Organic Chem
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Organic Chem

Course: CHEM 101, Spring 2010

School: Duke

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2.1Give the direction of the dipole moment (if any ) for each of the following molecules: (a) HF, Answer: (b) IBr, (c) Br2, (d) F2. H (a) F ; (b) I Br 2.2 Boron trifluoride (BF3) has no dipole moment (=0). Explain how this observation confirms the geometry of predicted by VSEPR theory. Answer: B is SP2-hybridized, so all atoms on the same plane, the center of positive charge and the center of negative charge...

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the 2.1Give direction of the dipole moment (if any ) for each of the following molecules: (a) HF, Answer: (b) IBr, (c) Br2, (d) F2. H (a) F ; (b) I Br 2.2 Boron trifluoride (BF3) has no dipole moment (=0). Explain how this observation confirms the geometry of predicted by VSEPR theory. Register to View Answeris SP2-hybridized, so all atoms on the same plane, the center of positive charge and the center of negative charge coincide, and the molecule has no net dipole moment 2.3 Tetrachloroethene doesnt have a dipole moment. Explain this fact on the basis of the shape of Cl2C=CCl2. Answer: Cl Cl Cl Cl The shape of the tetrachloroethene is a planar, the vector sum of the dipole moment of each C-Cl bonds is zero. 2.4 Sulfur dioxide has a dipole moment, on the other had, carbon dioxide has no dipole moment. What do these facts indicate about the geometries of the two molecules? Answer: The fact that SO2 has a dipole moment indicates that the molecule is angular, not linear: S O O The fact that CO2 has no dipole moment indicates that its shape is linear, not angular: O C O 2.5 Using a three-dimensional formula, show the direction of the dipole moment of CH3OH. net dipole O H CH3 2.6 Trichloromethane (CHCl3, also called chloroform) has a larger dipole moment than CFCl3. Use three dimensional structures and bond moments to explain this fact. Answer: H C Cl Cl Cl Cl F C Cl Cl 2.7 Indicate the direction of the important bond moments in each of the following compounds (neglect C-H bonds). You should also give the direction of the net dipole moment for the molecule. If there is not net dipole moment, state that =0. (a) cis-CHF=CHF (b) trans-CHF=CHF Answer: (a) cis-CHF=CHF H C F C F H (c) CH2=CF2 (d)CF2=CH2 net dipole moment (b) trans-CHF=CHF H C F C H F (c) CH2=CF2 H C H C F F F =0. net dipole moment (d) CF2=CH2 F C F C F =0. 2.8 Write structural formulas for all of the alkenes (a) with the formula C2H2Br2 and (b) with the formulaC2Br2Cl2. In each instance designate compounds that are cis, trans isomers of each other. Predict the dipole moment of each one. Answer: (a) Br Br Br H Br H C H C H H C C Br =0 Br C C H cis-1,2-Dibromoethene (b) Br trans-1,2-Dibromoethene 1,1-Dibromoethene Br Br Cl C Cl C Cl Cl C C Br =0 cis-1,2-Dibromo-1,2-dichloroethene trans-1,2-Dibromo-1,2-dichloroethene Br Cl C Br C Cl 1,1-Dibromo-2,2-dichloroethene 2.9 Write structural formulas (a) for two constitutional isomeric primary alkyl bromides with the formula C4H9Br, (b) for a secondary alkyl bromide ,and (c) for a tertiary alkyl bromide with the same formula. Answer: Br (a) Br Br (b) Br (c) 2.10 Although we shall discuss the naming of organic compounds later when we discuss the individual families in detail, one method of naming alkyl halides is so straightforward that it is worth describing here. We simply name the alkyl group attached to the halogen and add the word fluoride, chloride, bromide, or iodide. Write formulas for (a) ethyl fluoride and (b) isopropyl chloride. What are names for (c) CH3CH2CH2Br, (d) CH3CHFCH3, and (e) C6H5I? (a) CH3CH2F (b) CH3CHCH3 Cl (c) propyl br omide (d) isopropyl fluoride (e) phenyl iodide 2.11 Write structural