$P$ and $Q$ are isomers of dicarboxylic acid $C_4H_4O_4$. Both decolorize $Br_2/H_2O$. On heating, $P$ forms the cyclic anhydride.
Upon treatment with dilute alkaline $KMnO_4, P$ as well as $Q$ could produce one or more than one from $S,~ T$ and $U.$

$ S $ $ T $ $ U $

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Solution

The correct option is A ~
~

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Q. Treatment of compound O with

K M n O_{4} / H^{+}

gave P, which on heating with ammonia gave Q. The compound Q on treatment with

B r_{2} / N a O H

produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of

K O H

followed by acidification, gave a compound T.

The compound T is:

Q. Treatment of compound O with

K M n O_{4} / H^{+}

gave P, which on heating with ammonia gave Q. The compound Q on treatment with

B r_{2} / N a O H

produced R. On strong heating, Q gave S, which on further treatment with ethyl 2-bromopropanoate in the presence of

K O H

followed by acidification, gave a compound T.

The compound T is:

Compound $A (C_{10} H_{6})$ liberates 2 mole of $C H_{4}$ when treated with MeMgBr. On heating with $K M n O_{4}$ solution, “A” produces benzene dicarboxylic acid which on mononitration produces only one product and no other isomers. The compound “A” can be:

Q. Compound A

A (C_{10} H_{6})

liberates 2 mole of

C H_{4}

when treated with MeMgBr. On heating with

K M n O_{4}

solution, “A” produces benzene dicarboxylic acid which on mononitration produces only one product and no other isomers. The compound “A” can be:

Q. Among P, Q, R, S, T and U, P is taller than R and S but not shorter than T and U and Q is taller than R, S, T and U but not as tall as P. Who amongst them is the tallest?

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\(P\) and \(Q\) are isomers of dicarboxylic acid \(C_4H_4O_4\). Both decolorize \(Br_2/H_2O\). On heating, \(P\) forms the cyclic anhydride. Upon treatment with dilute alkaline \(KMnO_4, P\) as well as \(Q\) could produce one or more than one from \(S,~ T\) and \(U.\) \( S \) \( T \) \( U \)

The correct option is A ~~

\(P\) and \(Q\) are isomers of dicarboxylic acid \(C_4H_4O_4\). Both decolorize \(Br_2/H_2O\). On heating, \(P\) forms the cyclic anhydride.
Upon treatment with dilute alkaline \(KMnO_4, P\) as well as \(Q\) could produce one or more than one from \(S,~ T\) and \(U.\)

\( S \) \( T \) \( U \)

The correct option is A ~
~