Question

Define saponification equivalent `(S.E)` of an ester.
(b) When is `S.E` equal to molecular weight of an ester
( c) What is the `S.E` of diethyl fumarate ?
(d) Identify `(A)` as methyl or ethyl cinnamate from the following data :
`0.75 gm` of `(D)` is refluxed with `100.0 mol` of `125 M KOH`, and the excess `KOH` is neutralised by back titrating with `27.5 ml` of `0.3 M HCl`.

Answer 1

(a) `S.E` is the weight of an ester that reacts with `1 mol` of `KOH`.
(b) If the ester is derived from monocarboxylic acid, its `Mw` and `S.E` are identical.
( c) It it is diester, then
`S.E = (Mw)/(2) = (172.0)/(2) = 86.0 gm eq.^-1`
Diethyl fumarate,
`C_8 H_12 O_4 =96 +12 + 64` .
(d) Here, `m` moles of `KOH` used `= 100 xx 0.125 = 12.5`
Excess of `KOH = 27.5 xx 0.3 = 8.25`
`KOH` used `= 12.5 - 8.25 = 4.25 m mol`
=`(4.25)/(1000) = 0.00425 mol`.
Equivalent weight if ester `= 0.00425 mol`
`(("Since is mono"-),("carbo xylic ester" )) = 0.00425 xx 0.75`
=`176 gm mol^-1`
`Ew` of ester = `Mw` of ester = `176`
So `(A)` is ethyl cinnatmate `(PhCH=CHCOOEt)` (since its `Mw = 176`) rather than methyl cinnamate `PhCH=CHOOMe` whose `Mw = 162`.