1:14 Knowing that one of the starting materials in the (E)-stilbene synthesis re

Question

1:14 Knowing that one of the starting materials in the (E)-stilbene synthesis reaction of chloride, which combination of alkyl halide (reacted with compound (and base) would be best for making (E)-4-methylhept-3-ene by the Wittig reaction? Hint: Which combination would be best for the S2 Please note that Wittig reactions done in aprotic solvents and by removing a salt byproduct actually can produce Z 2 a. 1-iodopropane and pentan-2 b. 1-iodobutane and butanone C. 2-iodobutane and butanal d. 2-chlorobutane and butanal f. 1-chlorobutane and butanone g. 1-chloropropane and pentan- 2-one h. 2-iodopentane and propanal

Explanation / Answer

The correct answer is option g.

Explanation: There are two steps in the formation of the alkene. In the first step, 1-chloropropane is more reactive in SN2 reaction towards PPh3 than 1-iodopropane. For SN2 reaction, the alkyl halide should be a less hindered one.

Hinderance: primary alkyl halide > secondary alkyl halide

Therefore, the correct answer is 1-chloropropane and pentan-2-one (option g).

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