4
$\begingroup$

I would like to know the method of syntheses of making 1,1-diphenyl-1-butene $(\ce{C16H16})$ from benzophenone $(\ce{(C6H5)2CO},$ generally abbreviated $\ce{Ph2CO})$ and 1-bromopropane $(\ce{CH3CH2CH2Br}).$

I think that the carbonyl group which is a functional group composed of a carbon atom double-bonded to an oxygen atom $(\ce{C=O})$ of benzophenone will react to the alkyl halide.

Because oxygen is more electronegative than carbon, carbonyl compounds often have resonance structures which affect their reactivity. This relative electronegativity draws electron density away from carbon, increasing the bond's polarity, therefore making carbon an electrophile (i.e. slightly positive).

But I have no idea how it reacts and proceeds to work. Can anyone help me understand the detail mechanisms and reaction procedures to make 1,1-diphenyl-1-butene from benzophenone and 1-bromopropane?

$\endgroup$
2
$\begingroup$

I present another (almost) two-step synthesis, involving the famous Wittig reaction. The first step details the preparation of a phosphonium ylide, which subsequently is allowed to react with the carbonyl compound (benzophenone in our case) to yield the final product, 1,1-diphenyl-1-butene.

Wittig: benzophenone and n-propyl triphenyl phosphonium ion

For mechanistic details, visit NotEvans.'s answer to Which is the currently accepted mechanism of a Wittig reaction?.

$\endgroup$
  • $\begingroup$ Thank you for useful information about Wittig reaction. $\endgroup$ – Jan Jul 17 at 15:47
  • $\begingroup$ But I can’t understand well about this action procedures. $\endgroup$ – Jan Jul 17 at 15:48
  • $\begingroup$ @Jan Do you mean mechanism? $\endgroup$ – William R. Ebenezer Jul 17 at 15:49
  • $\begingroup$ Try this in that case: masterorganicchemistry.com/2018/02/06/wittig-reaction $\endgroup$ – William R. Ebenezer Jul 17 at 15:51
  • $\begingroup$ Thank you so much for your information! I can understand the mechanism of Wittig reaction that ylide attacks aldehyde or ketone and makes alkene and triphenylphosphine oxyde. $\endgroup$ – Jan Jul 20 at 14:40
6
$\begingroup$

Two steps:

  1. Form the Grignard reagent with 1-bromopropane and magnesium metal. This can be down in a variety of ethereal solvents, THF or $\ce{Et2O}$ are most commonly used. This species is nucleophilic through carbon and will add to the carbonyl group.

    Cool the Grignard solution in an ice bath under nitrogen with stirring, slowly add a solution of 0.9 eq of benzophenone. Allow it to stir until TLC shows consumption of the benzophenone starting material.

    Add a solution of aqueous ammonium chloride to quench residual Grignard. Add organic solvent $(\ce{Et2O}$ or $\ce{EtOAc})$ and collect the organic phase, dry and concentrate. This will form 1,1-diphenyl-butan-1-ol.

  2. Take the crude product from step 1 and dissolve in toluene. Add a catalytic amount of p-toluene sulfonic acid (PTSA) and heat under Dean-Stark reflux to remove the water. The PTSA will protonate the tertiary alcohol you have formed in step 1. This gives a cation stabilised by the two phenyl groups so it forms readily. This eliminates water to give the required product cleanly.

    When the system has ceased to generate water allow it to cool, wash with diluted aqueous sodium carbonate solution, dry and concentrate. This should result in fairly clean 1,1-diphenyl-1-butene.

    To produce cleaner material use chromatography on silica with hexane/ethyl acetate (0 to 10%).

Further reading on Grignard reagents: Master Organic Chemistry.

$\endgroup$
  • 1
    $\begingroup$ Step 2, why is conc H2SO4 not preferred? $\endgroup$ – thewitness Jul 13 at 7:37
  • 2
    $\begingroup$ PTSA is more soluble in Toluene and not oxidising $\endgroup$ – Waylander Jul 13 at 7:41
  • $\begingroup$ I can understand fairly well about Grignard agent and its detail reaction procedures $\endgroup$ – Jan Jul 14 at 9:36
  • $\begingroup$ Thank so much for your great teaching! I can study from this site organic chemistry fairly well! $\endgroup$ – Jan Jul 14 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.