| ⇦ | 
| ⇨ |
Acetylene, on reacting with formaldehyde under high pressure, gives
Options
(a) propanol
(b) propyl alcohol
(c) allyl aldehyde
(d) butyndiol
Correct Answer:
butyndiol
Explanation:
HC ≡ CH + HCHO (Acetylene) → HC ≡ C – CH₂OH (Formaldehyde)
HC ≡ C – CH₂OH + H – C = O → HOCH₂C ≡ CCH₂OH ( But – 2- yn-1,4-diol)
Related Questions: - The number of water molecules in gypsum and in plaster of paris are respectively
- 5 mL of N HCl, 20 mL of N/2 H₂SO₄ and 30 mL of N/3 HNO₃ are mixed together
- Which one of the following compounds exhibit both schottky and frenkel defects
- How many grams of dibasic acid (mol.wt.200) should be present in 100 mL.
- At room temperature the eclipsed and staggered forms of ethane can not be isolate
Topics: Hydrocarbons
(84)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The number of water molecules in gypsum and in plaster of paris are respectively
- 5 mL of N HCl, 20 mL of N/2 H₂SO₄ and 30 mL of N/3 HNO₃ are mixed together
- Which one of the following compounds exhibit both schottky and frenkel defects
- How many grams of dibasic acid (mol.wt.200) should be present in 100 mL.
- At room temperature the eclipsed and staggered forms of ethane can not be isolate
Topics: Hydrocarbons (84)
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
C2H2+2HCOH=OHCH2CCCH2OH