Do alcohols react with alkyl halides?
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:

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

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



The reaction proceeds through the ionization of phenol:

$$\ce{\underset{phenol}{PhOH} -&gt;[K2CO3] \underset{phenoxide}{PhO-}}$$

The next step is an SN$2$ step:

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



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

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

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