Kolbe reaction class 12

Tamilnadu State Board Solutions for Class 12th Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers Questions and Answers will help you to improve the complete subject knowledge. Every student must look at the single concept included in Students can easily avoid the struggle to get the best book for Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers Book Solutions Answers Guide learning. You have to go through Chapterwise Samacheer Kalvi Class 12th Textbook Solutions for Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers Book Solutions for better practice. Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers Students who wish to have the strong basics of Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers can use Tamilnadu State Board Class 12th Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers Questions and Answers Guide pdf. Enhance your knowledge by referring to the Samacheer Kalvi Solutions pdf. We provided a free pdf of Tamilnadu State Board Class 12th Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers Book Solutions Answers Guide material and textbook for students. Check it out now and start preparing for the exam immediately. Score maximum marks in the exam by referring to Samacheer Kalvi Class 12th Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers Book Solutions Answers Guide Solutions Pdf. Samacheer Kalvi 12th Chemistry Chapter 11 Hydroxy Compounds and Ethers Textua...

Substitution reaction: A substitution reaction is that which involves the direct replacement of hydrogen atoms of some other groups of a molecule by suitable atoms or groups without changing the structure of remaining part of the molecule. The product obtained is known as substitution product and new atom or group which enters the molecule is known as a substituent. Halogenation: It involves the replacement of one or more hydrogen atom(s) of alkane by the corresponding number of a halogen atom(s). (i) Chlorination: Chlorination of alkanes is carried out by treating alkane with chlorine in the presence of ultraviolet light or at 523-673 K temperature. (ii) Bromination: Bromination takes place similarly but not so rapidly. (iii) Iodination: Iodination is extremely slow and reversible due to the reducing nature of HI. Thus, iodination is carried out in the presence of some oxidising agent like iodic acid (HIO 3 ) of nitric acid (HNO 3) which converts HI to iodine and pushes the reaction in the forward direction. Halogenation of higher alkanes (ethane, propane etc), yields a mixture of all possible isomerism products. (a) Nitration of alkane: The process which involves the replacement of hydrogen atom of alkanes by nitro group (-NO 2 ) is known as nitration of an alkane. It can be carried out in two ways: (i) Liquid phase nitration; In this method higher alkane is heated with fuming HNO 3 at 413 K under pressure. (ii) Vapour phase nitration: Lower member of alkanes can be nitr...

NCERT Solutions for Exam Preparation Class 12 Chemistry Chapter 11 Alcohols, Phenols And Ethers are provided here with simple step-by-step explanations. These solutions for Alcohols, Phenols And Ethers are extremely popular among Exam Preparation Class 12 students for Chemistry Alcohols, Phenols And Ethers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Exam Preparation Class 12 Chemistry Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Exam Preparation Class 12 Chemistry are prepared by experts and are 100% accurate. Answer: (i) 2, 2, 4-Trimethylpentan-3-ol (ii) 5-Ethylheptane-2, 4-diol (iii) Butane-2, 3-diol (iv) Propane-1, 2, 3-triol (v) 2-Methylphenol (vi) 4-Methylphenol (vii) 2, 5-Dimethylphenol (viii) 2, 6-Dimethylphenol (ix) 1-Methoxy-2-methylpropane (x) Ethoxybenzene (xi) 1-Phenoxyheptane (xii) 2-Ethoxybutane Page No 344: Question 11.2: Write structures of the compounds whose IUPAC names are as follows: (i) 2-Methylbutan-2-ol (ii) 1-Phenylpropan-2-ol (iii) 3,5-Dimethylhexane −1, 3, 5-triol (iv) 2,3 − Diethylphenol (v) 1 − Ethoxypropane (vi) 2-Ethoxy-3-methylpentane (vii) Cyclohexylmethanol (viii) 3-Cyclohexylpentan-3-ol (ix) Cyclopent-3-en-1-ol (x) 3-Chloromethylpentan-1-ol. Answer: (i) The structures of all isomeric alcohols of molecular formula, C 5H 12O are shown below: (a) Pen...

