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The following question was asked in an exam today:

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I chose the option (a), but the correct answer given was (b). Following is the explanation why I chose (b):

Step 1 (Reaction with $\ce{PBr_3}$): $\ce{Br}$ substitutes $\ce{OH}$. Hence, the product is $\ce{CH(CH_3)_2Br}$

Step 2 (Reaction with $\ce{Mg}$ in the presence of dry ether): $\ce{Mg}$ gets inserted between $\ce{CH(CH_3)_2}$ and $\ce{Br}$. Hence, the product is $\ce{CH(CH_3)_2MgBr}$

Step 3 (Opening of epoxide and attacking of grignard reagent): First, the $\ce{CH(CH_3)_2MgBr}$ will break into $\ce{CH(CH_3)_2^-}$ and $\ce{MgBr^+}$. Then, $\ce{MgBr^+}$ will attack $\ce{O}$ of the epoxide. After that, $\ce{O}$ will take its electron away from 2nd $\ce{C}$ (referring to the catalyst used) due to the greater stability of secondary carbocation.

Further, $\ce{CH(CH_3)_2^-}$ will attack the carbocation and hence form the product $\ce{CH(CH_3)_2-CH(CH_3)-CH_2-OMgBr}$

Final step (Hydrolysis): On hydrolysis, $\ce{CH(CH_3)_2-CH(CH_3)-CH_2-OH}$ will be our product. Phew!

Where I am wrong and what should be the correct mechanism?

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    $\begingroup$ Steric hindrance is the major factor $\endgroup$
    – Waylander
    Commented Jun 17, 2018 at 15:57
  • $\begingroup$ The carbocation stability argument isn't so valid under basic conditions as the extent to which the C-O bond is elongated in the TS is less significant. Sterics tend to dominate $\endgroup$
    – NotEvans.
    Commented Jun 17, 2018 at 16:51
  • $\begingroup$ @NotEvans., How do you know that the medium is basic? $\endgroup$
    – Rahul Verma
    Commented Jun 17, 2018 at 17:00
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    $\begingroup$ Grignard reagents are pretty basic... $\endgroup$
    – NotEvans.
    Commented Jun 17, 2018 at 17:13

3 Answers 3

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As you are dealing with Grignard reagents, the carbocation argument is inadequate since the reaction is in basic media (carbocations are brønsted acids, so they would be easily deprotonated to produce an alkene). In fact, this mechanism is concerted (everything happens simultaneously, not in 2 distinct steps such as in the formation of a carbocation followed by a nucleophilic attack).

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The Grignard reagent could attack either the methylene $\ce{(CH_2)}$ carbon or the methylidyne $\ce{(CH(CH_3))}$ carbon. The methylene carbon is attacked preferentially because its substituents are smaller than the methylidyne ones (2 hydrogens against one hydrogen and a methyl group). So steric hindrance is the key factor for predicting the product correctly in this question, not carbocation stability.

ps.: I drew a specific enantiomer just for visualization, but in this case, the absolute configuration doesn't matter.

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When you are using Grignard reagent for opening of epoxides, the Grignard reagent will attack less sterically hindered carbon(i.e. SN2 attack will take place). This is because when we use Grignard reagent, the reaction takes place in basic medium. Therefore in step 3 it will attack the 1st carbon instead of the 2nd one.

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Grignard reagents are generally bulky in nature therefore they choose to attack the less sterically hindered carbon which in this case is the methylene carbon . Furthermore you should remember that presence of acidic media will destroy the Grignard reagent . The carbocation stability is not considered only because it is a bronsted lowry acid.

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