Aldehydes And Ketones Question 7
Question 7 - 30 January - Shift 1
A trisubstituted compound ’ $A$ ‘, $C _{10} H _{12} O_2$ gives neutral $FeCl_3$ test positive. Treatment of compound ‘A’ with $NaOH$ and $CH_3 Br$ gives $C _{11} H _{14} O_2$, with hydroiodic acid gives methyl iodide and with hot conc. $NaOH$ gives a compound B, $C _{10} H _{12} O_2$. Compound ’ $A$ ’ also decolorises alkaline $KMnO_4$. The number of $\pi$ bond/s present in the compound ’ $A$ ’ is
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Answer: (4)
Solution:
Formula: $ FeCl_3$ Test
$(C _{10} H _{12} O_2)$
(or)
$(C _{10} H _{12} O_2)$
$CH=O+C_3 H_7 \stackrel{O}{CH_3 I} \xrightarrow{NaOH} CH=O+C_3 H_7$
