/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Cc1ccc(O)cc1C#N (B) (D) O=C1CC(C... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 24

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

Expert verified
Compound D has an odd degree of unsaturation.

Step by step solution

01

(Write the title here)

(Write the content here) First, we will draw the molecular structures for the given SMILES strings:
02

Compound A

\(N\equiv C-c1ccc(O)cc1C\equiv N\)
03

Compound B

\(C_{2}H_{5}OH\)
04

Compound D

\(O=C1CC(C(=O)O)c2ccccc21\) Now, let's calculate the degrees of unsaturation for each compound:
05

Degree of Unsaturation Formula

Degree of Unsaturation = \(\frac{(2 \times Number\: of\: Carbons + 2 - Number\: of\: Hydrogens - Number\: of\: Halogens + Number\: of\: Nitrogens)}{2}\)
06

Degree of Unsaturation for Compound A

For Compound A: 2(8) + 2 - 8 - 0 + 2 = \(\frac{12}{2} = 6\)
07

Degree of Unsaturation for Compound B

For Compound B: 2(3) + 2 - 8 - 0 + 0 = \(\frac{-2}{2} = -1\)
08

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:
09

Odd Degree of Unsaturation

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

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