Draw the two major organic products of the reaction shown: Select Draw OH conc HBr
The Correct Answer and Explanation is:
Thank you for the image. The compound shown is an allylic alcohol:
3-buten-2-ol derivative with methyl substitutions — specifically, 3-methylbut-2-en-1-ol.
You’re reacting this with concentrated HBr, which will convert the –OH into a good leaving group (H₂O), enabling carbocation formation, and then substitution by Br⁻.
🔬 Step-by-Step Mechanism:
Step 1: Protonation of the alcohol
The –OH group is protonated by HBr → becomes H₂O, a good leaving group.
Step 2: Formation of carbocation
Loss of water leads to a primary carbocation at the terminal carbon — but this is allylic, so resonance stabilization allows delocalization of the positive charge over two positions:
CH2=CH–CH(+)-CH3 ⇌ CH2(+)-CH=CH–CH3
Step 3: Attack by Br⁻
The bromide ion (Br⁻) can attack at either resonance-stabilized carbocation center:
- The primary carbon (original position)
- The tertiary carbon after resonance
✅ Major Organic Products:
- 1-Bromo-3-methylbut-2-ene
(Br attaches to the original –OH position) - 3-Bromo-3-methylbut-1-ene
(Br attaches to the tertiary carbon via resonance)
🧪 Structures:
Product 1:
CH₂=CH–C(CH₃)₂–Br
(Allylic substitution at the terminal CH₂)
Product 2:
CH₃–C(Br)=CH–CH₃
(Br attaches at the tertiary carbon, double bond shifts)
Explanation:
When 3-methylbut-2-en-1-ol reacts with concentrated HBr, the alcohol undergoes substitution by a bromine atom. This transformation proceeds through a two-step mechanism: first, protonation of the hydroxyl group by HBr forms water, a good leaving group. Once water departs, a carbocation intermediate is formed. In this case, the resulting carbocation is allylic, meaning the positive charge is adjacent to a double bond. This allows for resonance stabilization, where the positive charge can delocalize across two carbon atoms.
Because of this resonance, the Br⁻ nucleophile has two electrophilic centers to attack: the carbon where the OH originally was (a primary position), and the more substituted tertiary carbon formed via resonance. Both of these pathways lead to major organic products, because both are resonance contributors of a stable allylic carbocation.
The first product results from bromide attacking the primary position, giving 1-bromo-3-methylbut-2-ene. This is considered major because it follows direct substitution. The second product forms when bromide attacks the tertiary resonance form, leading to 3-bromo-3-methylbut-1-ene. This product is also major due to the stability of the tertiary carbocation.
The reaction showcases two fundamental concepts in organic chemistry: carbocation resonance stabilization and nucleophilic substitution at allylic positions. These allow for multiple favorable pathways, yielding more than one major product. Such outcomes are especially common with allylic systems where resonance and substitution can compete or cooperate
