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In response to this diagram (Organic Chemistry By David Klein 3rd Edition Page 364):

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The textbook stated

A comparison of the transition states for hydroboration via Markovnikov addition or anti-Markovnikov addition. The latter will be lower in energy because of decreased steric crowding.

Is it supposed to mean 'former', as the diagram obviously shows higher crowding in the latter transition state?

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  • $\begingroup$ Yes, you are correct. This would also be more helpful to future readers if you were to cite the textbook and a page number $\endgroup$ – orthocresol Apr 14 '18 at 0:22
  • $\begingroup$ @orthocresol Done $\endgroup$ – George Tian Apr 14 '18 at 0:42
  • $\begingroup$ I would argue that bond polarity has more to do with it. In Markovnikov's studies, he dealt with HX addition where H was always more positive than X. In the B-H bond, boron is more positive than H. Thus, I would classify the lefthand TS as Markovnikov-type addition. After the alkylborane oxidation, where O replaces B, the net result is anti-Markovnikov addition of water. $\endgroup$ – user55119 Apr 25 '18 at 18:30

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