Number of isomeric compounds with molecular f | vedclass

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(A) The molecular formula $C_9H_{10}O$ has a degree of unsaturation of $6$. Given conditions: $(i)$ Does not dissolve in $NaOH$ (not a phenol). $(ii)$

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$2$

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(A) The molecular formula $C_9H_{10}O$ has a degree of unsaturation of $6$. Given conditions: $(i)$ Does not dissolve in $NaOH$ (not a phenol). $(ii)$ Does not dissolve in $HCl$ (not an amine). $(iii)$ Does not give orange precipitate with $2,4-DNP$ (no carbonyl group,i.e.,no aldehyde or ketone). $(iv)$ On hydrogenation,it gives an identical compound $C_9H_{12}O$. Since it is not a carbonyl compound and not a phenol,it is likely an ether. The structure $Ph-CH=CH-O-CH_3$ (methyl styryl ether) satisfies these conditions. It exists as $cis$ and $trans$ isomers. Both isomers on hydrogenation yield $Ph-CH_2-CH_2-O-CH_3$ $(C_9H_{12}O)$. Thus,there are $2$ such isomers.

Column-$I$ (Reaction)	Column-$II$ (Condition)
$(A)$ $CH_3-CH(OH)-CH_3 \rightarrow CH_3-CH=CH_2$	$(i)$ $20\% \ H_3PO_4, 358 \ K$
$(B)$ $CH_3-C(CH_3)_2-OH \rightarrow CH_3-C(CH_3)=CH_2$	(ii) $Cu, 573 \ K$
$(C)$ $CH_3CH_2OH \rightarrow CH_2=CH_2 + H_2O$	(iii) $85\% \ H_3PO_4, 440 \ K$
$(D)$ $R-CH(OH)-R' \rightarrow R-CO-R'$	(iv) $H_2SO_4, 443 \ K$

Number of isomeric compounds with molecular formula $C_9H_{10}O$ which $(i)$ do not dissolve in $NaOH$ $(ii)$ do not dissolve in $HCl$ $(iii)$ do not give orange precipitate with $2,4-DNP$ $(iv)$ on hydrogenation give identical compound with molecular formula $C_9H_{12}O$ is $...........$

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