what is the major product if 2,3,3-trimethylpentane was mixed with IBr?
To determine the major product when 2,3,3-trimethylpentane reacts with IBr, we need to consider the possibility of substitution reactions. In this case, IBr is a strong electrophile, meaning it has a positive charge and tends to attack nucleophiles (electron-rich regions).
To analyze the reaction, the first step is to identify the groups that can act as potential nucleophiles. In 2,3,3-trimethylpentane, there are three methyl groups (CH3) attached to the carbon chain. The presence of bulky methyl groups makes it difficult for the nucleophile to attack directly on that carbon atom.
Next, we need to identify the possible sites of attack in the molecule. In this specific case, the tertiary carbon atom (marked with *) adjacent to the three methyl groups is the most likely site for substitution.
CH3
|
CH3* - C - CH(CH3)2
|
CH3
Since IBr is a halogen, it can dissociate into I+ (iodonium ion) and Br- in a polar solvent such as acetic acid or acetone. The nucleophile, in this case, would be the Br- ion.
The Br- ion will attack the tertiary carbon atom, leading to the displacement of one of the methyl groups:
CH3
|
CH3 - C - CH(CH3)2 + Br- --> CH3 - C - CH(CH3)2 + Br
As a result, the major product in this reaction would be 2-bromo-2,3,3-trimethylpentane.
It's important to note that this explanation is based on the assumption that iodination proceeds via a substitution mechanism rather than a radical mechanism. Reaction conditions and different reactants may influence the specific outcome of the reaction.