JEE Main 2022 · 27 Jul · Shift 2 · Chemistry Q73

← JEE Main 2022 · 27 Jul · Shift 2

ChemistrySingle correct+4 / -1Question 73 of 90

Major product 'B' of the following reaction sequence is :

$\mathrm{CH_3-\underset{\displaystyle CH_3}{\underset{|}{C}}=CH-CH_3\xrightarrow[\ CH_3OH\ ]{Br_2}A\xrightarrow{HI}\underset{\text{(major product)}}{B}}$

\mathrm{(CH_3)_2C(OH)-CHBr-CH_3}

\mathrm{(CH_3)_2CI-CHBr-CH_3}

\mathrm{(CH_3)_2CBr-CH(OH)-CH_3}

\mathrm{(CH_3)_2CBr-CHI-CH_3}

Answer and worked solution

Official answer: B

Step 1. $\mathrm{Br_2/CH_3OH}$ forms a bromonium ion, and methanol attacks the more substituted carbon: A is $\mathrm{(CH_3)_2C(OCH_3){-}CHBr{-}CH_3}$ .

Step 2. HI cleaves the tertiary ether by $S_N1$ , putting I on the tertiary carbon: B is $\mathrm{(CH_3)_2CI{-}CHBr{-}CH_3}$ .

(B)

Topics:Alkenes Ethers

← Previous question Next question →

JEE Main 2022 · 27 Jul · Shift 2, Chemistry question 73