0.44 g of a monohydric alcohol when added to | vedclass

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Question

(A) At $S.T.P.$,$22400 \ cm^{3}$ of methane corresponds to $1 \ mol$.Therefore,$112 \ cm^{3}$ of methane corresponds to $\frac{112}{22400} = 0.005 \ m

Vedclass · Accepted Answer

$(CH_{3})_{3}CCH_{2}OH$

Vedclass · Answer

(A) At $S.T.P.$,$22400 \ cm^{3}$ of methane corresponds to $1 \ mol$.Therefore,$112 \ cm^{3}$ of methane corresponds to $\frac{112}{22400} = 0.005 \ mol$.Since $1 \ mol$ of monohydric alcohol reacts with $1 \ mol$ of methylmagnesium iodide to produce $1 \ mol$ of methane,the moles of alcohol used is $0.005 \ mol$.The molar mass of the alcohol is $\frac{	ext{Given mass}}{	ext{moles}} = \frac{0.44 \ g}{0.005 \ mol} = 88 \ g/mol$.Among the options,$CH_{3}CH(OH)CH_{2}CH_{3}$ (butan$-2-$ol) has a molar mass of $74 \ g/mol$,while $C_{5}H_{12}O$ isomers have $88 \ g/mol$. The alcohol must be a primary alcohol to form an aldehyde with $PCC$ that gives a positive silver mirror test. The structure $CH_{3}CH(CH_{3})CH_{2}OH$ ($2$-methylpropan$-1-$ol) has a molar mass of $74 \ g/mol$. Re-evaluating the molar mass $88 \ g/mol$ corresponds to $C_{5}H_{12}O$. The alcohol $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH$ or similar isomers fit. However,based on the requirement of forming an aldehyde that gives a silver mirror test,the alcohol must be primary. The correct structure is $CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH$ or similar. Given the options provided,the calculation leads to $88 \ g/mol$.

$0.44 \ g$ of a monohydric alcohol when added to methylmagnesium iodide in ether liberates $112 \ cm^{3}$ of methane at $S.T.P.$ With $PCC$,the same alcohol forms a carbonyl compound that answers the silver mirror test. The monohydric alcohol is:

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