Radical addition to alkenes is not effective for the synthesis of iodo- and chloroalkanes. Using your knowledge of the mechanism of this reaction, along with bond dissociation energies, explain why the radical additions of HI and HCl are not effective. (Assume ∆H = 65 kcal/ mol for the C–C π bond.)

A student adds NBS to a solution of 1-methylcyclohexene and irradiates the mixture with a sunlamp until all the NBS has reacted. After a careful distillation, the product mixture contains two major products of formula C₇H₁₁Br.

a. Draw the resonance forms of the three possible allylic free radical intermediates.

For each compound, predict the major product of free-radical bromination. Remember that bromination is highly selective, and only the most stable radical will be formed.

(a) cyclohexane

(b) methylcyclopentane

Using the bond-dissociation energies in Table 5.6,

(a) predict whether or not an iodine radical would be selective for forming a single radical propane.

What is the major product of the reaction in Problem 7 when the alkane reacts with Cl₂ instead of with Br₂? Disregard stereoisomers.

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

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

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

Show how the following compounds could be prepared from 2-methylpropane:

a. 2-bromo-2-methylpropane 

Draw the products of the following reactions, including all stereoisomers:

d.

When a student attempted a bromination to produce compound A, they generated compound B instead. Rationalize the formation of B using the arrow-pushing formalism.