Why i n a magneti c fi el d al l h exi st as h h 11

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Unformatted text preview: c. d H a H bd C Cl C O q t Cll C CH 2 CH 3 Why ? Why I n a magneti c fi el d, al l H exi st as H :H ~ 1:1. OCH 2CH 3 H a “ sees” two types of H b n ei ghbor s (thr ough bond): H b and H b (thr H b str engthens H 0 ar ound H a : deshi el di ng str d eshi H b weakens H 0 ar ound H a : shi el di ng w eakens shi Resul t: two l i nes (1:1 r ati o, d ) i nstead of one. Resul t wo D i stance between them i n H z i s the coupl i ng constant J . Same val ue Same H a and H b ar e sai d to “spl i t” each other w i th a J of 7 H z. “ spl J i s f i el d i ndependent: Same at 90, 300, etc. M H z 300, Range: 1-18 Range: F or H z. e than one n ei ghbor : H “sees” all l mor t han m or a α, β combi nati ons of nei ghbor s. Thus, two n ei ghbor i ng H s: t wo αα, αβ, βα , ββ: 1:2:1 tr i pl et αα αβ βα ββ :1 t r Thr ee n ei ghbor i ng H s: ααα, ααβ, αβα, βαα , αββ, βαβ, ββα , ααβ αβα βαα αββ βαβ ββα βββ βββ 1:3:3:1 quar tet :1 quar Summary Points Summary 1. Equi val ent pr otons show no spl i tti ng. 1. E qui no 2. J i s i ndependent of H 0. 3. Nonadjacent H s J ~ 0. 3. H CX H C 4. (N+1) Rul e. Rati o fr om Pascal ’s tr i angl e. 4. (N+1) 5. Spl i tti ng i s mutual . I f ther e i s one spl i t peak, 5. Spl t her e has to be (at l east) one other . 1:6:15:20:15:6:1 qui ntet t O O s d (CH 3)3CH (CH decet t H Al l –CH 2- h ydr ogens ar e hydr equi val ent: equi Four mi r r or pl anes tr i decet CH 3CH 3 CH 3CH 2Br + CH 3CH Br 2 + h q dq Br CH 2CH 2Br t Br Br 2 si ngl et A 3.427 B 1.679 p pm http:/ / www.ai st.go.jp/ RI ODB/ SDBS/ cgi -bi n/ cr e_i ndex.cgi A 5.842 B 2.458 p pm A 3.654 p pm Pr obl em: Pr 2H 1H 4H 3H The two butanol s: CH 3CH 2CH 2CH 2OH CH 3CH 2CH CH 3 OH Whi ch i s whi ch? 1H 3H 3H 2H 1H...
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This note was uploaded on 11/11/2009 for the course CHEM 368537 taught by Professor Voldhart during the Spring '09 term at Berkeley.

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