The reaction of C2H5Br with alcoholic KOH (potassium hydroxide) results in an elimination reaction, specifically a dehydrohalogenation. Here are the steps:

1. C2H5Br reacts with KOH. The KOH, being a strong base, will abstract a proton (H+) from the ethyl group (C2H5).

2. This results in the formation of water (H2O) and a carbocation intermediate (C2H4+ and Br-).

3. The negatively charged bromide ion (Br-) then leaves, creating a double bond between the two carbon atoms.

4. The final product is ethene (C2H4).

So, C2H5Br + alc.KOH gives C2H4 + H2O + KBr.

Question

The reaction of C2H5Br with alcoholic KOH (potassium hydroxide) results in an elimination reaction, specifically a dehydrohalogenation. Here are the steps:

1. C2H5Br reacts with KOH. The KOH, being a strong base, will abstract a proton (H+) from the ethyl group (C2H5).

2. This results in the formation of water (H2O) and a carbocation intermediate (C2H4+ and Br-).

3. The negatively charged bromide ion (Br-) then leaves, creating a double bond between the two carbon atoms.

4. The final product is ethene (C2H4).

So, C2H5Br + alc.KOH gives C2H4 + H2O + KBr.

Knowee AI · Accepted Answer

The reaction of C2H5Br with alcoholic KOH (potassium hydroxide) results in an elimination reaction, specifically a dehydrohalogenation. Here are the steps:

1. C2H5Br reacts with KOH. The KOH, being a strong base, will abstract a proton (H+) from the ethyl group (C2H5).

2. This results in the formation of water (H2O) and a carbocation intermediate (C2H4+ and Br-).

3. The negatively charged bromide ion (Br-) then leaves, creating a double bond between the two carbon atoms.

4. The final product is ethene (C2H4).

So, C2H5Br + alc.KOH gives C2H4 + H2O + KBr.

C2H5Br+alc.KOH gives ?

Question

Solution

Similar Questions

Upgrade your grade with Knowee