Question

Question: Which is a more stable intermediate in each pair?

a.

b.

Short answer

The compound for graph a is 2,2-dimethyl-1-phenylpropane.

The compound for graph b is 1-tert-butyl-4-methylbenzene.

Step-by-step solution

Activator or deactivator

The groups which are electron donating are called activators and the groups which are electron withdrawing are called deactivators.

Activators increase reactivity and deactivators decrease reactivity.

Effect of substituents on orientation

All activating substituents and the weakly deactivating halogens present on ring are ortho–para directors, and all substituents that are more deactivating than the halogens are meta directors for electrophile.

Part ‘a’

The two molecules given are only different for one substitution i.e OH and CH₃. The methyl group donates electron by + I effect and is a weak activator while OH donate electron by resonance and is a strong activator.

As NO₂ is present on ring, which is an electron withdrawing group so ring is electron deficient and for it to be stable more electron donating or activator group is needed.

So the more stable intermediate is:

More stable intermediate

Part ‘b’

Here NO₂ group is present in both molecules given are only difference is in substitution position. As we know NO₂ is meta-director so it doesn’t show its electron withdrawing effect on meta-position.

Here OH and Cl are present on same carbon so it can’t participate in resonance as carbon has complete octet and atoms can’t change position in resonance so the negative charge must be present at para or ortho position for stabilizing NO₂ .

So the more stable intermediate is:

More stable intermediate