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draw the major elimination product formed in the reaction. a 4 carbon chain where carbon 1 is bonded to benzene, carbon 2 has a bromide substituent, and carbon 3 has a methyl substituent. this reacts with potassium methoxide in d m s o to give the product.
draw the major elimination product formed in the reaction. a 4 carbon chain where carbon 1 is bonded to benzene, carbon 2 has a bromide substituent, and carbon 3 has a methyl substituent. this reacts with potassium methoxide in d m s o to give the product.
Theproductwill have a 3-carbon chain attached to thebenzenering. The methyl substituent remains unchanged.The reaction you described involves the elimination of a bromide substituent from a 4-carbon chain attached to a benzene ring. The reaction takes place withpotassium methoxide(KOMe) in DMSO (dimethyl sulfoxide).To draw the major elimination product, we need to consider the elimination reaction, which is typically an E₂ (bimolecular elimination) reaction. In this reaction, the bromide (Br) is eliminated along with a proton (H) from the neighboring carbon.Benzene|1-Carbon|2-Bromide|3-MethylWhen treated withpotassium methoxidein DMSO, the bromide is eliminated, and a double bond is formed between the neighboring carbons (1 and 2).The product will have a 3-carbon chain attached to the benzene ring. The methyl substituent remains unchanged.The major elimination product can be drawn as follows:Benzene|1-Carbon|2-Double bond|3-MethylThe specific reaction conditions and reaction mechanism can affect the outcome.To know more aboutbenzene, visit:brainly.com/question/31837011#SPJ11...