For reductive conjugative carbonil compounds (as chalcone) use hydride reagents ( NaBH4, LiAlH4, L-Selectride)

Yes, we can reduce carbonyl group selectively without intact of alkene by using NaBH4 in Et2O/THF. When use the NaBH4-CeCl3 (Luche Condition), we may reduce both as well. Hope this information is useful.

thaks Ananda sir

i tried many condition like Luche condition, NaBH4 bronsted acid in methanol at rt to reflux , NaBH4 with cat guanidine hydrochloride in MeOH but no any change in TLC also i tried NaBH4 in meoh at rt to reflux also but no change

Chalcone we have only one functional group i.e. ketone.So we can easily reduce to allylic alcohol by using LiAlH4,NaBH4,DIBALH or Luche reduction(NaBH4 with CeCl3.7H2O).These reagents wont effect double bond.Another case is if you have 2 double bonds in one molecule,if one is alpha beta unsatured double bond anoother one is normal double bond.If u want reduce only alpha beta unsaturated double bond we need to use Pd(PPH3)4 with n-Bu3SnH or Ph2SiH2,CHCl3 at rt.But if we use palladium or platinum reagnets both will reduce.This is good reaction. A colleague got 100% conversion within 30 mins

Hydride reagents are best for reduction of carbonyl group of chalcones

USE NaBH4 in ether

LAH in ether at a low temperature should do the reduction of the carbonyl group.

Thanks kamakshi

Simple by NaBH4 reduction.

Hi Mukund- Thanks for your feed back. If, still you are struggling to get read of this funny reaction, you could also try this technique- along with conditions/reagents described above add catalytic amount of DMF solvent to the reaction mixture. It may helpful to accelerate the reaction. Try your luck..)