formulas for (a) two primary alcohols , tertiary alcohol all having the molecular formula C4H10O. Answer: (b) (a) CH3CH2CH2CH2OH & (CH3)2CHCH2OH; (b) a secondary alcohol, and (c) a CH3CH2CH(CH3)OH; (c) (CH3)3COH. 2.12 One way of naming alcohols is to name the alkyl group that is attached to the OH and add the word alcohol. Write the structures of (a) propyl alcohol and (b) isopropyl alcohol. Answer: propyl alcohol isopropyl alcohol CH3CH2CH2OH CH3CHOHCH3 2.13 Write structural formulas for the materials below. (a) diethyl ether CH3CH2OCH2CH3 (b) ethyl propyl ether CH3CH2OCH2CH2CH3 (c) ethyl isopropyl ether CH3CH2OCH(CH3)2 What name would you give to (d), (e) and (f)? (d) CH3OCH2CH2CH3 methyl propyl ether (e) (CH3)2CHOCH(CH3)2 diisopropyl ether (f) CH3OC6H5 methyl phenyl ether 2.14 One way of naming amines is to name in alphabetical order the alkyl groups attached to the nitrogen atom, using the prefixes di- and tri- if the groups are the same. An example is isopropylamine for (CH3)2CHNH2. Write formulas for (a) propylamine, (b) trimethylamine, and (c) ethylisopropylmethylamine. What are names for (d) CH3CH2CH2NHCH(CH3)2, (e) (CH3CH2CH2)3N, (f) C6H5NHCH3, (g) C6H5N(CH3)2. Answer: (a) (b) (c) CH3CH2CH2NH2 (CH3)3N H3CH2C N CH3 H C CH3 CH3 (d) (e) (f) (g) isopropylpropylamine tripropylamine methylphenylamine dimethylphenylamine 2.15 Which amines in Problem 2.14 are (a) primary amines, (b) secondary amines, and (c) tertiary amines? Answer: Among the amines in Problem 2.14, (a). (b). (c). propylamine CH3CH2CH2-NH2 is a primary amine; isopropylpropylamine CH3CH2CH2NHCH(CH3)2 and methylphenylamine C6H5NHCH3 are secondary amines; trimethylamine (CH3)3N,ethylisopropylmethylamine CH3(CH3CH2)NCH(CH3)2, tripropylamine (CH3CH2CH2)3N, and dimethylphenylamine C6H5N(CH3)2 are all tertiary amines. 2.16 Amines are like ammonia in being weal bases. They do this by using their unshared electron pair to accept a proton. (a) Show the reaction that would take place between trimethylamine and HCl. (b) What hybridization state would you expect for the nitrogen atom in the product of this reaction? (a) (CH3)3N + HCl [(CH3)3NH]+Cl(b) SP3 hybridization. 2.17 The compounds in each part below have the same (or similar) molecularweights. Which compound in each part would you expect to have the higher boiling point? Explain your answers. (a) CH3CH2CH2CH2OH or CH3CH2OCH2CH3. (b) (CH3)3N or CH3CH2NHCH3 (d) CH3CH2CH2CH2OH or HOCH2CH2CH2OH. Answer: In group (a) I think CH3CH2CH2CH2OH have the higher boiling point .In group (b) is CH3CH2NHCH3. In group (c) is HOCH2CH2CH2OH. The reason is that those compounds all have a strong hydrogen bonding which is limited to molecules having a hydrogen atom attached to an O, N, or F atom. 2.18 Which compound would you expect to have the higher melting point, propane or cyclopropane? Explain your answer? Answer: The cyclopropane has the higher melting point because it is more symmetrical and more compact. 2.19 Classify each of the following compounds as an alkane, alkene, alkyne, alcohol, aldehyde, amine, and so forth. (a) O (b) H 3C (c) OH C CH (d) O H (e) (f) OH CH3(CH2)7 (CH2 )12CH3 H H Answer: (a) (d) Ketone Aldehyde (b) (e) Alkyne Alcohol (c) (f) Alcohol Alkene 2.20 Identify all of the fuctional groups in each of the following compounds: (a) OH functional groups: HO- ,aromatic ring (b) O OH C CH2 C H2N O C N H H CH2 CH C O O CH3 O C O C O CH3 functional groups: OH (c) C- , -O- , -NH2, aromatic ring, H2 C H C CH3 NH2 functional groups: aromatic ring , (d) NH 2 HO functional groups:-OH ,double bond (e) O C OCH2CH2 N CH3 O functional groups: C (f) OCH2CH2 ,aromatic ring , O H O functional groups: double bond , H (g) O O O O O functional groups: O C ,double bond 2.21 There are four alkyl bromides with the formula C4H9Br. Write their structure