Draw the major product formed in the reaction. Be sure to draw a hydrogen on any heteroatoms, where applicable. The problem asks to draw the major product formed in the reaction of a given epoxide with sodium cyanide (NaCN) in an ethanol and water solvent.**Reactants:**- A three-membered cyclic ether (epoxide) with an ethyl chain attached.**Reagents:**- NaCN (sodium cyanide)- Ethanol and water as solvents**Reaction Mechanism:**The epoxide undergoes nucleophilic ring-opening in the presence of NaCN. The cyanide ion (CN⁻) attacks the less sterically hindered carbon atom in the epoxide, leading to the opening of the ring.**Product:**The major product is a β-hydroxy nitrile. It consists of:- An open chain with a hydroxyl (OH) group.- A nitrile group (C≡N) attached to the former epoxide carbon.This transformation is a typical example of how epoxides react with nucleophiles in an acidic or neutral aqueous medium.

Answered: Image /qna-images/answer/968bd032-8253-4f33-ab01-0181d17fc357.jpgNote- In this reaction epoxide ring opening take place by cynide ion(ion) . Epoxide ring opening take place via SN2 reaction. CN- attack from less hindered side of the ring. Note -Always remember if any ring opening reactions are goes via both Sn1 as well as SN2 reaction. Depending upon the given conditions. If acidic medium is given then ring opening is take place from more hindered side of the ring that is SN1and acidic medium is not give the ring opening take place from less hindered side of the ring that is SN2 as given in the question.