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CH10136 > Nucleophilic Substitution-Leaving groups > Flashcards

Nucleophilic Substitution-Leaving groups Flashcards

1
Q

What factors affect leaving group ability

A
  1. Strength of C-X bond

2. Stability of the leaving group anion (X-)

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2
Q

What is needed from electrophilic substrates

A
  1. Need to be stable enough to make but reactive enough when treated with the required nucleophile
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3
Q

How can the stability of an anion be estimated

A
  1. Consider the pKA of the corresponding conjugate acid
  2. Strong acids readily dissociate to form stabilised anions
  3. A good leaving group is the conjugate base of a strong acid
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4
Q

What is the conjugate acid of X-

A
  1. HX
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5
Q

What is the conjugate base of H3O+

A
  1. H2O
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6
Q

Which of the halides is the best leaving group and why

A
  1. I- >Br->Cl->F-
  2. HF has the strongest C-X bond strength
  3. This means it is the least acidic
  4. HI has the weakest C-X bond strength so is easier to break= stronger acid
  5. HI has the lowest pKa
  6. HF has the highest pKa
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7
Q

Why can’t alcohols undergo direct substitution reactions

A
  1. The conjugate acid of the leaving group is H2O
  2. Water has a high pKa of 14
  3. It is a weak acid
  4. Therefore hydroxide is a poor leaving group
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8
Q

How can alcohols undergo nucleophilic substitution

A
  1. Must be activated by protonation using a strong acid e.g. HBr
  2. Now the leavin group is water and the conjugate acid is H3O+
  3. H3O+ has a low pKa and is a strong acid
  4. Therefore water is a good leaving group
  5. Useful for installing halides using HCl, HBr, HI
  6. Use of other nucleophiles is more difficult as they must be compatible with strong acid
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9
Q

What is another method that can activate alcohols

A
  1. Through the formation of a sulfonate ester

2. The sulfonate ester is a good leaving group that can undergo nucleophilic reactions

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10
Q

How can a sulfonate ester be formed

A
  1. Alcohols react with sulfonyl chlorides in the presence of base to form sulfonate esters
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11
Q

How can alcohols be converted to alkyl chlorides

A
  1. Using thionyl chloride (SOCl2)

2. OH attacks S and joins

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12
Q

Describe what a curly arrow mechanism would look like for the formation of an alkyl chloride from an alcohol

A
  1. Two sequential SN2 substitution reactions
  2. One at sulfur and the other at carbon
  3. OH group attacks S and curly arrow from s-cl bond to Cl
  4. Then in next step arrow from Cl- to carbon
  5. Curly arrow from C-O bond to O-S bond forming a double bond
  6. Curly arrow from S-Cl bond to Cl
  7. Forms R-C-Cl, SO2 and HCl
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13
Q

Why is the formation of alkyl chlorides from alcohols favourable

A
  1. Thermodynamically favourable due to formation of a strong S=O bond
  2. Entropically favourable due to formation of SO2 gas- increase entropy
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14
Q

Do ethers undergo substitution reactions

A
  1. No
  2. Alkoxides are poor leaving groups
  3. Conjugate acid of leaving group would be alcohol
  4. Weak acid so poor leaving group
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15
Q

How can ethers be made to undergo substitution

A
  1. The oxygen atom can be protonated with strong acid to make a better leaving group
  2. The leaving group is now an alcohol so conjugate acid is very acidic
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16
Q

What type of ethers can react without further activation

A
  1. Epoxides- cyclic ether
17
Q

Why can epoxides react without further activation

A
  1. Relief of ring strain makes epoxide ring-opening favourable
  2. Epoxide has- 3 x60 degree bond angles instead of the ideal 109.5
18
Q

Describe the mechanism for the epoxide ring opening

A
  1. Nucleophile attacks carbon next to oxygen
  2. curly arrow from C-O bond to O
  3. That opens the ring
  4. Next step is hydrogen is added to the O to form OH
19
Q

What happens with unsymmetrical epoxides

A
  1. Undergo regioselective ring-opening

2. Good nucleophiles at least sterically hindered end by SN2 mechanism

20
Q

Describe the stereoselectivity of epoxide ring-opening

A
  1. Inversion of stereochemistry as nucleophile approaches at 180 degrees from sigma*c-o
21
Q

How can you find the stereochemistry of epoxide ring-openign

A
  1. Draw the epoxide in plane of the paper
  2. Ring opening- inversion of symmetry- opposite to O
  3. Redraw product with longest carbon chain in the plain of the paper

CH10136 (30 decks)

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