The benzene ring alters the reactivity of a neighboring

QUESTION:

The benzene ring alters the reactivity of a neighboring group in the benzylic position much like a double bond alters the reactivity of groups in the allylic position.

\(H_{2} C=C H-C H_{2}-R\)

\(\mathrm{CH}_{2}\)

\(C H_{2}-R\)

Benzylic cations, anions, and radicals are all more stable than simple alkyl intermediates.

(a) Use resonance forms to show the delocalization (over four carbon atoms) of the positive charge, unpaired electron, and negative charge of the benzyl cation, radical, and anion.

(b) Toluene reacts with bromine in the presence of light to give benzyl bromide. Propose a mechanism for this reaction.

\(C H_{3}+B r_{2}\)

\(\mathrm{CH}_{2} \mathrm{Br}+\mathrm{HBr}\)

(c) Which of the following reactions will have the faster rate and give the better yield? Use a drawing of the transition state to explain your answer.

\(\mathrm{CH}_{2} \mathrm{Br}\)

\(\frac{\mathrm{NaOCH}_{3}}{\mathrm{CH}_{3} \mathrm{OH}}\)

\(\mathrm{CH}_{2} \mathrm{OCH}_{3}\)

Equation transcription:

Text transcription:

H{2} C=C H-C H{2}-R

{CH}{2}

C H{2}-R

C H{3}+B r{2}

{CH}{2}{Br}+{HBr}

{CH}{2}{Br}

frac{{NaOCH}{3}}{{CH}{3}{OH}}

{CH}{2}{OCH}{3}