Provide the products for this reaction: CH3O- +CH3SH -> ????
Exchange the acidic proton.
The reaction between CH3O- and CH3SH can produce multiple products. Here's a step-by-step breakdown of the possible reaction products:
Step 1: The nucleophile (CH3O-) attacks the electrophile (CH3SH), resulting in the formation of a new bond:
CH3O- + CH3SH → CH3S-CH3 + OH-
The primary product formed is called dimethyl sulfide (CH3S-CH3). Additionally, hydroxide ion (OH-) is also produced as a result of the reaction.
So, the products of the reaction between CH3O- and CH3SH are dimethyl sulfide (CH3S-CH3) and hydroxide ion (OH-).
To determine the products of the reaction between CH3O- and CH3SH, we need to consider the reaction type and the properties of the reactants.
The reaction appears to be a nucleophilic substitution reaction (SN2) because a strong nucleophile (CH3O-) is reacting with a primary alkyl halide (CH3SH). In an SN2 reaction, the nucleophile attacks the substrate and displaces the leaving group simultaneously.
Now, let's break down the reaction step by step:
1. The CH3O- nucleophile attacks the carbon atom attached to the sulfur atom in CH3SH.
2. The sulfur atom, being less electronegative than oxygen, donates its electron pair to the carbon atom.
3. The leaving group (-SH) is pushed off, carrying the pair of electrons with it.
4. The nucleophile (-CH3O) replaces the leaving group in the molecule.
Based on this mechanism, the products of the reaction are CH3OCH3 (methyl methyl ether) and HS- (hydrogen sulfide).