An organic compound (X) has an empirical form | vedclass

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(B) The empirical formula is $C_4H_8O$. The compound $(X)$ gives a pale yellow precipitate with iodine in $NaOH$ solution,which indicates the presence

Vedclass · Accepted Answer

$CH_2=CHCH(OH)CH_3$

Vedclass · Answer

(B) The empirical formula is $C_4H_8O$. The compound $(X)$ gives a pale yellow precipitate with iodine in $NaOH$ solution,which indicates the presence of a methyl ketone group or a secondary alcohol that can be oxidized to a methyl ketone (iodoform test). Among the options,$CH_2=CHCH(OH)CH_3$ is a secondary alcohol. Upon oxidation,it forms $CH_2=CHCOCH_3$ (methyl vinyl ketone),which contains a methyl ketone group $(CH_3CO-)$ and thus gives a positive iodoform test. The reaction is: $CH_2=CHCH(OH)CH_3$ $\xrightarrow{	ext{Oxidation}} CH_2=CHCOCH_3$ $\xrightarrow{I_2/NaOH} CHI_3 + CH_2=CHCOO^-Na^+$.

An organic compound $(X)$ has an empirical formula $C_4H_8O$. This gives a pale yellow precipitate with iodine in $NaOH$ solution. What is $X$?

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