I got the following exercise question on configurations, which the textbook does not have an explanation that matches this question, so could someone please tell me whether my attempt is correct or not:

My attempt:

A) both the (3R, 4R) and (3S, 4R) products are chiral since for each of them, there are chiral centers and they are not meso (no plane of symmetry observed).

B) since the alkyne anion preferentially attack carbonyl at re face i.e. attack from the bottom viewing from the figure of the question (top view) since the assigned rotation is anticlockwise, I get something like this after attacking (my thinking when drawing this: attack from 'bottom', then rotate the entire molecule such that the carbon linked to the bulky group is on the same plane as the alkyne group): the configuration of this (major product) is S. Moreover, the mixture is optically active since S:R = >50:<50 (not racemic mixture).

May I ask whether this approach is correct or not? And for B), is the reason that re face is attacked preferentially is because of the steric hindrance by CH3, which is out of the plane i.e. on the si face?

  • $\begingroup$ I don't think this is a real reaction. The pKa of the benzylic proton is around 15, it will just quench the acetylide anion (pKa 25), forming the enolate and racemising the chiral centre $\endgroup$
    – Waylander
    Oct 3 at 10:38
  • $\begingroup$ You need to consider the actual conformation of the aldehyde that reacts. You've assumed that the reacting conformation is the same as the one that is drawn in the textbook, but that's not necessarily true. Look up the Felkin–Anh model, for example. $\endgroup$
    – orthocresol
    Oct 3 at 12:06
  • $\begingroup$ @Waylander Oh you mean the acetylide anion is going to just grab the H+ bonded to the carbon which also bonds to CH3? (Is it called alpha-H?) And after this base-catalyzed racemization (I just known), and the reaction still take place by forming (3S, 4S) (3R, 4R) (3S, 4R) (3R, 4S) products, or any acetylene added will just grab the H+ again? $\endgroup$
    – Question
    Oct 3 at 14:32
  • $\begingroup$ Any acetylide will grab the H+ and form acetylene so you will have no nucleophilic attack on the aldehyde. When the reaction mixture is worked up the chiral centre will have racemised $\endgroup$
    – Waylander
    Oct 3 at 14:46
  • $\begingroup$ @Waylander's point is that racemic aldehyde will be the product should deprotonation prevail. No acetylene anion can add to the resultant enolate. Should addition occur as proposed, then follow orthocresol's lead. $\endgroup$
    – user55119
    Oct 3 at 15:05

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