| ⇦ | 
| ⇨ |
When ethyl alcohol (C₂H₅OH) reacts with thionyl chloride in the presence of pyridine the product obtained is
Options
(a) CH₃CH₂Cl + HCl
(b) CH₃CH₂Cl + H₂O + SO₂
(c) CH₃CH₂Cl + Cl₂ + SO₂
(d) C₂H₅Cl + HCl + SO₂
Correct Answer:
C₂H₅Cl + HCl + SO₂
Explanation:
C₂H₅OH + SOCl₂ → C₂H₅Cl + HCl + SO₂.
Related Questions: - The difference between ΔH and ΔE at 300 K for the reaction
- Naphthalene is a
- The product formed on reaction, of ethyl alcohol with bleaching powder is
- Which of the following elements are analogous to the lanthanides
- Ksp of CaSO₄.5H₂O is 9 ˣ 10⁻⁶, find the volume for 1g of CaSO₄ (M.wt, = 136)
Topics: Haloalkenes and Haloarenes
(78)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The difference between ΔH and ΔE at 300 K for the reaction
- Naphthalene is a
- The product formed on reaction, of ethyl alcohol with bleaching powder is
- Which of the following elements are analogous to the lanthanides
- Ksp of CaSO₄.5H₂O is 9 ˣ 10⁻⁶, find the volume for 1g of CaSO₄ (M.wt, = 136)
Topics: Haloalkenes and Haloarenes (78)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply