On the topic on neigbouring group participation, it is mentioned in Carey & Sundberg (2007)[1] that the effectiveness of the participation is dependent on on the ease with which the molecular geometry required for the participation can be attained. The example of the cyclization of ω-hydroxyalkyl halides was then given:

The rate of cyclization of ω-hydroxyalkyl halides, for example, shows a strong dependence on the length of the chain separating the two groups.

To prove the point, they showed the data for the solvolysis rates for ω-chloro alcohols with various chain lengths. Clearly, when a 5-membered ring is form, the rate of reaction is fastest. This is reasoned with the fact that a 5-membered ring has relatively little ring strain and thus, the rate of forming it would be relatively higher. However, this analysis does not seem to be convincing. If this reasoning was indeed correct, then wouldn't the next member in the series, with the capability of forming the 6-membered ring, have a higher solvolysis rate?

\begin{array}{|c|c|c|c|} \hline \text{Substrate} & k_\mathrm{rel}{}^\text{[2,3]} \\ \hline \ce{Cl(CH2)2OH} & 2000 \\ \hline \ce{Cl(CH2)3OH} & 1 \\ \hline \ce{Cl(CH2)4OH} & 5700 \\ \hline \ce{Cl(CH2)5OH} & 20 \\ \hline \end{array}

In general, why doesn't the rate increase as the alkyl chain increases in length, based on the reasoning using ring strain in the intermediate? We see that 3-chloropropanol is also another outlier.

Additional information

It seems that the relative solvolysis rates of ω-methoxyalkyl p-bromobenzenesulfonates also follow the same trend, with the rate peaking with the formation of the 5-membered ring.

\begin{array}{|c|c|c|c|} \hline \text{Substrate} & k_\mathrm{rel}{}^\text{[4]} \\ \hline \ce{CH3(CH2)2OSO2Ar^*} & 1 \\ \hline \ce{CH3O(CH2)2OSO2Ar} & 0.28 \\ \hline \ce{CH3O(CH2)3OSO2Ar} & 0.63 \\ \hline \ce{CH3O(CH2)4OSO2Ar} & 657 \\ \hline \ce{CH3O(CH2)5OSO2Ar} & 123 \\ \hline \ce{CH3O(CH2)6OSO2Ar} & 1.16 \\ \hline \end{array}

*The first entry shows the rate of solvolysis without any neighbouring group participation from the methoxy group.


  1. Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.

  2. Capon, B. Neighbouring group participation. Q. Rev., Chem. Soc. 1964, 18 (1), 45–111. DOI: 10.1039/QR9641800045.

  3. Richardson, W. H.; Golino, C. M.; Wachs, R. H.; Yelvington, M. B. Neighboring oxide ion and fragmentation reactions of 1,3-chlorohydrins. J. Org. Chem. 1971, 36 (7), 943–948. DOI: 10.1021/jo00806a019.

  4. Winstein, S.; Allred, E.; Heck, R.; Glick, R. Neighboring methoxyl participation in solvolytic nucleophilic substitution. Tetrahedron 1958, 3 (1), 1–13. DOI: 10.1016/S0040-4020(01)82605-3.

  • 1
    $\begingroup$ I think for 2-chloroethanol, the rate is high since only a single bond rotation is required for achieving the configuration favourable for NGP while for 4-chlorobutanol the 5mem ring formation increases the rate but when we go to 5-chloropentanol, the rate drops probably due to the low probability of achieving the right configuration required for NGP... however there might be other factors operating here too... $\endgroup$
    – Carrick
    Dec 24, 2018 at 6:51
  • $\begingroup$ Longer the chain the two reacting sites can span over a bigger volume therefore have littler chance to meet and react. In other words is an entropy effect. $\endgroup$
    – Alchimista
    Dec 24, 2018 at 11:58
  • $\begingroup$ This is a standard case of balancing entropy of activation (as Alchimista mentioned) vs enthalpy of activation (as mentioned in your question body) for ring closure reactions. Clayden 2ed pp 805–807 covers it quite well. And for the second table, the authors explain that NGP is not competitive for the 3-membered ring, so the rate is about the same as in the first entry. (I don't immediately see why this is the case, admittedly..) $\endgroup$ Dec 25, 2018 at 4:15
  • $\begingroup$ @orthocresol The gibbs free energy of activation has to do with kinetics. Thus, perhaps, the rate of formation of the 5-membered ring is the greatest. However, if we look at thermodynamic constants, would the equilibrium constant for the formation of the 6-membered ring always dominate? Under the thermodynamic conditions, what would be the major change? $\endgroup$ Dec 25, 2018 at 13:46
  • $\begingroup$ @TanYongBoon, under thermodynamic control it's difficult to get anything apart from 5- and 6-rings, and all else being equal the 6-membered ring would be preferred, as you said. That said, it will probably also depend on the exact substitution pattern; if you have substituents which can stabilise or destabilise the preferred conformations of the 5- or 6-ring, then things can change. I can't think of a good example – maybe the hemiacetal forms of hexose sugars, where you only really see furanoses (5-ring) and pyranoses (6-ring). $\endgroup$ Dec 25, 2018 at 13:53

