Question

can someone please explain how to do this with as much detail as possible. someone tried to explain it earlier in relation to (a,e) and (e,a) but im not sure even what that means

Bccurance of Methyl shift o. The following two elimination reactions are found to have very different rates of reaction. The cis compound produces two products and it reacts much faster than the trans isomer, which produces only one product. Explain these observations (the erence in rate and the number of products formed) by making appropriate drawings of transition state structures to support your argument. You can only receive minimal credits if you do not show key structures. a) C(CH3)3 C(CH3)3 CH3CH ON CH3CH2OH C(CH3)3 fast b)

Answer 1

or As we can see from the chair conformation, then is proper stereochemiel environment lies froms -acial C-H bond to that of c-Bn bond present. So elimination here is very fast and as the t-Bu- group is in equitorid position , so, is is most estable conformation also. H4 (6) Now, this is the most ustasce confennection of the molecute. But in this confumetion Bn is in equebarial position which is not favourable fur eliminelia reactin. Now, to bring on to axial position are have to flip the ring and which is energelically unfavourable as in Iut care t-Bu group will be in arial position.

Bm quins i unstabben highly will be very very Glow. so, flipping Cell) eta Here only which is one c-H trans-axial tress writt bord present to (- Bn. so only one Product will form.