My book says that the reaction of

$$\ce{Cl-CH2-O-CH2-CH3}$$ with ethanol to form $$\ce{CH3-CH2-O-CH2-O-CH2-CH3}$$ occurs through $\mathrm{S_N1}$ mechanism, but shouldn't it actually be $\mathrm{S_N2}$?

My understanding of this statement is that the substrate is a $1^\circ$ halide and on top of that ethanol is available to us as $\ce{C2H5-O-}$, which is a considerably strong nucleophile. All possible conditions are pointing towards an $\mathrm{S_N2}$ reaction but then why does the reaction proceed through $\mathrm{S_N1}$ mechanism?

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    $\begingroup$ The presence of the oxygen in chloromethylether stabilises +ve charge so Cl- can leave in an SN1. $\endgroup$ – Waylander May 29 '20 at 21:26
  • $\begingroup$ Ok, so basically what you are trying to say is that the carbocation formed when chlorine leaves overpowers the benefits of SN2 in this case , right? $\endgroup$ – TheChemist May 29 '20 at 21:35
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    $\begingroup$ yes, essentially H2C=OEt+ is the intermediate formed $\endgroup$ – Waylander May 29 '20 at 21:37
  • $\begingroup$ I still don't by Sn1 even with the alpha hydrogen. The bimolecular pathway also sees a significantly increased rate. $\endgroup$ – Zhe May 30 '20 at 0:50

It is well-established that alkyl chlorides where derivatization involves the introduction of a substituent atom such as oxygen, sulfur, or nitrogen on the α-carbon give, on ionization, resonance stabilized carbocations as depicted in equation $(1)$ (Ref.1):

$$\ce{R-X-CH(Cl)-R’ ->[-Cl] R-X -CH^+-R’ <-> R-X+=CH-R’} \\\quad \text{where } \ce{X = O, S, N} \tag1$$

The formation of the carbocations, with an appreciable amount of delocalized charge, is relatively rapid (Ref.1-Ref.5). This behavior of chloromethyl ethyl ether (where $\ce{R = CH3CH2 \!-}$, $\ce{X = -O \!-}$, and $\ce{R' = -H}$) is well explained in the page 289 of Ref.4:

The rate constant for the solvolysis of chloromethyl ethyl ether, chloromethyl octyl ether, and chloromethyl methyl sulfide have been determined in several pure and binary solvents. Aplication of the extended Grunwald-Winstein equation, $\log (k/k_\circ) = \ell \mathrm{N_T} + m\mathrm{Y} + c$, gave appreciable “$\ell$” values (0.55-071) for the three substrates indicating that there is significant nucleophilic salvation of the developing carbenium ion in the transition state of these reactions. The $k_\ce{Cl}/ k_\ce{F} = 1.2 \times 10^{5}$ found for the hydrolysis of chloromethyl methyl ether in water is virtually identical to that observed for the unimolecular solvolysis of t-butyl chloride and trityl halides confirming the unimolecular mechanism for these reactions.

The effect of the adjacent lone pairs of electrons can be dramatic (Ref.1). For example, the solvolysis of chloromethyl ethyl ether ($\ce{CH3CH2OCH2Cl}$) in $36\%$ dioxane proceeds at a rate of about $10^9$ times that of butyl chloride (Ref.3, p. 103). Also, Ballinger, et al. (Ref.6) has presented the evidence showing the solvolysis of related chlorodimethyl ether $(\ce{ClCH2OCH3})$ is unimolecular in ehanol and mixtures of ethanol/diethyl ether system, rates of which are at least about $ 10^{14}$ faster than the corresponding reactions of methyl chloride. With the same substrate, the ethanolysis rate is about $10^{13}$ times faster than that of propyl chloride (Ref.3, p.103). Note that it is of interest that when the nucleophile is ethoxide ion, the reaction undergoes in bimolecular mechanism.

