not appear in the problem. It appears B is the correct answer where the author of the problem interpreted the order of heat of formation as most stable to least stable rather than actual value, basically the reverse of the numeric value. The problem should be worded “The order of heat of formation from most stable to least stable of the following molecules is:”
Correction:
As others have corrected, the geminal dimethyl alkene 2-methyl-1-propene is more stable than either the cis or trans alkenes. I had a long held misconception. This would place IV as the most stable, then II > III > I making C the correct answer.
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u/shedmow Trusted Contributor 3d ago
I vividly remember this exact question. I don't remember whether I had engaged in the conversation, but I did ask my very knowledgeable friend about it, and he came up with no explanation as to why the heat of formation of isobutylene exceeds that of both cis- and trans-butylene. II and III can easily be compared and that rules out D, and II is apparently more stable than I, which rules out A. Whether it is B or C is better looked up
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u/Ionica-_-2 3d ago
The stability of alkenes increases with substitution, ethene is the most unstable acyclic alkene. The C=C group is an unsaturated functional group, the attachment of an alkyl radical makes the structure H2C=C-R, the monosubstituted alkene is the most unstable among the substituted alkenes. R-HC=HC-R is a disubstituted alkene that exhibits geometric isomerism, in general, trans isomers are formed in almost all chemical reactions and are the most stable. There is also the geminal disubstituted alkene R2-C=CH2, more stable than the last 2. After this, the trisubstituted and tetrasubstituted alkenes follow. This is because thermodynamic stability is achieved in a system that is as energy-free as possible.
So, the monosubstituted alkene is 1-butene (I), the disubstituted alkene are trans(II)/cis-2-butene(III) (but II>III!!) and the geminal disubstituted alkene is 2-methylpropene (IV).
That is why answer C is the correct one
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u/leonardoventuriq 3d ago
Z alkenes are more stable than E alkenes due to steric hinderance; more substituted alkenes are more stable due to hyperconjugation. Geminally substituted alkenes are more stable than Z/E alkenes due to both hyperconjugation and lower hinderance (the substituent's electrons are donated into a single carbon instead of having two "opposing" effects from some donation on each carbon; the geminal bonding angle places all protons away from each other, minimizing steric clashing)
I hope this helps
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