Wurtz Reaction with acidic alkyl halides
The Wurtz reaction involves the reaction of alkyl halide with sodium in dry ether. In this reaction, two molecules of alkyl halide react to form a alkyl-alkyl chain and sodium halide is also produced -

$$\ce{R-X + 2Na + X-R -&gt;R-R + 2NaX} \tag{$\ce{X}$ = halogen}$$ 

Now, sodium is a strongly electropositive metal, and is above hydrogen in the electrochemical series. Thus it replaces acidic hydrogen wherever it occurs. For example,

$$\ce{CH3COOH + Na -&gt;CH3COO- Na+ + 1/2 H2}$$

So, sodium reacts with the acidic hydrogen present in the starting alkyl halide.

Also, acid-base reactions are, in general, faster than a substitution reaction. If there was no acidic hydrogen, alkyl sodium intermediate would have formed and an SN2 reaction would have taken place.

However, in this case, if any alkyl sodium intermediate is formed, it would accept the acidic hydrogen and convert into an alkane, stopping the reaction.

$$\ce{R- Na+ + H-M -&gt; R-H + NaM}$$

(R = alkyl group, M represents the functional group or chain attached to the acidic hydrogen)