Chemistry Hydrocarbons MDCAT 2019
PMDC Verified Question 36 of 65
In the reaction sequence:
\( \text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{Br} + \text{Alc.KOH} \rightarrow \text{C} \xrightarrow{\text{H}_2\text{SO}_4 / \text{H}_2\text{O}} \text{D} \)
Product D will be
A
Mixture of methanol and ethanol
B
1-Propanol
C
Propanoic acid
D
2-propanol
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: 2-propanol
Concept:

Elimination of a primary alkyl halide yields a terminal alkene. Hydration of that terminal alkene follows Markovnikov's rule, yielding a secondary alcohol.

Solution:

  • Step 1: 1-bromopropane (\( \text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{Br} \)) is treated with alcoholic KOH. This is a dehydrohalogenation (elimination) reaction, yielding the alkene propene (Compound C): \( \text{CH}_3-\text{CH}=\text{CH}_2 \).


  • Step 2: Propene is hydrated using \( \text{H}_2\text{SO}_4 \) and water. This is an electrophilic addition of \( \text{H}_2\text{O} \).


  • Following Markovnikov's rule, the H adds to the terminal carbon (\( \text{CH}_2 \)) and the OH adds to the central carbon (\( \text{CH} \)).


  • This produces \( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3 \), which is 2-propanol.


Why other options are incorrect:

  • 1-Propanol would be the product if anti-Markovnikov hydration (hydroboration-oxidation) were used instead of acid-catalyzed hydration.


  • Propanoic acid requires an oxidation reaction.

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