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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