When does elimination and substitution happen please explain using OH reacting with 2-bromo pentane?
Always there is a competition between elimination and substitution reaction when an alkyl halide reacts with KOH. For example ethyl chloride (CH3CH2Cl) will react with potassium hydroxide(KOH) to give either ethanol (CH3CH2OH) or ethene (CH2=CH2). The competition between elimination Vs substitution is decided by the following factors.
Nature of alkyl halide: Relative primary carbocations are unstable and tertiary carbocations are stable. So elimination will preferably takes place for tertiary halide and substitution will take place for primary alkyl halide.
Nature of solvent: Water is polar solvent and will stabilise carbocation. So it will stabilise the tertiary carbocation and will favour elimination.
Nature of base: A stable nucleophile will prefer substitution only. For example comparing with OH- (hydroxide) and CH3CH2O-(ethoxide) hydroxide is stable so it will give substitution products. But ethoxide is unstable and it will give elimination products.
In our case we are taking secondary alkyl halide (2-bromo pentane). So there is 50% chance of elimination or substitution. But the reagent is OH- which will give only substitution products. So we will be getting substitution products.