Question

1. In the first step of the mechanism for this process, a phenoxide anion is generated. This phenoxide anion goes on to act as a nucleophile via an SN2 mechanism, displacing the chloride on 3-chloro-1,2-propanediol. Why doesn’t the phenoxide anion act as a base to deprotonate one of the alcohols on 3-chloro-1,2-propanediol? Write a brief, specific explanation (1-2 sentences).

Answer 1

The phenoxide anion prefers acting as a nucleophile in an SN2 mechanism on 3-chloro-1,2-propanediol because the electronegative chloride creates an electrophilic carbon that is more reactive towards nucleophilic attack than the deprotonation of an alcohol.

Step-by-step explanation:

The phenoxide anion does not act as a base to deprotonate one of the alcohols on 3-chloro-1,2-propanediol because the presence of the chloride makes that carbon a more electrophilic center, which is highly susceptible to nucleophilic attack.

In an SN2 mechanism, the nucleophile favors attacking an electrophilic carbon, here significantly activated by the chloride leaving group, rather than deprotonating an alcohol which is a less electrophilic and less favorable process.

This is especially true when considering that alcohols are not particularly acidic, and thus their protons are not as easily abstracted by a base as compared to more acidic hydrogens (e.g., hydrogens adjacent to carbonyl groups).

The SN2 reaction is characterized by the simultaneous bond formation by the nucleophile and bond breaking by the leaving group, typically observed in primary alkyl halides. The phenoxide anion is a good nucleophile due to its negative charge, making it highly reactive towards electrophilic carbons, particularly against an atom that bears a good leaving group like chloride.

Answer 2

See the explanation

Step-by-step explanation:

In this case, in order to get an elimination reaction we need to have a strong base. In this case, the base is the phenoxide ion produced the phenol (see figure 1).

Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).

Finally, the phenoxide will attack the primary carbon attached to the Cl. The C-Cl bond would be broken and the C-O would be produced at the same time to get a substitution (see figure 1).