A medicinal chemist needed to deprotonate acetylene ( HC≡CH )&...

Question

A medicinal chemist needed to deprotonate acetylene ( HC≡CH ) for use in a coupling reaction. Among the options given, which base(s) could be used for this process?

Accepted Answer

All right. Hi, everyone. So this question says that a student aims to use propine or a coupling reaction. The first step is to detonate propine, which of the following base can be used for this process. And here we have five different compounds. In order from one through five, we have sodium iso valerate, sodium hydride, potassium hydroxide tri methyl amine and potassium amide. Option A says one and four, option B says two and five, option C says 13 and four and option D says 23 and five. So let's go ahead and get started, right? Because this question is specifically discussing the detonation of propine. We can see that this is going to be an acid base reaction. More specifically propine is going to be the acid in this particular reaction. Now, recall that in the case of propine, which is a terminal all kind, the most acidic proton in propine. And therefore the one that gets dep proton is this proton here on the sp hybridized carbon of propine. So it's on the outermost carbon here of our triple bond. Recall that this particular proton on the sp hybridized carbon of the all kind has an approximate PK value of about 25 recall that Pka is a measure of acidity for a specific proton. It follows the same logic as the Ph skill. The lower the PK value is the stronger that acid happens to be. And therefore the weaker its conjugate base. Similarly, the higher the PK A value, the weaker the acid happens to be and therefore the stronger the conjugate base is. So recall also that in an acid base reaction, the equilibrium is always going to favor the formation of the weaker acid. Now because they're starting materials or our second reagents. Or rather our options here for the second reagent are all bases. We have to consider the conjugate acids of all five compounds provided. In other words, the species that forms after each of these bases receives a proton. So from there, after finding the conjugate acids of all five compounds, we have to find the relative acidities or PK values of those conjugate acids. Now, to be more specific, what we're looking for would be the conjugate acids that have a higher PK value than that of our all kind proton because the idea here is to find which of these compounds produces the weaker acid compared to propine. So let's go ahead and find the conjugate acids of all five compounds starting off with sodium iso valerate. The conjugate acid is going to be the proton version of itself. So we can see here that the sodium counter ion is hovering next to the oxygen atom of the carboxylate group. This means that the conjugate acid is going to be a carboxylic acid, meaning that iso valerate is the conjugate base of iso auric acid. And because it is a carboxylic acid, its relative PK value is going to be approximately five. Now notice here that iso VALIC acid has a lower PK value than the terminal all kind proton of propine. This means that the carboxylic acid is a stronger acid than are all kind here. Meaning it is not suitable rather sodium iso valerate is not suitable to detonate because it would form the stronger acid. So moving on to the second compound, we have sodium hydride. Once again, the presence of sodium as a counter ion indicates that the hydrogen atom has a negative charge. So when hydride gains an additional proton, the resulting conjugate acid is simply going to be hydrogen gas or H two. Recall that H two has an approximate PK value of about 35. So now hydrogen is a weaker acid than are all kind here with a PK of about 25. This means that compound two is suitable to deproteinate propine because it would form the weaker acid. So here we can see that sodium hydride compound two is going to be a suitable base here. Moving on to potassium hydroxide, the conjugate base of potassium hydroxide would simply be water or H two up if you recall, water has an approximate PK of 15.7. Now, because the PK of water is smaller than that of the most acidic proton of propine, it is not going to be suitable. Rather hydroxide would not be suitable to detonate it because hydroxide after it gets pronated would produce the stronger acid moving on to. Number four, we have tr methyl amine. Now recall that the nitrogen atom of amine of the amen I should say is what's going to be proton in this case. So here I would add a proton through the nitrogen of tri methyl amine. And because Ive gained this proton, the charge is going to increase by one. So I would go from a neutral charge to a positive charge, meaning that the conjugate acid of triethyl amine is the triethyl ammonium ion. Now, the relative PK of an ammonium ion is about 10, which would make it a stronger acid than propine. Meaning trime amine is not suitable to deproteinate propine last but not least for compound five, we have potassium amide. Now potassium is a counter ion, meaning that the amide ion would have a negative charge. So the conjugate acid of the amide ion would be ammonia or NH three with NH three having a relative PK value of about 38. So once again, the conjugate acid of potassium amide is weaker than propine. This means that potassium amide would be suitable to detonate propine. So after all is said and done. We can see that only compounds two and five can deproteinate propine, which means our answer is going to be option B in the multiple choice and there you had it if you watch this video all the way to the end. Thank you so very much for watching and I hope you found this helpful.

A medicinal chemist needed to deprotonate acetylene ( HC≡CH ) for use in a coupling reaction. Among the options given, which base(s) could be used for this process?

Key Concepts

Acidity and Basicity

Deprotonation

Strong Bases in Organic Chemistry

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