Question

Determine the structure of the following given the spectrum and molecular formula

CzH02 11 10 9 8 7 6 HSP-03-871 ppm

Answer 1

Rules to be used in this question

1) Rule of thirteen: It says that the formula of a compound is a multiple n of 13 (the molar mass of CH) plus a remainder r.

Base formula : C​​​​​​n H​​​​​​n+r

2) index of Hydrogen deficiency : = [2( number of carbon atoms) -( number of monovalent atoms ) + ( number of trivalent atoms) +2] /2

3) Factors affecting chemical shift

a) Electronegativity: An increase in electronegativity of the surrounding groups will decrease the electron density and increase the chemical shift value due to the deshielding.

If electropositive groups are present in molecules they increase the electron density on the adjacant atoms and decrease the chemical shift due to shielding

For example let's take an example :

CH​​​​​3​​​​ - Cl , CH​​​3 - H , CH​​​3 - C​​​​​​2 H​​​​​​5

Cl is more electronegative than H and C will decrease the electron density on adjacent CH3 and inturn increasing CH3 protons chemical shift or making them deshielded. Whereas C2H5 shows inductive effect (+I) which increases the electron density on CH3 inturn decreases the chemical shift of the protons attached.

chemical shift values for CH3Cl protons will be higher than CH4 which will be in turn higher than CH3-C2H5

b) Magnetic anisotropy : When molecules having pi electrons such as benzene, alkenes, alkyne etc are placed in a magnetic field then these mobile pi electrons circulate around and produces a new magnetic field which has two regions shielded or deshielded.

Aromatic protons come under deshielded region.

c) Hydrogen bonding
Hydrogen bonding results from the presence of electronegative atoms such as Fluorine, Nitrogen, Oxygen. Hydrogen bonds forms between F, N, O with Hydrogen of F, N , O. The interactions thus forms leads to deshielding to higher values of chemical shifts. This confirms the presence of Hydrogen bonding.

4) Spilting

NMR spectra provides information about a the neighbouring hydrogens present for a particular hydrogen or group of equivalent hydrogens.

An NMR spectral peak will be split into N + 1 peaks where N = number of hydrogens on the adjacent atom.

Equivalent hydrogens means  atoms that are completely interchangeable as to their role in the molecule. In the case of HNMR, the two Hydrogens which are on different atoms but have same chemical environment. They will show same chemical shift. As in the above question in benzene both protons which are labelled 'a' have same Chemical environment. Hence show same chemical shift.

5) Integration

It is the measurement of peak areas which corresponds to the amount of energy Absorbed or released by all nuclei participating in chemical shift during the nuclear spin flip. It also helps us to determine the ratio of hydrogens that correspond to the signal.

Now using all these information let's solve the questions

Ques E)

- ③ Cz Hq Og Index of Hydrogen deficiency = 3x2-6+2 2 2 - i double bond present en IR spectron a peak at nitro un shows the presence of c o group .

③ e peaks are in Proton NMR, the three as follows - a) triplet 34 [CH] triplet shows it is attached to two Hydrogen carbon. (2+ 15 b quartet - 24 CH2] peaks shows it is attached to (3 ta lains three hydrogen carbon () singlet 14 can be offor O CHO on coon group.

E ☺ compiling information till now o c=0 group double bond → CH₂ 4 Hz. as fragments attached ... to each other. So the probable structure comes out to be CH₂ CH₂COOH. As nopeak is seen. at 3300-3500 cant for - on & no peaks at 2750 & 2850 cont for CH! whereas a peak at 3000cm for coon acid is observed en addition, there is a peak at 1150 cm' for c-o stretching

O IR spectrum solved З ооо ам" — соон амоир – 0-И stretching : 1700 ing of coon group 1150 cmt :-o stretching of . NMR i singlet 8ppm cool froton (14) & quartet - 4.1ppm CH2 (24). triplet 1.2 ppm CH₂ (34) acid 6 .

Ques F)

104 27 . 2 0 Cq H2O : Index of hydrogen deficiency = 2X9 – 12 12 degree of unsaturation means 4 double bond or mixture of double bonds tring - (2) IR spectrum shows distinct peaks → : a) 3 300 cm' due to NH or on & stretching . But as in the formula o is present... Hence On' stretching b) peaks b/w 1400-1500 m shows presence I of benzene ring which is also satisfied by degree of unsaturation & signal al- I ppm in Proton NMR...

