Question

what is product 1?

A NaOH G Heated KOH Reagent 1 Reagent 2 Product 1 B HgSo. / H2SO4 H Br2/hv с 1.) Na NH2/ heat 2.) H20 LDA OH C1 D J Br E K Cl

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Answer #1

Reagent 1 - Br2/h\nu

There are possibilities of formation of two radical. But 2° radical is more stable than 1°. Hence we get more substituted bromoalkane as product.

Product 1 - E

E undergoes elimination by strong bulky base LDA, so we get the terminal alkene as the product.

Alkene on treatment with Cl2, gives dihalides.

Dihalides on double elimination by the reagent c (NaNH2/H2O), we get the alkyne. At first we get alkene with 1 equivalent of reagent. Then on the second equivalent of reagent we get the alkyne

Reagent 3 : 1.NaNH2/heat , 2. H2O

Product 2 : F

F on hydroboration oxidation reaction gives alcohol. The alcohol tautomerise to give aldehyde as product.

Br + Bralho LDA > E2 (majoo) Product I 외고 이. J A, NANA/D 2HO Nanh2Ja 2, H20 HO 대고대 Side BH H2O2/raoff H Produed 2

As you wanted to know only product 1, so that corresponding mechanism is shown below -

** Hopefully this would be helpful. If you don't understand, comment below.

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