Question

Identify the incorrect reaction from the following.

• A. \[ CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3CHO + HCHO \]
• B. \[ CH_3CH=CH_2 \xrightarrow[\;273\,K\;]{dil.\,KMnO_4} CH_3CH(OH)CH_2OH \]
• C. \[ CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3 \;(\text{Major}) + CH_3CH_2CH_2Br \;(\text{Minor}) \]
• D. \[ CH_3CH=CH_2 + HCl \xrightarrow{\text{Peroxide}} CH_3CHClCH_3 \;(\text{Major}) + CH_3CH_2CH_2Cl \;(\text{Minor}) \]

Hint:

\[ \begin{aligned} HBr &: \text{Shows peroxide effect} \\ HCl &: \text{Does not show peroxide effect} \\ HI &: \text{Does not show peroxide effect} \end{aligned} \] \[ \boxed{\text{Peroxide effect is observed only with } HBr} \]

Answer 1

The Correct Option is D

Concept: Alkenes undergo ozonolysis, hydroxylation and electrophilic addition reactions. The peroxide (Kharasch) effect is observed only with HBr and not with HCl or HI.

Step 1: Check reaction (A). Propene undergoes ozonolysis followed by reductive workup to give ethanal and methanal. \[ CH_3CH=CH_2 \xrightarrow[Zn/H_2O]{O_3} CH_3CHO + HCHO \] Hence, reaction (A) is correct.

Step 2: Check reaction (B). Cold dilute potassium permanganate converts alkenes into vicinal glycols. \[ CH_3CH=CH_2 \xrightarrow[\;273\,K\;]{dil.\,KMnO_4} CH_3CH(OH)CH_2OH \] Hence, reaction (B) is correct.

Step 3: Check reaction (C). Addition of HBr in the absence of peroxide follows Markovnikov's rule. \[ CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3 \] Major product is 2-bromopropane. Hence, reaction (C) is correct.

Step 4: Check reaction (D). Peroxide effect is exhibited only by HBr. \[ HCl \quad \text{does not show peroxide effect} \] Therefore, anti-Markovnikov addition of HCl is not possible. The reaction shown in option (D) is incorrect. \[ \begin{aligned} (A) &: \text{Correct} \\ (B) &: \text{Correct} \\ (C) &: \text{Correct} \\ (D) &: \text{Incorrect} \end{aligned} \] \[ \boxed{\text{Reaction (D) is incorrect}} \] Hence, option \(\mathbf{(D)}\) is correct.