...U2 mod 1chem. Halogenoalkanes page1 of 4 Introduction Halogenoalkanes or alkyl halides are derivatives of alkanes. General formula = CnH2n+1X or simply R—X where R represents an alkyl group They can be classified as primary, secondary or tertiary alkyl halides. Primary alkyl halides have either 1 or no alkyl groups attached to the central carbon atom. Secondary alkyl halides have 2 alkyl groups (circled) attached to the central carbon atom. Tertiary alkyl halides have 3 alkyl groups (circled) attached to the central carbon atom Aryl halides are compounds in which the halogen atom is attached directly to the aromatic ring. U2 mod 1chem. Halogenoalkanes page2 of 4 Reactions of alkyl halides The point of attack would be the electron deficient carbon atom (due to the difference in electronegativity in the C-X bond) attached to the halogen atom , thus the attacking species would be a nucleophile. All reactions need the present of heat i.e. reflux 1. Alcohol formation (R represents any hydrocarbon chain) R—X + OH- R—OH + XPrimary and secondary alkyl halides undergo alkaline hydrolysis to the alcohol while tertiary alkyl halides can be hydrolysed by water. Aryl halides do not undergo hydrolysis except under VERY SEVERE conditions. Industrially chlorobenzene is reacted with NaOH(aq) at 360 °C at a high pressure then adding dilute HCl. 2. Amine formation R—X + NH3(alc) heat in a sealed tube R—NH2 + HX Note (alc) means an alcoholic solution as opposed to an aqueous...
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...ELIMINATION VERSUS SUBSTITUTION IN HALOGENOALKANES This page discusses the factors that decide whether halogenoalkanes undergo elimination reactions or nucleophilic substitution when they react with hydroxide ions from, say, sodium hydroxide or potassium hydroxide. Details for each of these types of reaction are given elsewhere, and you will find links to them from this page. The reactions Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution. Nucleophilic substitution The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic substitution reaction. In the example, 2-bromopropane is converted into propan-2-ol. Note: If you want to read about nucleophilic substitution in this reaction in detail, follow this link. Elimination Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide. The 2-bromopropane has reacted to give an alkene - propene. Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons. Note: If you want to read about elimination in this reaction in detail, follow this link. What decides whether you get substitution...
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...Between Elimination and Substitution in Haloalkane 1.0 ABSTRACT The main focus of this study is to study the competition between substitution and elimination of haloalkanes. Substitution of haloalkanes This report will explain the two types of SN1 and SN2 reactions. Later, the factors which influence both SN1 and SN2 reactions will be explained and competition between the two different types of substitutions will be further elaborated to see which conditions favour each type of reactions. Next, the elimination process of haloalkanes will be discussed and similar to the substitution reaction, the elimination process is also comprised of two types, namely the E1 and E2 reactions. The factors that influence E1 and E1 reactions will be listed and competition between the two types of reactions will be discussed to see which conditions favour which type of elimination reaction. Later on, to ease the process of determining which reaction is favoured on the haloalkane, we will divide the process to see if the reaction is favoured on SN1/E1 or SN2/E2 reaction as each pairs of reactions are favoured by the same conditions. Then, the primary, secondary or tertiary structure of the haloalkane will further determine whether the major product of each reaction is a substitution product, elimination product or both. As a conclusion, the details of each reaction need to be taken into account to determine the product of the reaction of haloalkanes. Many factors are taken into account, namely...
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...substitution and elimination of haloalkanes. Substitution of haloalkanes This report will explain the two types of SN1 and SN2 reactions. Later, the factors which influence both SN1 and SN2 reactions will be explained and competition between the two different types of substitutions will be further elaborated to see which conditions favour each type of reactions. Next, the elimination process of haloalkanes will be discussed and similar to the substitution reaction, the elimination process is also comprised of two types, namely the E1 and E2 reactions. The factors that influence E1 and E1 reactions will be listed and competition between the two types of reactions will be discussed to see which conditions favour which type of elimination reaction. Later on, to ease the process of determining which reaction is favoured on the haloalkane, we will divide the process to see if the reaction is favoured on SN1/E1 or SN2/E2 reaction as each pairs of reactions are favoured by the same conditions. Then, the primary, secondary or tertiary structure of the haloalkane will further determine whether the major product of each reaction is a substitution product, elimination product or both. As a conclusion, the details of each reaction need to be taken into account to determine the product of the reaction of haloalkanes. Many factors are taken into account, namely the leaving group, the nucleophilicity, type of solvent, steric effect and structure of haloalkane. Thus, the steps mentioned in...
