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`overset(NaBH_(4))(leftarrow) PhCH=CH-CHOunderset(2. H_(3)O^(+))overset(1. LAH, ether)(rarr)(A)`
The products `(A)` and `(B)` are:
A. a.image
B. b.image
C. c.image
D. d.image

1 Answer

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Best answer
Correct Answer - D
`LAH` reduces `(CHO)` group to `(CH_(2)OH)` and also reduces double bond when `Ph` group is attached to `beta`-position of double bond, whereas `NaBH_(4)` reduces only `(-CHO)` group to `(CH_(2)OH)` group. So the answer is `(d)`.

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