Indicate which of the metallation reactions given below would be thermodynamically favourable on the basis of the pKa data in non-aqueous media provided.
What is the underlying principle that determines whether the reaction occurs?
(a) C6H5Li + C6H5CH3 → C6H5CH2Li + C6H6
(b) LiCH2CH=CH2 + C6H5CH3 → C6H5CH2Li + CH3CH=CH2
(c) CH3Li + HC≡CH → HC≡CLi + CH4
(d) LiH + CH3CH=CH2 → LiCH2CH=CH2 + H2
(e) LiCH3 + CH3CH=CH2 → LiCH2CH=CH2 + CH4
Table pka values
HC≡CH 24
C6H6 37
CH4 48
C6H5CH3 35
H2 35
CH3CH=CH2 43
I don't understand anything in this question.. Can someone please give me a little help?
Here is a site you can read. It may not be the BEST but if you follow it carefully you will see how it's done.
http://www.masterorganicchemistry.com/2010/09/29/how-to-use-a-pka-table/
Basically, if the acid produces a WEAKER acid the reaction will proceed. If not, the reaction will not proceed. Here is the first one.
pKa = 37 for C6H6 (the acid product)
pka = 35 for C6H5CH3 (the acid reactant)
Since 35 is a stronger acid than 37 the reaction will proceed.
I'm assuming you know how to determine which is the acid and which is the base.
Of course! I'll walk you through the question.
In this question, you are given several metallation reactions, along with the pKa values of the compounds involved. You need to determine which of these reactions would be thermodynamically favorable based on the provided pKa data.
Thermodynamic favorability is determined by comparing the acidity/basicity of the reactants and products. The more acidic (lower pKa) a compound is, the more likely it is to give up a proton. The more basic (higher pKa) a compound is, the more likely it is to accept a proton.
To analyze each reaction, you need to compare the pKa values of the compounds involved. Let's go through each reaction one by one:
(a) C6H5Li + C6H5CH3 → C6H5CH2Li + C6H6
In this reaction, both C6H5Li and C6H5CH3 are neutral compounds. Comparing their pKa values, C6H5Li has a pKa of greater than 37, while C6H5CH3 has a pKa of 35. Since C6H5Li is less basic than C6H5CH3, it is not favorable for C6H5Li to accept a proton from C6H5CH3. Therefore, this reaction is not thermodynamically favorable.
(b) LiCH2CH=CH2 + C6H5CH3 → C6H5CH2Li + CH3CH=CH2
In this reaction, LiCH2CH=CH2 is the neutral compound and C6H5CH3 is the acidic compound. Their respective pKa values are 43 and 35. Since C6H5CH3 is more acidic (lower pKa) than LiCH2CH=CH2, it is favorable for C6H5CH3 to donate a proton to LiCH2CH=CH2. Therefore, this reaction is thermodynamically favorable.
(c) CH3Li + HC≡CH → HC≡CLi + CH4
In this reaction, both CH3Li and HC≡CH are neutral compounds. Comparing their pKa values, CH3Li has a pKa greater than 48, while HC≡CH has a pKa of 24. Since HC≡CH is more acidic (lower pKa) than CH3Li, it is favorable for HC≡CH to donate a proton to CH3Li. Therefore, this reaction is thermodynamically favorable.
(d) LiH + CH3CH=CH2 → LiCH2CH=CH2 + H2
In this reaction, both LiH and CH3CH=CH2 are neutral compounds. Comparing their pKa values, LiH has a pKa greater than 35, while CH3CH=CH2 has a pKa of 43. Since CH3CH=CH2 is more acidic (lower pKa) than LiH, it is favorable for CH3CH=CH2 to donate a proton to LiH. Therefore, this reaction is thermodynamically favorable.
(e) LiCH3 + CH3CH=CH2 → LiCH2CH=CH2 + CH4
In this reaction, LiCH3 is the neutral compound and CH3CH=CH2 is the acidic compound. Their respective pKa values are greater than 43 and 35. Since CH3CH=CH2 is more acidic (lower pKa) than LiCH3, it is favorable for CH3CH=CH2 to donate a proton to LiCH3. Therefore, this reaction is thermodynamically favorable.
To summarize, reactions (b), (c), (d), and (e) would be thermodynamically favorable based on the given pKa data.