formulas and classify each as to whether it is a primary, secondary or tertiary alkyl bromide. Br primary alkyl bromide secondary alkyl bromide Br Br primary alkyl bromide tertial alkyl bromide Br 2.22 There are seven isomeric compounds with the formula C4H10O. Write their structures and classify each compound according to its functional group. Answer: (a) OH Alcohol (b) OH Alcohol (c) OH Alcohol (f) O Ether (d) O Ether (e) O Ether (g) OH Alcohol 2.23 Write structural formulas for four compounds with the formula C3H6O and classify each according to its functional group. H H C H H C H C H H C H O H Alkene&Alcohol O C H C H H Aldehyde H C H C H H Alcohol C H O H H C H C H O H C H H Ether & Alkene 2.24 Classify the following alcohols as primary, secondary or tertiary (a) (CH3)3CCH2OH (b) CH3CH(OH)CH(CH3)2 (c) (CH3)2C(OH)CH2CH3 primary secondary tertiary OH (d) tertiary OH (e) secondary Classify the following amines as primary ,secondary,or tertiary: 2.25 (a) CH3NHCH(CH3)2 (b) CH3CH2CH(CH3)CH2NH2 (c) (CH3CH2)3N (d) (C6H5)2CHCH2NHCH3 (e) HN (f) N Answer: (b) is primary, (a), (d), (e) are secondary and (c), (f) are tertiary amine. 2.26 Write structural formulas for each of the following: (a) Three ethers with the formula C4H10O. (b) Three primary alcohols with the formula C4H8O. (c) A secondary alcohol with the formula C3H6O. (d) A tertiary alcohol with the formula C4H8O. (e) Two esters with the formula C3H6O2. (f) Four primary alkyl halides with the formula C5H11Br. (g) Three secondary alkyl halides with the formula C5H11Br. (h) A tertiary alkyl halide with the formula C5H11Br. (i) Three aldehydes with the formula C5H10O. (j) Three ketones with the formula C5H10O. (k) Two primary amines with the formula C3H9N. (l) A secondary amine with the formula C3H9N. (m) A tertiary amine with the formula C3H9N. (n) Two amides with the formula C2H5NO. Answer: (a) O O O (b) OH OH OH (c) OH (d) OH CH3 (e) O O O O (f) Br Br Br Br Br (g) Br Br Br (h) Br (i) O O O O (j) O O O (k) NH2 NH2 (l) N H (m) N (n) O NH2 NH O 2.27 Which compound in each of the following pairs would have the higher boiling point? Explain your answers. (a) CH3CH2CH2OH or CH3CH2OCH3 (b) CH3CH2CH2OH or HOCH2CH2OH O OH (c) O OH or (d) or NH N CH3 (e) F or F F (f) F or O O (g) OH or O (h) Hexane CH3(CH2)4CH3 or nonane CH3(CH2)7CH3 O (i) or Answer: (a) CH3CH2CH2OH because it has a hydrogen bonding. (b) HOCH2CH2OH because it has a stronger hydrogen bonding. (c) OH OH Because it has a hydrogen bonding. (d) Because it has hydrogen bonding. NH (e) F Because it has hydrogen bonding. F (f) Because it has a Dipole-Dipole Forces. O (g) OH Because it has a hydrogen bonding. (h) nonane CH3(CH2)7CH3 because it has larger molecular weight and size. O (i) Because it has a stronger van der Waals Forces. 2.28 Predict the key IR absorption bands whose presence would allow each compound in pairs a, c, d, e, g and i from Problem 2.27 to be distinguished from each other. (a) CH3CH2CH2OH or CH3CH2OCH3 O OH (c) O (d) or or OH (e) NH or N CH3 O O (g) OH or O (i) or O Answer: (a) The typical OH adsorption would be around 3200-3500 cm-1; (c) The typical ketone adsorption wpuld be around 1700 cm-1; (d) Same reason for (a); (e) The secondary amine would have the adsorption around 3300-3500 cm-1; while the tertiary amine wouldnt have the adsorption around this region; (g)The carboxylic acid would have the adsorption around 2500-3500 cm-1 due to the OH stretch; (i) The ketone would have a strong adsorption around 1700 cm-1. 2.30 Cyclic compounds of the general type shown here are called lactones. What functional group dose lactone contain? O O Answer: Ester group. 