1 Answer 1


Thanks to orthocresol's suggestion to look at Clayden's (2nd ed. p. 805-807), I now have a satisfactory answer which I can now present.

We can explain these trends using the Gibbs free energy equation. Except that now, we are now interested in the quantities: Gibbs free energy of activation, enthalpy of activation and the entropy of activation, since we are analysing rate.

$$\Delta G^\ddagger = \Delta H^\ddagger - T\Delta S^\ddagger $$

Explaining the decreasing of $\ce {\Delta H^\ddagger}$ as the ring size increase

$\Delta H^\ddagger$ is associated with the ring strain of the resultant ring formed. The larger the ring strain, the larger the value of the enthalpy of activation. Clayden, Greeves, & Warren (2012) explains$\ce{^1}$:

Firstly, small rings form slowly because forming them introduces ring strain. This ring strain is there even at the transition state, raising its energy and slowing down the reaction. The activation energy for forming a three-membered ring is very high, due to strain, but decreases as the ring gets larger.

The data (from pg. 108 of ref. 2) presented here of the strain energy determined for cycloalkanes of varying sizes also supports this trend.

\begin{array}{|c|c|c|c|} \hline \text{Ring size} & \text{Strain energy}^\text{2,*}/ \text{kJ/mol} \\ \hline \ce{3} & 115 \\ \hline \ce{4} & 110\\ \hline \ce{5} & 26 \\ \hline \ce{6} & 0 \\ \hline \end{array}

*Reported to the nearest whole number.

Explaining the increasing of $\Delta S^\ddagger$ as the ring size increases

Think of it this way: a long chain has a lot of disorder, and to get its ends to meet up and react means it has to give up a lot of freedom$\ce {^1}$. However, for shorter chains, it is far easier for their ends to meet since these chains have fewer degrees of freedom and thus, the loss of entropy upon cyclisation is less negative.

Consolidating the trends

Clayden, Greeves, & Warren (2012) reasons that based on the above trends for $\Delta H^\ddagger$ and $\Delta S^\ddagger$, the activation energy for the formation of the 4-membered ring is the largest. This is because the formation of the 4-membered ring likely has similar $\Delta H^\ddagger$ to that for the 3-membered ring while its $\Delta S^\ddagger$ is relatively more negative.

This seems to agree with the solvolysis rates I have presented:

For the ω-chloro alcohols, we see that the relative solvolysis rate is lowest for $\ce {Cl(CH2)3OH}$. For the ω-methoxyalkyl p-bromobenzenesulfonates, the formation of the 4-membered ring also has the 2nd lowest rate ($\ce {k_{rel} = 0.68}$), with the rate for the 3-membered ring being the lowest ($\ce {k_{rel} = 0.28}$).

Why does the solvolysis rate not peak at the formation of the 6-membered ring and instead, peak for the formation of the 5-membered ring? This is probably a coincidence. There are 2 opposing trends here so it may be the case that the combined effects of the two trends give different results under different circumstances.

ron has provided a graphical representation of the data for strain energy and the entropy of activation here. We do see that in general, there is a peak in the rate of formation of the 5-membered ring. In other words, from the formation of 5-membered ring to formation of 6-membered ring, $\Delta S^\ddagger$ becomes significantly more negative.


  1. Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

  2. Anslyn, Eric V., and Dennis A. Dougherty. (2006). Modern Physical Organic Chemistry. Sausalito, CA: University Science.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.