Question

What is the major product of the reaction between 2-methyl butane and bromine in the presence of UV light?

• A. 1-bromo-2-methyl butane
• B. 1-bromo-3-methyl butane
• C. 2-bromo-2-methyl butane
• D. 2-bromo-3-methyl butane

Hint:

Bromine radicals are very selective and prefer to abstract the hydrogen that gives the most stable radical.

Answer 1

The Correct Option is C

Step 1: Bromination with Br2 under UV light proceeds by a free radical mechanism. Unlike chlorination, bromination is highly selective and strongly favors substitution at the most stable radical position, mainly a tertiary carbon if one is available. Step 2: 2-Methylbutane has the structure $CH_3-CH(CH_3)-CH_2-CH_3$. Carbon 2 is attached to three other carbon atoms, two methyl groups and one ethyl group, making it the only tertiary carbon in the molecule, while C1 and the methyl branch are primary and C3 is secondary. Step 3: Abstraction of the hydrogen at C2 by a bromine radical generates a tertiary free radical, far more stable than the primary or secondary radicals that would form at other positions, because of greater hyperconjugation and inductive donation from the surrounding alkyl groups. Step 4: This tertiary radical reacts rapidly with Br2 to place the bromine at C2 and regenerate a bromine radical to continue the chain. \[ (CH_3)_2CHCH_2CH_3 \xrightarrow[UV]{Br_2} (CH_3)_2CBrCH_2CH_3 \] Step 5: The product with bromine on the tertiary carbon, C2, bearing the methyl branch, is named 2-bromo-2-methylbutane. \[\boxed{\text{2-bromo-2-methylbutane}}\]