# Reaction involving opening of epoxides

The following question was asked in an exam today:

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?

• Steric hindrance is the major factor – Waylander Jun 17 '18 at 15:57
• 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 – NotEvans. Jun 17 '18 at 16:51
• @NotEvans., How do you know that the medium is basic? – Rahul Verma Jun 17 '18 at 17:00
• Grignard reagents are pretty basic... – NotEvans. Jun 17 '18 at 17:13

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.