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Explain the formation of `(A)` and `(B)` in the following reaction. `Oct-1-en eoverset(NBS+C C I_(4))underset(hv)rarrunderset((80%))((A))+underset((20%))((B))` |
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Answer» `C_(5)H_(11)-CH=CH_(2)overset (NBS//C C I)rarr` `underset(3-Bromo-1-oct a n e(20%))(C_(5)H_(11)-CH(Br)-CH)=underset(1-Bromo-2-octen e(80%))(CH_(2)+C_(5)+C_(5)H_(11)CH=CH-CH_(2)Br)` `[C_(5)H_(11)dot(C)H-CH=CH_(2)leftrightarrowC_(5)H_(11)CH=CH-dot(C)H_(2)]` The major brominated product results from the introduction of `Br` on methylene `C`, so as to get more `-` substituted brominated alkene. The shift of the double bond is an example of allylic rearrangement. |
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