Animated Solution for Chemistry - Organic Chemistry: Which of the following reactions produce(s) propane as a major product?

Welcome to a classic organic chemistry puzzle! Our mission is straightforward but requires a sharp eye: we need to identify which of the four given chemical reactions will yield propane ($C_3{H}_8$) as the major product. Propane is a simple, three-carbon alkane. To solve this, we must act as chemical detectives, analyzing the starting materials and the specific reagents in each option to predict the final outcome.

The Dimerization Trap (Kolbe Electrolysis)

Let's begin with option A, where we are presented with sodium butyrate ($C{H}_{3}C{H}_{2}C{H}_{2}COONa$) undergoing electrolysis in an aqueous solution. This is the famous Kolbe Electrolysis.

When the current flows, the carboxylate ion migrates to the anode and loses an electron, followed by the rapid loss of a carbon dioxide ($C{O}_2$) molecule. This leaves behind a highly reactive propyl radical ($C{H}_{3}C{H}_{2}C{H}_2^{\bullet }$).

Because radicals are incredibly unstable and eager to pair up their unpaired electrons, two propyl radicals will quickly find each other and dimerize.

The result is a six-carbon chain: n-hexane. Since our target is the three-carbon propane, option A is a trap and is incorrect.

The Carbon Chopper (Decarboxylation)

Moving on to option B, we again start with sodium butyrate, but this time it is heated with soda lime (a mixture of $NaOH$ and $CaO$).

Soda lime is a notorious "carbon chopper." It facilitates a decarboxylation reaction, which literally rips the entire carboxylate group ($-COONa$) off the molecule, releasing it as sodium carbonate ($N{a}_{2}C{O}_3$). In its place, a single hydrogen atom is attached to the remaining alkyl chain.

Our four-carbon starting material loses exactly one carbon, leaving us with a pristine three-carbon alkane. Yes, that is propane! Option B is a correct answer.

The Hydrogen Swap (Reduction)

In option C, we shift gears to an alkyl halide: 1-chloropropane ($C{H}_{3}C{H}_{2}C{H}_{2}Cl$). The reagent here is a mixture of zinc ($Zn$) and dilute hydrochloric acid ($HCl$).

This combination is a classic and powerful reducing agent. The reaction between zinc and acid generates nascent hydrogen, which aggressively attacks the carbon-chlorine bond. The chlorine atom is kicked out as a chloride ion, and a hydrogen atom takes its place.

The three-carbon skeleton remains completely intact during this substitution. The result is, once again, propane. Therefore, option C is also a correct answer.

The Double Bond Forger (Dehalogenation)

Finally, let's examine option D. We are given 1,2-dibromopropane ($C{H}_{3}CH(Br)C{H}_{2}Br$), which is a vicinal dihalide (halogens on adjacent carbons). The reagent is zinc dust ($Zn$).

Zinc has a strong affinity for halogens. When heated with a vicinal dihalide, zinc will grab both bromine atoms to form stable zinc bromide ($ZnB{r}_2$).

This removal of two adjacent atoms is an elimination reaction. To satisfy their valency, the two carbons that lost their bromines must form a double bond. The product is propene ($C_3{H}_6$), an alkene, not our desired alkane. Thus, option D is incorrect.

By carefully tracing the mechanism of each reaction, we've successfully navigated the traps. Decarboxylation (Option B) and reduction (Option C) both lead us straight to propane. Kolbe electrolysis gave us a larger alkane, and dehalogenation gave us an alkene. The correct choices are (B) and (C).