The reactions among the following which results in the formation of a pair of diastereomers are:

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I am aware of the fact that $\ce{HBr}$ gives anti addition product and $\ce{BH3}$ forms syn addition product. In part (a) since the back side is less sterically hindered ( due to presence of only $\ce{H}$ ) $\ce{BR-}$ preferentially attacks from the back side.Similar is the case with (b).c) proceeds with a free radical mechanism.d) however occurs via anti-markinicoff addition so that $\ce{OH}$ adds to the primary carbon.Since there are two possible ways in which the $\ce{OH}$ group may add( either from the back side or the front side)my idea is that it may result in a pair of diastereomers. However I am not quite sure about it. Thanks.

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    $\begingroup$ Consider that, in order to form diastereoisomers here, you need to form a new chiral center. So which of these reactions will generate a new chiral center in the product? $\endgroup$ – MarcoB Jan 4 '17 at 8:48
  • $\begingroup$ @MarcoB that implies that the correct answer should be b). $\endgroup$ – Pink Jan 4 '17 at 8:53
  • $\begingroup$ @aniline To quote the Lion King: ‘Look harder!’ $\endgroup$ – Jan Jan 4 '17 at 23:29

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