2-Bromopen tane is heated with potassium ethoxide in ethanol. The majo...
Potassium ethoxide is a strong base, and 2-bromopentane is a 2º bromide, so elimination raction predominates
Since trans- alkene is more stable than cis.thus trans-pentene -2 is the main product.
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2-Bromopen tane is heated with potassium ethoxide in ethanol. The majo...
Reaction: 2-Bromopentane + Potassium ethoxide → trans-2-pentene + Potassium bromide + Ethanol
Explanation:
The reaction involves the elimination of a bromide ion (Br-) from 2-bromopentane, resulting in the formation of an alkene. The reaction is carried out in the presence of potassium ethoxide (C2H5OK) in ethanol as the solvent.
Step 1: Formation of Potassium Bromide
- The reaction starts with the dissociation of potassium ethoxide in ethanol solvent:
C2H5OK → C2H5O- + K+
- The potassium ion (K+) acts as a catalyst in the reaction.
- The bromide ion (Br-) is displaced from 2-bromopentane and reacts with the potassium ion to form potassium bromide (KBr):
C5H11Br + K+ → C5H11 + KBr
Step 2: Elimination of Bromide Ion
- The alkoxide ion (C2H5O-) acts as a strong base and abstracts a proton from the carbon adjacent to the bromine atom in 2-bromopentane.
- This results in the formation of an alkene (trans-2-pentene) and the elimination of a bromide ion:
C5H11 + C2H5O- → C5H10 + C2H5OH
Step 3: Formation of trans-2-pentene
- The elimination of the bromide ion from 2-bromopentane leads to the formation of trans-2-pentene.
- In trans-2-pentene, the double bond is present between the second and third carbon atoms, and the two methyl groups are on opposite sides of the double bond.
- This is the major product obtained in the reaction.
Conclusion:
Upon heating 2-bromopentane with potassium ethoxide in ethanol, the major product obtained is trans-2-pentene. This is due to the elimination of a bromide ion from 2-bromopentane, resulting in the formation of the alkene trans-2-pentene.