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0.218 gm of the acetyl derivative of a p...

`0.218` gm of the acetyl derivative of a polyhydric acohol (molecular mass = 92) requires `0.168` gm of KOH for hydrolysis. Calculate the number of `(--OH)` groups in the alcohol.

Text Solution

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Let the formula of alcohol is `R(OH)_(n)`, and the formula of its actyl derivative is `R(OCOCH_(3))_(n)`.

Molecular mass of `R(OCOCH_(3))= (M + 42n)` M is the molecular mass of alcohol. Molecular mass of `(CH_(3) - CO -)` group = 43. One H atom is replaced by (OH) group of `CH_(3)COOH` group. Therefore, molecular mass of `R(OCOCH_(3))`
`= [M + (43 - 1)n]`
`= (M + 42n)`
`R(OCOCH_(3))_(n)` `overset (nKOH) rarr` `R(OH)_(n) + nCH_(3)COOH`
Molecular mass of `nKOH = 56n`
`0.218` gm of acteyl derivative requires
`=(0.168(M + 42n))/(0.218) = 56n`
On solving, we get n = 3
[Use the formula: `(w(M + 42n))/(W) = 56n` w = Weight of KOH, W = weight of acetyl derivative, M = Molecular mass of alcohol, n - number of `(-OH)` groups]
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