The reaction of 1-bromobutane C4H9Br with sodium hydroxide NaOH in aqueous solution is a nucleophilic substitution reaction. In this case, it is an SN2 reaction, where the hydroxide ion OH- acts as a nucleophile and attacks the electrophilic carbon atom bonded to the bromine atom. The bromine atom is a good leaving group, and it leaves as a bromide ion Br- . The products of this reaction are 1-butanol C4H9OH and sodium bromide NaBr . The overall reaction can be represented as:C4H9Br + NaOH C4H9OH + NaBrThe reason for this reaction is that the hydroxide ion is a strong nucleophile and can easily attack the electrophilic carbon atom in 1-bromobutane. The bromine atom is a good leaving group due to its relatively large size and ability to stabilize the negative charge when it leaves as a bromide ion. This results in the formation of 1-butanol and sodium bromide as the products.