Q.
CH3CH2COOHX →1 moleBr2 / Red P A →alc. KOHB →HBr C. Most acidic compound in this conversion is
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a
A
b
X
c
B
d
C
answer is A.
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Detailed Solution
A=CH3−CH|Br −COOH; B=CH2=CH−COOH;B → CCH2=CH−COOH→H+ CH3 −CH+−COOHLess Stable+C+H2−CH2−COOHMore Stable -COOH group shows negative inductive effect and destabilizes the positive charge As distance increases stability increases C+H2−CH2−COOH→Br− CH2|Br −CH2−COOHCAcidic strength ofα–halo carboxylic acid > HCOOH > β–halo Carboxylic acid > C6H5COOH > CH2=CH-COOH > C6H5CH2COOH > CH3COOH > CH3CH2COOH∴A > C > B > X
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