Kolbe's reaction is a type of addition reaction named after Sir Hermann Kolbe and Rudolf Schmitt. It is also known as Kolbe Schmitt's reaction. When phenol reacts with sodium hydroxide, phenoxide ion is generated. This generated phenoxide ion is more reactive as compared to phenol towards electrophilic aromatic substitution reactions. The phenoxide ion reacts with carbon dioxide (a weak electrolyte) and undergoes an electrophilic substitution reaction. Ortho-hydroxybenzoic acid (salicylic acid) is formed in the reaction as the primary product. This reaction is called Kolbe's reaction. An Example of Kolbe's Reaction is Given below: (Image to be Updated Soon) Details About Kolbe - Schmitt Reaction In the earlier classes, students must have learnt about the transformation of one type of chemical substance or compounds to some other form of compounds or substances. This process is commonly known as the process of chemical reaction that occurs due to the inherent chemical properties of that substance or by some external arrangements. The electrons present in the atoms of the molecules play the role of carrying out such reactions. The chemical bonds break or get exchanged with the transfer of electrons. In the chapters of organic chemistry students also must have got acquainted with the carboxylic acid group of hydrocarbons. These chains of carbon atoms have one Oxygen atom and an OH formed out of oxygen and hydrogen atoms, molecules attached to one carbon atom. Carboxylic acid ...

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ISC Chemistry Semester-1 Specimen Paper Solved Class -12 for practice. Step by step solutions of ISC Class-12 specimen model sample paper . During solutions of semester-1 Chemistry specimen paper we explain with figure , graph, table whenever necessary so that student can achieve their goal in next upcoming exam of council. Visit official website ISC Chemistry Semester-1 Specimen Paper Solved Class -12 Board ISC Class 12th (XII) Subject Chemistry Topic Semester-1 ISC Specimen Paper Solved Syllabus on bifurcated syllabus (after reduction) session 2021-22 Question Type MCQ/ Objective (as prescribe by council) Total question Total=50 with all parts Max mark 70 Colligative properties depend on: (a) The nature of solute particles in solution (b) The number of solute particles in solution (c) The nature of solute and solvent particles (d) The physical properties of solute particles in solution Answer- (b) The number of solute particles in solution Question 3 On dilution, the specific conductance of a solution: (a) Remains unchanged (b) Increases (c) Decreases (d) First increases then decreases Answer- (a) Remains unchanged Question 4 The flux used in the extraction of iron from haematite ore is: (a) Limestone (b) Silica (c) Coke (d) Calcium phosphate Answer- (a) Limestone Question 5 Which of the following xenon fluoride of xenon cannot be formed? (a) XeF2 (b) XeF4 (c) XeF6 (d) XeF3 Answer- (d) XeF3 Question 6 The gas obtained on heating iodoform with silver powder is: (a) Propan...

Hint: Aspirin is 2-Acetoxybenzoic acid which is prepared from Salicylic acid (2-Hydroxybenzoic acid) as the starting material. Aspirin is a non-narcotic analgesic which is used as antipyretic and as a pain reliever. Complete step by step answer: Aspirin is also called 2-acetoxy benzoic acid. It acts as an analgesic. An analgesic is a drug that reduces or relieves pain. They are classified as non-narcotic or narcotic pain relievers. Aspirin is non-narcotic i.e. non-opioid and is used to reduce pain and inflammation. It is also antipyretic. We can easily prepare aspirin by acetylating salicylic acid using acetic anhydride in the presence of concentrated $ $. For synthesising aspirin, we first need to prepare salicylic acid. Salicylic acid is ortho-hydroxybenzoic acid. It can be synthesised using the Kolbe-Schmitt process. The first step involves the reaction of phenol and sodium hydroxide in order to produce sodium phenoxide. This phenoxide is then reacted with carbon dioxide to form sodium salicylate which on acidification gives salicylic acid. The reactions are given below: So, the correct answer is “Option (B) $$”. Note: All the above reactions are taking place in an aqueous medium. The end product i.e. Aspirin is also crystallised from the aqueous medium. The Kolbe-Schmitt process is a very old process, though still in use, many new efficient methods are used for the manufacture of Aspirin now.