Knowing that one of the starting materials in the (E)-stilbene synthesis was mad

Question

Knowing that one of the starting materials in the (E)-stilbene synthesis was made from the SN2 reaction of triphenylphosphine and benzyl chloride, which combination of alkyl halide (reacted with triphenylphosphine) and carbonyl 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 SN2 reaction?

Please note that Wittig reactions done in aprotic solvents and by removing a salt byproduct actually can produce Z isomers.

Select one:

a. 2-chlorobutane and butanal

b. 2-iodopentane and propanal

c. 1-iodobutane and butanone

d. 2-chloropentane and propanal

e. 2-iodobutane and butanal

f. 1-iodopropane and pentan-2-one

g. 1-chloropropane and pentan-2-one

h. 1-chlorobutane and butanone

Explanation / Answer

Order of alkyl halide towards SN2 :1°>2 °>3 °

Hence primary halide are better for wittig reagent and Iodine is good leaving group in SN2 reaction.

Hence option F is suitable. Since it has 1-iodopropane and 2-pentanone gives E-4-methyl-3-heptene

Eventhough aldehyde is more reactive than ketone in B, but it is 2°halide, less reactive in SN2 reactions

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