Question

A hydrocarbon 'P' (\(C_4H_8\)) on reaction with HCl gives an optically active compound 'Q' (\(C_4H_9Cl\)) which on reaction with one mole of ammonia gives compound 'R' (\(C_4H_{11}N\)). 'R' on diazotization followed by hydrolysis gives 'S'. Identify P, Q, R and S.

• A. \(P=CH_3-CH_2-CH=CH_2\), \(Q=CH_3-CH_2-CH_2-CH_2Cl\), \(R=CH_3-CH_2-CH_2-NH_2\), \(S=CH_3-CH_2-CH(OH)CH_3\)
• B. \(P=CH_3-CH=CH-CH_3\), \(Q=CH_3-CH_2-CH_2-CH_2-Cl\), \(R=CH_3-CH_2-CH(NH_2)CH_3\), \(S=CH_3-CH_2-CH_2-CH_2OH\)
• C. \(P=CH_3-CH=CH-CH_3\), \(Q=CH_3-CH_2-CH(Cl)CH_3\), \(R=CH_3-CH_2-CH(NH_2)CH_3\), \(S=CH_3-CH_2-CH(OH)CH_3\)
• D. P=Cyclobutane, Q=Chlorocyclobutane, R=Cyclobutylamine, S=Cyclobutanol

Hint:

In multi-step organic synthesis problems, use key information like "optically active" to eliminate possibilities early on. Working forwards or backwards through the reaction sequence can help confirm the structure of all intermediates. Remember the stereochemical implications of reactions.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question describes a sequence of reactions starting from a hydrocarbon P (C₄H₈). We need to deduce the structures of P, Q, R, and S based on the reaction outcomes, with a key clue being the optical activity of compound Q.
Step 2: Detailed Explanation:
Reaction 1: P (C₄H₈) + HCl → Q (C₄H₉Cl, optically active)
\begin{itemize} \item P (C₄H₈) is an alkene. Possible isomers are But-1-ene, But-2-ene (cis/trans), and 2-Methylpropene.
\item The product Q must be optically active, meaning it must have a chiral carbon (a carbon atom attached to four different groups).
\item Let's check the addition of HCl to each isomer of P: \begin{itemize} \item But-1-ene (CH₃CH₂CH=CH₂) + HCl → CH₃CH₂CH(Cl)CH₃ (2-Chlorobutane, major product) and CH₃CH₂CH₂CH₂Cl (1-Chlorobutane, minor). 2-Chlorobutane is chiral at C2. So, P could be But-1-ene.
\item But-2-ene (CH₃CH=CHCH₃) + HCl → CH₃CH₂CH(Cl)CH₃ (2-Chlorobutane). This product is chiral. So, P could also be But-2-ene.
\item 2-Methylpropene ((CH₃)₂C=CH₂) + HCl → (CH₃)₃C-Cl (tert-Butyl chloride). This product has no chiral center. So P cannot be 2-Methylpropene.
\end{itemize} \item So, P is either But-1-ene or But-2-ene, and Q is 2-Chlorobutane, CH₃CH₂CH(Cl)CH₃.
\end{itemize} Reaction 2: Q + NH₃ → R (C₄H₁₁N)
\begin{itemize} \item This is a nucleophilic substitution reaction (ammonolysis), where the -Cl group is replaced by an -NH₂ group.
\item Q (2-Chlorobutane) + NH₃ → R (Butan-2-amine).
\item Structure of R: CH₃CH₂CH(NH₂)CH₃.
\end{itemize} Reaction 3: R --(diazotization)---- S
\begin{itemize} \item Diazotization of a primary aliphatic amine (with NaNO₂/HCl or HNO₂) followed by hydrolysis converts the amine group (-NH₂) into an alcohol group (-OH).
\item R (Butan-2-amine) will form S (Butan-2-ol).
\item Structure of S: CH₃CH₂CH(OH)CH₃.
\end{itemize} Step 3: Matching with Options:
Let's check the options with our deduced structures:
P = But-1-ene or But-2-ene
Q = CH₃CH₂CH(Cl)CH₃ (2-Chlorobutane)
R = CH₃CH₂CH(NH₂)CH₃ (Butan-2-amine)
S = CH₃CH₂CH(OH)CH₃ (Butan-2-ol)
\begin{itemize} \item Option (A) is incorrect because it shows Q as 1-Chlorobutane.
\item Option (B) is incorrect because it shows Q as 1-Chlorobutane and S as Butan-1-ol.
\item Option (C) shows P=But-2-ene, Q=2-Chlorobutane, R=Butan-2-amine, and S=Butan-2-ol. This sequence is entirely consistent with our analysis.
\item Option (D) starts with Cyclobutane, which has the formula C₄H₈, but its reaction with HCl would be a ring-opening reaction under specific conditions, not a simple addition to form chlorocyclobutane. The proposed sequence is less likely and doesn't fit the typical reactions described.
\end{itemize} Step 4: Final Answer:
The correct reaction sequence is described in option (C).