Chapter 2.6, Problem 2.22P

Interpretation Introduction

Interpretation:

The given below compound has to be matched with the appropriate spectrum.

Concept Introduction:

To analyze an IR spectrum certain tools are needed.  The first step is to draw a line at $1500c{m}^{-1}.$  Then focus on any signals to the left of this line.  This is called the diagnostic region.  Try to identify the following regions:

Double bonds: $1600-1850c{m}^{-1}$

Triple bonds: $2100-2300c{m}^{-1}$

$X-H$ bonds: $2700-4000c{m}^{-1}$

While looking for $X-H$ bonds, draw a line at $3000c{m}^{-1}$ and look for signals that appear to the left of the line.

Explanation of Solution

Consider spectrum A.

There are no signals in the triple-bond region, but there is one signal in the double-bond region.  The signal at $1650c{m}^{-1}$ is narrow and weak, consistent with a $C=C$ bond.  Next, look for $X-H$ bonds.  Draw a line at $3000c{m}^{-1}$ and identify if there are any signals to the left of this line.

This spectrum exhibits one signal just above $3000c{m}^{-1}$, indicating a Vinylic $C—H$ bond.  The identification of a Vinylic $C—H$ bond is consistent with the observed $C=C$ signal present in the double-bond region $(1650c{m}^{-1}).$  There are no other signals above $3000c{m}^{-1}$, so the compound does not possess any $OH$ or $NH$ bonds.  Thus, spectrum A corresponds to the given below compound.

Consider spectrum B.

There are no signals in the triple-bond region, but there are signals in the double-bond region.  The signals between $1450-1600c{m}^{-1}$ are narrow and weak, consistent with aromatic $C=C$ bond.  The signal between $1700-1750c{m}^{-1}$ is strong, consistent with a carboxylic $C=O$ bond.  There is a broad signal between $2400-3600c{m}^{-1}$ which indicates the presence of carboxylic $-OH$ group.  Thus, spectrum B corresponds to the given below compound.

Consider spectrum C.

There are no signals in the triple-bond region, but there are signals in the double-bond region.  The signal at $1650c{m}^{-1}$ is narrow and weak, consistent with a $C=C$ bond.  The signal at $1720c{m}^{-1}$ is strong, consistent with a $C=O$ bond.  Draw a line at $3000c{m}^{-1}$ and identify if there are any signals to the left of this line.  There are no other signals above $3000c{m}^{-1}$, so the compound does not possess any $OH$ or $NH$ bonds.

The little bump between $3400$ and $3500c{m}^{-1}$ is not strong enough to be considered a signal.  These bumps are often observed in the spectra of compounds containing a $C=O$ bond.  The bump occurs at exactly twice the wavenumber of the $C=O$ signal, and is called an overtone of the $C=O$ signal.  Thus, spectrum C corresponds to the given below compound.

Consider spectrum D.

There are no signals in the triple-bond region, but there are signals in the double-bond region.  The signal at $1650c{m}^{-1}$ is narrow and weak, consistent with a $C=C$ bond.  .  Draw a line at $3000c{m}^{-1}$ and identify if there are any signals to the left of this line.  There a broad signal which has two peaks above $3000c{m}^{-1}$, so the compound possess a $N{H}_2$ group.  Thus, spectrum D corresponds to the given below compound.

Consider spectrum E.

There is a signal in the triple-bond region, but there are no signals in the double-bond region.  The signal at $2100c{m}^{-1}$ is narrow and weak, consistent with a $C\equiv C$ bond.  Draw a line at $3000c{m}^{-1}$ and identify if there are any signals to the left of this line.  The signal at $3300c{m}^{-1}$ is narrow and weak, consistent with a terminal alkyne hydrogen $\left(C\equiv C-H\right)$.  Thus, spectrum E corresponds to the given below compound.

Consider spectrum F.

There are no signals in the triple-bond region and also in the double-bond region.  Draw a line at $3000c{m}^{-1}$ and identify if there are any signals to the left of this line.  There is a broad signal between $3200-3600c{m}^{-1}$ which indicates the presence of an $-OH$ group.  Thus, spectrum F corresponds to the given below compound.