0.1 mole of compound 'S' will weigh ......... | vedclass

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(A) $1$. The starting material is $2$-(hydroxymethyl)cyclopentanol. Oxidation with excess $CrO_3$ gives $2$-oxocyclopentanecarboxylic acid $(P)$.$2$.

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) $1$. The starting material is $2$-(hydroxymethyl)cyclopentanol. Oxidation with excess $CrO_3$ gives $2$-oxocyclopentanecarboxylic acid $(P)$.$2$. Protection of the ketone with ethylene glycol gives $(Q)$.$3$. Reaction with $CH_3MgBr$ followed by acid workup removes the protecting group and gives $2$-oxocyclopentanecarboxylic acid $(R)$ (as the Grignard reagent reacts with the acid group,but the overall sequence regenerates the keto-acid).$4$. Reduction of $(R)$ with $NaBH_4$ reduces the ketone to an alcohol,yielding $2$-hydroxycyclopentanecarboxylic acid $(S)$.$5$. The molecular formula of $(S)$ is $C_6H_{10}O_3$.$6$. Molar mass of $(S) = (6 	imes 12) + (10 	imes 1) + (3 	imes 16) = 72 + 10 + 48 = 130 \ g \ mol^{-1}$.$7$. Weight of $0.1$ mole of $(S) = 0.1 \ mol 	imes 130 \ g \ mol^{-1} = 13 \ g$.

$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.
(Given molar mass in $g \ mol^{-1} \ C: 12, H: 1, O: 16$)

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$0.1$ mole of compound $'S'$ will weigh $.......... \ g$.(Given molar mass in $g \ mol^{-1} \ C: 12, H: 1, O: 16$)