Characteristic organic reactions — practice questions

This is an electrophilic addition reaction. The HBr molecule is polar, $H^{\delta+}-Br^{\delta-}$.

Step 1: The electron-rich π-bond of propene attacks the $\delta+$ hydrogen atom of HBr. The H-Br bond breaks heterolytically.

Step 2: A carbocation intermediate is formed. The hydrogen can add to either carbon of the original double bond.

• Path A (forms secondary carbocation): H adds to C1, forming $CH_3\overset{+}{C}HCH_3$. This is a secondary (2°) carbocation, which is relatively stable due to the positive inductive effect of two adjacent alkyl groups.
• Path B (forms primary carbocation): H adds to C2, forming $CH_3CH_2\overset{+}{C}H_2$. This is a primary (1°) carbocation, which is less stable.

Step 3: According to Markovnikov's rule, the major pathway is the one that forms the more stable carbocation (Path A). The bromide ion, $Br^-$, then acts as a nucleophile and attacks the positive carbon of the secondary carbocation.

Conclusion: The major product is 2-bromopropane ($CH_3CHBrCH_3$). The minor product is 1-bromopropane.

The reactant is 1-chlorobutane ($CH_3CH_2CH_2CH_2Cl$), a primary halogenoalkane. The reagent is aqueous sodium hydroxide, which provides the hydroxide ion ($OH^-$), a strong nucleophile.

• Organic Product: The $OH^-$ nucleophile will replace the chlorine atom. The product is butan-1-ol ($CH_3CH_2CH_2CH_2OH$).
• Mechanism: Since it is a primary halogenoalkane, the reaction proceeds via an Sₙ2 (Substitution, Nucleophilic, Bimolecular) mechanism.
• Overall Equation: $CH_3CH_2CH_2CH_2Cl + NaOH(aq) \rightarrow CH_3CH_2CH_2CH_2OH(aq) + NaCl(aq)$