On the other hand, the reaction mechanism is completely different when ethoxide ion is involved as a nucleophile in these reactions. For example, note that it is of interest that when the nucleophile is ethoxide ion (when the substrate is $\ce{ClCH2OCH3}$), the reaction undergoes in bimolecular mechanism. Similar to the shown rate enhancement in unimolecular reaction with ethanol, in this bimolecular reaction with ethoxide ion, the second-order rate coefficient is also considerably larger (by a factor of about 105) than that for the methyl chloride (Ref.6).

All of these findings suggest that an unimolecular $\mathrm{S_N1}$ mechanism is the most possible for ethanolysis of $\ce{ClCH2OCH2CH3}$, a reagent of which has commonly used in $\ce{R-OH}$ protection in synthetic procedures.


  1. Dennis N. Kevill, Young Hoon Park, Byoung-Chun Park, Malcolm J. D'Souza, "Nucleophilic Participation in the Solvolyses of (Arylthio)methyl Chlorides and Derivatives: Application of Simple and Extended Forms of the Grunwald-Winstein Equations," Current Organic Chemistry 2012, 16(12), 1502–1511 (DOI: 10.2174/138527212800672592).
  2. C. A. Bunton, In Reaction Mechanisms in Organic Chemistry Series, Volume 1: Nucleophilic Substitution at a Saturated Carbon Atom; Publishing Co.: New York, NY, 1963. pp. 50–51 (ISSN: 0079-9823).
  3. Andrew Streitwieser, Jr., In McGraw Hill Series in Advanced Chemistry: Solvolytic displacement reactions; McGraw Hill: New York, NY, 1964 (ISBN-13: 978-1258354817).
  4. K. C. Westaway, “Chapter 7: Nucleophilic Aliphatic Substitutions,” In Organic Reaction Mechanisms•2012: An annual survey covering the literature dated January to December 2012; A. C. Knipe, Ed.; John Wiley & Sons, Ltd.: Chichester, West Sussex, United Kingdom, 2015, pp.267-306 (ISBN: 978-1-118-36259-4).
  5. Andrew Streitwieser, Jr., “Solvolytic Displacement Reactions at Saturated Carbon Atoms,” Chem. Rev. 1956, 56(4), 571–752 (https://doi.org/10.1021/cr50010a001).
  6. P. Ballinger, P. B. D. de la Mare, G. Kohnstam, B. M. Prestt, "The reaction of chlorodimethyl ether with ethanol and with ethoxide ions," J. Chem. Soc. 1955, 3641–3647 (https://doi.org/10.1039/JR9550003641).
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    $\begingroup$ Thank you so much , cleared all my concepts regarding this topic. $\endgroup$ – TheChemist May 31 '20 at 4:57

Be careful! Ethanol is not a strong acid ($\mathrm{p}K_\mathrm{a} \approx 16$). So its conversion to ethoxide anion is not so easy. Rather, it would attack the $\ce{C}$ center by its lone pair and then it would release proton to form the product.

Actually here what I think is, the hetero atom $\ce{O}$ just near to the $\ce{C}$ undergoing substitution will assist by its lone pair to kick out chloride and then being planar ethanol can attack from either side. And finally, you get the product of $\mathrm{S_N1}$ type.

But don't mess up between ethanol and ethoxide ion. They are totally different in nucleophilicity. Treat them as two different nucleophiles.

  • $\begingroup$ So, if the reaction would have stated that ethoxide ion is present then would it have undergone SN2? $\endgroup$ – TheChemist May 29 '20 at 21:39
  • $\begingroup$ Yes I think so. Of course ethoxide ion is a strong nucleophile than ethanol and will more bend towards Sn2 $\endgroup$ – Nilarun Koley May 29 '20 at 21:41
  • $\begingroup$ Thanks and i had just one more doubt , if it would have been given aq. ethanol then what would we have considered it as , ethanol or ethoxide ion? $\endgroup$ – TheChemist May 29 '20 at 21:44
  • $\begingroup$ Aq.Ethanol means you have water as solvent which is polar. As polar solvent would bend towards Sn1(stabilize the carbocation by solvation) I think it should be Sn1. $\endgroup$ – Nilarun Koley May 30 '20 at 3:56
  • $\begingroup$ Ok , Thanks.... $\endgroup$ – TheChemist May 30 '20 at 4:27

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