Question: How do the IR spectra of the isomers cyclopentane and pent-1-ene differ?

IR analysis can be used for the identification of hydrocarbons.

Although all hydrocarbons have carbon-carbon single bonds, the peak due to absorption by a $\mathbf{C}\mathbf{=}\mathbf{C}$double bond or $\mathbf{C}\mathbf{\equiv }\mathbf{C}$triple bond can be used to distinguish between hydrocarbons.

Therefore, IR spectra of alkenes and alkynes will contain an extra peak for the double bond and the triple bond, respectively.

The position of the peak due to the absorption of the carbon-hydrogen bond depends on the hybridization of the corresponding carbon atom.

${\mathrm{sp}}^3$hybridized carbon-hydrogen bonds cause absorption at less than $3000{\mathrm{cm}}^{-1}$. These absorptions appear as broad peaks in the spectrum.

${\mathrm{sp}}^2$and sp hybridized carbon produces peaks absorbed at higher frequencies.

Cyclopentane and pent-1-ene are examples of structural isomers. Therefore, both are composed of $\mathrm{C}-\mathrm{C}$single bonds and $\mathrm{C}-\mathrm{H}$bonds.

Pent-1-ene has a $\mathrm{C}=\mathrm{C}$double bond. This will give the IR spectrum of pent-1-ene an additional peak at $1650{\mathrm{cm}}^{-1}$. Since there are no double bonds in cyclopentane, this peak will be absent in the IR spectrum of cyclopentane.

Pent-1-ene will also show a $\mathrm{C}-\mathrm{H}$absorptions for ${\mathrm{sp}}^2$hybridized carbon at role="math" localid="1649146543231" $3150-3000{\mathrm{cm}}^{-1}$ because the double-bonded carbon atoms are hybridized.