My book States part C as the answer of the following question.

But according to me, part D, that is soda lime decarboxylation can also be a perfect answer besides the Corey House synthesis( that is the part C )the answer given by my book. Please help in clearing my misconceptions.enter image description here

  • $\begingroup$ Decarboxylation is usually assisted by the presence of an electron withdrawing group at the β position, so it may not be particularly effective here. Plus, the nucleophile generated by is not going to cause much elimination, so the Corey House synthesis is a better choice here $\endgroup$ – Yusuf Hasan Nov 23 '18 at 14:07
  • $\begingroup$ Can u please elaborate your second point.Moreover, then according to your first point,it implies that decarboxylation without a beta keto acid is surely not productive,can we say this way? $\endgroup$ – chemophilic Nov 23 '18 at 14:20
  • $\begingroup$ @yusufHasan Moreover,if there would have been a carbonyl group at the beta position it would lead to the formation of ketone,that we do not want because we want formation of alkane to occur... $\endgroup$ – chemophilic Nov 23 '18 at 15:04
  • $\begingroup$ Precisely, so if you look at the normal decarboxylation of the given compound, there is no additional factor favouring the decarboxylation or literally, 'removal of CO2'. What I said about the (c) option was that the substrate is a primary haloalkane, so it will have minimum chance to follow the E2 pathway, and also, CH3- will not be that strong a base to cause elimination in comparison to, say, EtO- so it is the best method here $\endgroup$ – Yusuf Hasan Nov 23 '18 at 15:16
  • $\begingroup$ Note that the question is asking for the BEST preparation of propane, not "a preparation that will give some propane". (c) is the best reaction. $\endgroup$ – Waylander Nov 23 '18 at 15:22

Simple carboxylic acid or carboxylate ions (like CH3CH2CH2COO) do not undergo decarboxylation to yield corresponding alkanes because in this case the leaving group would be a carbanion (in this case CH3CH2CH2) which is very strong base and hence a very poor leaving group (the reason behind this is that strong base wants to give electrons, but leaving group has to take electrons and leave the compound, strong base is poor leaving group and weak base is good leaving group)

For decarboxylation to happen a keto group must exist to stabilise the negative charge, which is not present in salt of butanoic acid CH3CH2CH2COONa+

See this image for illustration

Yield of propane is very less if you chose decarboxylation, that is why Corey-House synthesis is best for preparation of propane.

I want to know why do you think that Option D is correct, mention in comment.

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  • $\begingroup$ Is the negative charge itself not very stable in ch3ch2ch2coo- because of equivalent resonating structure?when coo- posseses this kind of delocalization of e- then verily,it is a poor base and hence good leaving group? $\endgroup$ – chemophilic Nov 24 '18 at 14:08
  • $\begingroup$ @chemophilic yes negative charge stable on ch3ch2ch2coo- due to resonance but leaving group is not co2, leaving group is ch3ch2ch2- (I am mentioning again leaving group takes electron and leave the compound, that is why leaving group is ch3ch2ch2- which is strong base and very poor leaving group). $\endgroup$ – Saurabh Singh Nov 24 '18 at 14:59
  • $\begingroup$ @chemophilic Also see that the first reaction is highly endergonic (meaning energy of products is greater than that of products) as product ch3ch2ch2- is very unstable compared to reactant ch3ch2ch2coo- which is stabilized by resonance. Therefore the reaction requires huge amount of energy which may decompose the reactant compound, and may not yield desired products, which is not the case with second reaction due to presence of carbonyl group the oxygen takes negative charge and is stable and energy required for reaction to proceed is low. $\endgroup$ – Saurabh Singh Nov 24 '18 at 15:08

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