Felkin-Anh model: predicting diastereoselectivity at carbonyls
Learn a step-by-step method to predict the major diastereomer when a nucleophile adds to a chiral carbonyl, with two worked practice questions and a study pla
This post gives you a fixed routine to predict which diastereomer forms when a nucleophile adds to a carbonyl next to a stereocentre.
You also get two worked practice questions and a short study plan for CSIR-NET, GATE, IIT-JAM and CUET-PG.
The problem the model solves
Take an aldehyde or ketone whose alpha carbon (the carbon next to C=O) is already a stereocentre. A diastereomer is a stereoisomer that is not the mirror image of another. When a nucleophile adds, the new stereocentre can form in two ways. The two products are diastereomers, and they do not form in equal amounts.
The model tells you which product is major. A chemist named Hugh Felkin proposed it in 1967. It differs slightly from Cram's rule, and it is one of the most accepted rules for these reactions. Textbooks and question papers write it as the Felkin-Anh model.
Three ideas behind the model
- Staggered transition state. The forming bond and the neighbouring CāC bond should not eclipse each other. Eclipsed bonds cost energy, called torsional strain. Felkin's own early papers were titled around torsional strain involving partial bonds.
- Large group perpendicular. The bulkiest group on the alpha carbon turns until it is at right angles to the C=O plane. In that position it is as far as possible from the incoming nucleophile.
- Oblique attack. The nucleophile does not strike the carbonyl carbon head-on. It comes in at an obtuse angle to the C=O bond, leaning away from oxygen. So it passes close to the substituent on the carbonyl carbon, not close to oxygen.
Cram's older rule uses a different reactive conformation. The Felkin-Anh picture works from a staggered one. This is why your answer should begin with an end-on projection along one bond, not with a flat drawing.
The routine: five steps
First rank the three groups on the alpha carbon by size: L (large), M (medium) and S (small). Then follow these steps every time.
- Look along the C(carbonyl)āC(alpha) bond. The carbonyl carbon is in front. Its O and its other substituent lie in the horizontal plane.
- Put L on the back carbon, at the top or bottom, perpendicular to that plane.
- Place M next to the oxygen and S next to the carbonyl substituent. This is the arrangement that lets the nucleophile pass S.
- Send the nucleophile in from the face opposite L, leaning away from oxygen.
- Redraw the product and assign R/S using the priority rules. This is where most marks are lost.
The major product is the Felkin product. The minor one is the anti-Felkin product.
When the alpha group is electronegative
Suppose the alpha carbon carries OR or a halogen. The rule then changes slightly. The CāX bond has a low-energy antibonding orbital (Ļ*), so it accepts electrons well from the forming bond. That group goes perpendicular, even if another group is bulkier. This is the polar Felkin-Anh idea.
There is one more case. A metal ion can bind both the carbonyl oxygen and an alpha oxygen substituent, forming a ring. This is chelation. The molecule is now locked, so the staggered picture no longer applies. The nucleophile attacks the ring from its less hindered face. This is the Cram-chelate model, and it usually gives the anti-Felkin product.
| Model | When it applies | What decides the face | Usual major product |
|---|---|---|---|
| Felkin-Anh | Alkyl or aryl alpha groups, no chelation | Largest group perpendicular to C=O | Felkin |
| Polar Felkin-Anh | Alpha OR or halogen, no chelation | Electronegative group perpendicular | Felkin |
| Cram-chelate | Alpha OR with a metal that can chelate | Rigid chelate ring, less hindered face | Anti-Felkin |
Practice question 1: assigning the Felkin product
Practice question 1. The aldehyde (R)-2-phenylpropanal reacts with methylmagnesium bromide in dry ether, followed by aqueous work-up. Using the Felkin-Anh model, give the configuration of the major product, 3-phenylbutan-2-ol.
Solution.
- The alpha carbon carries phenyl, methyl and hydrogen. Sizes: L = phenyl, M = methyl, S = hydrogen.
- Look along the C1āC2 bond. Phenyl goes perpendicular to the C=O plane. Methyl sits next to the oxygen. Hydrogen sits next to the aldehyde hydrogen. For the (R) aldehyde, this layout fixes which side each group takes.
