The SN1 reaction yields Entry field with incorrect answer two (number) products. This is because ______________________. Entry f
ield with incorrect answer The carbocation that forms can react with CH3OH from the top side of the molecule. The carbocation that forms can react with H2O from either the top or bottom side of the molecule. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from the top side of the molecule. The carbocation that forms can react with H2O from the bottom side of the molecule. The carbocation that forms can react with CH3OH from either the top or bottom side of the molecule. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from the bottom side of the molecule. The carbocation that forms can react with H2O from the top side of the molecule. The carbocation that forms can react with CH3OH from the bottom side of the molecule. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule.
Four substitution products are obtained. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule
Explanation:
An SN1 reaction usually involves the formation of a carbocation in the slow rate determining step. This carbocation is now attacked by a nucleophile in a subsequent fast step to give the desired product.
However, the product is obtained as a racemic mixture because the nucleophile may attack from the top or bottom of the carbocation hence both attacks are equally probable.
The attacking nucleophile in this case may be water or CH3OH
When it is heated it decomposes into mercury and oxygen gas. The mercury oxide reactant becomes the silver color of mercury. Hence, a color change can be noticed throughout the reaction.
The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the remaining electron in the 3s. Therefore the sodium electron configuration will be 1s22s22p63s1.
