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CHAPTER 7 225
 
In the final step of the process, the primary alkyl iodide is then treated with triphenylphosphine (PPh3), which functions 
as a nucleophile, giving another SN2 reaction, to afford a phosphonium salt. 
 
 
 
The factors favoring this step are the leaving group and the nucleophile. Let’s explore each separately. The leaving 
group is one of the best leaving groups that could be used (because HI is one of the strongest acids, pKa = -10). The 
nucleophile is PPh3. Why is it such a powerful nucleophile? Phosphorus is in the same column of the periodic table as 
nitrogen (5A), but it is in the third row, rather than the second row. As such, phosphorus is larger and more polarizable 
than nitrogen, and therefore more strongly nucleophilic. This argument is similar to the argument we saw in the text 
when we compared sulfur and oxygen. Recall that sulfur is larger and more polarizable than oxygen, and therefore, 
sulfur is very strongly nucleophilic (even if it lacks a negative charge). Similarly, PPh3 is a powerful nucleophile, even 
though it lacks a negative charge. 
 
(b) In the second step, one leaving group (tosylate) is replaced with another (iodide). Iodide is a better leaving group 
than tosylate, which renders step 3 more favorable. 
 
 
7.90. 
(a) The proton connected to the oxygen atom is the most acidic proton in compound 1, so it is removed upon treatment 
with a strong base. 
 
 
We can justify that hydroxide is a suitable base to achieve the conversion of 1 to 2, with either a qualitative argument 
(based on structural comparisons) or with a quantitative argument (based on pKa values). Let’s start with the 
qualitative argument. Compare the structures of the anions on either side of the reaction. 
 
 
 
The negative charge in a hydroxide ion is localized on one oxygen atom, while the negative charge in the other anion is 
delocalized over one oxygen atom and five carbon atoms. As such, we expect the latter anion to be more stabilized (via 
resonance delocalization). 
 
 
 
The equilibrium will favor formation of the more stable anion. That is, hydroxide is a sufficiently strong base, because 
it is stronger (less stable) than anion 2. 
Alternatively, we can use a quantitative argument to justify why hydroxide is an appropriate base to use in 
this case. Specifically, we compare the pKa values of the acids on either side of the equilibrium. 
 
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