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: A practical approach for radical halogenation questions is to rank every distinct type of hydrogen in the molecule by the stability of the radical it would generate if removed, since bromine radicals are reactive but highly selective. Step 2: Label the carbons of 2-methylbutane: $C1(H_3)-C2(H)(CH_3)-C3(H_2)-C4(H_3)$, where C2 also carries a methyl substituent. This gives four kinds of hydrogens: those on C1 (primary), on the methyl branch off C2 (primary), on C2 itself (tertiary, since C2 touches three carbons), and on C3 (secondary). Step 3: The relative rate of hydrogen abstraction by a bromine radical roughly follows tertiary much greater than secondary much greater than primary, and bromine's selectivity, unlike chlorine, is large enough that this preference dominates the product distribution even though there are more primary and secondary hydrogens present statistically. Step 4: Since C2 offers the only tertiary hydrogen in the molecule, abstraction there is overwhelmingly favored kinetically, forming the tertiary radical $(CH_3)_2\dot{C}-CH_2CH_3$. Step 5: This radical combines with a bromine molecule to deliver the bromine atom back to C2, closing the catalytic cycle. \[ (CH_3)_2\dot{C}CH_2CH_3 + Br_2 \rightarrow (CH_3)_2CBrCH_2CH_3 + Br^{\bullet} \] Step 6: Naming this product from the longest chain through the substituted carbon gives 2-bromo-2-methylbutane, matching the position where the most stable radical had formed. \[\boxed{\text{2-bromo-2-methylbutane}}\]