Q.9. - Organic Chemistry

Question

Question: Estimate where the protons bonded to the $\mathit{s}{\mathit{p}}^{\mathbf{2}}$ hybridized carbons will absorb in the 1 H NMR spectrum of each compound.

Step-by-Step Solution

Answer

Answer

1. 6.5 - 8 ppm
2. 4.5 - 6 ppm
3. Aromatic ring proton - 6.5 - 8 ppm

             Double bond proton - 4.5 - 6 ppm

Step 1: De-shielded and shielded protons

Shielded protons refer to decreased chemical shift value due to magnetic induction. De-shielded protons refer to increased chemical shift value due to magnetic induction.

Step 2: Chemical shift value of protons

In aromatic hydrocarbons, protons on $s{p}^2$  hybridized carbons are highly deshielded and absorb at 6.5–8 ppm,

Unsaturated non-aromatic hydrocarbons have shielded protons and absorb at 4.5–6 ppm, typical of protons bonded to the benzene ring.

Step 3: Absorption in the 1 H NMR spectrum

a. Aromatic ring present - di-shielded protons

Therefore, absorption at 6.5 - 8 ppm.

                 (a)

 

b. Unsaturated alkenes - shielded protons

Therefore, absorption at 6.5 - 8 ppm.

                 (b)

 

c. 1. Protons in the aromatic ring are di-shielded

Therefore, absorption at 6.5 - 8 ppm.

 

2. Protons of double bond outside the ring are shielded,

  Therefore, absorption at 6.5 - 8 ppm.

                      (c)