4
$\begingroup$

I know that A is the product of a Knoevenagel condensation between the aldehyde and the diethyl malonate, due to the acidic protons of the latter, but I don't know what the reaction between that product and potassium cyanide would be.

enter image description here

$\endgroup$
  • 3
    $\begingroup$ The are only 2 possible sites of nucleophilic attack on A for CN-. One attack is completely reversible and represents a dead end, so it must be the other. $\endgroup$ – Waylander May 5 at 14:01
  • $\begingroup$ I'm guessing then that you're referring to a nucleophilic attack on the double bond, because the attack on the ester would be pointless. That was one of my first thoughts, but I discarded that option because I didn't know where to get the proton that was missing, maybe from the solvent? Thanks for your answer! $\endgroup$ – shadowsilvergold May 5 at 14:25
  • 1
    $\begingroup$ The 1,4 attack on the double bond is the pointless one. Think through attack on one of the ester carbonyls $\endgroup$ – Waylander May 5 at 14:30
  • 1
    $\begingroup$ @Zenix not as I see it. C is the cinnamic acid and D is the mixed anhydride. $\endgroup$ – Waylander May 5 at 14:47
  • 1
    $\begingroup$ I would suggest OP to edit the title of question. As it stands, the title looks vague. Be specific with the title. A title should always reflect the body of question. Check this meta post to see what to write in a title. $\endgroup$ – Nilay Ghosh May 6 at 15:07
5
$\begingroup$

Here is what I think is going on:

  1. $Step$ $1$: As the OP correctly identified is a Knoevenagel condensation to give A diethyl 4-cyclohexyl-benzalmalonate.

  2. $Step$ $2$ is an example of little-used ester hydrolysis using cyanide ion to give initially the acyl cyanide which is unstable in aq EtOH and gives the diacid B

  3. $Step$ $3$ is loss of one carboxy group by acid catalysed decarboxylation to give the cinnamic acid C

  4. $Step$ $4$ formation of the mixed anhydride D

  5. $Step$ $5$ cyclisation through formation of an aluminium-acyl complex literature example here to give the final product 6-cyclohexyl-Inden-1-one E

enter image description here

| improve this answer | |
$\endgroup$
  • $\begingroup$ Mechanism of step 5? $\endgroup$ – Zenix May 5 at 15:03
  • 1
    $\begingroup$ Formation of an acyl-aluminium complex similar to acyl halide/AlCl3 cyclisation. I've edited to include a reference. Not the way I would choose to do it (PPA would be my choice), but I think this question is trying to show alternative ways this could be done $\endgroup$ – Waylander May 5 at 15:13
4
$\begingroup$

@Waylander has provided a well-reasoned solution to the post. I offer a different interpretation. Knoevenagel product 1a has been converted in high yield to cyano ester 2a and subsequently hydrolyzed to phenylsuccinic acid (3a) 1. In the addition of cyanide to the double bond of 1a and protonation, one of the labile ester groups is cleaved by cyanide. (e.g., Diethyl malonate is readily monosaponified by base in the cold.) Cyclization of anhydride 4a has been accomplished by standard Friedel-Crafts methods but, like Waylander, I am a big fan of polyphosphoric acid (PPA). The cyclohexyl series b (R = c-C6H11) is likely to lead to ketoacid 5b.

enter image description here

1) C. F. H. Allen and H. B. Johnson, Organic Synthesis, Coll. Vol. 4, 804, 1963.; Vol. 30, 83, 1950.

| improve this answer | |
$\endgroup$

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.