The central idea
Classify the reaction before recalling a named rule. A nucleophile donates an electron pair; an electrophile accepts one. Compare possible intermediates using resonance, inductive effects and the reaction conditions. Solvent, leaving group and steric access can change the preferred pathway.
A worked example
For an SN1 comparison with the same leaving group in a suitable polar protic solvent, a tertiary substrate usually ionises more readily than a primary substrate because the resulting carbocation is more stable. This argument is about the ionisation mechanism, not a universal rule for every substitution.
A common mistake
Do not transfer SN1 trends directly to SN2. In SN2, steric obstruction to the incoming nucleophile is central.
Try the idea
These original practice questions also appear in the foundation mock. Attempt them before opening the explanation.
Which species is the most acidic in water under ordinary conditions?
Answer: C
Ethanoic acid forms a carboxylate ion with negative charge delocalised over two electronegative oxygen atoms. Its pKa is much lower than those of phenol, ethanol and ethane, so it is the strongest acid of this set.
Which substrate generally reacts fastest by an SN1 mechanism in a polar protic solvent?
Answer: D
The rate-determining SN1 step forms a carbocation. The tertiary carbocation from tert-butyl bromide is stabilised by alkyl groups more than the secondary or primary alternatives. Methyl carbocations are especially unfavourable.
How many stereogenic carbon centres are present in CH₃CH(OH)CH₂CH₃?
Answer: B
The carbon carrying OH is attached to H, OH, CH₃ and CH₂CH₃: four different groups. It is stereogenic. None of the other carbons has four different substituents.
Reductive ozonolysis of an alkene produces only ethanal. Which alkene fits?
Answer: D
Cut the C=C bond and replace each alkene carbon's double bond by C=O. CH₃CH=CHCH₃ gives two CH₃CHO molecules. Both cis- and trans-but-2-ene give this same constitutional product.
Addition of HBr to propene in the presence of a radical initiator under suitable peroxide conditions gives predominantly:
Answer: A
In the radical pathway, Br· adds to the terminal carbon so that the intermediate carbon radical is secondary. That radical takes H from HBr, giving CH₃CH₂CH₂Br. This is the peroxide effect for HBr.
Which statements are correct?
Answer: A, B, D
Enantiomers are non-superimposable mirror images. Internal symmetry can make a molecule with stereogenic centres achiral, as in a meso compound. A racemic mixture contains equal amounts of enantiomers whose rotations cancel. Therefore the presence of a stereogenic carbon alone does not guarantee a rotating bulk sample.
What to do next
Explain your answer aloud, then find a related question in an official paper. If you can only follow a worked solution but cannot start a fresh question, revisit the central idea and try again without the notes.
Apply it to an official paper