0.44 g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP 112 cm^3 of methane with PCC the same alcohol form a carbonyl compound that answers silver mirror test. Identify the compound.

0.44 g of a monohydric alcohol when added to methyl magnesium iodide i

1.	0.44 g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP 112 cm^3 of methane with PCC the same alcohol form a carbonyl compound that answers silver mirror test. Identify the compound.
Answer» Solution :0.44g of a monohydric alcohol liberates `112 cm^3` of methane. `underset(1 "MOLE")(R- OH + CH_3MgI to underset(1" mole")(CH_4) + mgI(OR)` `:.` ass of monohydric alcohol which gives `22400 cm^3` of methane `=(22400 xx 0.44)/112= 88` `C_5H_12O` 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, compound is, neopentyl alcohol (or) 2, 2-dimethyl propan-1-ol. `CH_3- underset(CH_3) underset(\|)OVERSET(CH_3) overset(\|)C - CH_2OH`