NMR spectrim 4H (2+2) 7 8ppin shains benzene ning protons. In addition to this information, it is also clear the ring is disubstituted because only 4 protons are giving signals. The spilling of benzene proton is a characteristic feature shown by fara substituted benene. 4.5ppm an singlet shows CH2 fragment with no hydrogen on adjacant carbon 4ppm 24 quartet CH2 fragment but has 3 (3+1) protons in near vicinity

18ppas 14 singlet or proton. 1ppm 34 triplet shows it is a CH3 group attached to CH2 fragment: Both LH2 protons are highly deshielded suggesting their presence near election- a withdrawing oxygen. compiling all information o on group 2 - CH2 fragment ② CH3 fragment @ Para subsituted Benzene

o the structure came out to be . си, он си, осны ол, Ба ин О but the structure is this as phenolic - on comes at 5-6ppm whereas alcoholic comes b/w 2-3ppm: Proving stuchire a to be our structure o

© IR spectrum 3300 cmt → 04 broad peak 1400-1500 cm! Benzene cacpeaks . Tosocm c-o stretching o Proton WMR d 45ppm ... tesppm 40ppm Tore Y CH2 04 18ppm CH₂ - CH2 - 0 I singlet do → per .. R X-4 I ppm Briti) doublet 7.4 ppm . (2+1=3) ter (317 quartet (1+1) ... doublet V

Note : alcoholic proton is labile in nature. This means it readily exchanges with the solvent. Hence doesnot spilt the adjacant present protons of CH_{​​​​​2.}

Ques G)

A o CGhio Oa Degree of unsaturation = 6x2-1012 2 ... means a double bond @ In IR spectrum signal above 1700cm (17 25 cm) suggests presence of carbonyl stretching In addition to this peak, a peak at 1640 cm! suggests the presence of C= C Amefening

( NMR → peaks at 4, 5, 7 ppm of T4 each suggests that they are attached to etc that's why dishielded to this extent. I 2.4ppm - triplet (24) means I He is attached to ä carbon with Id.pmtoms (2+1= 3 ü miplet) This carbon may be is attached to carbonyl carbon that's has this value of chemical shift.

- 0 Isppm multiplet (24) CH2 fragment which is attached to many protons. 1 ppm triplet (34) > CH3 fragment which is attached to c having at's. compiling all the information o ce of c= c 6 Chaq 3 CH, - cha (6 H-, * ,c=H - The structure will be H₂ C = CHTG CH2=CH2 CH3 The three protons spilt each other.

(4) Now solving IR spectra > . 1725cm' 1640 cm - c=0 stretching c = c stretching ③ NMR spectrum 9 - 4. C = Ć H2CH2 .ppm 4 - D'old) 2 Uppm los por

Ques H)

(I. Ha & Ho are in different environment Hence will spilt each other. also, in addition to the proton . → He 7.1ppm → dd doublet of doublet - He is spilted by the than by Ha. B (141)=2 (1+1)=2 same will be with Ha & Hb But the peaks get merged to doublet, Proton d deshielded due c=0 group + Proton ex (2+3+1=6) u multiplet 1.8 ppm Proton of (2+1) - triplet - 1ppm will triplet. show 2+1=3 →

7 Culoo degree of unsaturation=4x2-672, 2 a double bonds on rings. 2. In IR spectrum → : peaks at too cm > c=0 i at too cm.. stretching " 275 of 2850 cm! (84) de ch stretching V 1650 cm v c retching

③ NMR spectra - aroma aldehydic proton appm- deshielded due to Magnetic anisoropy 6ppm 4 7ppm (m) c=c-y deshielded due to magnetic . anisotropy TO 2ppm (d) CH3 (34)

4 Compiling all the information б-сно c=c ③ - CH₂ fragment. The structure possible will be ch- c— съено и.

6 IR spectra a) 275042850 cm 0-c-12 stretching 6 1700 cm Lc=0 stretching @ 1600 cmt c =( stretching

© NMR NMR it 115 multiplet CH3 - CH=CH - 64 2ppm oppen tippm ppm en la dolet" utan doublet /