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प्रश्न
0.44 g of a monohydric alcohol, when added to methyl magnesium iodide in ether, liberates at STP 112 cm3 of methane with PCC the same alcohol form a carbonyl compound that answers the silver mirror test. Identify the compound.
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उत्तर
0.44g of a monohydric alcohol liberates 112 cm3 of methane.
\[\ce{\underset{(1 mole)}{R - OH} + CH3MgI -> \underset{(1 mole)}{CH4} + MgI(OR)}\]
Mass of monohydric alcohol which gives 22400 cm3 of methane = `(22400 xx 0.44)/112` = 88
C5H12O molecular formula has mass number 88 and it shows eight possible isomers. But neopentyl alcohol reacts with PCC to form neopentyl aldehyde, which shows positive silver mirror test.
Therefore, the compound is neopentyl alcohol (or) 2, 2-dimethyl propan-1-ol.
\[\begin{array}{cc}
\ce{CH3}\phantom{.}\\
|\phantom{....}\\
\ce{CH3 - C - CH2OH}\\
|\phantom{....}\\
\ce{CH3}\phantom{.}
\end{array}\]
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संबंधित प्रश्न
Among the following ethers which one will produce methyl alcohol on treatment with hot HI?
Identify the product(s) is/are formed when 1-methoxy propane is heated with excess HI. Name the mechanism involved in the reaction.
Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent of HI.
Write the chemical equation for Williamson's synthesis of 2-ethoxy-2-methyl pentane starting from ethanol and 2-methyl pentan-2-ol.
Complete the following reaction.
\[\ce{C6H5 - CH2CH(OH)CH(CH3)2 ->[conc. H2SO4]}\]
What will be the product (X and A)for the following reaction
acetylchloride \[\ce{{acetylchloride}->[i)CH3MgBr][ii)H3O+]X ->[acidK2Cr2O7] A}\]
Predict the major product, when 2-methyl but -2-ene is converted into an alcohol in each of the following method.
Acid catalysed hydration
Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent of HI.
Predict the major product, when 2-methyl but – 2 – ene is converted into an alcohol in each of the following method.
Acid catalysed hydration.
The correct IUPAC name of the compound,
\[\begin{array}{cc}
\ce{CH3}\phantom{......}\\
|\phantom{........}\\
\ce{H3C - CH - CH - CH - CH2 - OH}\\
|\phantom{............}|\phantom{........}\\
\ce{Cl}\phantom{...........}\ce{CH3}\phantom{......}
\end{array}\]
