JEE Main 2022 · 24 Jun · Shift 2 · Chemistry Q89

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ChemistryInteger+4 / -1Question 89 of 90

$\text{(structure)}\xrightarrow[\mathrm{H_2O}]{\mathrm{NaOH}}\underset{\text{(Major Product)}}{\mathrm P}$
Consider the above reaction. The number of $\pi$ electrons present in the product 'P' is ______.

Reactant: 4-chloro-3-methylpent-1-ene, CH3-CHCl-CH(CH3)-CH=CH2, drawn as a skeletal formula, treated with NaOH/H2O to give P.

Your answer

Answer and worked solution

Official answer: 2

Substrate. $\mathrm{CH_3CHCl\text{-}CH(CH_3)\text{-}CH=CH_2}$ , a secondary chloride.

Hydrolysis goes through the cation: the secondary cation takes a 1,2-hydride shift to the tertiary, allylic position, and water attacks the allyl cation. Whatever the site of attack, the product alcohol keeps exactly one C=C.

$\pi$ electrons $2$ .

Answer: 2

Topics:SN1 and SN2 Reactive intermediates

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JEE Main 2022 · 24 Jun · Shift 2, Chemistry question 89