Animated Solution for Chemistry - Organic Chemistry: Which of the following compounds does not exhibit resonance ?

Welcome to the fascinating world of organic chemistry, where electrons don't always stay put! Today, we are tackling a classic problem: identifying which molecule out of a given set does not exhibit resonance.

Resonance is a stabilizing phenomenon where $\pi$ electrons are delocalized across multiple adjacent atoms. But for this magical delocalization to occur, the molecule must possess a specific structural feature known as a conjugated system.

Spotting Conjugation

What exactly is a conjugated system? Think of it as a continuous highway of overlapping p-orbitals. If the highway is unbroken, electrons can travel freely. You can easily spot this highway if you see alternating double and single bonds (like in a benzene ring), or a double bond separated by exactly one single bond from a lone pair of electrons, a radical, or a positive charge.

If an atom in the chain is $s{p}^3$ hybridized (meaning it has only single bonds and no unhybridized p-orbitals), it acts as a massive roadblock, completely shattering the conjugation chain.

Options A and B

Let's examine our first contender, option (a): ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OCH}={\mathrm{CH}}_2$. Look closely at the oxygen atom. It possesses lone pairs of electrons. Right next to it, separated by just one single bond, is a carbon-carbon double bond. This is a perfect lone pair-$\sigma$-$\pi$ conjugated system! The p-orbitals overlap beautifully, meaning this molecule definitely exhibits resonance.

Now for option (b), which is benzyl alcohol. While the $-{\mathrm{CH}}_2\mathrm{OH}$ group itself isn't conjugated with the ring (thanks to that $s{p}^3$ hybridized $-{\mathrm{CH}}_2-$ carbon), the benzene ring itself is the absolute poster child for resonance! It features a continuous loop of alternating single and double bonds. So, resonance is undeniably happening within the ring.

Options C and D

Moving on to option (c): ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CONH}}_2$. Here we have an amide group. The nitrogen atom has a lone pair, and it is separated by exactly one single bond from the carbon-oxygen double bond ($\mathrm{C}=\mathrm{O}$). Once again, the p-orbitals can overlap, allowing the lone pair to delocalize towards the electronegative oxygen. This molecule exhibits resonance.

Finally, let's look at option (d): ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CH}={\mathrm{CHCH}}_2{\mathrm{NH}}_2$. We have a carbon-carbon double bond, and we have a nitrogen atom with a lone pair. But wait! Look at what lies between them. There is a $-{\mathrm{CH}}_2-$ group. That means there are two single bonds separating the double bond and the lone pair.

Because of that $s{p}^3$ hybridized $-{\mathrm{CH}}_2-$ carbon in the middle, the p-orbital highway is broken! The electrons on the nitrogen cannot delocalize into the double bond. Therefore, this compound lacks a conjugated system and does not exhibit resonance.

Option (d) is our correct answer! Always remember to scan your molecules for those sneaky $s{p}^3$ roadblocks when hunting for resonance.