CHE 230

University of Kentucky

Draw the product of substitution or elimination

A common question on an organic chemistry exam is, "Draw the product of this reaction." The following instructions should help you figure out how to solve this kind of problem when it involves substitution or elimination.

1. Identify the electrophilic C atom.

   An electrophilic C atom is one that is attached to a leaving group. Typical leaving groups are:

   • most often, I, Br, or Cl;
   • a sulfonate such as tosylate (OTs), mesylate (OMs), or triflate (OTf);
   • if the conditions are acidic (see below), OH, OR (R = Me, Et, etc.), or O₂CR such as OAc;
   • occasionally, a positively charged group that can break off to make a small, neutral molecule such as N₂, SMe₂, NMe₃, or OMe₂.

   The OH group is sometimes converted into a better leaving group in a first step before the substitution or elimination is executed:

   • treatment with TsCl, MsCl, or Tf₂O and a base (often pyridine) gives OTs, OMs, or OTf;
   • treatment with SOCl₂ or PBr₃ gives Cl or Br.

2. Identify the nucleophilic or basic atom, if there is one.

   • An atom that is attached to a metal is likely to be the nucleophilic or basic atom.

   • If no metal is present, a negatively charged atom that bears a lone pair is likely to be the nucleophilic or basic atom.

   • If no negatively charged atom is present, a neutral atom that bears an acidic H atom may be the nucleophilic atom, especially if it is the solvent or the conditions are basic. (E.g., NH₃, HBr, HCl, H₂O, EtOH, RSH, RCO₂H such as AcOH, RC≡CH, and a C atom adjacent to a C=O group.)

   • If no neutral heteroatom bearing an acidic H atom is present, a neutral heteroatom that bears a lone pair may be the nucleophilic or basic atom. (E.g., Ph₃P, Me₂S, Et₃N.)

   • A species that is present in only catalytic amounts cannot contain the nucleophilic or basic atom.

   • The O atoms in H₂SO₄ and in sulfonic acids such as TsOH, MsOH, and TfOH do not act as nucleophiles.

   • Solvents such as benzene, ether, THF, and CH₂Cl₂ are not usually nucleophilic.

3. Determine whether the reaction conditions are acidic or basic. (They may be neutral, but for purposes of this exercise, we count them as acidic or basic.)

   • Look for a good acid. The acid may be:
     • indeterminate, such as H⁺ or H₃O⁺;
     • a strong mineral acid such as HCl, HBr, or H₂SO₄;
     • a carboxylic acid such as AcOH, perhaps written as RCOOH or RCO₂H;
     • a sulfonic acid such as TsOH, MsOH, or TfOH;
     • an ammonium salt of any of the above, such as pyrH^{+ –}OTs or Et₃NH^{+ –}OAc.

     Note that neither water nor an alcohol ROH counts as an acid for the purposes of determining what the reaction conditions are!

   • Look for the presence of any of the following indicators of basic conditions:
     • Anions such as MeO^–, EtS^–, F^–, etc.;
     • Salts of alkali metals such as Li, Na, or K, or nonhalide salts of Mg such as CH₃MgBr;
     • Amines such as NH₃, NEt₃, i-Pr₂NEt, or pyridine;
     • Typical nonnucleophilic bases that may fall into any of these three categories, such as DBU (an amine), LDA (contains an N-Li bond), or t-BuOK.

     (Note: Alkali metal salts of the heavier halogens such as NaCl, LiI, and KBr can exist under acidic as well as basic conditions. However, if no strong acid is present, then you can act as if the conditions are basic.)

   • If your electrophile is an alkyl halide or alkyl sulfonate and it can generate a fairly stable carbocation (usually tertiary, sometimes secondary, or even primary if it is stabilized by resonance), the conditions are "stealth acidic". Strong acid is generated as the reaction proceeds, regardless of whether it is substitution or elimination.

   • If you see a molecule that contains a neutral, heavy congener of N or O with a lone pair, such as Ph₃P or Me₂S, the conditions can be considered basic.

4. Predict whether substitution or elimination will occur.

   • If the conditions are basic:

     A nucleophile or base is classified as:

     • a good nucleophile and a poor base if it is in the 2nd row of the main group or below (P, S, Cl, Br, I);
     • a good nucleophile and a good base if it is in the 1st row of the main group and fairly unhindered;
     • a poor nucleophile and a good base if it is in the 1st row of the main group and it is very hindered (t-BuOK, i-Pr₂NEt, LiN(i-Pr)₂).

     	good nuc, poor base	good nuc, good base	poor nuc, good base
     1° elec	SN2	SN2	E2, rarely SN2*
     2° elec	SN2	SN2 < E2**	E2
     3° elec	No reaction	E2	E2

     Notes:

     If elimination is to occur, there must be at least one H-bearing C adjacent to the electrophilic C.

     *Some very unhindered electrophiles, especially those that lack H-bearing C atoms adjacent to the electrophilic C (e.g., CH₃X and PhCH₂X), may undergo S_N2 reactions even with poor nucleophiles.

     **In the middle square, the balance of S_N2 and E2 can easily be tilted in one direction or the other depending on various factors.

     • Polar aprotic solvents such as DMSO and DMF promote substitution.
     • The less basic the nucleophile, the larger the proportion of substitution. So AcO^– gives more substitution product than HO^–, HC≡C^– more than H₂C=CH^–, and (EtO₂C)₂CH^– more than EtO₂CCH₂^–. The less basic forms of the nucleophile can often be converted into the more basic forms after the substitution reaction has been executed.
     • Compounds in which the electrophilic C atom is adjacent to a π bond, such as PhCH(CH₃)X or H₂C=CHCH(CH₃)X, tend to give more substitution.
     • Cycloalkyl halides tend to give more elimination.

