A trisubstituted compound 'A',C_{10}H_{12}O_2 | vedclass

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(B) $1$. Compound '$A$' is trisubstituted and gives a positive neutral $FeCl_3$ test,indicating it is a phenol derivative.$2$. The formula $C_{10}H_{1

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$4$

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(B) $1$. Compound '$A$' is trisubstituted and gives a positive neutral $FeCl_3$ test,indicating it is a phenol derivative.$2$. The formula $C_{10}H_{12}O_2$ with a benzene ring ($6$ carbons) leaves $4$ carbons for substituents. The degree of unsaturation is $10 - 12/2 + 1 = 5$. Since the benzene ring accounts for $4$ degrees of unsaturation ($3$ $\pi$ bonds + $1$ ring),there is $1$ additional $\pi$ bond or ring present in the side chains.$3$. The compound decolorises alkaline $KMnO_4$,confirming the presence of an alkene or an oxidizable side chain.$4$. The reaction with $HI$ gives methyl iodide,indicating the presence of a methoxy group $(-OCH_3)$ or a similar ether linkage. However,the initial test with $NaOH$ and $CH_3Br$ suggests the presence of a phenolic $-OH$ group which gets methylated.$5$. Given the structure is trisubstituted on a benzene ring,the total $\pi$ bonds are $3$ (from the benzene ring) + $1$ (from the alkene side chain) = $4$ $\pi$ bonds.

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