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8.8 Identify the compounds of the given molecular formulas that show a. One 'H NMR signal: CHO CH C.H.Br, b. Two 'H NMR signals: CH.CL CHO CHO, CHOCI 8.9 8.10 How many 'H NMR signals will each of the compounds below show? a. (CH, CHỊCH, d. CH.COCH,CH.COCH, b. CH,COCOOCH, e. CH.COCH.CH.COOCH, c. CH,COCH,COCH, L. CHCOCH CH(CI)COCH, Suggest structures consistent with the following 'H NMR data: а. Сно CHCI triplet (1.0 ppm, 3H) singlet (1.6 ppm, 6H) singlet (2.1 ppm, 3H) singlet (3.1 ppm, 2H) quartet (2.4 ppm, 2H) multiplet (7.3 ppm, 5H) b. C,H,O, 1, C,H, 0, triplet (1.2 ppm, 3H) triplet (1.2 ppm, 3H) singlet (2.1 ppm, 3H) singlet (2.4 ppm, 3H) quartet (4.1 ppm, 2H) quartet (4.2 ppm, 2H) c. C,H,O j. CHCIO singlet (2.4 ppm, 3H) triplet (1.4 ppm, 3H) multiplet (7.5 ppm, 5H) quartet (3.8 ppm, 2H) singlet (5.6 ppm, 2H) d. C,H,BRO k. CHBr singlet (3.3 ppm, 3H) quintet (2.2 ppm, 2H) singlet (4.9 ppm, 2H) triplet (2.8 ppm, 2H) triplet (3.4 ppm, 2H) multiplet (7.2 ppm, 5H) есно 1. CH singlet (3.7 ppm, 3H) doublet (0.9 ppm, 6H) multiplet (7.0 ppm, 5H) multiplet (1.9 ppm, 1H) doublet (2.4 ppm, 2H) singlet (7.2 ppm, SH) m. CH20, Singlet (1.3 ppm, 9H) triplet (1.3 ppm, 3H) singlet (7.3 ppm, 5H) quartet (3.7 ppm, 2H) singlet (4.2 ppm, 2H) singlet (11.0 nm. 1H) CH

Answer 1

Answer 8 lo c) MF- C8H8O iDegee o Unsalurahom = (No os c +1)-No ob H 2 E 5 The 5 clegeee ok unsakusahom indlicate the poesence os benzene sing in given Compound iiThe NmR Singlet at S 2.4ppm due to 3H atoms încicate the Presence os 96oup and it has no psotons on adjacent Casbon The doon sield S 'ndicate CHa 8oup attached to the carbonyl group Thus present in structuse i The NmR muikplet at 7-5pem due 5H indrcate +he i os monosubshtu tedl benzene sing. pfesence The possible stou ctuv e is 2.4(s) 7.5(m)

Answer Neglect ' 8.lo M-F GH8O e) С7н8 i Degsee o unsatysahcm = (No. ob Ct1)- H (+1)) 8- 4 The given Compound has Four degsee os unsatusahom and it îndicate the presence ob monosubshtutecl benzene ving in structuse NmR singlet at 3.7ppm due to SH atoms indicate no peotens c the odjacent The pre sence ols CH3 goup and i+ hos carhon. The downlseld & Tepresent c group a tached to electro negahve oxygen atom. Thus 'c-o' present in structuse . The NMR malh'plet a t presence op monosubehtuted benzene. 7.o Ppr due 5i atom indicate the The possible staucture an the bais ob NmR data is 8-10) F). mF CloH14 iDegree o unsatearakcon = No ob o +1)-No.ob= (l0+1)-(G 1-T = 4 The given Compound hove 4 degvee ob unsatusahan, it indicatle the presence os benzene a'ng in Compound ii The peesence ob three equivalent CH5 9soup and all no prolor on adjacent carboan . Thu NMR Cinglet at S J.3 pm due to9t atom indicate the gsoup has C roup present i) The NmR RRtpo singlet at To3S due to sH atom indfcate the presence os monosubshtuted benzene eing. on the bosi's ob NmR possible steucture i's CH3 кE Č

add 1H Fo c m.F CloHi3 cl 8-10 ) New lmulq i Degree os ansakunab'om No ob c+1)- (No ob ) (Io+1) he given Compound has 4 sites os unsaterrahcn, it indi cate the Peesence ob benzene oing în stsucterse NmR Singlet at S l-6Ppm due to 6H fndicate The presence o tuoo equivalent CHgoup and both has no Ptoton on adjacent carbon Thus group present lii) The NmR Singlet at S 3-1ppm dlue tu 2H atom îndicate Presence os -CH- gtoup and i hau no protun om odiacent carbom The downbield S sepoesent r attoched to electsonega tve cl gtom - ii) The NmR mulhplet at T-3 pem due to H atom îndicate the pre sence ob monosubshituded benzene ring in srackure Onthe bosis ob NmR the possible stuctuse is Thw, -сH,-cl mF C2H4B 8.80 C) DUO 1t show one Nme signal and then possible structuse is Br-CHCBr 8-8 b) c) m.F Cg Hg02 4it shouw thwo NMR signals D.U o Possible stucture is -o-c-o--CH