Chapter 15, Problem 15.32P

Interpretation Introduction

Interpretation:

Four isomers consistent with the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}$ and longest UV-vis. absorption at $\text{215 nm}$ are to be drawn.

Concept introduction:

The IHD accounts for the number of rings or multiple bonds the structure could possess.

${\lambda }_{\max }$ depends heavily on the extent of conjugation. In a molecule contain only one sigma bond do not absorb in the UV or visible regions- their ${\lambda }_{\max }$ is too much smaller. On the other hand a molecule that contain at least one pi bond have a greater ${\lambda }_{\max }$ than about nearly $\text{160 nm}$.

Number of conjugated C=C double bond are predicted using following table.

Conjugated $\text{C=C}$ bonds	UV ${\text{λ}}_{\text{max}}$	Change in ${\text{λ}}_{\text{max}}$
$\text{1}$	$\text{161}$ nm	-
$\text{2}$	$\text{217}$ nm	$\text{56}$ nm
$\text{3}$	$\text{274}$ nm	$\text{57}$ nm
$\text{4}$	$\text{290}$ nm	$\text{16}$ nm

Answer to Problem 15.32P

Four isomers consistent with the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}$ and longest UV-vis. absorption at $\text{215 nm}$ are as shown below:

Explanation of Solution

The given molecular formula is ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}$. A completely saturated molecule with $\text{5C}$ would have $\text{12H}$, but the given molecular formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}$ has $\text{6H}$, thus have IHD of $3$. The compound should have two conjugated bond, from the table $\text{15}\text{.1}$ compound with two conjugated $\text{π}$ bonds absorbs at $\text{217 nm}$ which is very close to the wavelength of given unknown compound.

From the table $\text{15}\text{.1}$ trend can be seen as follows:

Conjugated $\text{C=C}$ bonds	UV ${\text{λ}}_{\text{max}}$	Change in ${\text{λ}}_{\text{max}}$
$\text{1}$	$\text{161}$ nm	-
$\text{2}$	$\text{217}$ nm	$\text{56}$ nm
$\text{3}$	$\text{274}$ nm	$\text{57}$ nm
$\text{4}$	$\text{290}$ nm	$\text{16}$ nm

The two conjugated $\text{π}$ bonds account for as IHD $2$. To account IHD $1$, the compound must have another $\text{π}$ bond or a ring. If a third bond $\text{π}$ bond is added, it must not be conjugated to the other two; the added $\text{π}$ bond present as triple bond, nonconjugated double bond or a ring.

Therefore four isomers consistent with the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}$ and longest UV-vis. absorption at $\text{215 nm}$ are as shown below:

Conclusion

Isomers consistent with the formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}$ are drawn using given conditions.