Question

1) excess 03 2) Mes 1) NaOEtZn(Hg) E) MeBr HCI
can someone explain how ro generate this?

Answer 1

Dear Candidate,

Here first stage is ozonolysis. O3 will react with carbon carbon double bond to give carbonyl compound. ie Aldehyde or Ketone.

In stage 2 we have CH group sandwiched in between two carbonyl group. It will form carbanion ion. This carbanion ion will attack on methyl group to give product as shown in attached image.

Now lets lets understand whats happening in last step.

Last step is nothing but reduction of carbonyl group by using Clemmensen Reduction. In Clemmensen Reduction Zinc amalgam in presence of concentrate HCl will reduce carbonyl group alkane.

Please go through attached image for more details.

stage-1 1,3 cyclo C Do addition or similarly another double bond breaks to give this product I Stage 2 age-1 NUO ER Resonance Stabilisation stage- (HLTBV STG-02

stage-3 410 -zano 7 In at This zo et with realt with eo im to form Lo el2