When ethyl bromide reacts with sodium hydroxide, the primary outcome is a nucleophilic substitution reaction that yields ethanol (CH₃CH₂OH) and sodium bromide (NaBr). This reaction is a fundamental process in organic chemistry, demonstrating the conversion of an alkyl halide to an alcohol.
The Primary Reaction: Nucleophilic Substitution (SN2)
The main pathway involves the hydroxide ion (OH⁻) from sodium hydroxide acting as a strong nucleophile. It attacks the electrophilic carbon atom of ethyl bromide, which is bonded to the bromine. Simultaneously, the bromine atom, an excellent leaving group, departs as a bromide ion (Br⁻). This concerted, single-step process is known as a bimolecular nucleophilic substitution (SN2) reaction.
The overall chemical equation for this reaction is:
CH₃CH₂Br (aq) + NaOH (aq) → CH₃CH₂OH (aq) + NaBr (aq)
- Ethyl Bromide (CH₃CH₂Br): A primary alkyl halide, where the carbon atom attached to the bromine is slightly positive due to bromine's higher electronegativity, making it susceptible to nucleophilic attack.
- Sodium Hydroxide (NaOH): In an aqueous solution, it dissociates into Na⁺ and OH⁻ ions. The hydroxide ion acts as a potent nucleophile.
- Ethanol (CH₃CH₂OH): The desired alcohol product, formed by replacing the bromine atom with a hydroxyl (-OH) group.
- Sodium Bromide (NaBr): An ionic salt formed from the sodium cation and the bromide anion.
Kinetics of the Reaction
The rate at which this reaction proceeds is dependent on the concentrations of both reactants. Specifically, the reaction is first order with respect to ethyl bromide and first order with respect to sodium hydroxide. This means the overall reaction is second order. This kinetic profile is a hallmark of an SN2 mechanism, indicating that both the ethyl bromide and the hydroxide ion are involved in the rate-determining step.
Potential Side Reaction: Elimination (E2)
While SN2 is generally favored for primary alkyl halides like ethyl bromide when reacting with strong nucleophiles in polar protic solvents (like water), an elimination reaction (E2) can also occur to some extent, particularly under certain conditions.
In an E2 reaction, the hydroxide ion acts as a strong base, abstracting a proton from a carbon atom adjacent to the one bearing the bromine (a beta-carbon). This leads to the simultaneous formation of a carbon-carbon double bond and the departure of both the bromide ion and a water molecule.
CH₃CH₂Br (aq) + NaOH (aq) → CH₂=CH₂ (g) + NaBr (aq) + H₂O (l)
- Ethene (CH₂=CH₂): An alkene, which is typically a gaseous product.
- Water (H₂O): Formed from the abstracted proton and the hydroxide ion.
Factors Influencing Reaction Pathway
The specific conditions under which the reaction is carried out significantly influence whether substitution (SN2) or elimination (E2) predominates:
- Temperature: Higher temperatures generally favor elimination (E2) reactions due to the increased entropy associated with forming multiple product molecules, including a gas. Lower temperatures tend to favor substitution (SN2).
- Solvent: Polar protic solvents (like water, denoted by 'aq') stabilize the transition state of SN2 reactions, but strong bases/nucleophiles like hydroxide can still facilitate both pathways.
- Nature of Alkyl Halide: For primary alkyl halides such as ethyl bromide, SN2 is typically the dominant pathway due to minimal steric hindrance around the carbon atom that the nucleophile attacks.
Summary of Reaction Outcomes
| Aspect | Description |
|---|---|
| Main Reaction Type | Nucleophilic Substitution (SN2) |
| Reactants | Ethyl Bromide (CH₃CH₂Br) and Sodium Hydroxide (NaOH) |
| Primary Products | Ethanol (CH₃CH₂OH) and Sodium Bromide (NaBr) |
| Secondary Products | Ethene (CH₂=CH₂) and Water (H₂O) (from E2 elimination, favored at higher temperatures or if NaOH acts more as a base) |
| Dominant Mechanism | SN2: A concerted, single-step process where the hydroxide nucleophile attacks the carbon bearing the bromine, and the bromine simultaneously departs. This is supported by the observed kinetics. |
| Rate Law (for SN2) | Rate = k [CH₃CH₂Br]¹ [NaOH]¹ (The reaction is first order in both reactants, indicating a bimolecular rate-determining step). |
| Role of NaOH (OH⁻) | Primarily acts as a strong nucleophile to displace the bromide ion. It can also act as a strong base to abstract a proton, leading to elimination, particularly under elevated temperatures. |
Practical Insights and Applications
The reaction of ethyl bromide with sodium hydroxide holds significant practical and educational value:
- Organic Synthesis: This reaction is a classic laboratory method for synthesizing alcohols from alkyl halides. Ethanol, a key industrial solvent, fuel additive, and chemical intermediate, can be produced this way, although industrial synthesis often uses fermentation or hydration of ethene.
- Mechanism Understanding: Studying this reaction provides critical insights into the principles of SN2 kinetics, the role of nucleophiles and leaving groups, and the competition between substitution and elimination pathways in organic chemistry.
- Chemical Selectivity: By carefully controlling reaction conditions (e.g., temperature, solvent polarity, concentration), chemists can influence the selectivity of the reaction, favoring either the desired alcohol (ethanol) or the alkene (ethene). This control is crucial in the synthesis of more complex organic molecules.
