Chapter 11, Problem 11.1P

Neopheliosyne B is a novel acetylenic fatty acid isolated from a New Caledonian marine sponge. (a) Label the most acidic H atom. (b) Which carbon-carbon $\text{σ}$ bond is shortest? (c) How many degrees of unsatuartion does nepheliosyne B contain? (d) How many bonds are formed from ${\text{C}}_{sp}-{\text{C}}_{s{p}^3}$ ? (e) Label each triple bond as internal or terminal.

Interpretation Introduction

(a)

Interpretation: The most acidic H atom present in nepheliosyne B is to be labeled.

Concept introduction: The most acidic hydrogen $\left(\text{H}\right)$ atom or proton is the one which can be easily eliminated from the molecule. This higher acidity of $\text{H-atom}$ is dependent upon the electronegativity of the functional groups that are bonded with carbon atom containing hydrogen. The elimination of a proton leads to the formation of a conjugate base. The higher stability of the conjugate base corresponds to the higher acidity of the proton.

Answer to Problem 11.1P

The labeling of the most acidic H-atom present in nepheliosyne B is shown below.

Explanation of Solution

The labeling of most acidic H-atom is shown as,

Figure 1

The stability of conjugate base of carboxylic acid is more than the stability of alcohol’s conjugate base. Thus, the acidity of hydrogen atom present in the carboxylic acid is more than that of alcohol because resonance stabilized conjugate base results in an increase in the acidity of a compound.

Conclusion

The labeling of most acidic H-atom present in nepheliosyne B is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation: The shortest carbon-carbon $\text{σ}$ bond present in nepheliosyne B is to be labeled.

Concept introduction: The higher value of s-character corresponds to the higher electronegativity and lesser bond strength The highly electronegative carbon that is $sp$ carbon will attract more bonded electrons than $s{p}^2$ and $s{p}^3$ carbons that leads to the high concentration of negative charge and decrease in the bond strength. The lesser amount of bond strength corresponds to long bond length and vice-versa.

Answer to Problem 11.1P

The shortest carbon-carbon $\text{σ}$ bond present in nepheliosyne B is bond (a) as shown below.

Explanation of Solution

The carbon-carbon $\text{σ}$ bonds present in nepheliosyne B are shown as,

Figure 2

The bond (a) in the given compound is formed by $s{p}^2$ and $sp$ carbon atoms. The end-on-overlapping of $s{p}^3$ hybridized carbon atom and $sp$ hybridized carbon atom produces a single bond (b). Thus, due to increasing bond strength, the s-character of bond (a) is less than the s-character of bond (a). Therefore, the shortest bond $\text{σ}$ bond present in nepheliosyne B is bond (a).

Conclusion

The shortest carbon-carbon $\text{σ}$ bond present in nepheliosyne B is bond (a) as shown in Figure 2.

Interpretation Introduction

(c)

Interpretation: The number of degree of unsaturation contained by nepheliosyne B is to be stated.

Concept introduction: The degree of unsaturation states the total number of double bonds, triple bonds or rings present in a compound. It is calculated by the molecular formula of compound containing carbon and hydrogen atom.

Answer to Problem 11.1P

There are total $13$ degrees of unsaturation contained by nepheliosyne B.

Explanation of Solution

The molecular formula of nepheliosyne B is given by, ${\text{C}}_{\text{48}}{\text{H}}_{\text{72}}{\text{O}}_{\text{11}}$. Thus, the maximum number of hydrogen for $48$ carbon atoms are calculated by the formula,

$\text{Number}\text{of}\text{hydrogen}\text{atoms}=2n+2$

Where

• $n$ is the number of carbon atoms.

Substitute the value of $n$ in the above formula.

$\begin{array}{rcl}\text{Number\hspace{0.17em}of\hspace{0.17em}hydrogen\hspace{0.17em}atoms}& =& 2\left(48\right)+2\\ & =& 98\end{array}$

As the molecular formula of compound contains $72$ hydrogen atoms, so the actual number of hydrogen atoms which are lesser than maximum number is $=98-72=26$. Thus, two hydrogen atoms will be removed for each degree of unsaturation.

The degree of unstauration is calculated by the formula,

$\begin{array}{rcl}\text{Degree\hspace{0.17em}of\hspace{0.17em}unstauration}& =& \frac{{\text{26\hspace{0.17em} H}}^{\text{'}}\text{s\hspace{0.17em}fewer\hspace{0.17em}than\hspace{0.17em}the\hspace{0.17em}maximum}}{{\text{2\hspace{0.17em} H}}^{\text{'}}\text{s\hspace{0.17em}removed\hspace{0.17em}for\hspace{0.17em}each\hspace{0.17em}degree of\hspace{0.17em}unsaturation}}\\ & =& 13\end{array}$

Hence, the total number of degree of unstauration present in nepheliosyne B is $13$.

Conclusion

The total number of degree of unstauration present in nepheliosyne B is $13$.

Interpretation Introduction

(d)

Interpretation: The number of bonds formed by ${\text{C}}_{sp}-{\text{C}}_{s{p}^3}$ is to be identified.

Concept introduction: The procedure of intermixing of the atomic orbitals in an atom is known hybridization. This intermixing of the atomic orbitals is used to form a set of new atomic orbital with different geometry.

Answer to Problem 11.1P

The number of bonds formed by ${\text{C}}_{sp}-{\text{C}}_{s{p}^3}$ is six.

Explanation of Solution

The number of bonds formed by ${\text{C}}_{sp}-{\text{C}}_{s{p}^3}$ is shown as,

Figure 3

Thus, there are six bonds formed by ${\text{C}}_{sp}-{\text{C}}_{s{p}^3}$ in nepheliosyne B.

Conclusion

The total number of bonds formed by ${\text{C}}_{sp}-{\text{C}}_{s{p}^3}$ in nepheliosyne B is $6$.

Interpretation Introduction

(e)

Interpretation: Each of the triple bond present in nepheliosyne B is to be labeled as internal or terminal.

Concept introduction: The triple bond or alkyne which contains carbon substituents over acetylenic carbon is known as internal triple bond or alkyne. In case of terminal triple bond or alkyne, there is at least one acetylenic carbon that contains H-atom linked to it.

Answer to Problem 11.1P

The labeling of the triple bonds present in nepheliosyne B as internal or terminal is shown as,

Explanation of Solution

The labeling of the triple bonds present in nepheliosyne B as internal or terminal is shown as,

Figure 4

In Figure 4, there are three internal triple bonds and one terminal triple bond present.

Conclusion

The labeling of the triple bonds present in nepheliosyne B as internal or terminal is shown in Figure 4.