JEE Main 2019 · Chemistry · Alcohols, Phenols, Ethers and Haloalkanes · medium

Propene is treated with B2H6 followed by H2O2/OH^- to give X; X is oxidised with PCC to Y; Y is treated with CH3MgBr and then hydrolysed to give Z. Z…

Single correct option

Propene is treated with

\mathrm{B_2H_6}

followed by

\mathrm{H_2O_2/OH^-}

to give X; X is oxidised with PCC to Y; Y is treated with

\mathrm{CH_3MgBr}

and then hydrolysed to give Z. Z is warmed with

\mathrm{I_2}

in aqueous NaOH. Besides iodoform, the organic product formed is

Show answer and solution

Answer: (C) $\mathrm{CH_3CH_2COONa}$

Hydroboration-oxidation gives anti-Markovnikov propan-1-ol (X); PCC stops at propanal (Y);

\mathrm{CH_3MgBr}

adds to the carbonyl to give butan-2-ol,

\mathrm{CH_3CH(OH)CH_2CH_3}

(Z). Z contains the

\mathrm{CH_3CH(OH)-}

unit, so

\mathrm{I_2/NaOH}

first oxidises it to butan-2-one and then cleaves the

\mathrm{CH_3CO-}

group:

\mathrm{CH_3COCH_2CH_3 + 3I_2 + 4NaOH \to CHI_3 + CH_3CH_2COONa + 3NaI + 3H_2O}

. Sodium propanoate is formed.

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