The mechanism of the given reaction, RCH2CH2-ONa + CS2 + CH3-I → ∆ → R-CH=CH2, involves several steps that lead to the formation of an alkene, R-CH=CH2. Let's break down the reaction into its individual steps and explain each of them in detail.

1. Formation of Sodium Dialkyl Dithiocarbamate:
The first step involves the reaction between RCH2CH2-ONa (sodium alkoxide) and CS2 (carbon disulfide). This reaction forms a sodium dialkyl dithiocarbamate intermediate, which can be represented as RCH2CH2-N(CS2)SR, where R is an alkyl group.

2. Formation of Alkyl Isocyanide:
In the next step, the sodium dialkyl dithiocarbamate reacts with CH3-I (methyl iodide). This reaction is an S-alkylation process, where CH3-I replaces the sulfur atom in the dithiocarbamate, forming an alkyl isocyanide intermediate. The reaction can be represented as RCH2CH2-N(CS2)SCH3.

3. Elimination of Sulfur:
The alkyl isocyanide intermediate undergoes thermal decomposition (∆) to eliminate sulfur, resulting in the formation of the desired alkene, R-CH=CH2. This elimination reaction is known as the von Richter reaction.

Overall Reaction:
RCH2CH2-ONa + CS2 + CH3-I → RCH2CH2-N(CS2)SCH3 → ∆ → R-CH=CH2

To summarize, the reaction involves the formation of a sodium dialkyl dithiocarbamate intermediate by reacting sodium alkoxide with carbon disulfide. The intermediate then undergoes S-alkylation with methyl iodide to form an alkyl isocyanide. Finally, the alkyl isocyanide undergoes thermal decomposition to eliminate sulfur and yield the desired alkene.

RCH2CH2ONA+CS2+CH3---(Mec-zn/na)---∆---(mec.-deep HCL)---CH2CH3