2.31 Hydrogen fluoride has a dipole moment of 1.82 D; its boiling point is 19.34C. Ethyl fluoride (CH3CH2F) has an almost identical dipole moment and has a larger molecular weight, yet its boiling point is -37.7C.Explain. Answer: The two molecules have almost the same dipole moment. That means they have almost the same London force. However, the major molecular interaction between hydrogen fluoride is the hydrogen bonding, it makes hydrogen fluoride a higher boiling point. 2.32 Which of the following solvents should be capable of dissolving ionic compounds? (a) liquid SO2 (b) liquid NH3 (c)benzene (d) CCl4 Register to View Answerand B are polar, and hence are capable of dissolving ionic compounds. 2.33 Write a three-dimensional formula for each of the following molecules using the wedge-dashed wedge-line formalism. If the molecule has a net dipole moment, you should so state. (You may ignore the small polarity of C-H bonds in working this and similar problems.) (a) CH3F (b) CH2F2 (c) CHF3 (d) CF4 (e) CH2FCl (f) BCl3 (g) BeF2 (h) CH3OCH3 (i) CH3OH (j) CH2O (a) F C H H (b) F H F C H H (c) H C F F (d) F F F C F F (e) F C Cl H (f) H Cl Cl Cl B (g) F Be F (h) H C H H H C O H H (i) H H O C H (j) O C H H H 2.35 Analyze the statement: For a molecule to be polar, the presence of polar bonds is necessary, but it is not a sufficient requirement. Answer: Its right. For a polar molecule polar bond is necessary, but its not sufficient. For instance CO2, it has polar bonds but it is not polar molecule. 2.36 Identify all of the functional groups in Crixivan (the structure of Crixivan is shown in the chapter-opening vignette). C6H5 N H N N H HN C(CH3)3 O O OH H H N OH H Answer: hydroxy, phenyl, carbonyl, amide, amine. 2.37 The IR spectrum of propanoic acid indicates that the absorption for the O-H stretch of the carboxylic acid functional group is due to a hydrogen bonded form. Draw the structure of two propanoic acid molecules showing how they could dimerize via hydrogen bonding. Answer: CH3 CH2 C O H O H O O C CH2 CH3 2.38 Tow isomers having molecular formula C4H6O are both symmetrical in structure. In their infrared spectra, neither isomer when in dilute solution in CCl4 (used because it is nonpolar) has absorption in the 3600-cm-1 region. Isomers A has absorption bands at approximately 3080,1620,and 700cm-1. Isomers B has bands in the 2900-cm-1 region and at 1780cm-1 .propose a structure for A and two possible structure for B Register to View Answer O B; O C O and H 2.39 When two substituents are on the same side of a ring skeleton they are said to be cis, and when on opposite sides, trans (analogous to use of those terms with 1,2-disubstituted alkene isomers). Consider stereoisomeric forms of 1,2-cyclopentanediol (compounds having a 5-membered ring and hydroxyl groups on two adjacent carbons that are cis in one isomer and trans in the other). At high dilution in CCl4 both isomers have an infrared absorption band at approximately 3626 cm-1 but only one isomer has a band at 3572 cm-1. (a) Assume for now that the cyclopentane ring is coplanar (the interesting actuality will be studied later) and then draw and label the two isomers using the wedge-dashed method of depicting the OH groups. (b) Designate which isomer will have the 3572-cm-1 band and explain its origin. ANSWER: (a) 1. 2. HO OH HO HO (b) The cis isomer will have the 3572 cm-1 band because only in it are the two hydroxyl groups close enough to form the intramolecular hydrogen bonding. The intermolecular hydrogen bonding couldnt be formed in the dilute carbon tetrachloride solution. 2.40 Compound C is asymmetric, has molecular formula C5H10O, and contains two methyl groups and a 30 functional group. It has a broad infrared absorption band in the 3200-3550-cm-1 region and no absorption in the 1620-1680-cm-1 region. (a) Propose a structure for C. (b) Is your suggested structure capable of stereoisomerism? If so, draw the stereo isomers using the wedge-dashed wedge method. Register to View AnswerH3 C H H OH H CH3 (b) Yes H3C H OH C H3 H OH H3C CH3 HO H3C H CH3 HO H3C H CH3

