H-Nmr Spectroscopy

- H-NMR Spectroscopy 0 > 9 ppm . - Examples Rz - CHz 1 : . 115 - 2 DPM 4 5-6 : < -- ppM . -H R-CH3 : 2 5 : H = C3 . : 2 : 2 . 5 ppm 3 PpM 1 ppM . & one show50 signal R connecting as Rs- . X 5 : = 8 CH-CH L D R *H Proton 0 B: 1 - 3 2 Proton C Proton : 9ppM . . 2 2ppM i th 5-4ppM . ItII : D: 1 0 PPM ↓ ⑧ idential Is IIid ↓ Triplet 3 = -adjacent peaks Br I CH-cHz CH3-2 A - - * adjacent has two n+1 will = hydrogens 2 +1 - be carbon a signal triplet * , so --T Br T proton b to a = 5 , I' 5-4ppm = 2(4) + DUS te =no H !* 2 = Hy CH I - & adjacent so n+1- (quintett - proton C is to , III a CH3 en C adjacent total of 5 So , St1 = 0 , illl , hydrogens -> 8 CHz-cHz + - * proton d to = C3 adjacent total of one a hydrogen H is 2 , so It + 0-9-1 0 benzene E - total of 4 hydrogens 4+1 is DUS iH . . RCH2X if bott II 3 . 3ppM : 2 . 5-4PPM Br , I Cl , F , [TT* CyAqBr Proton A : = . zJ ppM 9-10ppm : : RCA2X 7ppm 0 I B 1 <3 R-0-CH 10-1ppm : NH2 2 Br A 3 R -- * proton"d" is closer to the bromine* CHz -CH2 * I ppM . - willdie - 1 5 : H: ** up ↓ CH ving : 3 1-5 . ppm 5-8ppM Ha A B I c CH3 - I -CHz - 0 - 2 H adjacent carbon next to three have Of = - c 243 Proton the - chemical shifts 0 - the no neighboringso hydrogens methyl groups no hydrogen so Oth = 1 singlet , 1 - neighboring carbon has a total of 3H 3 = hydrogens 4 -quartet -Singlet - , so neighboring has 2 2 +1 3 Carbon hydrogens = B I CH3 - - , Ha ~ 9H - signal CHz - et 0 - 2 3 . =triplet . Examples ↓· 1. 30 0 A (2CHz si ② 0 B W D A has C ⑧ I C Il a total -D L & ..... OCHz CH3 _ & = signals CH2 -C 2 (H3 - - =4 Signals 2 sing a A B CHz/-> , - O total signals · ↳ hydrogens = 3 Bio jo ⑧ W A , ⑧ ⑧ 0 equivalent = / RCH3 is ⑰ W 28 so ppm =0 9 - , C3 D Proton D - carbony a to i a attached is Proton C - A 0 so . R "A Ha - 2 illl Proton A A , Oppon ppm 3 5 around D Proton C Proton . ProtonB Come e - B 1 Signal CH3-10 - i it 1 / or I - is eachhydroge methyl 2 groups z signall ↓ to d'b's ! Proton A2 Singlet Br 5-4PpM Proton 2-2 B 2pp -quartet . ⑯, - 3 g to 0 9-1 be a L CH3 - RCH - 0ppM -triplet - I neighbor has . . to be a citz) WAS wrong Signal choice I Answer I Choices CHz-CHz CHz = - CH3 the 1 = - B - 4 I CH3 B Proton C * - A Proton -Proton G B CH3 CH2 A I . 9ppm 1 2 . 2 = ppM -proton B -proton C = . 2 . care / S I I ~ I i 3 ppm 1ppon would be i your singlet proton B and high around 9ppm 1 2 . = a has so 0 = CHz = 2 1 & Answer attached electronegativity chlorine 5-4ppmn . RCH2X Proton A This 0 = = to directly the carbonyl so it will Pull the hydrogens further away B CHz-CH-CH3 - Cl L Br . -only have 2 I - A b answer A answer is 2 3 ppon ? POOTOR C - ! CH I think C-CH2 my - A (nel nbor was - Br I - - A tz / CHi =RCH2 X . Examples Work Inneigh e 54PA . 2 Ti ils ' I ! 