Question

draw structure of C8H10 and C8H8O2 based on the 1HNMR AND 13C-NMR shown in the pictures provided.

any tips of steps will help! thank you!

C-NMR data given. D ons: Solve the structures for C8H10 and C8H8O2 using only the molecular formula and 'H- and R data given. DO NOT WORK IN GROUPS. You may ask your TA for minor assistance. You may use your tab manual (pages 31-39). Please write your final structures in the space provided below for each This is due before you leave and will not be accepted at another time. C8H8O2 = 5 DOU C8H10 C&H10 =4DOU 5 DOU UN= [(2C+2) - (H +X-N)] [(a+8+2)-8), 10 = 5 DOU 2 Catio 18 [(a)(2)+2) -(10) ? 4 DOV

$PROTON CDC3 (C:\Bruker|TOPSPIN) masterson 1 10 total H 3 wrique H C8H10 PRCITON COOL (C:\ B UTOPSPIN) masterson 1 -60000 -500$

$C13CPD32 CDC3 (C:\Bruker|TOPSPIN) masterson 1 130.00 -1400 -1300 -1200 C8H10 5 unquie carbons -1100 400 CDC13 -300 Sp² 200190$

$PROTON CDC (C:\BrukerTOPSPIN) masterson 1 55000 C8H802 -50000 45000 PROTON CDG (C:XBrukerTOPSPIN masters 40000 40000 - 30000$

$C13CPD32 CDCI (C:\BrukerTOPSPIN) masterson 1 CBHBO2 Peak shifts are on bottom cpoly omso-do 160 140 120 100 80 190 170 150 13$

please scroll for all lictures!

Answer 1

Solution for Ce Ho Double bind equivalent 5 Carbons – Hydrogens +1 8-10 + = 4. 2. DBU = 4. li je to from NMR Data given :- The structure of the compound C8 Mio is CH3 CH3 HH S13/ Descaiption of HIMMR: H (6) CH3 (d) CH3L 13 didy HT There are named as Hay l eyen four types of Hydrogen atoms abic& d.

HACIH) It gives signal at 8= 7.00pm 1.1 and multiplet due to long range coupling Hb CIH) - gt gives signal at se 7.00ppm to 6.96 pbm approacimately. 97 LB memo to CHg groups l at ortho positions. So Due to election releasing nature of CH3 groups, there is sh Slightly decrease in s-value because of shielding. So 8-value is less thom Ha. He (2H):- gt also gives siqnals near 846.94 bbm as these two hydoogens are Ortho - to CH3 group. Hd 16 HD!These protons gives singlet at 8= 2.3-2.4 ppm. These protons' C- atoms are directly attached to phenyl ring. So deshielding occur and gives high S-value for both CHz (allylic protons).

Describtion of 13C-NMR: - CHE sf ; CH3 There are five types of carbon atoms denoted by 1, 2, 3,4 $ 5. Gi>(20)→ 6 = 21-22 bpm. Highly shielded e due to protons and benzene ring. G s (2.) -= 126 127 þþm. Ootho to dectoon dona ting (CH3) groups. - Czy (IC) meta to C13 groups, so it has I'll negligible inductive effect. So 8 = 128 bbm. . CA C 1-6) - Although highly shielded due to two CH3 groups. But gives signal at 6= 129-130 ppm, higher thon C2$ (3. C5 (20)- 8= 137 bbm. Due to hyper conjugation, electron density away from CS. da so s-value is highest

S toucture for Cg Hg 02 :- Double bond equivalenta C-H+ → 8-8 +1 3 5 DBU = 5. The structure of the compound is :- crohia) (C) 1 CH₂ (d) botoluic acid. Description of 1H-NMR :- There are four types of proton named as a, b, c & dan Ha (1H) :- I gives a broad bignal at 8 = 11.7 ppm due to highly election withdrawing nature of — Coo@gooup. This proton is exchangible. So gives broad signal. Hb ( 2H): - Cool has +R effect due to which these protons are in deshielded environment. So & value = 7.9-8.0 ppm.

-HC (2 H): These protons are meta to coooh! group. So there is no + R effect on ortho position. So 8-value = 7.2-7.3 bbm lesser thom Hb protons. Hd. (3H):- Benzylic (CH3) group generally gives a sharp singlet at 8 = 2.4 ppm always due to allylic nature of 3 protons in CH3- Describtion of 13C NMR :- ) ) W slo @cool CH3 0 There are six type of carbon atoms mim bered as 1, 2, 3, 4,5$ 6. C-1:-(CH₃)= 8= 21.48 ppm, normal value for benzylic CHZ, C-2:- (1.Cls S= 127.70 þbm less s-value due to tI effect of CH3 grous

C3- (20948= 128.81 → Slightly away from -CH3 group. Su lesser +I effect. meta to cool group, nu tr effect cq (20):+ 8 = 129.68 bbm northo to cool group having +R effect. C5 CICI -> 8= 143.28 ppm-j directly attached carbon atom with electron with drawing cool group Go C1 CJ Carbonyl carbon of coon group gives asignal at 8= 165-175pbm. Here 8-value => 188.98 ppm, Please like