This is based in the underlying redox rection. If we take a hexose e.g., mannose and attempt to oxidize that, we will obtain a corresponding hexuronic acid (mannuronic acid in this case). The (unbalanced) half-reaction we need is the following:

$$\ce{C6H12O6 +[O]-> C6H12O7 + 2e-}\tag{Ox1}$$

To balance it, we either need to add protons to the right-hand side (acidic medium) or negative charges to the left-hand side. Since we are asked about alkaline solutions, let’s do the latter:

$$\ce{C6H12O6 + 2 OH- -> C6H12O7 + 2 e- + H2O}\tag{Ox2}$$

As you can see, the oxidation of sugars consumes hydroxide ions. If you increase the concentration of hydroxide, that will typically increase the reaction rate.

Of course, one should always check what the reduced compound is. In your case, it is always a complexated copper(II) ion and is reduced by the following equation:

$$\ce{2Cu^2+ + 2 e- + 2 OH- -> Cu2O v + H2O}\tag{Red}$$

Even here, if we set up the half-reaction, we notice that we are consuming hydroxide. Thus, the overall redox reaction is:

$$\ce{C6H12O6 + 2 Cu^2+ + 4 OH- -> C6H12O7 + Cu2O v + 2 H2O}\tag{Redox}$$

If a reaction consumes hydroxide ions, it is enhanced by basic conditions. Likewise, if a redox reaction consumes protons, it is enhanced by acidic conditions.

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