The Shapiro Reaction involves the base-mediated decomposition of sulfonyl hydrazones like 14 below. Methyl lithium (CH3Li) is a strong base; Ar = p-tolyl (i.e. para-methylphenyl, -C6H4CH3).

i. Under conditions A, doubly-deuterated product 15 is formed. Draw a mechanism to explain this outcome. 3 marks 

ii. Under conditions B, the mono-methylated product 16 is formed. Draw a mechanism to explain this outcome. What does the position of methylation tell you? 2 marks 

iii. Under conditions C, the decomposition product 17 is formed. Draw a mechanism to explain this outcome. Why does the change in reaction conditions (relative to A) open this alternative pathway? 4 marks

Transcribed Image Text:

Os.Ar N. N D A i. 2x CHLI, -78 °C. ii. D20 Ph. D 15 O- Ar в Ar N. NH i. 2 x CH3LI, -78 °C ii. 1x CH3I iii. H20 (work-up) Ph. N° Ph. 14 16 D i. 2x CHLI, 25 °C ii. D20 Ph. 17 z= Os.Ar N. N D A i. 2x CHLI, -78 °C. ii. D20 Ph. D 15 O- Ar в Ar N. NH i. 2 x CH3LI, -78 °C ii. 1x CH3I iii. H20 (work-up) Ph. N° Ph. 14 16 D i. 2x CHLI, 25 °C ii. D20 Ph. 17 z=

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Answer Shapiro reaction is the conversion of tosyl hydrazones of carbonyl compounds into olefins wit... View the full answer