a) Give the major elimination product derived from each of the following elimination reactions. Pay attention to stereochemistry where appropriate.
2-Bromo-3-methylbutane
(H3CH2C-O(-)Na(+))-> ethanol
[Ethanol is below the arrow while (H3CH2C - ONa) is on top. the plus sign is above the N and the minus sign is above the O]
b) CH2CH2CH2CH2CH2CCH2I (H3C)3C-O(-)K(+) /(H3C)3C-OH -->
[The (H3C)3C-OK is on top of the arrow while the (H3C)3C-OH ]
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To determine the major elimination product derived from each of the given elimination reactions, we need to identify the leaving group and the strongest base present.
a) In the first reaction, the leaving group is the bromine atom (Br) attached to the 2nd carbon (counting from the end that yields the highest priority substituent). The strongest base present is the negatively charged oxygen atom (O-) in ethanol.
To get the major elimination product, we need to eliminate the leaving group (Br) with the base (O-) to form a new double bond. The elimination reaction occurs by removing a hydrogen atom from the β-carbon and forming a new bond between the β-carbon and the α-carbon.
The major elimination product from 2-bromo-3-methylbutane using ethanol as the base is 2-methyl-2-butene.
b) In the second reaction, the leaving group is the iodine atom (I) attached to the 6th carbon. The strongest base present is the negatively charged oxygen atom (O-) in (H3C)3C-O(-)K(+).
To obtain the major elimination product, we need to eliminate the leaving group (I) with the base (O-) to form a new double bond. The elimination reaction occurs by removing a hydrogen atom from the β-carbon and forming a new bond between the β-carbon and the α-carbon.
The major elimination product from CH2CH2CH2CH2CH2CCH2I using (H3C)3C-O(-)K(+) as the base is 3-hexyne.
Remember, when drawing elimination products, it is important to pay attention to stereochemistry when appropriate. However, stereochemical information is not provided for the given reactions, so we cannot determine the stereochemistry of the products.