0.44 g of a monohydric alcohol when added to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

The product obtained is

Identify the final product in the following reaction sequence: $CH \equiv CH$ $\xrightarrow[H_2SO_4]{HgSO_4}$ $\xrightarrow[H_2O]{CH_3MgBr}$ $\xrightarrow{P/Br_2}$

Consider the following reaction sequence:
$\text{Cyclohexanol}$ $\xrightarrow{CrO_3} A$ $\xrightarrow{CH_3MgBr / H^+, H_2O} B$ $\xrightarrow{H_2SO_4, \Delta} C$ $\xrightarrow{B_2H_6, H_2O_2, OH^-} D$
Identify the final product $D$.

The iodoform reaction is given by all of the following,except:

Vedclass Test Series

Exam Paper Generator

Online Exam Module