# How to determine the reactivity of a set of compounds that may undergo nucleophilic substitution?

1. Which of the following compounds is most reactive under $$\mathrm{S_N1}$$ conditions? 1. Arrange the following compounds in order of decreasing rate of $$\mathrm{S_N1}$$ hydrolysis: A. II > III > IV > I
B. IV > III > II > I
C. III > IV > II > I
D. I > II > III > IV

Both these questions are based upon the rate of reactivity of $$\mathrm{S_N1}$$ reaction.

1) In this I thought the answer should be B because the rate of $$\mathrm{S_N1}$$ reaction depends on the ease of formation of carbocation. Being more electronegative, the electron density between carbon and chlorine bond is quite shifted towards chlorine, so it should be easier for chlorine to leave, hence forming the carbocation. Options A, B, and C allow the carbocation to be resonance stabilized so D cannot be the answer.

But the answer given is A.

Where am I wrong with this logic? I guess it might be because of more polarizing power of iodine, if it is this, then how are we supposed to figure out which one will be the dominating factor?

2) I think we check this on the basis of carbocation stability as well. Carbocation will be formed as $$\ce{Br}$$ leaves. Alkyl chain is electron releasing, so it must increase the electron density on the benzene ring which should be better for the stability of carbocation (as there is more electron density on the benzene ring to stabilize the positive charge). Hence the answer should be I > II > III > IV

Where am I wrong with my logic in both?

• 1) Think about leaving group ability. 2) draw resonance forms, see where the charge appears, and especially consider how that charge is going to to be stabilised by the differing alkyl substituents when it appears at their position. Sep 27, 2016 at 6:00
• @getafix We know poor bases are good leaving groups. Hence iodide ion being the worse base amongst other halide ions here must be the best leaving group. But what if I think in terms of electronegativity as in chlorine is able to hold the electron pairs better than others( higher electronegativity) Sep 27, 2016 at 6:16
• @getafix Sorry I sniped you … but you’ve had your fair share of sniping what I wanted to answer, so … ;)
– Jan
Sep 27, 2016 at 12:03
• @Jan Oh no it is perfectly alright, your answer is probably better than anything that I could've come up with. (up voted it myself). Excellent job, like always! Sep 27, 2016 at 12:05

In question 1), you are wrong about the tendency of leaving groups. While you are correct that the carbon-chlorine bond is more polarised towards chlorine, iodide is still a better leaving group. This is because it is much larger and can thus stabilise the resulting negative charge much better.

You can observe this in $\mathrm{S_N2}$ reactions, where iodides are much more reactive than bromides which are more reactive than chlorides (and fluorides are basically unreactive) even though the same polarisation argument could have applied.

In question 2), the stabilisation that alkyl groups give carbocations, or the electron density they donate into a phenylic ring actually comes from $\ce{C-H}$ bonds. Thus, the para-xylene derivative can stabilise a cation better since there are three $\ce{C-H}$ bonds it can use for stabilisation. No matter which way the methyl is oriented, it can stabilise. para-ethylbenzyl bromide can only be stabilised in two out of three orientations (when a $\ce{C-H}$ bond is orthogonal to the phenyl ring) and para-isopropylbenzyl bromide only in one. Thus, the rate of hydrolysis is accordingly.

• I do understand that iodide ion is a better leaving group because of its large size but I don't understand why we cannot extend the electronegativity reasoning to predict the better leaving group Sep 27, 2016 at 15:41
• In question 2, I was supposed to account for hyperconjugation since it is more dominant than inductive effect of alkyl chain. Right? Sep 27, 2016 at 15:46
• Is it that the net effect of size and electronegativity factors make iodide ion the best leaving group? Sep 27, 2016 at 15:50
• @Blue Re 2nd comment: Hyperconjugation is how methyl groups exercise their $+I$ effect.
– Jan
Sep 28, 2016 at 14:10
• @Blue Re 1st and 3rd comments: It’s size and thus charge distribution. Electronegativity doesn’t play too much of a role when determining nucleofug…acity?
– Jan
Sep 28, 2016 at 14:11