Question

6. Explain the basis for the selective formation of the product shown over the alternative product. CH CO2CH3 C150ccEncuzcn1.01 CH,O,CCCH2CH2CH2OH not - CIL o 2.7.1 (b) H C CH3 H3CCH; H:C, CH, RCO3H I versus Versus a yo (88%) (12%) (c) H2B2 Н4С., HC-N ) + BH preferred to V GR- HC" H Hº CH3 CH3 CH3

explain the basis for selective formation of the product

Answer 1

co,CH3 not coscly :-- CH2-CH2-CH,0H CHe To undorsland tbis let's fris see the mechanfsm of paoduct. -сH -сH -СИ -> 0-C-C 8o th9s 93 9ntermediate for both the reactHon , Now lets consider Now, G B more eJectoph?l?c conter than the &. Go the nucleophile s îs less electoophil?c ,so the there che hoo centers for attack of Ocide anion.at C, or G. 09 attack on G e not on Q, lefs see mechansm NO loesn't aHack on that canbon, CH CH2 CH,-CH, -CH,-0 CH2 CH3-O CHy-O CH снz So, we get ths as major pooduct, :CH2 cogCHg Mpnor pooduct.

RCO3H Vs 88.1. In thi3 case, there are two methylene goups are present on bridlged carbon which does not allow the peracid to come foom the above stde the exo epoxide major poduct. So we don't get But foom the endo face attack of peracid 93 Favoured and we get endo poduct as major product. exo face cmore hindeged to attack CHz RCQ3H -> endo face c Dess hindened) So , we get this poduct as major procduct . ИС + BHz Вн. prefored tof To undesstand this deaction, we should draw thPs shucture in chalr form As we can see CH is above the plane in chaso form, So we should put 9t above +'8H3 CH3 As we can see the Jone par of ni hugen attack from amfal side els, which 9s Jes hindeIed, the equatorial side 9s hindered by-CH, group

So auttack of Jone palrs does not takes place foom equatorfal side. CH3 CH3 CHy Ha d] LIAIHY rathes han CH3 EHO (1) the bychide 9on whpch acts as reclucing approach Aom the opposfto sfde of aready In this case, cagent which Stuaded -CH3 goup. So thats why we get CI) as major pooduct nor CII). CH3 CH3