Question

If ethanol will be used in elimination reaction of halogenoalkanes it will produce

Answer 1

Answer:

Explanation:

If a halogenoalkane is heated under reflux with a solution of sodium or potassium hydroxide, the halogen is replaced by -OH and alcohol is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent the loss of volatile substances from the mixture.

The solvent is usually a 50/50 mixture of ethanol and water, because everything will dissolve in that. The halogenoalkane is insoluble in water. If you used water alone as the solvent, the halogenoalkane and the sodium hydroxide solution wouldn't mix and the reaction could only happen where the two layers met.

For example, using 1-bromopropane as a typical primary halogenoalkane:

You could write the full equation rather than the ionic one, but it slightly obscures what's going on:

Answer 2

If ethanol will be used in the elimination reaction of halogenoalkanes, it will produce an alkene.

Explanation:

• Halogenoalkanes or alkyl halides undergo an elimination reaction in the presence of a strong base like an alkoxide ion ($RO^{-}$).
• This reaction results in the formation of alkene, alcohol and hydrogen halide.
• The reaction can be given as:

        $\[C{H_3}C{H_2}Cl + {C_2}{H_5}OH \to C{H_2} = C{H_2} + \[{C_2}{H_5}OH\] + HCl\]$

• Thus, the product of the reaction is an alkene.
• Weak bases yield poor elimination products.
• Also, the elimination reaction of alkyl halides competes with the substitution reaction in the case of nucleophiles.

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