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Duke - CHEM - 101
4.1 Write condensed structural formulas for all of the constitutional isomers with the molecular formula C7H16 (There is a total of nine constitutional isomers). Answer:Heptane2-methylhexane3-methylhexane2,2-dimethylpentane2,3-dimethylpentane2,4-dim
Duke - CHEM - 101
5.1 Classify each of the following objects as to whether it is chiral or achiral: (a) A screwdriver achiral (b) A baseball bat achiral (c ) A golf club chiral (d) A tennis shoe chiral (g) A car chiral (e) An ear chiral chiral (h) A hammer achiral (f) A wo
Duke - CHEM - 101
7.1 Using the (E)-(Z) designation [and in parts (e) and (f) the (R-S) designation as well], give IUPAC names for each of the following.Br H C Cl C CH2CH2CH3H3C CH2CH(CH3)2 C H C CH3(a)Cl(b)BrCl(c)ICH3 C C CH2CH3Cl C CCC CH2 CH3H CH3(d)IBr
Duke - CHEM - 101
8.1 Give the structure and name of the product that would be obtained from the ionic addition of IBr to propene. Answer:BrII I Br Br8.3 Provide mechanistic explanations for the following observations a addition of hydrogen The chloride to 3-methyl-1-b
Duke - CHEM - 101
10.1 Calculate the heat of reaction, H, for the following reactions:(a) H2 +F2 2HF (b) CH4+F2 CH3F+HF (c) CH4 +Cl2 CH3Cl+HCl (d) CH4+Br2 CH3Br+HBr (e) CH4 +I2 CH3 I+HI (f) CH3CH3+Cl2 CH3 CH2Cl+HCl (g) CH3CH2CH3+Cl2 CH3CHClCH3+HCl (h) (CH3)3 CH+Cl2 (CH3 )
Duke - CHEM - 101
11.1 What is wrong with the use of such names as isopropanol and tert-butanol? Answer: We should say them as isopropyl alcohol and tert-butyl alcohol. 11.2 Give bond-line formulas and appropriate names for all of the alcohols and ethers with the formulas
Duke - CHEM - 101
12.1 One method for assigning an oxidation state to a carbon atom of an organic compound is to base that assignment on the groups attached to the carbon; a bond to hydrogen for (anything less electronegative than carbon) make it 1, a bond to oxygen, nitro
Duke - CHEM - 101
13.1 (a) What product(s) would you expect to obtain if propene labeled with subject to allylic chlorination or bromination? (b) Explain your answer.H214C CHCH314C at C1 were+X2high temperature or low conc. of X2?(c) If more than one product would
Duke - CHEM - 101
14.1 Listed below are four compounds that have the molecular formula C6H6. Which of these compounds would yield only one monosubstitution product, if, for example, one hydrogen were replaced by bromine?(a) H3CCC C CCH3(b)(c)(d)Answer:(a) H3CC(a) a
Duke - CHEM - 101
16-1.(a)Give IUPAC substitutive names for the seven isomeric aldehydes and ketones with the formula C5H 8O (b).Give structures and names (common or IUPAC substitutive names)for all the aldehydes and ketones that contain a benzene ring and have the formula
Duke - CHEM - 101
17.1 For all practical purposes, the compound 2,4-cyclohexadien-1-one exists totally in its enol form. Write the structure of 2,4-cyclohxandien-1-one and of its enol form. What special factor account for the stability of the enol form? Answer:O OHBecaus
Duke - CHEM - 101
19.1 (a) Write a mechanism for all steps of the Claisen condensation that take place when ethyl propanoate reacts with ethoxide ion. (b) What products form when the reaction mixture is acidified? Answer: (a)O H3CHC H COC2H5 + OC2H5 H3CHC O COC2H5 + C2H5O
Duke - CHEM - 101
20.1 Outline a procedure for separating hexylamine from cyclohexane using dilute HCl, aqueous NaOH, and diethyl ether. Answer:organic phase: CH3(CH2)5NH2 dil HCl ether NaOH ether organic phase dry distill CH3 (CH2 )5NH2 dry distillaqueous phase20.2 Out
Duke - CHEM - 101
22.1 How many stereocenters are contained in the (a) aldotetrose and (b) ketopentose just given? (c) How many stereoisomers would you expect from each general structure?O CH CHOH CHOH CH2OHCH2OH C CHOH CHOH CH2OH OAn aldotetroseA ketopentoseThe answe
