Question

4) Chemical Formula: C4H100 IR: strong broad peak at 3300cm-1 6H 2H 1Η 1Η 3 PPM 5) Chemical Formula: C&H11N IR: Two weak peaks at 3350cm-1 ЗН Doublet, 11 Triplet, 1H Triplet, 1H Doublet, 1H 2H он PPM

Please explain how to read and draw the Chemical structure for the following questions depending on the information provided: Chemical Formula, IR and NMR.

Final answer should be skeletal structure, thank you.

Answer 1

Dear candidate,

Any problem related to 1H NMR can be solved in following steps.

a) If molecular formula is given, Go through molecular formula and find out Degree of Unsaturation or it is also known as sites of unsaturation. Formula is given in image.

Where C= no of carbon atom

N= no of nitrogen atom

X= no of Halogen

H= no of Hydrogen atom

present in given formula.

b) Go through IR data, we will get information about functional group present in the molecule.

c) Collect data from 1H NMR as shown in Image.

Answer to problem 4)

a) Degrees of unsaturation came zero, it means their is no double bond present in the molecule.

b) By IR we came to know presence of alcoholic OH group which gives absorption peak at 3300cm^-1.

C) By NMR data,

i) 6H doublet at 0.9ẟ indicates two CH₃ groups are attached to CH.

ii) Signal at 1.25ẟ 1H shows multiplate means CH group is surrounded by carbon having more Hydrogens(More carbon and more hydrogens).

iii) Signal at 3.5ẟ 2H shows doublet indicating adjusting carbon has one hydrogen atom.

iv) Signal at 4.5ẟ 1H Singlet Indicates OH group.(This point not Indicated in image 1)

So according to above data(Go through image 1 ) we got some fragments, If we join those fragment we will get structure as shown in image 2

4 Chemical formula cahoo @ Degrees of 20+2+N-X-H unsaturation - 2 x 4 +2 +o-o-10 = 10-10 2 = 0 sites of unsaturation or Degrees of un saturation is zen. 6 IR: strong broad Indicates peak at 33co cont presence of –OH group. © HUMR: 0.95 C6H,d). CMg-ca-cha 1.258 (1H,m) - 3.56 (2H, d) = Chy-dy-chy CH – CH₂-OH

(IHM) 1.258 CH - Co - OH = 4.38 (14,5) (24, d) 8.55 (6H, d) o.gd

5)

a) Here we got Degrees of unsaturation 4, it means benzene ring must be present in the molecule.

b) By IR we came to know presence of NH₂ group which gives absorption peak at 3350cm^-1.

C) By NMR data,

i) Signal at 1.05ẟ 3H shows triplet means CH₃ group is surrounded by carbon having two Hydrogens.

ii) Signal at 2.6ẟ 2H shows quartet indicating adjusting carbon has three hydrogen atom.

iii) Signal at 7.5ẟ 2H singlet indicate NH2 Protons.

iv) Signal at 6.5-7.0ẟ 4H , It means di substituted benzene ring.

So according to above data(Go through below image) we got some fragments,

5) Chemical formula @ Degrees of in sathulim. C&HIN 2XC + 2 +N-X-H 2 2 x 8+2 +1-0-1) 1 2 - 19-11 1 1 1 As degrees of unsaturation is 4 aromatic benzene ring must be present 6 IR: Two weak peaks at 3350 cm Indicatos presence of Niy group ề 1H NMR: 1:058(3H, t) = crt3 - CHE 2.66 (2H, t) = -ch City 7.58(2H s broad) = - Hly 6.5 to tod(44) = Di substituted benzene. ging

If we join those fragment we will get three possible structures as shown below

ΖΗΝ

But according to above NMR spectra shared by you we are having four different hydrogen atom signals, so possibility of C is zero. Because Structure C will give two aromatic protons and in image we are having four aromatic signals.

So most probable structure are A and B. If we get more information we can surely tell its A or B.

more information means zoomed NMR at aromatic region.

Thanks.