An alkene X with formula C_4H_8 does not exhi | vedclass

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(A) Since $X$ $(C_4H_8)$ does not exhibit geometrical isomerism,it must be $2$-methylpropene,$CH_2=C(CH_3)_2$.For $Y$ to give the iodoform test,it mus

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$BH_3, H_2O_2 / OH^{-}, \quad PCC$

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(A) Since $X$ $(C_4H_8)$ does not exhibit geometrical isomerism,it must be $2$-methylpropene,$CH_2=C(CH_3)_2$.For $Y$ to give the iodoform test,it must contain a $CH_3CH(OH)-$ group or a $CH_3CO-$ group.Using reagent $A = BH_3, H_2O_2 / OH^-$ (hydroboration-oxidation) on $CH_2=C(CH_3)_2$ gives anti-Markovnikov addition of water,resulting in $2$-methylpropan-$1$-ol,$(CH_3)_2CHCH_2OH$.Using reagent $B = PCC$ (pyridinium chlorochromate) oxidizes the primary alcohol to an aldehyde,$2$-methylpropanal,$(CH_3)_2CHCHO$,which does not give the iodoform test.Using reagent $A = H_2O / H^+$ (acid-catalyzed hydration) on $CH_2=C(CH_3)_2$ gives Markovnikov addition of water,resulting in $2$-methylpropan-$2$-ol,$(CH_3)_3COH$.Using reagent $B = Cu / 573 \ K$ on a tertiary alcohol leads to dehydration to the alkene,not a ketone.However,if we use $A = BH_3, H_2O_2 / OH^-$ to get $(CH_3)_2CHCH_2OH$ and then oxidize it,we don't get a methyl ketone. Let's re-evaluate: $2$-methylpropan-$2$-ol does not give the iodoform test. $2$-methylpropan-$1$-ol does not give the iodoform test. The only way to get a methyl ketone from $C_4H_8$ is to form butan-$2$-ol and oxidize it. But $X$ is $2$-methylpropene. Wait,$CH_3CH_2CH=CH_2$ (but$-1-$ene) shows geometrical isomerism. $CH_3CH=CHCH_3$ (but$-2-$ene) shows geometrical isomerism. Thus $X$ must be $2$-methylpropene. The only alcohol from $2$-methylpropene is $2$-methylpropan-$2$-ol. This does not give iodoform. There might be a typo in the question options,but based on standard reagents,$A$ followed by $B$ in option $A$ is the most common sequence for alcohol synthesis and oxidation.

An alkene $X$ with formula $C_4H_8$ does not exhibit geometrical isomerism. In the conversion of $X$ to $Y$,the correct sequence of reagents $A$ and $B$ used are ($Y$ gives iodoform test):
$X$ $\xrightarrow{A} \text{Intermediate}$ $\xrightarrow{B} Y$

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An alkene $X$ with formula $C_4H_8$ does not exhibit geometrical isomerism. In the conversion of $X$ to $Y$,the correct sequence of reagents $A$ and $B$ used are ($Y$ gives iodoform test): $X$ $\xrightarrow{A} \text{Intermediate}$ $\xrightarrow{B} Y$