Stereoisomerism

Working backwards:
The final product is 1-bromo-1-methylcyclopentane (Tertiary bromide).
This is formed by the addition of HBr to an alkene. Following Markovnikov's rule, the alkene must be 1-methylcyclopentene.
1-methylcyclopentene is formed by the reaction of (P) with Zn/ .
Zn dust acts as a debrominating agent for vicinal dihalides.
Therefore, (P) must be 1,2-dibromo-1-methylcyclopentane.

Step 1: Understanding the Question:
The question requires us to identify the correct statements about different types of isomerism for given pairs of organic compounds. We need to evaluate each statement based on the definitions of functional isomerism, metamerism, optical isomerism, and chain isomerism.
Step 2: Detailed Explanation:
Let's analyze each statement:
A. Propanal and propanone are functional isomers.
\begin{itemize} \item Propanal (CH₃CH₂CHO) is an aldehyde.
\item Propanone (CH₃COCH₃) is a ketone.
\item The molecular formula for both is C₃H₆O.
\item Isomers with the same molecular formula but different functional groups are called functional isomers.
\item Conclusion: Statement A is correct.
\end{itemize} B. Ethoxyethane and methoxypropane are metamers.
\begin{itemize} \item Ethoxyethane is C₂H₅-O-C₂H₅.
\item Methoxypropane is CH₃-O-C₃H₇.
\item Both are ethers with the molecular formula C₄H₁₀O.
\item Metamers are isomers having the same functional group but different alkyl groups attached to the functional group. Here, the alkyl groups attached to the ether oxygen are different (ethyl/ethyl vs. methyl/propyl).
\item Conclusion: Statement B is correct.
\end{itemize} C. But-2-ene shows optical isomerism.
\begin{itemize} \item The structure of but-2-ene is CH₃-CH=CH-CH₃.
\item For a molecule to be optically active, it must be chiral, i.e., non-superimposable on its mirror image. This typically requires a chiral center (a carbon atom bonded to four different groups).
\item But-2-ene does not have any chiral carbon. It shows geometrical isomerism (cis and trans), but neither the cis nor the trans isomer is chiral.
\item Conclusion: Statement C is incorrect.
\end{itemize} D. But-1-ene and but-2-ene are functional isomers.
\begin{itemize} \item But-1-ene is CH₂=CH-CH₂-CH₃.
\item But-2-ene is CH₃-CH=CH-CH₃.
\item Both compounds have the same functional group (alkene, C=C double bond). They differ only in the position of the double bond.
\item Isomers with the same functional group but differing in its position are called positional isomers, not functional isomers.
\item Conclusion: Statement D is incorrect.
\end{itemize} E. Pentane and 2, 2-dimethyl propane are chain isomers.
\begin{itemize} \item Pentane (n-pentane) is CH₃CH₂CH₂CH₂CH₃.
\item 2,2-dimethylpropane (neopentane) is C(CH₃)₄.
\item The molecular formula for both is C₅H₁₂.
\item They differ in the arrangement of the carbon skeleton (carbon chain). Pentane has a straight chain of 5 carbons, while 2,2-dimethylpropane has a parent chain of 3 carbons with two branches.
\item Isomers with the same molecular formula but different carbon chains are called chain isomers.
\item Conclusion: Statement E is correct.
\end{itemize} Step 3: Final Answer:
The correct statements are A, B, and E. Therefore, option (C) is the correct choice.

Step 1: Analyzing the statements.
- (A) Propanal and Propanone are functional isomers since they differ in functional groups (aldehyde and ketone).
- (B) Ethoxyethane and methoxypropane are metameres because they are ethers with different alkyl groups on either side of the oxygen atom.
- (C) But-2-ene doesn’t show optical isomerism as it contains a plane of symmetry and no chiral center.
- (D) But-1-ene and But-2-ene are position isomers, not functional isomers.
- (E) Pentane and 2, 2-dimethylpropane are chain isomers since they differ in the carbon chain arrangement.
Step 2: Conclusion.
The correct answer is (3) as statements A, B, and E are true.