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...Apa Ertinya Saya Menganut Islam__________________________________________________________ َﺎ ﹶﺍ َﻌﻨﻲ ﺍﻨﺘ َﺎﺌﻲ ﻟﻺﺴ ﹶﻡ؟ ﻼ ِﻤ ﻤﺫ ﻴ Apa Ertinya Saya Menganut Islam Fathi Yakan 1 Apa Ertinya Saya Menganut Islam__________________________________________________________ BAHAGIAN PERTAMA Apa Ertinya Saya Menganut Islam 1. Saya Mestilah Muslim Di Sudut Akidah. 2. Saya Mestilah Muslim Di Sudut Ibadat. 3. Saya Mestilah Muslim Di Sudut Akhlak. 4. Saya Mestilah Muslim Di Sudut Berkeluarga. 5. Saya Mestilah Mampu Mengawal Diri. 6. Saya Mestilah Yakin Bahawa Masa Depan Di Tangan Islam. 2 Apa Ertinya Saya Menganut Islam__________________________________________________________ Mukadimah Bahagian Pertama Bahagian pertama buku ini bertajuk "Apa Ertinya Saya Menganut Islam" membentangkan sifat-sifat penting yang wajib ada pada seseorang bagi membolehkan ia menjadi seorang Muslim dalam erti kata yang sebenarnya. Penggabungan diri dengan agama Islam bukanlah secara warisan, bukan secara hobi malah ia juga bukan penggabungan secara zahir sahaja. Sebenarnya penggabungan yang dimaksudkan ialah penggabungan dengan ajaran Islam itu sendiri dengan cara berpegang teguh dengan seluruh ajaran Islam serta menyesuaikan diri dengan Islam di segenap bidang kehidupan dengan penuh kerelaan. Seterusnya kami akan menerangkan secara ringkas sifat-sifat yang wajib dimiliki oleh setiap muslim untuk memastikan penggabungan dengan agama ini merupakan penggabungan yang sah dan benar. Firman Allah Subhanahu...
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...Chemistry Unit 2 AS Group 7, the halogens - Physical properties, fluorine is a pale yellow gas, chlorine a greenish gas, bromine a red-brown liquid and iodine a black solid o Fluorine is small and leads to the repulsion between non-bonding electrons because they are so close together o The atoms get bigger as we go down the group because each element has one extra filled main level of electrons compared with the one above it |Halogen |Atomic number |Electron Arrangement |Electronegativity |Atomic radius nm |Melting point K |Boiling Point K | |Fluorine |9 |[He]2s22p5 |4.0 |0.071 |53 |85 | |Bromine |35 |[Ar]3d104s24p5 |2.8 |0.114 |266 |332 | Electronegativity - This is a measure of the ability of an atom to attract electrons, or electron density towards itself within a covalent bond o This depends on the attraction between the nucleus and bonding electrons in the outer shell, which depends on the balance between the nuclear charge and the distance between the nucleus and the bonding pairs plus the shielding effect of inner shells of electrons Melting and boiling points - Melting and boiling points involve weakening and breaking...