Duke - CHEM - 101
23.4 Give structural formulas for the products that you would expect from the following reactions:(a) -Pinene+ hot KMnO4(b) ZingibereneH+ H2PtH(c) CaryonphylleneCH 3+ HCl(d) -Selinene Answer:OH+ 2THF:BH3 (2)H2O2,OH(a)H(b)CH3ClHHCl(c
Duke - CHEM - 101
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Duke - STAT - 101
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Duke - STAT - 101
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Duke - STAT - 101
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Duke - STAT - 101
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Duke - STAT - 101
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Duke - STAT - 101
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Duke - STAT - 101
Chapter 3: Data Descriptions In statistics, we like to describe what the data is telling us. It is important to distinguish between a statistic and a parameter. A statistic is obtained from a sample. A parameter is obtained from a population. We use bot
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Duke - STAT - 101
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Duke - STAT - 101
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Duke - PHIL - 101
Theatre 101I.Worksheet 1general terms/conceptsArt: makes order out of chaos; man-made, therefore artificial. Fine Art: art made for entertainment, culture; brings us pleasure, sharpens our perception of life. All art requires knowledge, study and expe
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Duke - CHEM - 201
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Duke - CHEM - 202
Organic Chemistry IDr Luke A. BurkeOrganic chemistry is the chemistry of Carbon and its compounds. For many scientists, the study of Organic Chemistry will be their only opportunity to explore synthesis. All sciences use analysis where ideas or things a
Duke - CHEM - 202
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Duke - CHEM - 202
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Duke - CHEM - 202
The Study of Chemical Reactions Mechanism: The complete, step by step description of exactly which bonds are broken, formed, and in which order. Thermodynamics: The study of the energy changes that accompany chemical and physical transformations. It allow
Duke - CHEM - 202
Stereochemistry This is study of the 3 dimensional arrangement in space of molecules. In organic chemistry, subtle differences in spatial arrangements can give rise to prominent effects. E.g. the isomers of butenoic acid:H HO2C CO2H H HO2C CO2Htrans iso
Duke - CHEM - 202
Alkyl Halides Alkyl halides are a class of compounds where a halogen atom or atoms are bound to an sp3 orbital of an alkyl group. CHCl3 (Chloroform: organic solvent) CF2Cl2 (Freon-12: refrigerant CFC) CF3CHClBr (Halothane: anesthetic) Halogen atoms are mo
Duke - CHEM - 202
Structure and Synthesis of Alkenes Alkenes (olefins) are hydrocarbons which have carboncarbon double bonds.H H CC H HA double bond is a bond and a bond. Double bond B.D.E. bond B.D.E. = 146 kcal/mol = 83 kcal/molTherefore B.D.E. must = 63 kcal/mol. A b
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Reactions of Alkenes Since bonds are stronger than bonds, double bonds tend to react to convert the double bond into bonds+ X-Y XYThis is an addition reaction. (Other types of reaction have been substitution and elimination).Addition reactions are typi
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Alkynes Alkynes or acetylenes are compounds that contain a carboncarbon triple bond. E.g.HCCH acetylene CH3CH2 C C H H3C C C CH3The triple bond results in a molecular formula of CnH2n-2 Ethane Ethene Ethyne C2H6 C2H4 C2H2 0 elements of unsaturation 1 el
