The reaction between ethyl bromide CH3CH2Br and sodium hydroxide NaOH is a nucleophilic substitution reaction, specifically an SN2 reaction Substitution Nucleophilic Bimolecular . The mechanism for this reaction is as follows:1. Nucleophile approach: The hydroxide ion OH- from sodium hydroxide, which acts as a nucleophile, approaches the ethyl bromide molecule. The oxygen atom in the hydroxide ion has a lone pair of electrons, making it electron-rich and attracted to the partially positive carbon atom in the ethyl bromide molecule. This carbon atom is partially positive because the bromine atom is more electronegative, pulling electron density away from the carbon.2. Transition state formation: As the hydroxide ion gets closer to the carbon atom, a transition state forms. In this transition state, the carbon atom is partially bonded to both the bromine atom and the oxygen atom of the hydroxide ion. The bond between the carbon and bromine is in the process of breaking, while the bond between the carbon and oxygen is in the process of forming.3. Bond formation and leaving group departure: The bond between the carbon atom and the oxygen atom of the hydroxide ion becomes stronger, while the bond between the carbon atom and the bromine atom weakens. Eventually, the bond between the carbon and bromine breaks completely, and the bromine atom leaves as a bromide ion Br- . The bond between the carbon and oxygen is now fully formed, resulting in the formation of ethanol CH3CH2OH .4. Overall reaction: The overall reaction can be represented as follows:CH3CH2Br + OH- CH3CH2OH + Br-In summary, the mechanism of the nucleophilic substitution reaction between ethyl bromide and sodium hydroxide involves the hydroxide ion acting as a nucleophile, attacking the partially positive carbon atom in the ethyl bromide molecule, forming a transition state, and ultimately resulting in the formation of ethanol and a bromide ion.