# 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 toward 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.

Where am I wrong with this logic? The answer given is A. 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 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 but the answer is A.

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. – getafix Sep 27 '16 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) – Arishta Sep 27 '16 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 '16 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! – getafix Sep 27 '16 at 12:05

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.
• @Blue Re 2nd comment: Hyperconjugation is how methyl groups exercise their $+I$ effect. – Jan Sep 28 '16 at 14:10