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Synthesis and Structure of Alcohols Alcohols can be considered organic analogues of water.HOH ROHAlcohols are usually classified as primary, secondary and tertiary.H R H primary OH R R secondary H OH R R tertiary phenol R OH OHAlcohols with the hydrox
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Reactions of Alcohols Alcohols are versatile organic compounds since they undergo a wide variety of transformations the majority of which are either oxidation or reduction type reactions. Oxidation is a loss of electrons Reduction is a gain of electrons.
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Infrared Spectroscopy and Mass Spectrometry Introduction It is fundamental for an organic chemist to be able to identify, or characterize, the new compound that he/she has just made. Sometimes this can be achieved by a chemical means, such as determining
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Nuclear Magnetic Resonance (NMR) Spectroscopy NMR is the most powerful analytical tool currently available to an organic chemist. NMR allows characterization of a very small amount of sample (10mg), and does not destroy the sample (non-destructive techniq
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Organic Chemistry II with Prof. BurkeLecture Notes Email Room Office Hours Texthttp:/camchem.rutgers.edu/~burke burke@camden.rutgers.edu Sci 114B Half-hour before and after each lecture or by appt. (a) Organic Chemistry Wade, 4th ,5th, or 6th Edition (b
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Conjugated Systems, Orbital Symmetry and UV Spectroscopy Introduction There are several possible arrangements for a molecule which contains two double bonds (diene): Isolated: (two or more single bonds between them)Conjugated: (one single bond between th
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Aromatic Compounds Historically, benzene and its first derivatives had pleasant aromas, and were called aromatic compounds. Structure of Benzene Kekul Structure Kekul (1866) bravely proposed that benzene had a cyclic structure with three alternating C=C d
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Reactions of Aromatic Compounds Just like an alkene, benzene has clouds of electrons above and below its sigma bond framework.Although the electrons are in a stable aromatic system, they are still available for reaction with strong electrophiles. This ge
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Ketones and Aldehydes The carbonyl group is of central importance in organic chemistry because of its ubiquity. Without studying the carbonyl group in depth we have already encountered numerous examples of this functional group (ketones, aldehydes, carbox
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Amines Amines are derivatives of ammonia with one or more alkyl groups bonded to the nitrogen. Amines can be classified as primary, secondary or tertiary, meaning one, two and three alkyl groups bonded to the nitrogen respectively. E.g.Prim ary Am ines N
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Carboxylic AcidsWhen a carbonyl carbon also bears a hydroxyl group, then these compounds are appreciably acidic, and are called carboxylic acids.O R C O-HRCO2HRCOOHCarboxylic acids are classified according to the substituent that is bonded to the car
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Carboxylic Acid DerivativesCarboxylic derivatives are described as compounds that can be converted to carboxylic acids via simple acidic or basic hydrolysis. The most important acid derivatives are esters, amides and nitriles, although acid halides and a
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Duke - CHEM - 202
Duke - CHEM - 202