# Question #8786b

##### 1 Answer
Apr 29, 2017

Well the aldehyde is oxidized to a carboxylic acid......

#### Explanation:

$R \stackrel{+ I}{C} \left(= O\right) H + {H}_{2} O \rightarrow \text{R} \stackrel{+ I I I}{C} \left(= O\right) O H + 2 {H}^{+} + 2 {e}^{-}$ $\left(i\right)$

Now I used standard assignments for oxidation numbers here. As always the sum of the formal oxidation numbers equals the charge on, the molecule; and for a neutral organic molecule this charge is ZERO. Again, as always, the oxidation reaction balances mass and charge.

Now the $\text{Benedict's reagent}$ is aqueous $\text{copper sulfate}$, which forms the ${\left[C u {\left(O {H}_{2}\right)}_{6}\right]}^{2 +}$ ion in aqueous solution, and this has a truly beautiful blue colour. It is reduced to $\text{cuprous oxide}$, an insoluble reddish salt in aqueous solution.

And we can represent the reduction reaction as:

$\text{Cu"^(2+) + H_2O + e^(-) rarr stackrel(+I)"Cu"_2"O} + 2 {H}^{+}$ $\left(i i\right)$

We add the two redox reactions together to eliminate the electrons. And thus we add $\left(i\right) + 2 \times \left(i i\right)$ to give.............

$2 C {u}^{2 +} + R C \left(= O\right) H + 2 {H}_{2} O \rightarrow 2 C {u}_{2} O + R C \left(= O\right) O H + 4 {H}^{+}$

Which (I think!) is balanced with respect to mass and charge, as is required for any chemical reaction. What do we see in this reaction? The deep blue colour of ${\left[C u {\left(O {H}_{2}\right)}_{6}\right]}^{2 +}$ dissipates, and a deep red precipitate of the insoluble copper oxide deposits. This is the macroscopic change we observe.

Sometimes, Benedict's reagent can oxidize primary alcohols (and sugars) all the way up to the acids. With all these discriminating reactions, these qualitative tests, it is a good idea to perform the test on a reagent whose identity you know, so that you can appreciate the results of a positive test.

Claro?