Acetone reacts with excess formaldehyde in presence of Alkali condition, then how many number of “-OH” groups are present in reduced form of product

Acetone + Formaldehyde (Alkali) -OH Groups in Product

Organic Chemistry: Reaction of Acetone with Excess Formaldehyde in Alkali

Solution:

• Acetone (CH₃COCH₃) reacts with 2 moles of formaldehyde (HCHO) in base (Cross Cannizzaro/Aldol reaction, then reduction).
• The reduced product is pentane-1,3,5-triol:
• Structural formula: HOCH₂-CH₂-C(OH)(CH₃)-CH₂OH
• The number of -OH groups in the reduced product is 3.

Final Answer:

There are 3 "-OH" groups in the reduced form of the product.