Question

12. Draw all major products for the following reactions. Indicate any region specificity Mark, Non Mark, syn or anti addition were necessary. (36 points) HgSO H SO. Na NH 1.2 HB 2. cther R BH THE NaOH, HOA Bry CCL OH NaOCI HOAC o E

Answer 1

[Markovnikov addition] secondary incipient carbocation more stable Hg2+ [Hg2+1 [Hg2+1 protodemetallation H Hg2 Hg2+ SO42- {proton abstraction) O-H ' trans-addition Hg2+ enol-form less stable keto-form more stable (product) [anti-Markovnikov addition] 4-06 H primary incipient carbocation less stable Na Na® + e H H ee H-NH2 If the electron attacks at any of the carbons of the akynes, it will give the same product H (product) H secondary incipient carbocation more stable [Markovnikov addition) HO HO H* Br Н H H H -Br Br Br H Br H Br (product) trans-addition * Hot Brº primary incipient carbocation less stable [anti-Markovnikov addition]

R hydroboration via 4-membered cyclic Transtition state [TS] due to the bulk of borane this TS is more stabler favoured path R R R-B-H R-B OH B-OH H R H H-OHO BO R-B keto more stable R R-B-H R BH OH enol (product) R R-B- * sterically hindered less stable TS less favoured path OH H нө R Br Bre Br-Br R HO CI. *** 10 O trans-diromo HO OC CI но HCIO OH Br Br anti-attack (product) HCIO 1 attack at any position (carbon) gives the same product R H R (product)

TA H-Br Br Br hv H-Br more stable secondary radical product RO-R-OR RO Br ROH Br less stable primary radical Usually traces amount of HBr is present in NBS which helps in catalyzing the reaction H-Br N-Br N-Br Br-Br + . less stable radical H Br hv Br-Br Br Br Br product more stable radical

Hg(OAC)2 Hg(OAc) + OAC NH2 Br H elimination reaction elimination reaction Br. H NH Me Me= Na SN2 attack inversion Me Br Me H NH3 + Br NH3 + Br NH3 M-OH нө ou vicinal dibromide NH Hg(OAC) (អ(0 ...Hg(OAc) IHO HOH Hg(OAc) Me H *Hg(OA) product OAC 220 Hg(OAc) I assume the reagent is LiAIH4 and not LiAIH2 HOH Mg-Coac) нө O-H H O! H-B-H H H-AICH I- H-OH HO H H 'Hg(OAc) H product ОАс H + Hg(0) H H-B-H H quo P2 ... + Hg(0) Chodac) P1 Both P1 and P2 are same! As these molecules are not chiral Thus we can write the product as: