substitution reactions summary with examples

      
                   

For each reaction below, determine whether the primary reaction is S_N1, S_N2, E1, or E2, and then draw the product.

Note: Me = methyl (CH₃)

problem set 1

SOLUTION

a) NaCN is charged! (Na⁺ and CN^-), so it's S_N2 or E2. CN is not a strong base, so it's S_N2.

b) KOtBu (potassium tert-butoxide) is charged, so it's S_N2 or E2. ^-OtBu is a strong base, so if anything is more bulky than 1º it will go E2. ^-OtBu is 3º, so it will definitely go E2 (KOtBu is a classic E2 reagent).

c) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. ^-OMe is 1º (actually, not even 1º), but the alkyl halide is 2º, so it will go E2.

d) NaOMe is charged so E2 or S_N2. NaOMe is a strong base, so if anything >1º it will go E2. But in this case there is no bulk whatsoever- nothing is >1º! NaOMe is 1 and the alkyl halide is also 1º, so it will go S_N2.

e) Methanol (MeOH) is neutral so probably E1 or S_N1. Methanol is a weak base and there's no bulk, so S_N1. In general water and alcohol do a mixture of S_N1 and E1 with alkyl halides (mostly S_N1).

f) H₂SO₄ is acidic so probably E1 or S_N1. Can't be S_N1 though because there is no nucleophile in H₂SO₄. (HSO₄^- is a very weak nucleophile). An alcohol with H₂SO₄ or H₃PO₄ is a dehydration reaction- E1.

g) H₂SO₄ is acidic so probably E1 or S_N1. In this case we have a nucleophile- Cl^-, so it will go S_N1.

h) Amines are neutral but they don't so S_N1/E1- they tend to go S_N2/E2, because they are basic (an amine solution has a basic pH). This amine is really bulky so it will go E2.

problem set 2

Rank the following compounds in order of decreasing reactivity with water (solvolysis). (1 = most reactive)

This is an S_N1 reaction. We know this because none of the reagents have charges (H₂O is neutral; if it were HO^-, it would probably be S_N2 or E2). So a carbocation will be formed in this reaction, and the compound whose carbocation is the most stable will react the fastest. So the 3º bromide will react fastest with water.

Problem # 3

Rank the following compounds in order of decreasing reactivity with NaI in acetone. (1 = most reactive)

This is an S_N2 reaction. We know this because NaI is an S_N2 reagent- charges give it away; when we see charges (Na⁺ is positive and I^- is negative) it's probably an S_N2 reaction. So we want reactants that are less substituted: methyl is more reactive than 1º, which is more reactive than 2º, etc. So methyl bromide reacts the fastest with NaI.

Problem #4

Rank the following electrophiles in order of decreasing reactivity with NaN₃ in DMF. (1 = most reactive)

Good leaving groups are stable. Larger ions tend to be more stable than smaller atoms (due to a smaller charge:size ratio. See problem 288), so when going down the periodic table, stability increases. I^- is more stable than Br^-, which is more stable than Cl^-, etc. So I^- is the best leaving group, and 2-iodobutane will react the fastest with a nucleophile.

Problem # 5

Indicate the reagents necessary to carry out each transformation.

For a), the wedge remains a wedge, so we have to do two S_N2 reactions (wedge to dash to wedge again). So we use PBr₃ to turn the OH into a Br. (Bromination of an alcohol with PBr₃ is an S_N2 reaction and so inverts stereochemistry).

For b), the wedge becomes a dash, so we can only do one SN2 reaction. So instead of using PBr₃ to make the OH a better leaving group, we use RSO₂Cl, which doesn't break the carbon-oxygen bond and so doesn't invert the stereochemistry.