8.8 g of monohydric alcohol added to ethyl m | vedclass

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Question

(B) The reaction of monohydric alcohol $(R-OH)$ with ethyl magnesium iodide $(C_2H_5MgI)$ is: $R-OH + C_2H_5MgI \rightarrow R-OMgI + C_2H_6 \uparrow$.

Vedclass · Accepted Answer

$2, 2$-dimethylpropan-$1$-ol

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(B) The reaction of monohydric alcohol $(R-OH)$ with ethyl magnesium iodide $(C_2H_5MgI)$ is: $R-OH + C_2H_5MgI \rightarrow R-OMgI + C_2H_6 \uparrow$. At $STP$,$1 \ mol$ of ethane occupies $22400 \ cm^3$. Given $2240 \ cm^3$ of ethane is liberated,which corresponds to $2240 / 22400 = 0.1 \ mol$ of ethane. Since $1 \ mol$ of alcohol produces $1 \ mol$ of ethane,$0.1 \ mol$ of alcohol has a mass of $8.8 \ g$. Thus,the molar mass of the alcohol is $8.8 / 0.1 = 88 \ g/mol$. The general formula for a saturated monohydric alcohol is $C_nH_{2n+1}OH$,so $12n + (2n+1) + 16 + 1 = 88$,which gives $14n = 70$,so $n = 5$. The alcohol is a pentanol isomer $(C_5H_{12}O)$. Since the alcohol forms an aldehyde upon oxidation with $PCC$ (which gives a positive Tollen's test),it must be a primary $(1^{\circ})$ alcohol. Among the options,$2, 2$-dimethylpropan-$1$-ol has the formula $(CH_3)_3CCH_2OH$,which is a primary alcohol with molar mass $88 \ g/mol$.

$8.8 \ g$ of monohydric alcohol added to ethyl magnesium iodide in ether liberates $2240 \ cm^3$ of ethane at $STP$. This monohydric alcohol when oxidized using pyridinium chlorochromate $(PCC)$,forms a carbonyl compound that answers the silver mirror test (Tollen's test). The monohydric alcohol is:

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