--- title: "Elimination reaction; dehydrohalogenation of tertiary halide" subject: "Engineering Chemistry" chapter: 10 key: "semester_2/engineering_chemistry:e1-tertiary-dehydrohal" source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/e1-tertiary-dehydrohal site: BCT Engineering --- # Elimination reaction; dehydrohalogenation of tertiary halide Engineering Chemistry · Chapter 10: Reaction Mechanism in Organic reactions - Asked in: 2073 Magh, 1+4 marks; 2069 Bhadra, 2+3 marks (asked 2 times) - Video: [Elimination Reaction of Alkyl Halide | Dehydrohelogenation |E1 And E2 Elimimation Reaction #bscnotes](https://www.youtube.com/watch?v=Dk4xhxXJioo) — Kanhaiya Patel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/e1-tertiary-dehydrohal - Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-10 ## Question What is elimination reaction? Explain the reaction mechanism for the dehydrohalogenation of tertiary alkyl halide. ## Short answer An elimination reaction removes two atoms or groups from adjacent carbons, forming a double bond. Dehydrohalogenation of a tertiary alkyl halide proceeds by E1: slow ionization to a carbocation, then fast base removal of a β-hydrogen yielding an alkene. Rate = k[RX] (first-order). ## Answer #### Elimination reaction An **elimination reaction** is a reaction in which two atoms or groups are removed from adjacent carbon atoms of a molecule, forming a double (or triple) bond, without being replaced by other groups. **Dehydrohalogenation** of an alkyl halide is the elimination of H and X (as HX) using a strong base such as alcoholic KOH to form an alkene. #### Dehydrohalogenation of a tertiary alkyl halide Tertiary halides eliminate by the E1 mechanism, in a polar protic solvent such as ethanol. **Mechanism: E1 (two steps).** ```text Step 1 (slow): ionisation to carbocation (CH₃)₃C—Br ---slow---> (CH₃)₃C⁺ + Br⁻ Step 2 (fast): base removes a β-hydrogen OH⁻ + H—CH₂—C⁺(CH₃)₂ ---fast---> H₂O + CH₂=C(CH₃)₂ ``` Overall: (CH₃)₃C–Br + KOH(alc) → (CH₃)₂C=CH₂ + KBr + H₂O (2-methylpropene) - Rate = k[RX]: first order (unimolecular), as only the halide takes part in the slow step. - Carbocation intermediate, so rearrangement is possible; competes with SN1. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens in an elimination reaction? A. A group is replaced by another group on the same carbon B. Two molecules add together to form a larger saturated molecule C. Two atoms or groups are removed from adjacent carbons, forming a double or triple bond D. A hydrogen is exchanged between two carbons without forming a pi bond Answer: C. Elimination removes two groups from neighbouring carbons without replacing them, so a C=C (or triple) bond forms. (Key point: Elimination reaction) 2. How does E1 dehydrohalogenation of (CH3)3C-Br proceed? A. One concerted step in which base removal of H and loss of Br happen together B. Slow ionisation to a carbocation, then fast removal of a beta-hydrogen by the base C. Fast attack of OH- on carbon, then slow loss of Br- D. Slow removal of H by the base, then fast loss of Br- Answer: B. E1 is two steps: the halide ionises slowly to a carbocation, then the base quickly removes a beta-H to form the alkene. (Key point: E1 mechanism steps) 3. What is the rate law for E1 elimination? A. Rate = k[RX][Base] B. Rate = k[Base] C. Rate = k[RX]^2 D. Rate = k[RX], independent of base concentration Answer: D. Only the halide takes part in the slow ionisation step, so E1 is first order in RX and independent of the base. (Key point: E1 rate law) 4. Which statement about E1 elimination is correct? A. A carbocation intermediate forms, so rearrangement is possible and SN1 competes B. There is no intermediate and no rearrangement C. It needs a strong, concentrated base D. It is favoured in non-polar solvents Answer: A. E1 goes via a carbocation, so rearrangement and competition from SN1 can occur. (Key point: Features of E1) ## Close variants set in other exams - [Dehydrohalogenation of 1° alkyl halide](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Adehydrohal-primary) — Engineering Chemistry, chapter 10, asked 4× ## Related in Engineering Chemistry - [Tertiary alkyl halide with aq. NaOH / alc. KOH](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-halide-naoh-koh) — Engineering Chemistry, chapter 10 - [Mechanism for primary alkyl halide in aqueous medium; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn2-primary-solvent) — Engineering Chemistry, chapter 10 - [Elimination reaction; dehydrohalogenation; Saytzeff's rule](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Aelimination-saytzeff) — Engineering Chemistry, chapter 10, asked 2× - [Dehydrohalogenation; mechanism and kinetics of E1 [partly illegible]](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ae1-kinetics) — Engineering Chemistry, chapter 10, asked 2× - [SN reactions; hydrolysis of 3° alkyl halide](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Asn-tertiary-hydrolysis) — Engineering Chemistry, chapter 10, asked 2× - [3° alkyl halide with alcoholic KOH; solvent](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Atert-alc-koh-solvent) — Engineering Chemistry, chapter 10