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a
H2/Pt ; PCC
b
NaBH4 ; PCC
c
NaBH4 ; KMnO4/H+,∆
d
H2/Ni ; KMnO4/H+,∆
answer is B.
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Detailed Solution
Step-I . . .Reduction of -CHO to -CH2OH without reducing C = C∴NaBH4 is preferred; H2/cat not only reduces -CHO to -CH2OH but also reduces C = C to C - C.Step-II: Oxidation of -CH2OH to -CHO without oxidizing C = C∴PCC is preferred; KMnO4/H+ oxidizes -CH2OH to -COOH and at the same time leads to oxidative cleavage of C = C