R_f value for 2-methylpropene in a solvent sy | vedclass

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Question

(D) The reaction of $2$-methylpropene with dilute $H_2SO_4$ is an acid-catalyzed hydration reaction which follows Markovnikov's rule to yield $2$-meth

Vedclass · Accepted Answer

$0.12$

Vedclass · Answer

(D) The reaction of $2$-methylpropene with dilute $H_2SO_4$ is an acid-catalyzed hydration reaction which follows Markovnikov's rule to yield $2$-methylpropan-$2$-ol (tert-butyl alcohol) as the major product $(X)$.In thin-layer chromatography $(TLC)$,the $R_f$ value is inversely related to the polarity of the compound when using a polar stationary phase like silica gel.$2$-methylpropene is a non-polar hydrocarbon,whereas $2$-methylpropan-$2$-ol is a polar alcohol capable of forming hydrogen bonds with the stationary phase.Due to stronger interactions with the polar stationary phase,the polar product $(X)$ will travel a shorter distance compared to the non-polar reactant.Therefore,the $R_f$ value of $(X)$ must be significantly lower than $0.42$.Among the given options,$0.12$ is the only value lower than $0.42$.

$R_f$ value for $2$-methylpropene in a solvent system (Ethyl acetate + ether) is $0.42$. $2$-methylpropene is treated with dilute $H_2SO_4$ to give major organic product $(X)$. $R_f$ value for $(X)$ in the same solvent system under identical condition will be:

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