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...CHAPTER 12 HYDROCARBONS 12.2 Alkenes Learning Outcomes At the end of this topic, students should be able to: a) Describe alkenes as unsaturated hydrocarbon with the general formula CnH2n, n≥2. (b) Draw the structures and name the compunds according to the IUPAC nomenclature for: i) straight chain and branched alkenes (parent chain ≤ C10) ii) cyclic alkenes (C3-C6) iii) simple dienes (C4-C6) 2 Learning Outcomes At the end of this topic, students should be able to: c) Show the preparation of alkenes through: i) dehydration of alcohols ii) dehydrohalogenation of haloalkanes d) Write the mechanism for c(i) e) State Saytzeff’s Rule. f) Deduce the major product of elimination reaction. 3 Alkenes - General formula CnH2n , n 2. - Functional group C=C double bond - Unsaturated hydrocarbon - C═C 1 σ bond and 1 π bond A B C=C A B Restricted rotation of carbon-carbon double bond causes cis-trans isomerism 4 Alkenes Cycloalkenes - General formula CnH2n-2 - Isomeric to alkynes CnH2n-2 Example : C4H6 CH2 CH CH2 Functional group isomerism CH CH3C – CCH3 5 Alkenes Common Names Of Alkenes Many older names for alkenes are still in use … example: CH2 CH3 CH2 ethene ethylene CH2 CH CH3 CH2 2–methylpropene isobutylene CH3 CH3 propene propylene C CH3 C CH CH2 2–methylbut-1,3-diene 6 isoprene Alkenes IUPAC...
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...Mathematics: Q1. If 1, , 2 are cube roots of unity, then the value of (1 2 )(1 2 ) is a) 4 b) c) 2 d) zero. Q2. The value of sin(cot -1 x) is a) x 1 x2 1 1 x2 b) c) 1 x2 x d) 1 x 2 Q3. Two circles x2 + y2 – 6x + 8 = 0 and x2 + y2 – 6 = 0 are given. The equation of the circle through their point of intersection and the point (1,1) is a) x2 + y2 -3x + 1 b) x2 + y2 – x + 5 = 0 c) x2 + y2 -8x + 6y = 8 d) x2 + y2 – 4x + 8y = 7 /2 Q4. The value of a) 1 0 1 tan x dx is b) /2 c) /4 d) 3/2 Q5. The volume of the parallelopiped whose edges are represented by -12i + k, 3j – k, 2i + j – 15k is 546. Then the value of is a) -5 b) 6 c) 1 d) -3 Physics Q1. A boat takes time t to go downstream from point A to point B and time 2 t to go upstream from point B to point A . If the speed of the stream is v and the speed of the boat with respect to the stream is u , then which of the following gives the correct relationship between u & v ? a) b) c) d) u u v v v 3v 2u 3u Q2. Which of the following is correct about the acceleration of a particle describing a uniform circular motion ? a) b) c) d) The acceleration is constant both in magnitude and direction The acceleration is constant in direction but not in magnitude The acceleration is constant in magnitude but not in direction The acceleration is neither constant in magnitude nor in direction Q3. Five identical resistors of resistance 1 each...
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...Centre for Foundation Studies, UTAR Chapter Scopes FHSC1124 Organic Chemistry Alkanes • IUPAC Nomenclature / naming of aliphatic alkanes and cycloalkanes • Physical properties • Combustion reactions • Free-radical substitution reactions & mechanism • Crude oil and “cracking” Chapter 2 Introduction to Alkanes IUPAC Nomenclature • Simplest member of hydrocarbon family • General formula of alkanes = CnH2n+2 • Alkanes = Paraffins (hydrocarbons with general formula CnH2n+2) • Aliphatic compounds: open chain / acyclic compounds • The names of alkanes end with suffix -ane. • Saturated hydrocarbon: only have C−C & C−H single bonds & contain the maximum possible number of H per C. 3 IUPAC Rules 1. Select the longest continuous C chain as parent chain (use root word for the no. of C) 2. Name each of the branch/substituents as an alkyl / aryl group 3. Number the C chain beginning from the end nearest to the branch ⇒ branch/substituents appear at the lowest no. possible FHSC1124 ORGANIC CHEMISTRY • IUPAC International Union of Pure & Applied Chemistry • The IUPAC nomenclature system is a set of logical rules devised and used by organic chemists to name the organic compounds. Prefix − Parent − Suffix What are the How many What family? substituents? carbons? IUPAC Rules 4. Name each substituent according to its chemical identity & the no. of the C atom to which it is attached ⇒ For identical substituent, use the prefix ...
