An organic compound "P" of molecular formula \(C_7H_{12}O\) (likely \(C_7H_{12}O_2\) or similar based on options), gives positive Iodoform test but negative Tollen's test. When "P" is treated with dilute acid, it produces "Q". "Q" gives positive Tollen's test and also Iodoform test. The structure of "P" is:
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Acetals are stable to base but hydrolyze in dilute acid to regenerate carbonyl compounds. This is a common method to "protect" aldehydes.
Step 1: Analyze Compound P
\begin{itemize}
\item Iodoform Test (+): P contains a methyl ketone (\(CH_3-C=O\)) or methyl carbinol (\(CH_3-CH(OH)-\)) group.
\item Tollen's Test (-): P is not an aldehyde.
\end{itemize}
Step 2: Analyze Reaction and Compound Q
\begin{itemize}
\item Reaction: P + dilute acid \(\rightarrow\) Q. This implies hydrolysis (likely of an acetal, ketal, or enol ether).
\item Q Properties:
\begin{itemize}
\item Tollen's (+): Q contains an aldehyde group (\(-CHO\)).
\item Iodoform (+): Q contains a \(CH_3-C=O\) group.
\end{itemize}
\end{itemize}
So, Q must be a compound like 3-oxobutanal (\(CH_3-CO-CH_2-CHO\)).
Step 3: Deduce P
P must be a protected form of Q that masks the aldehyde (causing Tollen's negative) but keeps the ketone available (or a group that becomes a ketone).
A likely structure is the dimethyl acetal of the aldehyde group:
\[ CH_3-C(=O)-CH_2-CH(OCH_3)_2 \]
\begin{itemize}
\item Check P: Contains Ketone (Iodoform +), Acetal (Tollen's -). Matches criteria.
\item Hydrolysis: \(CH_3-C(=O)-CH_2-CH(OCH_3)_2 \xrightarrow{H_3O^+} CH_3-C(=O)-CH_2-CHO + 2CH_3OH\).
\item Check Q: Contains Ketone (Iodoform +) and Aldehyde (Tollen's +). Matches criteria.
\end{itemize}
Option 2 typically represents this acetal structure (or a related enol ether).
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