The compound shown is subjected to different hydride reagents. Select ALL correct statements.

A $NaBH X_{4}$ in methanol reduces the ketone and leaves the nitrile unchanged.
BExcess $LiAlH X_{4}$ followed by work-up reduces both groups to give an amino alcohol.
C1 equivalent of $DIBAL - H$ at $- 7 8^{\circ} C$ reduces only the nitrile and leaves the ketone untouched.
DThe product of the $NaBH X_{4}$ reduction is formed as a racemic mixture (it contains one stereocentre).

Answer: A, B, D

The compound is 4-acetylbenzonitrile.

NaBH X_{4}

reduces ketones but not nitriles (statement 1 true); excess

LiAlH X_{4}

reduces C=O to CHOH and

C \equiv N

CH X_{2} NH X_{2}

(statement 2 true). DIBAL-H reduces a ketone far faster than a nitrile, so statement 3 is false. The

NaBH X_{4}

product 4-(1-hydroxyethyl)benzonitrile has one stereocentre and is formed as a racemate (statement 4 true).