Which of the following compounds is most reactive to an aqueous solution of sodium carbonate?

First we remind ourselves of a very common principle of acid-base chemistry: a base removes the most acidic hydrogen that is available. A hydrogen atom is called “acidic” when, after its removal as $$H^+$$, the conjugate base that remains is stabilised (generally by resonance or by an electron-withdrawing group). The better the stabilisation, the smaller the $$\mathrm{p}K_a$$ of the hydrogen-bearing compound and the more easily even a weak base can abstract that hydrogen.

In aqueous solution sodium carbonate provides the weak base $$CO_3^{2-}$$:

$$Na_2CO_3 \;\;\xrightarrow{H_2O}\;\; 2\,Na^+ + CO_3^{2-}$$

The carbonate ion itself is the species that can abstract an acidic hydrogen. Because $$CO_3^{2-}$$ is only a mild base, it can deprotonate a substrate only if the resulting conjugate base is appreciably stabilised. With this idea in mind we examine each of the four hydrocarbons offered in the question.

(i) Cyclopropane, $$C_3H_6$$
All six hydrogens are attached to ordinary sp3 carbon atoms. When one such proton is removed we would have a simple alkyl carbanion

$$C_3H_5^-$$

This carbanion has no resonance stabilisation and experiences strong angle strain; it is therefore highly unstable. The corresponding $$\mathrm{p}K_a$$ of cyclopropane is close to that of a normal alkane (well above 45). A base as weak as $$CO_3^{2-}$$ cannot remove such a proton.

(ii) Benzene, $$C_6H_6$$
All hydrogens in benzene are bound to sp2 ring carbons. Removal of one proton produces the phenyl carbanion

$$C_6H_5^-$$

Although this anion is conjugated with the ring, it is antiaromatic and hence severely destabilised. Consequently the $$\mathrm{p}K_a$$ of benzene is still extremely high (≈43). Again, $$CO_3^{2-}$$ is far too weak to effect deprotonation.

(iii) Toluene, $$C_6H_5CH_3$$
Toluene contains three types of hydrogen, but the ones of interest are the benzylic hydrogens on the $$CH_3$$ group. If one benzylic proton is abstracted we obtain the benzyl carbanion

$$C_6H_5CH_2^-$$

This anion is resonance-stabilised; the negative charge can be delocalised into the aromatic ring:

$$C_6H_5CH_2^- \;\;\rightleftharpoons\;\; C_6H_4CH_2 \; \text{(multiple resonance forms)}$$

Because of this stabilisation the $$\mathrm{p}K_a$$ of a benzylic hydrogen is appreciably lower (≈41) than that of a normal alkane and, more importantly, distinctly lower than the hydrogens in cyclopropane, benzene or naphthalene. Although $$CO_3^{2-}$$ is not strong enough to drive the equilibrium completely to the right, it can abstract a benzylic proton to a noticeable extent. Symbolically we can write

$$C_6H_5CH_3 + Na_2CO_3 \;\;\rightleftharpoons\;\; C_6H_5CH_2Na + NaHCO_3$$

The equilibrium lies far to the left, but the fact that it is accessible means that toluene is the most reactive of the four compounds toward aqueous sodium carbonate.

(iv) Naphthalene, $$C_{10}H_8$$
Like benzene, every hydrogen is attached directly to an sp2 carbon of the aromatic framework. Removal of a proton would break aromaticity in both fused rings, giving an extremely unstable anion. Its $$\mathrm{p}K_a$$ is even higher than that of benzene, so no reaction with $$CO_3^{2-}$$ occurs.

Putting all these points together, only toluene possesses hydrogens that are sufficiently acidic for even the mild base $$CO_3^{2-}$$ to abstract. Therefore toluene exhibits the greatest reactivity toward an aqueous solution of sodium carbonate, while the other hydrocarbons remain virtually inert.

Hence, the correct answer is Option C.