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

Step 1: Goal.
Find the one reaction written wrongly. Check each addition or oxidation against the standard rule.

Step 2: Reaction (A).
Ozonolysis of propene with reductive workup cleaves the double bond into ethanal and methanal. This is correct.

Step 3: Reaction (B).
Cold dilute $KMnO_4$ adds two $-OH$ groups across the double bond giving a glycol, here propane-1,2-diol. This is correct.

Step 4: Reaction (C).
$HBr$ adds by Markovnikov's rule, putting bromine on the more substituted carbon, so $CH_3CHBrCH_3$ is the major product. This is correct.

Step 5: Reaction (D).
The peroxide (anti-Markovnikov) effect works only with $HBr$. With $HCl$ there is no peroxide effect, so $HCl$ still follows Markovnikov and the named anti-Markovnikov outcome under peroxide is wrong.

Step 6: Conclusion.
Reaction (D) is the incorrect one.
\[ \boxed{\text{Reaction (D)}} \]