Column I (Reactions) | Column II (Products) | ||
I. | P. | ||
II. | Q. | ||
III. | R. | ||
IV. | S. |
A) I - R II - S III - Q IV - P
B) I - P II - Q III - S IV - R
C) I - Q II - P III - R IV - S
D) I - S II - R III - P IV - Q
Correct Answer: C
Solution :
[c] (I) (II) This produces \[C{{O}_{2}};\] the reaction is Kolbe reaction. (III) Carbene also converts \[(-N{{H}_{2}})\] group to \[(-\overset{\oplus }{\mathop{N}}\,\equiv \overset{\bigcirc -}{\mathop{C}}\,-)\]group (Carbylamine reaction) and also adds to \[(C==C)\]bond of cyclopentane ring and undergoes Reimer-Tiemann reaction at o-position w.r.t \[(-OH)\] group in benzene ring. (IV) \[CC{{l}_{2}}BrI\xrightarrow{4NaOH}2NaCl+NaBr+NaI+C{{(OH)}_{4}}\] \[\xrightarrow{-2{{H}_{2}}O}C{{O}_{2}}\]You need to login to perform this action.
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