Question

i need help with this HNMR practice problem. i obtained an IHD of 6 but im unsure what this means as well.

2964 1729 43 2 68 O 6E 7 113 52 37 42 4 2. DU (IHD): Show calculation(2 pts) Comments on IHD (1pt) 3. Report IR analysis here. Tabulate (5 pts) Region Functional group Base Value Absorption, cm-1 (Exact number read from IR Give as a range. Read from correlation chart ectra 4000-3000 cm 3000-2500 2500-1700 cm 1700-1500 cm 1500-400 cm 4 Explain any IR special features in this box (2 pts) (Symmetry, conjugation, resonance, H-bonding, Evidence for Ortho, para, meta disubstitution or mono substitution.), if applicable 5 Report your 1H NMR analyses here: (10 points) 6. Conclusion: (5 points) Draw Structure showing all carbons and hydrogens and match the peaks with the protons. 7. Explain any special features on the back of this HNMR: symmetry, Homotopic, heterotopic, constants, etc. (2 points) CH untroun 37 80 7.70 7.60 7.50 3.0 ppm 01)

Answer 1

Dear candidate,

please go through attached image.

Q2)

DU=C_nH_2n+2

Lets consider our MF: C₁₀H₁₀O₄

Now apply this value of 'n' in our formula,

10X2+2=22

our compound have 10 hydrogen atoms so,

(22-10)/2

=12/2

=6

Degrees of unsaturation is 6.

Q3)

IHD=(2n+2- no of hydrogen in given molecule)/2

=(2X10+2-10)/2

=(22-10)/2

=12/2

=6

Region	Absorption cm-1	Functional group	Fragments
4000-3000cm-1	3029	C-H stretch alkene	C-H stretch
3000-2500cm-1	2964	C-H stretch alkane	C-H stretch
2500-1700cm-1	1729	Ester	C=O
1700-1500cm-1	1600	C=C	C=C
1500-400cm-1

  

Sr No	Chemical Shift	Integration	Splitting	No of neighbours	Fragments
1	3.82	3H	Singlet	0	CH3-O-
2	7.56	2H	dd	1	Aromatic
3	7.66	2H	dd	1	Aromatic

7.66 3.82 7.56 7.56 3.82 7.66