| Directions (Q. Nos. 53): An organic compound [A] \[{{C}_{7}}{{H}_{6}}O\] gives positive test with Tollen's reagent. On treatment with alcoholic \[C{{N}^{\odot -}}\] [A] gives the compound [B]\[{{C}_{14}}{{H}_{12}}{{O}_{2}}\]. Compound [B] on reduction with \[Zn-Hg,\,\,HCl\] and dehydration gives an unsaturated compound [C], which adds one mole of\[B{{r}_{2}}/CC{{l}_{4}}\]. The compound [B] can be oxidized with \[HN{{O}_{3}}\] to a compound [D] \[{{C}_{14}}{{H}_{10}}{{O}_{2}}\]. Compound [D] on heating with KOH undergoes rearrangement and subsequent acidification of rearranged products yields an acidic compound (E) \[{{C}_{14}}{{H}_{12}}{{O}_{3}}\]. |
A) benzoin condensation
B) aldol condensation
C) Cannizzaro reaction
D) Perkin condensation
Correct Answer: B
Solution :
Compound [A] gives positive test with Tollen's reagent \[{{C}_{7}}{{H}_{6}}\] and undergoes reaction with \[C{{N}^{-}}\] giving condensation product. Hence, [A] is\[{{C}_{6}}{{H}_{5}}CHO\]. \[{{C}_{6}}{{H}_{5}}CHO\]. (Compound A) cannot undergo aldol condensation as it does not contain a-H atom.
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