Question

Draw the structure of the product of each step in the following three-step synthesis. Show the formal charges, if applicable. As a start, the benzene ring is drawn for you in each product.

Although the first step produces a mixture of isomers, the para isomer is isolated as the sole product of interest and used in the second step. Give only the major product for the second and third steps.

Answer 1

The chemical behavior of neopentyl bromide, 2,2-dimethyl-1-bromopropane, is an instructive place to begin this discussion. The very low SN2 reactivity of this 1º-bromide was noted earlier, and explained by steric hindrance to the required 180º alignment of reacting orbitals. Under conditions that favor SN1 reactivity, such as solution in wet formic acid, neopentyl bromide reacts at roughly the same rate as ethyl bromide. Both of these compounds are 1º-alkyl halides, and for an SN1 reaction the rate determining step requires ionization to a 1º-carbocation. As noted in the carbocation stability order shown below, such carbocations are relatively unstable and are formed slowly. The product from ethyl bromide is ethanol, the simple and direct substitution product, but neopentyl bromide yields 2-methyl-2-butanol instead of the expected neopentyl alcohol. A change in the way the five carbon atoms in this product are bonded to each other has clearly taken place.

Answer 2

(para-bromo toluene) +HNO3/H2SO4= (4-bromo-2-nitrotoluene)+Br2/FeBr3= 2,4-dibromo-6-nitrotoluene

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