Question

Please explain your reasoning behind the answers.

2.) Identify each reaction as either an 'oxidation' or a ‘reduction'. RE + TMSN+ TOISO,Na + TsN, 3a Ag3PO4(20 mol %) H2O (2.0 equiv) DMSO. 70 °C, 10 h R NHTS 1 a. J. Am. Chem. Soc. ASAP, Published online January 22, 2019 PdCl(Ph)(PPh3)2 (0.1 mol %) THF/toluene, rt, 1-3 h 0 R1R2 R1SP + BrZ-R2 c. 36 Org. Lett. ASAP, Published online January 18, 2019

Answer 1

a. Here the alkyne carbon gets oxidised from "0" to "+3" and azide nitrogen gets reduced from 0 / -1 to -3.

oxidation state = -1, as nitrogen is oxidation state = 0, as carbon is attached electropositive silicon attached to all four carbons oxidation state = -1, as nitrogen Me has one negative charge © R -H + Me—SN=NSEN Me oxidation state = -1, as carbon is attached to only one hydrogen oxidation state = 0, as nitrogen is attached nitrogen only oxidation state = -1, as nitrogen is attached electropositive sulphur oxidation state = -1, as nitrogen has one negative charge + Tolso Na + -S-NENEN oxidation state = 0, as nitrogen is attached nitrogen only H oxidation state = +3, as carbon is attached to one nitrogen by single bond and to another nitrogen by double bond R o xidations oxidation state = -3, as attached to electropositive atoms NHTS

b. Here the carbonyl carbon gets oxidised from "-3" to "-2" oxidation state.

NaCN Oxidation state of carbon -3, as carbon is attached to one chlorine by single bond and one oxygen by double bond Oxidation state of carbon = -2 as carbon is attached one oxygen by double bond

c. Here the carbonyl carbon gets reduced from "-1" to "-2" oxidation state.

Oxidation state of carbon = -2, as carbon is attached to one oxygen by double bond RSPB R2 R R2 Oxidation state of carbon = -1, as carbon is attached to one oxygen by double bond and one sulphur by single bond