Chemistry Hydrocarbons MDCAT 2013
PMDC Verified Question 55 of 65
What is the product formed when propene reacts with HBr?
A
\( \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} \)
B
\( \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \)
C
\( \text{BrCH}_2-\text{CH}=\text{CHBr} \)
D
\( \text{CH}_3-\text{C}(\text{Br})_2-\text{CH}_3 \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \)
Concept:

The electrophilic addition of an unsymmetrical hydrogen halide (HBr) to an unsymmetrical alkene (propene) follows Markovnikov's rule.

Solution:

  • Propene is \( \text{CH}_3-\text{CH}=\text{CH}_2 \).


  • According to Markovnikov's rule, the electrophile (\( \text{H}^+ \)) adds to the \( \text{sp}^2 \) carbon with the greater number of hydrogen atoms to form the most stable intermediate carbocation.


  • \( \text{H}^+ \) adds to \( \text{CH}_2 \), yielding a stable secondary carbocation (\( \text{CH}_3-\text{C}^+\text{H}-\text{CH}_3 \)).


  • The bromide ion (\( \text{Br}^- \)) then attacks the central carbon, resulting in 2-bromopropane: \( \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \).


Why other options are incorrect:

  • Option A (1-bromopropane) would require anti-Markovnikov addition (only occurs with HBr in the presence of peroxides).


  • Option C and D imply multiple additions or substitutions, which does not happen with a single equivalent of HBr on an alkene.

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