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MS 908 - SN1-SN2-E1-E2 Review

MS 908 - SN1-SN2-E1-E2 Review

Submitted by Matt on August 6, 2011.

Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 8

Keywords: leaving groups, nucleophilicity, SN1/SN2/E1/E2

Description: Goes through substitution and elimination competition reactions:

Nucleophile strength trends
Leaving group trends
Steric effects
How to tell the difference between SN1, SN2, E1, and E2 reaction conditions

Total Problems: 10

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Problem # 534

Using curved arrows, draw the mechanism for the S_N2 reaction below.
Solution
Arrows in organic chemistry always go from regions of high electron density to regions of low electron density. Most of the time this means arrows start from negative charges and go towards positive charges.

Because bromine is electronegative, the carbon directly bonded to it (also known as the alpha carbon) has a partial positive charge, and can be attacked by a nucleophile such as azide (N₃^-).

Because this is an S_N2 reaction, no carbocation is formed; as the nucleophile attacks the alpha carbon, the leaving group (Br^-) leaves.

Problem Details
MendelSet practice problem # 534 submitted by Matt on July 2, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Mechanism (show a mechanism using curved arrows..)
Keyword(s): SN2 mechanism
Problem # 535

Rank the following anions in order of decreasing stability (1 = most stable)
Solution
Charged molecules are generally less table than neutral ones, and each of the molecules below has a negatively charged oxygen.

But not all negative charges are equal; some oxygens are "closer to neutral" than others. How? Because resonance stabilizes charges by sharing electron density over multiple atoms (this is called electron delocalization).

Hydroxide (HO^-) doesn't have resonance, so the oxygen has 100% of the negative charge to itself. On the other hand, the sulfonate ester (SO₃R^-) has three resonance forms, so each oxygen only has ~33% of a negative charge. So the sulfonate ester is the most stable anion.

In general, the more resonance forms a molecule has, the more stable it is.

Problem Details
MendelSet practice problem # 535 submitted by Matt on July 2, 2011.
Subject: Organic Chemistry
Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): resonance
Problem # 536

Rank the following electrophiles in order of decreasing reactivity with NaN₃ in DMF. (1 = most reactive)
Solution
The only differences among these three molecules is their leaving groups, so whichever has the best leaving group will react fastest with the nucleophile (azide, N₃^-). Stable molecules make good leaving groups, so the best leaving group of the three is SO₃R^-, which has has three resonance forms (see problem 535). So the third compound will react the fastest with NaN₃.

Problem Details
MendelSet practice problem # 536 submitted by Matt on July 2, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): resonance, leaving groups
Problem # 537

Indicate the reagents necessary to carry out each transformation.
Solution
These are substitution reactions. In each case we're replacing the -OH with either -CN or -I, so the nucleophiles will be NaCN or NaI (S_N2 conditions).

Hydroxide (HO^-) is a poor leaving group, so the first step in each of these reactions will be to convert the alcohol into a better leaving group. Two good leaving groups are Br^- and SO₃R^-, but which one to use?

Every time an S_N2 reation takes place the wedge on the alpha carbon becomes a dash (and vice-versa).

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.

Problem Details
MendelSet practice problem # 537 submitted by Matt on July 2, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Reactions (predict the product, show the starting material..)
Keyword(s): SN2 mechanism, Walden inversion
Problem # 538

Rank the following electrophiles in order of decreasing reactivity with NaN₃ in DMF. (1 = most reactive)
Solution
This problem is similar to problem 536. The only difference among these compound is their leaving group, so the compound with the best leaving group will undergo substitution reactions most rapidly.

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 Details
MendelSet practice problem # 538 submitted by Matt on July 3, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): periodic trends, leaving groups
Problem # 539

Rank the following compounds in order of decreasing nucleophilicity. (1 = most nucleophilic)
Solution
Nucleophilicity increases with electron density; negatively charged molecules are more nucleophilic than neutral molecules. So HS^- is the best nucleophile in this group.

Nucleophilicity also increases with size. This is because the larger an atom, the more polarizable it is. Size increases as we go down the periodic table, so H₂S is a better nucleophile than H₂O.

Problem Details
MendelSet practice problem # 539 submitted by Matt on July 3, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): periodic trends, nucleophilicity
Problem # 540

Rank the following anions in order of decreasing stability (1 = most stable)
Solution
Each of these ions has a charge of -1, and none of them has any resonance forms. So the two things to consider are electronegativity and size. Electronegativity is relevant as we go from left to right across the periodic table. But here we are going up to down on the periodic table, so size is relavant. Larger ions are more stable than smaller ions (due to a smaller charge:size ratio), so I^- is the most stable of the group.

Problem Details
MendelSet practice problem # 540 submitted by Matt on July 3, 2011.
Subject: Organic Chemistry
Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): periodic trends, leaving groups
Problem # 541

Rank the following compounds in order of decreasing reactivity with NaI in acetone. (1 = most reactive)
Solution
The only differences among these compounds is the substitution of the alpha carbon (methly, 1º, 2º, or 3º).

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 Details
MendelSet practice problem # 541 submitted by Matt on July 3, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): SN2 mechanism
Problem # 542

Rank the following compounds in order of decreasing reactivity with water (solvolysis). (1 = most reactive)
Solution
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 Details
MendelSet practice problem # 542 submitted by Matt on July 3, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Concept (explain why this is so..)
Keyword(s): SN1 mechanism
Problem # 560

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)
Solution
Predicting S_N1/S_N2/E1/E2 competition reactions tends to drive students crazy, but it's not so bad once you notice the general pattern:

basic conditions (positive and negative charges) tend to go S_N2 or E2 (no carbocation)

neutral or acidic conditions tend to go S_N1 or E1 (carbocation is formed).

That's how you determine a SN1/E1 reaction from an SN2/E2 reaction. But how to decide between substitution or elimination? General things to watch for are bulk, nucleophilicity, and heat:

If you see heat (or Δ), the reaction will go elimination.

If you see a big, bulky compound, the reaction will go elmination.

If you see a strong base, the reaction will go elimination. Strong base is anything stronger than RO^-.

The exception: if everything is primary, it will probably go S_N2.

These rules probably seem confusing, so let's go through these eight examples and see how they apply.

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 Details
MendelSet practice problem # 560 submitted by Matt on July 7, 2011.
Subject: Organic Chemistry
Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions
Question Type: Reactions (predict the product, show the starting material..)
Keyword(s): SN1/SN2/E1/E2

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