On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is:

We start by recalling that when a primary aliphatic amine is heated with chloroform $$\left(CHCl_3\right)$$ in the presence of strong alcoholic potassium hydroxide, a characteristic reaction known as the Carbylamine reaction (or isocyanide test) takes place. The overall transformation produces an isocyanide (also called an isonitrile).

The general chemical equation for this reaction is first stated:

$$RNH_2 + CHCl_3 + 3KOH \;(\text{alc}) \;\rightarrow\; RNC + 3KCl + 3H_2O$$

Here $$RNH_2$$ stands for any aliphatic primary amine. Now let us see every algebraic-type step that leads to this summary equation.

1. The alcoholic $$KOH$$ provides hydroxide ions. The first step is deprotonation of the amine:

$$RNH_2 + OH^- \;\rightarrow\; RNH^- + H_2O$$

2. Simultaneously, hydroxide removes one hydrogen from chloroform, generating the powerful base chloride ion and forming the intermediate dichlorocarbene $$:CCl_2$$:

$$CHCl_3 + OH^- \;\rightarrow\; CCl_3^- + H_2O$$

$$CCl_3^- \;\rightarrow\; :CCl_2 + Cl^-$$

3. The anion $$RNH^-$$ formed in step 1 attacks the electron-deficient dichlorocarbene:

$$RNH^- + :CCl_2 \;\rightarrow\; RNH\!-\!CCl_2$$

4. Repeated action of hydroxide ions now removes two chlorides and two protons step by step, giving potassium chloride and water each time, and finally furnishing the isocyanide:

$$RNH\!-\!CCl_2 + 2OH^- \;\rightarrow\; RNC + 2Cl^- + 2H_2O$$

5. Collecting all chloride ions as potassium chloride and summing water molecules from the previous elementary processes, we arrive again at the net equation already written:

$$\boxed{RNH_2 + CHCl_3 + 3KOH \;\rightarrow\; RNC + 3KCl + 3H_2O}$$

The organic product contains the functional group $$-NC$$, which identifies an isocyanide (alkyl isocyanide).

Therefore, among the given choices, the compound formed is an alkyl isocyanide.

Hence, the correct answer is Option D.