I wonder if alkyl halides react with alcohols or not. I searched on Google but could only find reaction of alcohols with hydrogen halides


Not directly. As I hope you know, alkyl halides are electrophiles, accepting an electron pair at the electron-poor $\alpha$-carbon. Alcohols are nucleophiles, donating an electron pair from the electron-rich oxygen. So your question is a very reasonable one. However, both reagents are only mildly reactive, and the reaction will not proceed at useful rate.

We can deprotonate the alcohol $\ce{ROH}$ to form an alkoxide anion $\ce{RO^-}$. This requires a relatively strong base ($\ce{NaNH2}$ and $\ce{NaH}$ come to mind). The alkoxide is a much stronger nucleophile, and will attack the alkyl halide forming an ether. This usually occurs via the $\mathrm{S_N2}$ mechanism (unless the alkyl halide $\alpha$-C is tertiary). In fact, this reaction has a special name, the Williamson Ether Synthesis:

$$\ce{RO- + RX -> ROR + X-}$$

TL;DR Yes, if the alcohol is activated first.

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  • $\begingroup$ Can we say that they are less basic than amines because they don't react with alkyl halide to give addition product on these basis? $\endgroup$ – Saqeeb Jan 25 '17 at 18:39
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    $\begingroup$ Alcohols are indeed less basic than the corresponding amines, but no this substitution reaction would not be a good rationale for that. For one thing, this substitution requires the O (or N) to act as a nucleophile, which has subtle but important differences to acting as a base. I would rationalize nitrogen's greater basicity than oxygen by the fact that nitrogen is less electronegative, so it "holds its lone pair more loosely", making it more available to be shared. I think there is a more rigorous explanation using molecular orbitals, if you're interested in digging deeper. $\endgroup$ – electronpusher Jan 25 '17 at 21:26
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    $\begingroup$ Please note that although in general N is a stronger nucleophile than O and N is a stronger base than O, the nucleophilicity and basicity trends will not always be the same for any given pair of atoms. $\endgroup$ – electronpusher Jan 25 '17 at 21:28
  • $\begingroup$ And the base had better not be a good nucleophile for carbon. Putting a Grignard reagent in and having it react directly with the alkyl halide gets ugly. $\endgroup$ – Oscar Lanzi Jan 5 '18 at 23:35

I am assuming that you want the alcohol to react with the $\ce{OH}$ group and the alkyl halides to react with the $\ce{Cl}$ group.

Both groups like to withdraw electrons, so they tend to be negative, and if we take them away, the resulting ions would be positive so they cannot react.

This leaves us to the possibility of the $\ce{H+}$ being cleaved off from the $\ce{OH}$, producing an alkoxide anion, which can react with the cation resulting from the cleaving off of the halide.

However, to effectively cleave off an $\ce{H+}$, we would need an acidic alcohol, such as phenol, which does react with alkyl halides (taken from here):

In the presence of a base, such as potassium carbonate ($\ce{K2CO3}$), and heat, phenol reacts with an alkyl halide to form an ether.

For example:

$$\ce{\underset{phenol}{PhOH} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + HCl}$$

where $\ce{Ph}$ denotes the phenyl group.

The reaction proceeds through the ionization of phenol:

$$\ce{\underset{phenol}{PhOH} ->[K2CO3] \underset{phenoxide}{PhO-}}$$

The next step is an SN$2$ step:

$$\ce{\underset{phenoxide}{PhO-} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + Cl-}$$

For other alcohols, they are generally not acidic, so you would need the corresponding alkoxide by reacting with sodium:

$$\ce{\underset{alcohol}{R1-OH} ->[Na] \underset{alkoxide}{R1-O-} ->[R2Cl] \underset{ether}{R1-O-R2}}$$

This, as the comments have pointed out, is called the Williamson ether synthesis.

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