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This one was a bit involved. The alkyl halide turning into an alcohol with retention of stereochemistry suggested that an SN1 mechanism was at work. The addition of NaOH and the inversion of stereochem followed by "only" suggested again an SN2 mechanism. No E2 products; hydroxide isn't basic enough to deprotonate an alkane to a significant extent. The transformation of the alcohol into a nitrile group with inversion of stereochemistry is SN2; attacked with a non-protic solvent for good measure. The transformation into a nitrile group with retention of stereochem is SN1; treatment with tosyl chloride turns the -OH into a good leaving group.

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closed as unclear what you're asking by Martin - マーチン, tschoppi, ron, Klaus-Dieter Warzecha, Jannis Andreska Aug 27 '14 at 16:32

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Sorry, but I don't see what your question is? Are you asking whether your proposed reaction pathways are correct? $\endgroup$ – Philipp Aug 27 '14 at 2:34
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    $\begingroup$ @Dissenter Reactions involving an SN1 mechanism don't proceed with retention of configuration, rather both enantiomers are usually formed. $\endgroup$ – ron Aug 27 '14 at 2:46
  • $\begingroup$ Correct, I meant they proceed with both a retention and inversion. @Philipp - that is my question. $\endgroup$ – Dissenter Aug 27 '14 at 2:55
  • $\begingroup$ @Dissenter So in the TsCl cyanation shouldn't you also show the other enantiomer? $\endgroup$ – ron Aug 27 '14 at 2:59
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    $\begingroup$ Can you please rework your post to actually include a question and while you are doing that, give it a more descriptive title. $\endgroup$ – Martin - マーチン Aug 27 '14 at 3:27
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Since Greg E. focused on the first parts of the roadmap, I'll focus on the cyanation reactions.

What you have shown for the sequence from (R)-2-butanol to (R)-2-cyanobutane (retention of configuration), actually gives inversion of configuration. It is probably more clear if we look at this as two discrete steps. Step one is reaction with TsCl (and pyridine typically) to give the tosylate with retention. Step 2 is reaction with NaCN to give the cyano compound (S)-2-cyanobutane with inversion. Perhaps that can all be done in one step, but I've never seen it or tried it.

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Going from (R)-2-butanol to (S)-2-cyanobutane (inversion of configuration) by direct reaction with NaCN will not give the target compound. Hydroxide is not a suitable leaving group for SN2 reaction. The hydroxyl must be converted into a better leaving group first, such as the tosylate above.

That leaves the problem of how to synthesize (R)-2-cyanobutane (retention of configuration). Since the SN2 reaction with NaCN is going to invert configuration, you need a method of converting -OH into a leaving group with inversion. Two successive inversions give overall retention of configuration. The two most common ways taught for this transformation are SOCl2 (-OH to -Cl) and PBr3 (-OH to -Br). Both of these reagents install a halide with inversion of configuration, allowing SN2 reaction with NaCN to invert back to the original stereochemistry.

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    $\begingroup$ +1, very good answer, IMO. I do want to point out that if $\ce{SOCl2}$ is used, it's important to add pyridine, not just to neutralize evolving $\ce{HCl}$ but also because reaction of the pyridine with the $\ce{ROSOCl}$ intermediate prevents the reaction from taking place by $\mathrm{SN_i}$ instead. $\endgroup$ – Greg E. Aug 27 '14 at 16:18
  • $\begingroup$ @GregE. Thanks. I've never used SOCl2 for the reaction from an alcohol (only an acid), so I wasn't aware of that detail. $\endgroup$ – jerepierre Aug 27 '14 at 17:48
  • $\begingroup$ Yes, it's rarely mentioned in most textbooks I've seen and even many professional organic chemists I've encountered aren't aware of it. It's an uncommon mechanism; I'm aware of it only because I happened upon it mostly by accident while thumbing through March's Advanced Organic Chemistry some time ago. $\endgroup$ – Greg E. Aug 27 '14 at 17:57
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In theory, one can substitute a leaving group with retention of stereochemistry by performing two successive $\mathrm{S_N2}$ reactions. The first one inverts stereochemistry, and the second one inverts it again, thus restoring the original configuration. The Finkelstein reaction might be applicable: reaction of your 2-bromobutane with sodium iodide in acetone should yield 2-iodobutane via $\mathrm{S_N2}$, which can then be reacted with sodium hydroxide. Reaction of a secondary alkyl halide with $\ce{NaOH}$ will definitely yield competing elimination products, however. Note that, with secondary substrates, some competition will exist between $\mathrm{S_N1}$ and $\mathrm{S_N2}$, but the latter can be favored by use of an aprotic solvent and an excess of nucleophile.

Reaction with $\ce{H2O}$ is going to be slow and would yield a racemic mixture of $\mathrm{S_N1}$ products, as well as the competing $\mathrm{E1}$ product (with possibly some minor side-products from bimolecular pathways as well, depending on conditions). If you're satisfied with a racemic mixture, I would think elimination followed by acid-catalyzed hydration or oxymercuration-demercuration would be faster and more reliable in practice.

Cyanation with inversion can be done by direct reaction of your 2-bromobutane with, e.g., sodium cyanide. Cyanide is less basic basic than hydroxide, so competing eliminations are much less of a problem. If you want retention of configuration, replace the bromine first (as described above), then perform the substitution with $\ce{NaCN}$. The same ideas apply if starting from a tosylate instead.

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  • $\begingroup$ Oops, so the only part I erred on was thinking that hydroxide couldn't deprotonate the alkyl halide and give me some elimination products? $\endgroup$ – Dissenter Aug 27 '14 at 3:14
  • $\begingroup$ @Dissenter, yes, in general you would see a large proportion of elimination products. $\endgroup$ – Greg E. Aug 27 '14 at 3:26

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