1-bromobutane reacts with alcoholic Koh to give
potassium bromide and butanal.
When 1-bromobutane reacts with alcoholic KOH (potassium hydroxide), it undergoes an elimination reaction called the E2 (bimolecular elimination) reaction. During this reaction, the bromine atom is replaced by a hydrogen atom from the alcohol. Here is the step-by-step reaction:
Step 1: Deprotonation
The alcohol (ROH) reacts with KOH to form an alkoxide ion (RO^-) and water (H2O). This step involves the deprotonation of alcohol using a strong base (KOH). The reaction is as follows:
ROH + KOH ⟶ RO^- + H2O
Step 2: Nucleophilic attack
The alkoxide ion (RO^-) acts as a nucleophile and attacks the carbon atom bearing the bromine (C-Br) in 1-bromobutane. The bromine atom leaves as a bromide anion (Br^-). This step is known as the nucleophilic attack. The reaction is as follows:
RO^- + CH3(CH2)2CH2Br ⟶ CH3(CH2)2CH2OR + Br^-
Step 3: Elimination
In the final step, a proton from the adjacent carbon atom (beta carbon) is removed by the alkoxide ion (RO^-). Simultaneously, the bromide ion (Br^-) acts as a base and abstracts the acidic hydrogen, resulting in the formation of an alkene. This step is referred to as elimination due to the loss of a leaving group (Br^-) and a proton (H+). The reaction is as follows:
CH3(CH2)2CH2OR ⟶ CH3(CH2)2CH=CH2 + ROH
The final product of the reaction is an alkene, specifically 1-butene, and alcohol as a byproduct.