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my teacher told that major product would be when acetone will form nucleophile because it will attack more positive carbon of ethanal(less +i than acetone) however i read in my book that since formation of carbanion is rds ethanal should be nucleophile (more stable due to less +i)

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  • $\begingroup$ Remember the whole process is reversible $\endgroup$ – Waylander Jan 2 at 17:35
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    $\begingroup$ The anion of ethanal may form, and may react with acetone but it can just as easily break down reforming the starting materials if that is not the most energetically favourable pathway. Don't get hung up on which anion forms fastest, it is not the factor that determines the outcome. $\endgroup$ – Waylander Jan 2 at 18:02
  • $\begingroup$ ya i got it but why do we then in most cases decide major product using rds step? $\endgroup$ – Akshat Gautam Jan 2 at 18:07
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    $\begingroup$ Unclear what you mean by "most cases." I've got 15+ years of experience, and I'm not sure I've seen close to enough to count as having seen most cases. $\endgroup$ – Zhe Jan 2 at 19:44
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This is a question of thermodynamic vs kinetic control. For reactions that are irreversible, the kinetics will determine the product ratio. For reactions that are reversible, the product ratio will be determined by the stability of the products if you wait long enough before working up the reaction.

Highly reactive starting materials (e.g. Grignard reagents) typically are associated with kinetic control (you don't expect the Grignard reagent to reform).

Slow reactions that could go in either direction depending on conditions (e.g. esterification/saponification) typically are associated with thermodynamic control.

For some reactions, you can steer them in either direction. One example from molecular biology is taking single-stranded DNA and annealing it either by slow cooling (thermodynamic control, the best matches will be made) or fast cooling (kinetic control, the first matches made will be the products). This is used in the PCR process, where short primers will bind to long DNA because they are present at high concentration even though there is complementary long DNA in the mixture that would form a more stable product.

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As pointed out by Waylander, the question essentially wants students to go for the most stable product, or thermodynamic product. This is sort of like a well-understood thing for aldol product determination questions. Why? Well... Simply because there is usually only 1 thermodynamic product while there are several possible kinetic products. It is therefore hard to argue that the kinetic product drawn is really the major product under kinetic control. Under kinetic control, you may just get a mixture of products with nothing formed in significantly higher quantity.

Based on this understanding, we can go on to determine the thermodynamic product. This reaction is a very straightforward aldol reaction due to there being only two possible reaction pathways since one reactant is a symmetrical ketone while the other is an aldehyde. So there really are only two possible products...

One possible product comes from aldehyde as a the electrophile and acetone behaving as the nucleophile. Upon dehydration, we obtain A, which is a ketone and has a di-substituted conjugated $\ce {C=C}$ double bond. The other possible product comes from aldehyde as the nucleophile and acetone as the electrophile. Upon dehydration, we obtain B, which is still an aldehyde but it has a tri-subustituted $\ce {C=C}$ bond.

Structures of possible products

Which one is the thermodynamic product? Usually, we will just look at the substitution of the conjugated double bond formed. Very clearly, the kinetic product must be A since the more nucleophilic enolate from acetone attacks the more electrophilic aldehyde. However, is B really more stable? No, it isn't. The explanation presented below is based on qualitative molecular orbital theory.

Well... The key differences between A and B are:

  • A, being a ketone, has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=O \pi^*}$ MO, compared to the B

  • B instead has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=C \pi^*}$ MO, compared to A

The former hyperconjugative interactions are stronger since the $\ce {C=O \pi^*}$ MO is lower in energy and thus more energetically-accessible to the $\ce {C-H \sigma}$ MO. This is due to the higher electronegativity of the oxygen atom. Thus, we would expect the thermodynamic stability of A to be higher than that of B.

To provide even more concrete substantiation for the higher stability of A over B, Karsten Theis has kindly provided a source of data which indicates the Gibbs free energy of formation of the two compounds. As stated on chemeo, here and here, A has $\Delta G_f = \ce {-57.48 kJ/mol}$ while B has $\Delta G_f = \ce {-36.63 kJ/mol}$. This very clearly indicates that A has higher stability than B.

Thus, A is the answer.

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  • $\begingroup$ On chemeo, the Gibbs energy of formation is tabulated as -57.48 kJ/mol for the ketone A and as -36.63 kJ/mol for the aldehyde B. $\endgroup$ – Karsten Theis Jan 5 at 14:17
  • $\begingroup$ @KarstenTheis Thanks for your data. I will add it to my answer. $\endgroup$ – Tan Yong Boon Jan 5 at 14:19

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