JEE Advanced levelSingle correct optionDifficulty 3 of 3o4128
A tertiary alkyl bromide reacts in aqueous ethanol at 50 ∘C by a first-order process with half-life 3.0 h. Product analysis shows that 24% of the substrate consumed goes to alkene (E1) and the remaining 76% to substitution products (SN1). Take ln2=0.693. The first-order rate constant of the elimination pathway alone, in units of 10−5s−1, is closest to
A1.54
B4.88
C2.22
D6.42
Answer: A
The total rate constant is k=0.693/(3.0×3600s)=6.42×10−5s−1. The branching fractions are product ratios, so kE1=0.24k. Hence kE1=1.54×10−5s−1.