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Would H2, Pd/C be a sufficient reagent for this reaction to proceed?

The question is from a test which asked what would be the most suitable reagent for the above reaction

Patrick Schick
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    LiAlH4 plus a lanthanide salt is the way to go. I don't think any of these conditions are good options see https://pubs.rsc.org/en/content/articlelanding/1986/p1/p19860001929 – Waylander Nov 26 '23 at 18:41
  • I faced the same question today in the test! Think it should be option b. – Harikrishnan M Nov 26 '23 at 18:54
  • @HarikrishnanM Why would you guys use a reagent good for C=C when you got C=O reduction? – Mithoron Nov 26 '23 at 18:57
  • @HarikrishnanM I don't think simple hydrogenation with Pd/C will give you the allylic alcohol. See this answer https://chemistry.stackexchange.com/questions/19628/enones-and-hydrogenation – Waylander Nov 26 '23 at 19:12
  • I checked the answer key given by a famous coaching institute. It is given as 'b' option as $\ce{LiAlH4}$ is a very strong reducing agent and reduces everything it finds. Thus $\ce{H2 / Pd + C}$ is the way to go – Harikrishnan M Nov 27 '23 at 03:05
  • Also allylic alcohols are more stable because of the resonance effect and thus it is justified. – Harikrishnan M Nov 27 '23 at 03:05
  • Zn-Hg / HCl is the reagent for Clemmensen reduction and will remove the =O group altogether hence it's not suitable.

    Na / liq. NH₃ makes it undergo Birch reduction and reduces the aromatic ring

     – Harikrishnan M Nov 27 '23 at 03:08
  • By the way, this is the National Standard Examination in Chemistry, the 1st part of the qualifier to int'l olympiads in India. – Harikrishnan M Nov 27 '23 at 03:16

2 Answers2

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As with B option, H2/Pd will also reduce the double which we don't want.. C, LiAlH4 does not reduce double bond( exception incase of cinnamic acid as double bond is in conjugation with phenyl ring as well as carbonyl carbon) and will reduce ketone to alcohol. Hence Option 3 is most accurate.

Innocat
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Answer is option (c) according to the answer key uploaded by the IAPT agency.

$\ce{Na / liq. NH₃}$ is the reagent of Birch reduction and will reduce the aromatic ring. So, it's not the right way.

$\ce{Zn-Hg / HCl}$ will result in Clemmensen reduction and will completely remove the carbonyl oxygen. We don't want that.

$\ce{LiAlH4}$ looks a good enough reagent to go about this reaction as it correctly reduces the ketone group and also gives our desired product.

$\ce{H2 + Pd}$ will lead to hydrogenation, as pointed out rightly by some members.

Thus it is (c).

Harikrishnan M
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    Big difference between those people who write exam questions and those who actually run reactions every day. This answer is incorrect. Hydrogenation under these conditions reduces the double bond first. – Waylander Nov 27 '23 at 06:58
  • I don't think IAPT judges purely based on the answer key. Surely going to challenge this question if isn't already 'deleted' or given as a bonus – Patrick Schick Nov 27 '23 at 07:50
  • Let us wait then... – Harikrishnan M Nov 27 '23 at 07:58
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    @Harikrishnan M: I see that you are a 12th grader; I admire your enthusiasm. You have much to learn. From my personal experience, hydrogenation with Pd/C will give the saturated alcohol. $\ce{LiAlH4}$ is the best solution as long as it doesn't react like cinnamaldehyde with $\ce{LiAlH4}$ and give the saturated alcohol. ....continued.... – user55119 Nov 27 '23 at 20:54
  • ....continued....Na/$\ce{NH3}$ will afford the saturated ketone upon protonation of the enolate of the ketone. There will be no Birch reduction because there is no alcohol present. Too bad the constructor of the question didn't offer $\ce{NaBH4}$ or $\ce{LiA(O-t-But)3H}$ as. options. – user55119 Nov 27 '23 at 20:57
  • @user55119 I also was looking for $\ce{NaBH4}$... on not finding it I was pretty confused – Harikrishnan M Nov 28 '23 at 01:09
  • And the fact that it was emphasised to me lot of times to sparingly use $\ce{LiAlH4}$ as it was very strong and reduces everything – Harikrishnan M Nov 28 '23 at 01:28
  • The answer key has been released, and I have updated my answer accordingly. – Harikrishnan M Dec 19 '23 at 11:36