Reaction of CH3CH2CH2Br with aq.KOH

When CH3CH2CH2Br (1-bromopropane) is reacted with aqueous potassium hydroxide (KOH), it undergoes a nucleophilic substitution reaction known as an SN2 reaction. This reaction results in the formation of an alcohol, CH3CH2CH2OH (1-propanol), and the potassium salt of bromide, KBr.

SN2 Reaction Mechanism
The SN2 reaction mechanism involves a direct attack of the hydroxide ion (OH-) on the carbon atom bonded to the bromine atom. This attack occurs from the backside of the carbon-bromine bond, leading to the inversion of the stereochemistry.

Reaction Steps:
1. Nucleophilic Attack: The hydroxide ion acts as a nucleophile and attacks the carbon atom bonded to the bromine. This attack displaces the bromine atom from the carbon, forming a transition state.

2. Transition State: In the transition state, the hydroxide ion partially bonds to the carbon atom, while the bromine atom begins to detach. This transition state is a high-energy species.

3. Bond Formation: As the bromine atom fully detaches, a new bond forms between the carbon and the hydroxide ion. This leads to the formation of 1-propanol.

4. Product Formation: The final product is 1-propanol, an alcohol, in which the hydroxide ion has replaced the bromine atom. Additionally, potassium bromide (KBr) is formed as a byproduct.

Overall Reaction Equation:
CH3CH2CH2Br + KOH → CH3CH2CH2OH + KBr

The presence of aqueous KOH facilitates the reaction by providing hydroxide ions as the nucleophile. The potassium ion (K+) from KOH does not participate in the reaction but forms a salt with the bromide ion.

In conclusion, the reaction of CH3CH2CH2Br with aqueous KOH results in the substitution of the bromine atom by a hydroxide ion, leading to the formation of 1-propanol and potassium bromide. This reaction follows the SN2 mechanism, where the hydroxide ion attacks the carbon atom bonded to the bromine from the backside.