Chapter 18.15, Problem 37ATS

1. The sex attractant for the common housefly is a hydrocarbon with the molecular formula C23H46. On treatment with KMNO4, two products are formed: CH3(CH2)2CH(CH(CH3)2)- (CH2)6CO2H and (CH3)3C-(CH2)4-CO2H. Propose TWO structures for this sex attractant.

Which of the following are consistent with the requirements for aromaticity?I. A system with delocalized p electrons in a ring.II. 4n p electrons in the ring.III. All the ring atoms must be carbons.IV. (4n + 2) p electrons in the ring.  Give the answer why.

10. MM and NN are amines with molecular formula, C3H9N. Reaction of MM with sodium nitrite and HCl releases nitrogen gas and produces a mixture of XX and YY, while NN produces yellowish oily compound ZZ when reacts with the same reagent. Gives the structures of MM, NN, XX, YY and ZZ. Outline a synthesis of MM from a suitable alkene.

(10pts) Compound A, C10H16, was found to be optically active. On catalytic reduction over a palladium catalyst, 2 equivalents of hydrogen were absorbed, yielding compound B, CioH2o. On ozonolysis of A, two fragments were obtained. One fragment was identified as acetic acid (CHCOOH). The other fragment, compound C, was an optically active carboxylic acid, C8H14O2. Write reactions, and draw the correct structures for A-C, explain your answer in detail.

(1) (11) (111) (b)Give products and mechanisms for the following reactions. OH CH₂SO над сам _N_N_ но HO B

5) Limonene is a compound found in orange oil and lemon oil with molecular formula C₁0H16- When limonene is treated with 1 equivalent of mCPBA, the product is limonene oxide as depicted below. When limonene is treated with excess ozone (03) and then with dimethylsulfide (Me₂S), one of the organic products of the reaction is formaldehyde. Of the structures below, which correspond to the structure of limonene and/or the structure of the other major organic product when limonene is reacted with O3 and Me₂S? Please select any and all that apply. A) 'Н H B) ? H 1) 03 (XS) 2) Me₂S (xs) ? limonene, C10H16 D) mCPBA (1 equiv) E) limonene oxide None of the above

The treatment of (CH3)2C=CHCH2Br with H2O forms B (molecular formulaC5H10O) as one of the products. Determine the structure of B from its 1H NMR and IR spectra.

to to 1. N(CH3)3 HN(CH3)3 CI :0: Na 2. + NaCl Proton transfer = Electrophilic addition g = SN1 Nucleophilic substitution a = b = Lewis acid/base e = E1 Elimination h = SN2 Nucleophilic substitution c = Radical chain substitution f= E2 Elimination Identify the mechanism by which each of the reactions above proceeds from among the mechanisms listed. Use the letters a - i for your answWers. 1. 2.

Nucleophilic substitution happens on compounds having nucleophilic groups as leaving groups. The rule is, the weaker the basicity of a group of the substrate, the better is its leaving ability. In these substitution reactions, the basicity of leaving group must be less than the incoming nucleophilic group. Nucleophilic substitution reaction at sp3-hybridized carbon is either bimolecular (SN2) or unimolecular. Bimolecular reaction takes place in single step, involving transition state intermediate. In SN2 reaction, inversion in configuration occurs. In case of optically active alkyl halides, the inversion in configuration is called Walden inversion. SN2 reaction is preferred if the compound has less steric hindrance. On the other hand, unimolecular (SN1) reaction involves two steps and a carbonium ion intermediate. Optically active substrates give racemic mixture in these type of reactions. Which of the following will produce enantiomeric pair on treatment with HOH? " I ÇH, C,Hs-C-Br…