Question

Number of isomeric compounds with molecular formula \(C _9 H _{10} O\) 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 \(C _9 H _{12} O\) is _________

Hint:

Ethers are neutral compounds that do not react with acids, bases, or 2,4-DNP, and aromatic ethers retain their structure upon hydrogenation of the benzene ring.

Answer 1

Correct Answer: 2

The compound must have two fewer hydrogens compared to \(C_9H_{12}O\) for it to have one degree of unsaturation, indicating one possible double bond or ring. Hydrogenation converts this structure to the saturated compound \(C_9H_{12}O\). This degree of unsaturation points to the presence of an aromatic ring or a double bond. For solubility criteria:

• Not dissolving in \(NaOH\) implies no phenol or acidic functional group.
• Not dissolving in \(HCl\) implies absence of basic groups like amine.
• Not reacting with 2,4-DNP indicates absence of carbonyl groups (aldehydes or ketones).

Considering these clues, the functional group has to be non-reactive entities in electrophilic or nucleophilic substitution. Toluene derivatives without other reactive groups fit this criterion. Possible isomers are methyl phenyl ether (anisole) and its derivatives. Anisole does not react with alkali, acid, or 2,4-DNP, and forms the same compound by hydrogenation given the above conditions. Therefore, the two possible isomers are:

1. Anisole.
2. 0-Methyl anisole.

Thus, the number of such isomeric compounds is \(2\). This sits within the expected range of \(2,2\).