3.What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:
a)One equivalent of NaI in acetone?
b)One equivalent of Silver Nitrate in ethanol?
a) Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction.
Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide.
b) Silver ion tends to suck off a halide ion and leave a carbocation, which means E1 and SN1.
If there's only one equiv, then the tertiary bromide on C4 is the one that will go. The resulting carbocation can give:
E1 products 5-Br-2-Me-2-pentene (major, trisub) and 5-Br-2-Me-1-pentene (disub, minor).
SN1 product 5-bromo-2-ethoxy-2-methylpentane
To determine the major product of the reaction between 1,4-dibromo-4-methylpentane and different reagents, we need to consider the type of reaction that will occur. Let's analyze each case:
a) One equivalent of NaI in acetone:
In this case, we are performing a nucleophilic substitution reaction, or an SN2 reaction. Sodium iodide (NaI) acts as the nucleophile, and acetone is the solvent. The reaction proceeds by the nucleophile attacking the carbon attached to the bromine atom, leading to the replacement of the bromine with iodine.
To determine the major product, we need to identify the site where the nucleophile will attack. In this case, it will attack the primary carbon (where the bromine is attached) because SN2 reactions typically occur at this position. As a result, the major product will be 1-iodo-4-methylpentane.
b) One equivalent of Silver Nitrate in ethanol:
In this case, we are performing an elimination reaction, specifically an E2 reaction. Silver nitrate (AgNO3) acts as a source of the nucleophile, while ethanol is the solvent. The reaction proceeds by the nucleophile removing a proton, resulting in the elimination of a leaving group.
To determine the major product, we need to identify the site where the proton will be removed. In this case, it will be removed from a beta carbon adjacent to one of the bromine atoms. Since we have two bromine atoms, there are two possibilities for the elimination to occur. Therefore, we will get two major products: 1-bromo-4-methylpent-2-ene and 4-bromo-3-methylpent-2-ene.
Remember, in both cases, the major product is the one that forms most abundantly based on the reaction conditions and the preference of the reaction mechanism.