Question

4. What is the structure of the compound that gives the IR and proton NMR spectra? Examine the H NMR of each structure as a review. Note the chemical shift for each proton 100 MHz proton NMR spectrum CDC13 solution 2270 IR Spectrum (liquid film) 1745 - 4000 3000 1200 800 TMS 2000 1600 fem') oppm) O HƏNCH:C=c-c-OCH; NEC-CH:-0-OCH:CH; o 1,6--OCH HC --CHÚC H:C-C-OCH:CH.CN H;CCH: -2-0-CH:CN

Answer all

Answer all

Answer 1

Answer: From the H NMR data, the structure of the molecule is 1H-NMR data(ppm) 1 1.3(t, 3H) 2 H2 2 3.4(s, 2H) NEC-C CH3 H2 For-CH3 group bonded with a - CH2 group For-CH2 group bonded with a -CN and -CO group For-CH2 group bonded with oxygen atom and a-CH3 group. 3 4.3(q, 2H) From the IR spectra Functional Group Type of Vibration Characteristic Absorptions (cm) Intensity For CEO bond of ester stretching 1745 strong For CEN bond stretching 2270 medium-weak For C-H bond of alkyl group stretch 2950 strong Answer: H2 NEC-CLOCH2CH3

1. Answer: Ozonolysis is the rection where an alkene is treated with 03. After that it was treated with (CH3)2S and it cleave the double double bond and gives two carbonyl compound. In case of cyclic alkeke we got a compound which contains two carbonyl groups. Compound B is also an cyclic alkene because, when it was treated with O3 and (CH3)2S, it gives a compound which contains two carbonyl groups. The reaction is i. O3 Нас H ii. (CH3)2S Н. 1-methylcyclohex-1-ene (C7H12) 6-oxoheptanal On catalytic hydeogentation of compound B gives and alkene because the alkane get reduced. H2 H Pd/C 1-methylcyclohex-1-ene (C7H12) methylcyclohexane (C7H14) Here compound B absorbs one mole hydogen gas i.e. one carbon absorbs one mole hydrogen atom. Therefore, the structure of B is Н. 1-methylcyclohex-1-ene (C7H12)

2. Answer: The given reaction is an E2 elemination reaction, where a base abstract a B proton which is antiperiplaner with the leaving group. H H CH3 BICH CH3ONa Br B H H DH H In this reaction, the reactant contains two B position where it has two hydrogen and one deuterium. Among this three, only the deuterium atom is antiperiplaner with the leaving group (-Br), other two hydrogen are cis with the leaving group (-Br). So the base abstract the deuterium and form the obsarved product. H ...CH₃ Н. CH3 CBI CH3ONa H DH Н. Therefore, we donot observe the other two product. To form the other two products the base needs to take the cis hydrogens but in E2 elemiation it does not take place

3 Answer. The mechanism of the two reactions are shown below. - 4 OAC OAC -Hg -:Hg-OAC OAC CH ÖH2 Chamos H20: OH ОАс -Hg NaBH (Reduction) н il Methyl shift or when ge H Secondary carbocation Tertiary carbocation H20 H H. HO a