Question

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Problem IV The Elimination reaction allows the removal from a molecule of a Leaving group and a neighboring Hydrogen to generate a double bond. When that reaction is performed on a cyclohexane, the leaving group (or in the example below, and the must be Axial as shown below. When the reaction is performed on the two molecule below (A and B), it is found that the reaction works very well on molecule A but very slowly on molecule B. Propose an explanation for this observation

Answer 1

Iv Solution ginn artio is The B Y thea bromine and hydrugen 1- bromo teht cyclohex ane In the anti periplanar to each o ther The bromine at axial above anti-periplanar to axiul hyd rogen bo th ar and htdrogen axial beluw Axial Br So yeaction under go cycluhexene EL elimination and give Consider Now Br Chair conformation of structu A Br elimination E1 olefin In structux A Br and H anti-periplanar and the bully tertiarg a bu group at equitorial po sition So the reaction under elimination to gi olefin. goes Er chair conformathon of structure B BY BY E elimination steric hind rance

Br and H are anti-periplanar but the buky tertiary Because of steric hindrance of ter- In structuB bulyl group at avial position tiary bully group zeaction under goes E elimination ve slow NH H O The steptocentecrof the aboe compount ind ic at ed by The above compound contain 11 sterto centers OH Dot chiral chiral nat chirnd chiral centeY: Caybon atom having four diffent atom or 9YUuP Enantiomer ot the following compound is NH NI NIP4