21 R . . 50ppM Example quartet ↓ 2H !I is M ' singlet triplet ↓d 31t 3H I i R-0-cHz 3 = . A 3PPM cy-"a i - : . ! CH Cro I3 CH 0 - a ] " I Hz - R3CH-1 5 ppM C a : . R2CH2 -1 2 ppm O-CHz -(13 Bic O2 ener I RCHz-0 9ppM a b . ⑭ R CHz-2 1ppm . Aromatic cyclic cyclic Sp - - no hybridized sonly sp sp2) , - pi electrons must planar 4n+ 2 sp carbons conjugated no Huckels Aromaticity noncydic - Sp3 e -Non-planar /3D structure) -3 , 5 , 7 9 , N = 0 & 1) To ⑰ + -NO2 NOz - = H H i O / - - 1 I ⑰X - --M NOz - = / I /- ** Hot Natio T sa e Hs3H . + · r ⑦ + " HSO4 H /- -I I I - H - H - [ - ⑰ H sost sOzH / Is / - + ⑰ &H /W Hy I - - SOztI It - Aso ⑰ & It & /W↑ - II - Sub * + H2 SO3 It -> + H2SO4 + Agot Friedel's Craft ① Alkylation T*+ : ⑫ Acylation ② In I : ⑦ ① so :9 WAIIy A /--H I'll -/ M s + R Cl All - [R-9E (1 A - - [E * = -↑ - / - : AC3 a - E I 1 - -> 2 R R en I - II (l-Al-(13 ... p R + [ 2 - - HCl A11 methyl +x + Cs a I↓ / / - kYY it st ⑰ & + (1 - Al-(13 - and io halide -- H - + - S HzCHz -!: E -H 11 +AIS + ⑦ Al - - It e 3 x / / - ① & 0 I ⑰ +- -CE0"] - --↳ I - + ↓ = /-It / - W · E - - - + A1C3 - [I : ***Y * + / Reactions it s ① Zu - + HzCH , ( - C - Hz(Hy) & Allz -- All3 -> - / - -T as -> ↓ !] & - / - + HC1 + Ald3 OCH3 ↑ gety OCH3 /- --NOZ ↳ ~ Ex or Y - - on or Benzenes para OcoR= Ortho/para NH2 OR - examples H3C0 1310x Noz both * meta . I ......... 12 -- 0 - --- - ......... - istren will is beriche - X I OzN o ASCO "No** - E/ O ⑧ en Br i * position IY 3 -- - - - "I -- Ro Envacarea ra 0 0 C 0 - Ill - ↓ NO + ee ,T "Foz - NOZ - I electron donating =Less acidic the Iwatratio & - - - Ja rating - ioz meta Ortho activating most Substituted para electron withdrawing= stro a H u - --- -- -] ONH2 - (N - - - dangener SOzH WAzR t - NH3 , t - Meta NR3 t - , I NARZ stronatiga ! 0 Br · / Nucleophilic 0H ot ↑ - - i Aromatic - Noz - Bu / i-1) - substitution NOTE : * NOZ if the leaving group and windrowing group are meta is the reaction to Bu /Ill - LO AH 2) Azot , NO2 each other, y observed NO REACTION ① Example -[ *· I a ae-ol o Nut c W- Cl - -~ 1: ⑦ r an - ⑦ -/-I I -- S I - I I 1 ↳ - Cl ② ③ ↑ -- Ea T* replace do Cl G Not Chlorine -meta Nat + NaNHz N ~ NHz : 0H - - an ve 1 J- I- Oz " NaOCH + it * _ 0CH3 NOn this " /-H to ↳ - / Now Hs ]rn / NH2 Kil -NHZ - Nie * he dati ov -COUA My H2SO4 * - B nationAteent 1 wa *S _ 1 Il kMnO4 , Tai win I'll - OOH COOH / + I ⑧ /2-W / 1 ↑ rom - Reduction H2 , PAIC Fe+ HCl Ver it as L E * For carboitis I base , / I # bulky - " H2SO4 e acti + HCl I / - I In+HCl No reaction * varuyer Sn NH2 semmensen - o WOIF-kishner * waiff-Kishner/clemmensen /--R / In-Agal 1) -X- k 11 -R NHINA2 / KOI -> A 1 - / R I i / -E III : CI : - ALLS -> Al-C3 - - -- - -- /- 14 # ⑰ C1- AIC13 - Ed - ⑰ / - I1 I- + 1/ 2 ] · #Di I - I /- S -- --