Question

'HNMR - Problem Set-3 How would you distinguish between the compounds in each of the following pairs by 'H NMR? a. CH CH2CH2OCH; and CH3CH2OCH2CH3 b. BrCH CH CH Br and BrCH2CH2CH NO2 CH3 CH3 c. CH:CH-CHCH; and CH3 CH CH CH3 CH CH d. CH3-C-C-OCH; CH OCH and CH3-C-CH; OCH e. H3C CH; -C-CH3 CH; and -CHỤCCH CH 'HNMR-Problem Set-3 (Page 1 of 2)

Answer 1

mis titlt nics & Itlt dusel Ans. 2 -cal tit Initian - tltlt Itt it H-NMR @ equilibri 4 signals ww, debuiflet) on it Ellies It Elements $ P- b 15 1 1+1+ + 3.35 3.23 3 cych - o - th- chis (symmetric) htG THI Searntan ent only 2 signals due to synay Il triplet tly quartet it Elt 21+ 4 3.5 3. NMO

(6) Bre- cth eth-cth Br I Breth- ch-cth No2 (Syminettve) IR 4 (3 signals) (2 signals) friplet pentate pentale. Ariplet nel 43

at C-c-c-crb O the E-cth-clbze com (3signals) (2 signals) symmetric doublet tit singlet briplet multiplet queretet with - 3 2

2 . HI 2. Pauli atom 2 signals (symmetric) the sam singlet As an two elen spin 1-27 1 4 3.64 3 2 singo Position of the signals in both the compounds will be different Singlet + 1-20 1 N

> عسل عنهمطعمملممل нс a singlet doublet doublet, t enos H 1 Chho-c-eth nglet u unu inslet singlet

ons 2 CHIO O2 I 6 GHz Oz a this inco che a= triplet (t) = 1.25 ba quartet (9) = 2.33 quartet (a) = 4.13 a= singlet (6) = 1.5 .bz singlet (s) = 2:0 c= da triplet (t) = 1.15

(0) C Hq la atc ccth HD a doublet (d) = 1.60 be multiplet (m), z 4.27 ca multiplet (m 2 2015 da briplet (2) 23:72

89281 az singlet = 21 ob 2 quartet = 4.1 ca dhon triplet = 1.2 at

(K) CAHA Ha Az singlet brwad = 7.1 ba singlet =2)

(1) Cg Hz 10 yo b = pentate = 2.0 z triplet = 3.7 ca friplet = 3.8 da singlet = 2.8

(J) CG Hio (в на HeHc a= triplet = 1.2 b2 quartet = 2.6 c z broad singlet = 71 a= singlet = 3.7

A, F and G either number of carbons or number of hydrogens are not visible