An organic compound X with molecular formula | vedclass

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Question

(A) The molecular formula $C_6H_{10}$ corresponds to the general formula $C_nH_{2n-2}$,indicating an alkyne. Treatment with $HBr$ to form a $gem$-dibr

Vedclass · Accepted Answer

hex$-1-$yne

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(A) The molecular formula $C_6H_{10}$ corresponds to the general formula $C_nH_{2n-2}$,indicating an alkyne. Treatment with $HBr$ to form a $gem$-dibromide indicates that the alkyne is a terminal alkyne (Markownikoff addition). Acid-catalyzed hydration of a terminal alkyne using $HgSO_4$ and dil. $H_2SO_4$ produces a methyl ketone. $A$ methyl ketone $(CH_3CO-R)$ gives a positive iodoform test. Among the options,hex$-1-$yne $(CH_3CH_2CH_2CH_2C \equiv CH)$ is a terminal alkyne. Upon hydration,it forms hexan$-2-$one $(CH_3CH_2CH_2CH_2COCH_3)$,which is a methyl ketone and gives a positive iodoform test. Therefore,the compound $X$ is hex$-1-$yne.

An organic compound $X$ with molecular formula $C_6H_{10}$,when treated with $HBr$,forms a $gem$-dibromide. The compound $X$ upon warming with $HgSO_4$ and dil. $H_2SO_4$,produces a ketone,which gives a positive iodoform test. The compound $X$ is

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