The major monobromination product in the following reaction in UV light would be?
(CH3)3CCH2CH3 + Br2 -------------------> Product
a.1-Bromo 3.3 dimethylbutane
b.2-Bromo 2,2 dimethylbutane
c.1-Bromo 2, 2 Dimethylbutane
d. 2-Bromo 2 methylpentane
e. None are correct
c. 1-Bromo 2, 2 Dimethylbutane
To determine the major monobromination product in the given reaction, we need to consider the reactivity of the substrate and the stability of the resulting product.
In this case, we have (CH3)3CCH2CH3 as the substrate, which is a tertiary alkyl group. Tertiary carbons are more stable and less reactive compared to primary or secondary carbons due to the presence of three alkyl groups attached to the carbon.
When Br2 reacts with (CH3)3CCH2CH3 in the presence of UV light, a substitution reaction occurs where one hydrogen atom is replaced by a bromine atom. Considering the reactivity and stability, the major monobromination product would be the one that results in the most stable compound.
The product that would be formed is 2-Bromo 2,2-dimethylbutane, which is represented by option b. This product is more stable due to the presence of two methyl groups on the carbon attached to bromine, providing additional electron-donating groups that stabilize the positive charge on the carbon.
To determine the major monobromination product in the given reaction, we need to consider the reactivity and the regioselectivity of the reaction.
In general, free radical bromination of alkanes occurs via a three-step mechanism: initiation, propagation, and termination. In the initiation step, UV light is used to generate bromine radicals (Br•) from diatomic bromine (Br2). These bromine radicals react with the alkane (in this case, (CH3)3CCH2CH3) in the propagation step, leading to the formation of various brominated products. Finally, the termination step involves the combination of two radical species to form stable products.
In terms of regioselectivity, the major monobromination product will be the one where the bromine atom is bonded to the most substituted carbon atom. This is due to the relative stability of the resulting carbon-centered radical intermediate.
Now, let's analyze the given options:
a. 1-Bromo 3.3 dimethylbutane: In this product, the bromine is attached to one of the tertiary carbons. This is a possible outcome in the reaction, but we need to evaluate the other options to determine if it is the major product.
b. 2-Bromo 2,2 dimethylbutane: This product shows a bromine atom attached to one of the secondary carbons. Again, we need to evaluate the other options.
c. 1-Bromo 2,2 Dimethylbutane: Here, the bromine atom is bonded to the primary carbon. This can be a possible outcome, but it is less likely to be the major product due to the relatively unstable primary carbon radical intermediate.
d. 2-Bromo 2-methylpentane: This product has the bromine attached to one of the tertiary carbons. This is the same as option a, so we can exclude both options a and d.
e. None are correct: This option suggests that none of the given products are the major monobromination product.
Based on the analysis, the most likely major monobromination product in the given reaction would be option b. 2-Bromo 2,2 dimethylbutane, where the bromine is bonded to one of the secondary carbons.
It is worth noting that the regioselectivity of free radical reactions can be influenced by factors such as steric hindrance and stability of the intermediate radicals. Therefore, it is always crucial to consider these factors in determining the major product.