Basics Of Organic Chemistry Pdf 23

I began writing an organic chemistry textbook in 1992. The journey has been long and filled with unexpected twists and turns. I had hoped that I might some day see the book on shelves of campus bookstores, but I realized several years ago that this was unlikely.

As a result, I decided to post it on the internet. I wanted it to be available to students and faculty for studying and teaching organic chemistry. I have now posted 21 of the 23 chapters. I am currently working on Chapter 19 and plan to begin Chapter 6. The figures have been merged with the text in each posted chapter. With the exception of those in Chapter 20, they are presently all hand-drawn.You can access the book directly through the links on this page. They also lead to biographical information about me. The thumbnail version of my biography is "I retired from a 35 year position in the chemistry department at UC Riverside in 1998 and moved to Santa Barbara where I taught part-time at UCSB from 2000 to 2008".My greatest fan and supporter, through all of my efforts to be an author, has been, and continues to be, my wife Pat! Without her encouragement I would have abandoned all of this a long time ago!

basics of organic chemistry pdf 23

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The Author: I was born in Chicago, moved with my parents to Los Angeles at the age of 3, attended Dacotah St. Elementary School, and Gage Junior High School, moved to Whittier, CA, when I was 13, and graduated from Whittier High School in 1955. I did my undergraduate studies at UCLA, earned my Ph.D. at Caltech, was a post-doc at Columbia, and then joined the chemistry faculty at the University of California, Riverside. You can see my professional biography here - Robert Neuman.

The FULL COURSE Notes (1st and 2nd Semesters) of Organic Chemistry Notes is 367 pages in length (Section 1 through Section 23) and covers ALL lecture notes and topics discussed in your ENTIRE organic chemistry lecture course.

AP Chemistry is an introductory college-level chemistry course. Students cultivate their understanding of chemistry through inquiry-based lab investigations as they explore the four Big Ideas: scale, proportion, and quantity; structure and properties of substances; transformations; and energy.

Based on the Understanding by Design (Wiggins and McTighe) model, this course framework provides a clear and detailed description of the course requirements necessary for student success. The framework specifies what students must know, be able to do, and understand, with a focus on big ideas that encompass core principles and theories of the discipline. The framework also encourages instruction that prepares students for advanced chemistry coursework.

At the very beginning it was said that inspite of the huge variety of organic compounds, it is surprisingly simple to predict how they react using some basic principles. And we have already seen what those few fundamental tenets of organic chemistry are. But the reason why a huge molecule like cholesterol (a vital molecule in our body) would react mostly in the same way as a two carbon ethyl alcohol (a popular drink) is because they share the same functional group-an alcoholic OH group with which we will familiarize ourselves a bit later.

The knowledge of some basic functional groups and how they react would give us tremendous leverage to tackle the problem of predicting chemical reactivity in organic chemistry. Â Following is the table of the common functional groups you will encounter in organic chemistry.

General introduction, methods of purification, qualitative and quantitative analysis, classification and IUPAC nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.

Question 18. Give a brief description of the principles of the following techniques taking an example in each case: (a) Crystallisation (b) Distillation (c) ChromatographyAnswer: (a) Crystallisation: In this process the impure solid is dissolved in the minimum volume of a suitable solvent. The soluble impurities pass into the solution while the insoluble ones left behind. The hot solution is then filtered and allowed to cool undisturbed till crystallisation is complete. The crystals are then separated from the mother liquor by filtraration and dried.Example: crystallisation of sugar.(b) Distillation: The operation of distillation is employed for the purification of liquids from non-volatile impurities. The impure liquid is boiled in a flask and the vapours so formed are collected and condensed to give back pure liquid in another vessel. Simple organic liquids such as benzene toluene, xylene etc. can be purified.(c) Chromatography: Chromatography is based on the principle of selective distribution of the components of a mixture between two phases, a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. When the stationary phase is solid the basis is adsorption and when it is a liquid the basis is partition. Chromatography is generally used for the Reparation of coloured substances such as plant pigments or dyestuffs.

Question 23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.Answer: Estimation of halogens: It involves oxidising the organic substance with fuming nitric acid in the presence of silver nitrate. The halogen of the substance is thus converted to silver halide which is separated and weighed:1Weight of organic compound = W gmweight of silver halide = x g.Estimation of sulphur: The organic substance is heated with fuming nitric acid but no silver nitrate is added. The sulphur of the substance is oxidised to sulphuric acid which is then precipitated as barium sulphate by adding excess of barium chloride solution. From the weight of BaSO4 so obtained the percentage of sulphur can be calculated.Estimation of phosphorous: The organic substance is heated with fuming nitric acid whereupon phosphorous is oxidised to phosphoric acid. The phosphoric acid is precipitated as ammonium phosphomolybdate, (NH4)3 PO4 .12MOO3, by the addition of ammonia and ammonium molybdate solution which is then separated, dried and weighed.

Question 26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.Answer: Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na2S, NaX and Na3PO4. Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

Question 28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ? Answer: It is because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

Question 32. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is subjected to complete combustion.Answer:

Question 33. 0.50 g of an organic compound was Kjeldahlished. The ammonia evolved was passed in 50 cm3 of IN H2SO4. The residual acid required 60 cm3 of N/2 NaOH solution. Calculate the percentage of nitrogen in the compound.Answer:

Question 34. 0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.Answer: Mass of the compound = 0.3780 gMass of silver chloride = 0.5740 g

Question 35. In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.Answer: Mass of the compound = 0.468 gMass of barium sulphate= 0.668 g

Question 39. The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography.Answer: (d) is the correct answer.

Question 1. How will you separate a mixture of two organic compounds which have different solubilities in the same solvent?Answer: By fractional crystallisation.

Question 8. 0 .12 g of an organic compound containing phosphorous gave 0.22 g of Mg2 P2O7 by usual analysis. Calculate the percentage of phosphorous in the compound.Answer:

Question 2. (a) What is the basic principle involved in the estimation of nitrogen by Dumas method. (b) In a Dumas nitrogen estimation method, 0.30 g of an organic compound gave 50 cm3 of N2 collected at 300 K and 715 mm Hg pressure. Calculate the percentage composition of nitrogen in the compound. (Vapour pressure of water at 300 K is 15 mm Hg)Answer: (a) This method is based upon the fact that nitrogenous compound is heated withcopper oxide in an atmosphere of carbon dioxide yield free nitrogen.

Question 1. The large number of organic compounds is due to (a) the valency of carbon (b) a small size of carbon (c) a special property of carbon known as catenation Question 2. The IUPAC name of (a) 1, 2-dichloropropane (b) 3, 3-dichloropropane (c) 1, 1-dichloropropane (d) dichloropropane Question 3. The IUPAC name of (a) 2-methyl butanal (b) butan-2-aldehyde (c) 2-ethylpropanal (d) 3-methyl isobutraldehyde Question 4. The bond that undergoes heterolytic cleavage most readily is (a) C-C (b) C-O (c) C-H (d) O-H Question 5. The reaction (a)carbocation formation (b) free-radical mechanism (c) carbanion formation (d) none of these Question 6. The hybridization state of a carbocation is (a) sp4 (b) sp3 (c) sp2 (d) sp Question 7. Which of the following are electrophiles? (a) Dimethyl sulphide (b) Bromides (c) Carbon dioxide (d) Ammonia 2ff7e9595c