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...Radiation Atoms and Nuclei Electronic Devices CHEMISTRY S.No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. CLASS XI Some Basic Concepts of Chemistry Structure of Atom Classification of Elements and Periodicity in Properties Chemical Bonding and Molecular Structure States of Matter: Gases and Liquids Thermodynamics Equilibrium Redox Reactions Hydrogen s-Block Element (Alkali and Alkaline earth metals) Some p-Block Elements Organic Chemistry- Some Basic Principles and Techniques Hydrocarbons Environmental Chemistry CLASS XII Solid State Solutions Electrochemistry Chemical Kinetics Surface Chemistry General Principles and Processes of Isolation of Elements p- Block Elements d and f Block Elements Coordination Compounds Haloalkanes and Haloarenes Alcohols, Phenols and Ethers Aldehydes, Ketones and Carboxylic Acids Organic Compounds Containing Nitrogen Biomolecules Polymers...
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...CHEMISTRY CHEMISTRY (CLASSES XI–XII) Rationale Higher Secondary Stage is the most crucial stage of school education because at this stage specialised discipline based, content oriented courses are introduced. Students reach this stage after 10 years of general education and opt for Chemistry with a purpose of mostly for pursuing their career in basic sciences or professional courses like medicines, engineering, technology and studying courses in applied areas of science and technology at tertiary level. Therefore, at this stage, there is a need to provide learners with sufficient conceptual background of Chemistry, which will make them competent to meet the challenges of academic and professional courses after the higher secondary stage. National Curriculum Framework for School Education – 2005 recommends a disciplinary approach with appropriate rigour and depth with the care that syllabus is not heavy and at the same time it is comparable to the international level. It emphasizes a coherent focus on important ideas within the discipline that are properly sequenced to optimize learning. It recommends that theoretical component of Higher Secondary Science should emphasize on problem solving methods and the awareness of Syllabus for Secondary and Higher Secondary Levels 22 historical development of key concepts of science be judiciously integrated into content. The present exercise of syllabus development in Chemistry at Higher Secondary Stage is based on this framework...
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...CHAPTER 15 ORGANIC COMPOUNDS AND THE ATOMIC PROPERTIES OF CARBON 15.15 a) Octane denotes an eight carbon alkane chain. A methyl group (–CH3) is located at the second and third carbon position from the left. CH3 CH3 CH CH CH2 CH2 CH2 CH2 CH3 CH3 b) Cyclohexane denotes a six-carbon ring containing only single bonds. Numbering of the carbons on the ring could start at any point, but typically, numbering starts at the top carbon atom of the ring for convenience. The ethyl group (–CH2CH3) is located at position 1 and the methyl group is located at position 3. CH2 CH3 1 6 2 3 5 CH3 4 c) The longest continuous chain contains seven carbon atoms, so the root name is “hept-.” The molecule contains only single bonds, so the suffix is “-ane.” Numbering the carbon chain from the left results in side groups (methyl groups) at positions 3 and 4. Numbering the carbon chain from the other will result in side groups at positions 4 and 5. Since the goal is to obtain the lowest numbering position for a side group, the correct name is 3,4– dimethylheptane. Note that the prefix “di-” is used to denote that two methyl side groups are present in this molecule. d) At first glance, this molecule looks like a 4 carbon ring, but the two –CH3 groups mean that they cannot be bonded to each other. Instead, this molecule is a 4–carbon chain, with two methyl groups (dimethyl) located at the position 2 carbon. The correct name is 2,2–dimethylbutane...