   • If the conditions are acidic:

     The nucleophile has a high concentration if it is a protic solvent (H₂O, ROH, RCO₂H), if it is in the same molecule as the electrophile, or if it is noted as being in high concentration.

     	low [nuc]	high [nuc]
     1° elec	No reaction	SN2*
     2° elec	E1 (maybe rearrangement)**	SN1 (maybe rearrangement)**
     3° elec	E1	SN1

     Notes:

     If elimination is to occur, there must be at least one C adjacent to the electrophilic C that bears a H atom.

     *The S_N2 reaction occurs only under very strongly acidic conditions and very high concentrations of a nonbasic nucleophile, usually Cl^–, Br^–, or I^–. Typical conditions include 37% aq. HCl and 48% aq. HBr. The electrophile is usually an alcohol.

     **Whether rearrangements are expected to occur is difficult to predict. Normally you would not be expected to predict that a rearranged product is the major one.

5. Draw the substitution or elimination product, as appropriate.

   • If substitution occurs, erase the bond between the electrophilic C atom and the leaving group, and create a new bond between the electrophilic C atom and the nucleophilic atom.
     • If the nucleophilic atom bears a metal, the bond between the nucleophilic atom and the metal is also cleaved in the product, and the nucleophilic atom remains uncharged.
     • If the nucleophilic atom is negatively charged, it is neutral in the product.
     • If the nucleophilic atom bears an acidic H atom, the bond between the nucleophilic atom and the H atom is usually also cleaved in the product, and the nucleophilic atom remains uncharged. (Exception: if the nucleophilic atom is N, sometimes the N–H bond is retained in the product, and the N becomes positively charged.)
     • If the nucleophilic atom is a neutral, lone-pair-bearing atom that does not have an acidic H atom, it is positively charged in the product.
     • The electrophilic C atom's formal charge remains unchanged. The leaving group decreases its formal charge by one.
     • It is customary to draw only the product that contains the original electrophilic C atom.

     Furthermore, if the electrophilic C atom is stereogenic and its configuration is indicated with a bold or hashed bond:

     • If the reaction conditions are acidic (S_N1 mechanism), then the electrophilic C atom's configuration becomes scrambled in the product (squiggly bond).
     • If the reaction conditions are basic (S_N2 mechanism), then the electrophilic C atom's configuration becomes inverted in the product (bold to hashed or vice versa). If the nucleophile and the leaving group are assigned the same priority with respect to the other substituents on the electrophilic C atom, the configuration will change from R to S or vice versa.

   • If elimination occurs, draw in the H atoms on the electrophilic C atom and on all C atoms attached to it (Grossman's rule).

     • If only one C atom adjacent to the electrophilic C atom bears a H, then follow these steps:

       • Erase the bond between the electrophilic C atom and the leaving group, erase the bond between the adjacent H-bearing C atom and its H atom, and create a new π bond between the two aforementioned C atoms.

       • Does the alkene that you have drawn have the possibility of E/Z stereoisomerism? If so, then draw the lower-energy stereoisomer of the product unless all of these statements are true:
         • the conditions are basic (E2 mechanism);
         • the electrophilic C atom in the starting material is stereogenic; and,
         • the adjacent H-bearing C atom is also stereogenic.

         If all of these statements are true, determine whether the compound can achieve a conformation in which the C–X bond (X = leaving group) and the adjacent C–H bond are anti.

         • If so, draw the compound in this conformation. Groups that are gauche in this conformation become cis in the product.

         • If not, draw the compound in the conformation in which the C–X bond (X = leaving group) and the adjacent C–H bond are syn (eclipsed). Groups that are eclipsed in this conformation become cis in the product.

     • If more than one C atom adjacent to the electrophilic C atom bears a H:

       1. Choose which adjacent C atom will lose a H atom.

          • The most substituted product is usually obtained predominantly (Zaitsev's rule). Loss of H from the C atom with the fewest H atoms already attached will give the most substituted product.

          • If the leaving group is particularly poor (Me₃N⁺, OH, F, even Cl) and the base is strong and particularly hindered (t-BuOK, LDA), then the least substituted product may be obtained predominantly (Hofmann's rule). If so, then the adjacent C atom with the most H atoms already attached loses the H atom.

          • If an adjacent C atom is particularly acidic (for example, because it is adjacent to a carbonyl group), that C atom will lose the H atom.

       2. If any of these statements is false:
          • the conditions are basic (E2 mechanism);
          • the electrophilic C atom in the starting material is stereogenic; or,
          • the adjacent H-bearing C atom is also stereogenic;

          then erase the bond between the electrophilic C atom and the leaving group, erase the bond between the adjacent H-bearing C atom that you have chosen and its H atom, and create a new π bond between the two aforementioned C atoms. If there is a possibility of E/Z isomerism in the product, draw the lower-energy stereoisomer of the product.

          If, on the other hand, all of these statements are true, determine whether the compound can achieve a conformation in which the C–X bond (X = leaving group) and the adjacent C–H bond that you have identified are anti.

          • If so, draw the compound in this conformation. Groups that are gauche in this conformation become cis in the product, and groups that are anti become trans.
          • If the compound cannot achieve such a conformation, then identify the C atom adjacent to the electrophilic C that has the next fewest H atoms attached. If it is also stereogenic, go back to the three statements above.

     It is customary not to draw the product that contains the leaving group. If you do choose to draw it, the atom of the leaving group that was attached to the electrophilic C decreases its formal charge by one. However. if the conditions are acidic, and the leaving group is uncharged before it leaves, it may pick up an H atom before it leaves; if so, it is uncharged after it leaves.

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