An alcohol $(A)$ on treatment with conc. $H_{2} {SO}_{4}$ gave an alkene $(B)$ . The compound $(B)$ , on reacting with ${Br}_{2}$ water and subsequent dehydrobomination with ${NaNH}_{2}$ produced a compound $(C)$ . The compound $(C)$ with $dil . H_{2} {SO}_{4}$ in presence of ${HgSO}_{4}$ gave a compound (D). The compound (D) can also be obtained by oxidation of A by acidified ${KMnO}_{4}$ or from dry heating of calcium acetate. Calculate the mass of 0.1 mol of $D$ in gms. Round off your answer to next nearest integer.

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answer is 6.

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Detailed Solution

Reaction is,

$Ca {({CH}_{3} COO)}_{2} \overset{∆}{⟶}$ ${CH}_{3} {COCH}_{3} + {CaCO}_{3} ↓$

Here, D is Acetone CH₃COCH₃

In this,

Mass of the Carbon (C) = 12 g
Mass of the Hydrogen (H) = 1 g
Mass of the Oxygen (O) = 16 g

Molecular mass of the acetone,

$= 12 + 1 \times 3 + 12 + 16 + 12 + 1 \times 3 = 12 + 3 + 12 + 16 + 12 + 3 = 36 + 6 + 16 = 58 g / mol$

For 0.1 mol

$Molecular mass = 5.8 gm$

In round figure, it is 6.

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