JEE Challenger
More from Hydrocarbons
ChemMedAdv 2015 • Paper 1

Major Product Determination in Hydrobromination of Conjugated Diene

Determine the major organic product obtained in the following transformation:

Question Diagram 1

Options

A
Option A Diagram 1
B
Option B Diagram 1
C
Option C Diagram 1
D
Option D Diagram 1
Correct

Topics & Concepts

HydrocarbonsAlkanes, Alkenes, AlkynesElectrophilic Addition

Step-by-Step Solution

To determine the major product of the electrophilic addition of 1 equivalent1\text{ equivalent} of HBr\text{HBr} to 2-methylbuta-1,3-diene (isoprene), we analyze the reaction mechanism step-by-step:

Step 1: Protonation (Electrophilic Attack of H+\text{H}^+)

The starting diene, 2-methylbuta-1,3-diene, has two distinct double bonds: C1H2=C2(CH3)−C3H=C4H2\overset{1}{\text{C}}\text{H}_2=\overset{2}{\text{C}}(\text{CH}_3)-\overset{3}{\text{C}}\text{H}=\overset{4}{\text{C}}\text{H}_2

Protonation occurs to yield the most stable allylic carbocation intermediate:

  1. Attack at C-1\text{C-1}: Gives a 3∘3^\circ allylic carbocation: CH3−C+(CH3)−CH=CH2⟷CH3−C(CH3)=CH−C+H2\text{CH}_3-\overset{+}{\text{C}}(\text{CH}_3)-\text{CH}=\text{CH}_2 \longleftrightarrow \text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\overset{+}{\text{C}}\text{H}_2
  2. Attack at C-4\text{C-4}: Gives a 2∘2^\circ allylic carbocation: H2C=C(CH3)−C+H−CH3⟷C+H2−C(CH3)=CH−CH3\text{H}_2\text{C}=\text{C}(\text{CH}_3)-\overset{+}{\text{C}}\text{H}-\text{CH}_3 \longleftrightarrow \overset{+}{\text{C}}\text{H}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3

Since a tertiary allylic carbocation is significantly more stable than a secondary allylic carbocation, protonation preferentially takes place at C-1\text{C-1}. This generates the resonance-stabilized allylic intermediate: [(CH3)2C+−CH=CH2⟷(CH3)2C=CH−C+H2]\left[ (\text{CH}_3)_2\overset{+}{\text{C}}-\text{CH}=\text{CH}_2 \longleftrightarrow (\text{CH}_3)_2\text{C}=\text{CH}-\overset{+}{\text{C}}\text{H}_2 \right]

Step 2: Nucleophilic Attack of Br−\text{Br}^-

The bromide ion (Br−\text{Br}^-) can attack at two possible positions of the delocalized carbocation:

  1. 1,2-Addition (Kinetic Product): Attack of Br−\text{Br}^- at C-2\text{C-2} forms 3-bromo-3-methylbut-1-ene: (CH3)2C(Br)−CH=CH2(\text{CH}_3)_2\text{C}(\text{Br})-\text{CH}=\text{CH}_2 This product forms faster at lower temperatures due to lower activation energy (kinetic control), but it contains a mono-substituted alkene.

  2. 1,4-Addition (Thermodynamic Product): Attack of Br−\text{Br}^- at C-4\text{C-4} forms 1-bromo-3-methylbut-2-ene: (CH3)2C=CH−CH2Br(\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_2\text{Br} This product contains a tri-substituted, more highly substituted (and hence significantly more stable) double bond.

Step 3: Product Stability and Conclusion

Under standard reaction conditions, the reaction is thermodynamically controlled. The thermodynamically most stable alkene (1,4-addition product) predominates as the major product because it features a tri-substituted double bond rather than a mono-substituted double bond.

Thus, the major product is (CH3)2C=CH−CH2Br(\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_2\text{Br}.

Correct Option: (D)

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