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| 1. |
An alkyl chloride (RCI) reacts with aqueous KOH to form an alcohol predominantly while it reacts with alcoholic KOH to form an alkene predominantly. Explain these observations. |
| Answer» Solution :A STRONGER base and a less polar solvent favours ELIMINATION (E2) more than substitution `(S_(N)2)`. In such a condition, an alkene is formed predominantly. Alcohol is a much less polar solvent than water and the base `Oet^(Theta)` (formed by the reaction betweenEtOH and `OH^(-)`) PRESENT in alcoholic medium is stronger than `OH^(Theta)`. So, alkyl chloride reacts with alcoholic KOH to form an alkene predominantly by an `E_(2)` reaction. Since `OH^(Theta)` is a weaker base than `OEt^(Theta)` and water is more polar than alcohol, an alkyl chloride reacts with aqueous ALKALI to form an alcohol, predominantly by an `S_(N)2` reaction. | |