Why is base (NaOH or KOH) added as a second step in each of the reactions shown
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Why is base (NaOH or KOH) added as a second step in each of the reactions shown in Eqs. 23.23 and 23.24?
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Ph—C−H+H_N—C(CH3)3 benzaldehyde tert-butylamine + Me,NH dimethylamine cyclohexanone NaBH(OAC) 3 HOAC 1,2-dichloroethane (solvent) NaBH 3 CN HCl (1 equiv.) MeOH NaOH KOH NHC(CH3)3 Ph—CH—H N-tert-butylaniline (95% yield) Me₂N + H₂O N,N-dimethylcyclohexanamine (71% yield) (23.23) (23.24)
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