1‚6-Heptadiyne Common names: prefix the substituents on the triple bond to the name “acetylene” IUPAC name: Common name: 2-Butyne D imethylacetylene 1-Buten-3-yne Vinylacetylene Physical Properties Similar to alkanes and alkenes of comparable molecular weight and carbon skeleton 2-Butyne 1-Pentyne Melting Point Formula (°C) -81 HC CH CH3 C CH -102 CH 3 CH2 C CH -126 -32 CH3 C CCH3 CH 3 ( CH2 ) 2 C CH -90 1-Hexyne CH 3 ( CH2 ) 3 C CH 1-Octyne
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form‚ write the structure of any meso compounds below. If none would form‚ write the word “NONE” below. [pic] 3. (25 points) The following reaction produces a major alkene product. First‚ on the Newman template provided draw the conformation from which the raction will occur. Then‚ on the product alkene template provided‚ draw the product that will form from this reaction. Finally‚ indicate the stereochemical designation of the major product.
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I. Abstract The experiment: Classification tests on Organic Compounds‚ allows the students to be familiarized with different classification tests used for identifying the different classes of organic compounds; examine unknown compounds using appropriate tests; and identify functional group of an organic compound based on the tests performed. Several organic compounds with different functional groups were tested to identify the functional groups present in the compound. n-heptane‚ pentene
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Keghan Chapter 8 MULTIPLE CHOICE QUESTIONS Topic: Structure Elucidation 1. An alkene adds hydrogen in the presence of a catalyst to give 3‚4-dimethylhexane. Ozonolysis of the alkene followed by treatment with zinc and acetic acid gives a single organic product. The structure of the alkene is: CH3 A) CH3CH=C-CHCH2CH3 (cis or trans) CH3 CH3 B) CH3CH2C=CCH3 (cis or trans) CH2CH3 C) CH3 CH2=CCH2CHCH2CH3 CH3 CH2 D) CH3CH2CCHCH2CH3 CH3 CH3 E) CH3CH2CHCHCH=CH2
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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 more electronegative than carbon atoms‚ and so the C-Hal bond is polarized. H H μ C + C-l δ δ H The C-X bond is polarized in such a way that there is partial positive charge on the carbon and partial negative charge on the halogen. Dipole moment
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B) is a deactivator and an o‚p-director. C) is an activator and a m-director. D) is an activator and an o‚p-director. E) none of the above Answer: B 32) Which of the following is an intermediate in the bromination of toluene? A) 13 B) C) D) Answer: B 33) Provide the major organic product of the following reaction. Answer: 14 34) Provide the major organic product which results when PhCHOHCH3 is treated with a chromate (PCC- strong oxidant). Answer:
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4-bromoaniline Introduction Aromatic compounds tend to undergo electrophilic aromatic substitutions rather than addition reactions. Substitution of a new group for a hydrogen atom takes place via a resonance-stabilized carbocation. As the benzene ring is quite electron-rich‚ it almost always behaves as a nucleophile in a reaction which means the substitution on benzene occurs by the addition of an electrophile. Substituted benzenes tend to react at predictable positions. Alkyl groups and other electron-donating
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(CH3)2CHO- (d) (CH3)3CO- (Q.11) The correct order of decreasing dipole moment is: ( 1 mark ) (a) CH3Cl‚ CH3Br‚ CH3F (b) CH3Cl‚ CH3F‚ CH3Br (c) CH3Br‚ CH3Cl‚ CH3F (d) CH3Br‚ CH3F‚ CH3Cl (Q.12) HCl and HI do not give anti- markovnikov’s addition to alkenes in the presence of peroxides because: ( 1 mark ) (a) Both are highly ionic (b) One is oxidizing and other is reducing (c) One of the step is endothermic in both the cases (d) All the steps are exothermic in both the reactions.
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1. Which of the following are the products of a homolytic cleavage of a C-C bond of ethane? A) 2 B) CH3 CH3 C) H D) 2 + + CH3 CH2H3 CH3 Ans: A Chapter 11 Topic: Radicals Section: 11.1 Difficulty Level: Easy 2. Which of the following is the most stable radical? A) B) C) D) E) Ans: B Chapter 11 Topic: Radicals Section: 11.1 Difficulty Level: Easy 3. In the molecule shown below‚ determine which of the labeled bonds is
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from two factors: the presence of π electron density above and below the plane of the ring - making it nucleophilic‚ and the drive to regain the aromatic character by opting for substitution as opposed to a simple addition reaction. Preference towards addition reactions in the case of alkenes and substitution in the case of aromatic compounds becomes evident if we analyze the energy profiles of these reactions (Figures 1 and 2). Figure 1. Indian Institute of Technology Madras Engineering Chemistry
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