(A Level Chem) Nucleophilic substitutions and eliminations

Let's try to summarise nucleophilic substitutions (S_N1, S_N2) and eliminations.

In nucleophilic substitution reactions, the leaving group (e.g. Br) is replaced by a nucleophile (e.g. OH^-, CN^-, CH₃O^-, NH₃):

e.g. CH₃CH₂-Br + 2NH₃ --> CH₃CH₂-NH₂ amine + NH₄⁺Br^-

e.g. CH₃CH₂CH₂CH₂-Br + OH^- --> CH₃CH₂CH₂CH₂-OH alcohol + Br^-

e.g. CH₃CH₂-Br + CN^- --> CH₃CH₂-CN nitrile + Br^-

In S_N1 and S_N2, alkyl halides like R-Br are made into alcohols R-OH, amines R-NH₂ or nitriles R-CN...

Note: CH₃CH₂-CN can be hydrolysed by aqueous H₂SO₄ to CH₃CH₂-COOH or reduced by LiAlH₄ in dry ether to CH₃CH₂-CH₂NH₂. Therefore, to get CH₃CH₂-COOH or CH₃CH₂CH₂NH₂ from CH₃CH₂Br, you got to first add in one more C atom in the chain by heating with NaCN in ethanol. 

 

The above nucleophilic substitution examples involves S_N2 (bimolecular). The substrates are 1º implying less steric hindrance on nucleophile. S_N2 can also occur for 2º substrates. This allows backside attack of the nucleophile. S_N2 reactions are concerted, meaning it has only one step. This step involves a transition state with the nucleophile backside attacking the central carbon that's attached to the leaving group.

If the substrates are 2º or 3º, steric hindrance will prevent S_N2 and so S_N1 (unimolecular) occurs. S_N1 reactions involve two steps:

1. (This is the rate-determining step) Formation of a 2º or 3º trigonal planar carbocation intermediate (not a transition). (1º carbocation is unstable and so 1º alkyl halides don't undergo S_N1.)
   
2. Attack of the nucleophile on either side of the trigonal planar carbocation with equal probability. This results in a mixture of two enantiomers (mirror image stereoisomers). We say that racemisation has occurred.
   

e.g. C₂H₅CH₃CH-Br + CH₃O-Na⁺ --> C₂H₅CH₃CH-OCH₃ + Na⁺Br^-
(with two enantiomers)

In elimination reactions, the leaving group (e.g. Br) and ßH are removed and C=C π bond is formed. We get alkenes from halogenoalkanes! Bear in mind that the elimination of HBr from a halogenoalkane can result in a number of isomers including cis-trans isomers.

In elimination reactions, alkyl halides become alkenes...

Example 1 (GCE A Level N2008/P3)
The molecules of compound P C₇H₁₅Br are chiral. On treatment with NaOH(aq), P produces alcohol Q, C₇H₁₅OH, which does not react with hot acidified Na₂Cr₂O₇(aq). The elimination of HBr from P produces a mixture of four different isomeric alkenes with the formula C₇H₁₄. Only two are geometrical isomers of each other. Suggest the structural formulas of P and the four alkenes. 
Explanation
On treatment with NaOH(aq), C₇H₁₅Br (P) becomes C₇H₁₅OH (Q). This is a nucleophilic substitution.
P is chiral. ==> All the groups attached to the central carbon are different. P is either 2º or 3º.
Since Q can't be oxidized by Na₂Cr₂O₇/H⁺. ==> P is a 3º alcohol. The central carbon is attached to 3 other carbons.
Trial 1: CH₃-CH₂-CH₂-CBr(CH₃)-CH₂-CH₃ giving us 5 isomers with 2 pairs of geometric (cis-trans) isomers.
Trial 2: (CH₃)₂CHCBr(CH₃)CH₂CH₃ giving us 4 isomers with 1 pair of geometric (cis-trans) isomers.

Example 2 (GCE A Level N2009/P3)
The compound CH₂Br(CH₂)₃CHBr(CH₂)₂CH₃ is reacted with NaOH in ethanol. A mixture of four isomeric alkenes with the molecular formula C₈H₁₄ is formed. Suggest the structures of the four isomers. 
Explanation
This is 1,5-dibromo-octane. The bromine at C1 would be eliminated leading to H₂C=C-... The elimination of the bromine at the C5 will lead to the 4 different isomers with 2 pairs of cis-trans isomers. Either the ßH at C4 or the ßH at C6 is eliminated with the bromine at C5.