Lecture Outline More Applications of Nucleophilic Substitution Chemistry

Name ______________________________________ Text section 10.5 deals with conversion of alcohols to alkyl halides. 1 Write the mechanism of the reaction below. OH H–I I 2 In Figure 10.3 (p 360), there is a disturbing pair of curved arrows that are meant to show the key connection of the alcohol O with the S of thionyl chloride. Those arrows seem to imply an SN2 reaction. Although that sulfur is tetrahedral (there's a lone pair on S!), that reaction probably doesn't go by a direct displacement. Let's clean that up and write a more plausible mechanism. Write out the following mechanistic steps below, being careful to keep track of all electron pairs and formal charges. (a) the O of iBuOH attack the S of thionyl chloride, as shown in fig 10.3, but Cl– is not ejected in this step. Instead a pair of π-e–s is pushed onto O. (b) a pair of e–s shifts in from O to reform the SO π-bond and expel chloride. (c) some base present in the solution deprotonates the O that came from the alcohol. Lecture outline More applications of nucleophilic substitution chemistry Conversion of alcohols to alkyl halides (and related rxns) R–OH ———> R–X not directly via SN1 or via SN2 chemistry — an –OH is not a leaving group — hydroxide, HO–, is much too strong a base to be a leaving group. In all the methods for carrying out this conversion, the –OH is first converted to a better leaving group, either as the first step of the mechanism or as the first experimental operation. a. Rxns with HX — After protonation of the hydroxyl group to make it a better leaving group, 2° and 3° alcohols (and others that can form reasonably stable carbocations) react via an SN1 mechanism. Carbocation formation is fast in these cases because no separation of charge is involved. Write the mechanism. H–Cl (CH3 )3 C OH (CH3 )3 C Cl — After protonation of the hydroxyl group, 1° alcohols react (slowly) via an SN2 mechanism. (As mentioned in the text, ZnX2 is usually added to make this go faster, but this is a minor experimental detail that we can ignore.) Write the mechanism. OH H–Br Br — under these conditions, β-branched 1° alcohols often react with rearrangement, e.g., Br H–Br OH This can't go via a 1° cation; instead, concerted loss of the leaving group and hydride shift lead directly to the 3° cation. This is essentially an internal SN2 — a displacement of the leaving group by migration of a hydride "nucleophile". This type of process is most common when (1) unassisted departure of the l.g. would lead to an unstable 1° cation, and (2) SN2 attack of a halide ion is slow due to steric hindrance (β-branching). H OH2 b. Rxns with S- or P-based halide reagents — SOCl2 (thionyl chloride), PBr3, PI3, etc. These rxns involve an SN2 step, so they only work with 1° and 2° ROH. e.g., PBr3 R OH R Br Br2P–Br Br– Br R O P SN2 Br H Of course, the SN2ness has stereochemical consequences, e.g. OH SOCl2 c. Rxns via sulfonate ester intermediates (2-step method) — this allows introduction of any nucleophile, not just halogen. The reaction of alcohol and sulfonyl chloride was introduced back in Ch 8 (p 283), but was shown with a "don't take me seriously — I'm just a pointing to the important new connection, not representing a mechanism, so I shouldn't really be drawn to look like a curved arrow" arrow. This actually happens by a mechanism just like the one you wrote in the pre-lecture assignment. Because the reaction of an alcohol with a sulfonyl chloride generates HCl as a byproduct, a base (e.g., Et3N or pyridine) is usually added to soak up the HCl before it can do any mischief. Let's illustrate this 2-step conversion by writing (a) the overall reaction of (R)-4-methyl-2pentanol and methanesulfonyl chloride (mesyl chloride), followed by reaction of the mesylate with a nice nucleophile like NaCN. OH One more example: (S)-2-pentanol and p-toluenesulfonyl chloride (tosyl chloride); then NaOH in DMSO solvent. Synthetically useful hydride and carbanion nucleophiles (a) Nucleophilic hydride reagents The most obvious source of H:– is probably NaH or something similar. But NaH is insoluble in just about every organic solvent. Solubility is not an issue for acid-base chemistry — an acid can be deprotonated by NaH if it comes in contact with the surface — but the H:– has to be in solution to act as a nucleophile. Two nucleophilic hydride sources are LiAlH4 and NaBH4. Being sources of hydride, these reagents are strongly basic. Both react with water — LiAlH4 violently; NaBH4 more slowly — as well as alcohols and other moderately weak acids. As nucleophiles, they will deliver an "H:–" and displace a leaving group. Br O NaBH4 Et2O LiAlD4 OTs Et2O (b) Acetylide ions are easy to make by deprotonation of terminal alkynes (alkynes with an H on a triple-bond C). These have pKas of about 25, so a strong base is needed to deprotonate them. This SN2 reaction is a great way to make new CC bonds and as such is one important tool that synthetic chemists can use to build molecules from smaller, commercially available pieces. Ph C C H NaNH2 CH3—I liquid NH3 (solvent) Why is it important to do this in two separate steps rather than just tossing the base and alkyl halide in together? (hint: think chemistry, not alphabet soup) How would you use this chemistry to synthesize 7-methyl-3-octyne from acetylene, HCCH?