Question

Cc1ccc(O)cc1C#N (B) (D) O=C1CC(C(=O)O)c2ccccc21 # Which of the following has odd number of degree of unsaturation? (A) N#Cc1ccc(O)cc1C#N (B) (D) O=C1CC(C(=O)O)c2ccccc21

Short answer

Compound D has an odd degree of unsaturation.

Step-by-step solution

(Write the title here)

(Write the content here) First, we will draw the molecular structures for the given SMILES strings:

Compound A

\(N\equiv C-c1ccc(O)cc1C\equiv N\)

Compound B

\(C_{2}H_{5}OH\)

Compound D

\(O=C1CC(C(=O)O)c2ccccc21\) Now, let's calculate the degrees of unsaturation for each compound:

Degree of Unsaturation Formula

Degree of Unsaturation = \(\frac{(2 \times Number\: of\: Carbons + 2 - Number\: of\: Hydrogens - Number\: of\: Halogens + Number\: of\: Nitrogens)}{2}\)

Degree of Unsaturation for Compound A

For Compound A: 2(8) + 2 - 8 - 0 + 2 = \(\frac{12}{2} = 6\)

Degree of Unsaturation for Compound B

For Compound B: 2(3) + 2 - 8 - 0 + 0 = \(\frac{-2}{2} = -1\)

Degree of Unsaturation for Compound D

For Compound D: 2(14) + 2 - 12 - 0 + 1 = \(\frac{17}{2} = 8.5\) Now, we can check which of these values are odd:

Odd Degree of Unsaturation

In our calculations, Compound D has an odd degree of unsaturation (8.5). So, the answer is Compound D.