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...Introduction The transition metal catalyzed cross-coupling reactions such as Negishi,1 Heck,2 Suzuki-Miyaura,3 Stille,4 Kumada,5 Hiyama,6 and Sonogashira7 are extremely powerful tools for making C−C bonds. These methods have been extensively used in wide range of academic areas including material product synthesis, materials science, medicinal, biological, supramolecular chemistry, catalysis and coordination chemistry. Some of these reactions are also useful in pharmaceutical, agrochemical and fine chemical industries.8 Among these powerful transformations, the Suzuki-Miyaura cross-coupling (Scheme 1) (defined as transition-metal catalyzed cross-coupling between an organoboron compound and an organic halide or pseudo halide) is the most widely applied transition-metal catalyzed C−C bond forming reaction since its discovery in 1979.9 Scheme 1. Suzuki-Miyaura cross-coupling reaction. Suzuki reaction has become the most attractive approach for C−C bond formation due to many reasons. Firstly, the mild reaction conditions and broad functional group tolerance of this transformation. Secondly, the use of organoboron reagents has many advantages such as they are inexpensive, some are thermally stable and inert to water and oxygen, thus easy to handle (industry), they transmetallate with a variety of metal compounds under exceptionally mild reaction conditions, low toxicity, good availability of diverse organoboron reagent and easy separation of boron containing by-products from the reaction...
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...1 UTME 2001 Type R Questions Use the graph below to answer questions 1 and 2. 100 200 Time (mins) T(0C) 0 1 2 3 4 5 6 7 8 300 Liquid Vapour Solid 1. If the gas is cooled, at what temperature will it start to condense? (A) 1250C (B) 1500C (C) 1750C (D) 2500C 2. How long does it take all the solid to melt? (A) 2.5 mins (B) 6.0 mins (C) 1.0 mins (D) 3.0 mins 3. 25cm3 of a gas X contains Z molecules at 150C and 75mmHg. How many molecules will 25cm3 of another gas Y contain at the same temperature and pressure? (A) 2Y (B) 2Z (C) Z (D) Y 4. What mass of water is produced when 8.0g of hydrogen reacts with excess oxygen? (A) 36.0g (B) 8.0g (C) 72.0g (D) 16.0g [H = 1, O = 16] 5. A particle that contains 9 protons, 10 neutrons and 10 electrons is a (A) negative ion (B) positive ion (C) neutral atom of a non metal (D) neutral atom of a metal 6. Cancerous growth are cured by exposure to (A) γ -rays (B) β -rays (C) α -rays (D) x-rays 7. An oxide XO2 has a vapour density of 32. What is the atomic mass of X (A) 32 (B) 20 (C) 14 (D) 12 [O = 16] 8. Milikan’s contribution to the development of atomic theory is the determination of (A) charge on electron (B) positive rays (C) charge to mass ratio (D) cathode rays 9. Four elements W, X, Y, and Z have atomic numbers 2, 6, 16 and 20 respectively. Which of these elements is a metal? (A) X (B) W (C) Z (D) Y 10. An element X with relative atomic mass 16.2 contains two isotopes 16X 8 with relative abundance of 90% and X m8 with relative abundance of 10%...
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...Table of Contents Partial table of contents: Carbon Compounds and Chemical Bonds. Representative Carbon Compounds. An Introduction to Organic Reactions: Acids and Bases. Alkanes and Cycloalkanes: Conformations of Molecules. Stereochemistry: Chiral Molecules. Alkenes and Alkynes I: Properties and Synthesis. Alkenes and Alkynes II: Addition Reactions. Radical Reactions. Alcohols and Ethers. Conjugated Unsaturated Systems. Aromatic Compounds. Reactions of Aromatic Compounds. Aldehydes and Ketones I: Nucleophilic Additions to the Carbonyl Group. Aldehydes and Ketones II: Aldol Reactions. Carboxylic Acids and Their Derivatives: Nucleophilic Substitution at the Acyl Carbon. Amines. Carbohydrates. Lipids. Answers to Selected Problems. Glossary. Index. Solomons/Advices ADVICES FOR STUDYING ORGANIC CHEMISTRY 1. Keep up with your studying day to day –– never let yourself get behind, or better yet, be a little ahead of your instructor. Organic chemistry is a course in which one idea almost always builds on another that has gone before. 2. Study materials in small units, and be sure that you understand each new section before you go on to the next. Because of the cumulative nature of organic chemistry, your studying will be much more effective if you take each new idea as it comes and try to understand it completely before you move onto the nest concept. 3. Work all of the in-chapter and assigned problems. 4. Write when you study. over and over again. Write the reactions, mechanisms...
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