- The nucleophile (CH3ā from the organomagnesium reagent) comes from the face opposite phenyl. It leans away from oxygen, so it passes the small hydrogen.
- In the product, the new carbinol carbon holds OH, the old alpha carbon, the new methyl and H. Priorities: OH, then the CH(Ph)CH3 carbon, then CH3, then H. Reading the drawing gives R.
- The old alpha carbon is untouched. Its priority order (CH(OH)CH3, phenyl, CH3, H) matches the order in the aldehyde (CHO, phenyl, CH3, H). So it stays R.
Answer: the major (Felkin) product is (2R,3R)-3-phenylbutan-2-ol. The anti-Felkin minor product is (2S,3R).
Check yourself: if the old centre changed its descriptor and its priority order did not change, you made an R/S error, not a model error.
Practice question 2: Felkin versus chelation
Practice question 2. An aldehyde has an alkoxy group on the alpha carbon. An organomagnesium reagent adds to it in two experiments. In the first, no extra additive is used. In the second, a metal salt that can bind two oxygen atoms is added first. Which statement is correct?
- Both experiments give the same major diastereomer.
- Without the additive the Felkin product is major. With chelation the anti-Felkin product is major.
- Without the additive the anti-Felkin product is major. With chelation the Felkin product is major.
- The alkoxy group is always the largest group, so chelation has no effect.
Solution.
- Without chelation, the alkoxy group is electronegative. By the polar Felkin-Anh idea it sits perpendicular to C=O, and the nucleophile attacks opposite it. That gives the Felkin product.
- With the metal ion, the carbonyl oxygen and the ether oxygen bind to the same metal. This locks them into a ring. The groups can no longer rotate to the staggered arrangement the model needs.
- The nucleophile now attacks the ring from the face carrying the smaller alpha substituent. This gives the opposite diastereomer, the anti-Felkin product.
Answer: B. Option D fails because chelation changes the conformation, and the whole model depends on conformation. Option A fails for the same reason.
How questions are usually framed
A question seldom tells you which model to apply. Look for these patterns:
- A chiral aldehyde or ketone treated with an organometallic reagent or a hydride, with a request for the major product.
- Two sets of conditions, with and without a metal that can chelate.
- A question on which group sits perpendicular, or which conformer is reactive.
- A matching list of models and their key features.
The method stays the same. Rank the alpha groups, draw the end-on view, send the nucleophile in, assign descriptors. If an alpha oxygen is present, check for chelation first.
Study plan
- Day 1. Revise R/S priority rules and end-on projections. Without them the model earns no marks.
- Day 2. Learn the five steps. Redo practice question 1 on a blank page without the solution.
- Day 3. Study the polar and chelation variants. Redo practice question 2 and write one line on why each wrong option fails.
- Day 4 onwards. Solve stereoselectivity questions from the chemistry question bank. Draw the end-on view for every one, even when the answer looks obvious.
- Every week. Go through your wrong answers. Tag each as a size-ranking error, a drawing error or an R/S error. Fix the most frequent habit first.
Key takeaways
- Start from the end-on view along the C(carbonyl)āC(alpha) bond.
- The large group goes perpendicular to C=O. The nucleophile attacks opposite it and passes the smallest group.
- An alpha OR or halogen takes the perpendicular position because of its low-energy Ļ* orbital.
- Chelation locks the conformation and usually reverses the outcome to the anti-Felkin product.
- When assigning R/S, check the untouched alpha centre. Its descriptor stays the same if its priority order is unchanged.
Frequently asked questions
Is the Felkin-Anh model the same as Cram's rule?
No. Both predict the major diastereomer, but they differ slightly. The Felkin-Anh model uses a staggered transition state with the large group perpendicular to the C=O plane.
Does the model work for hydride reductions?
Yes. Felkin's early work studied the lithium aluminium hydride reduction of simple open-chain ketones. The hydride is the nucleophile, and the same steps apply.
Why does the nucleophile approach at an obtuse angle?
It must reach the C=O Ļ* orbital, and the best overlap comes from an oblique path, not a head-on one. This is why it passes the carbonyl substituent and not the oxygen.
How do I know if chelation applies?
Look for an alpha oxygen substituent together with a metal ion that can bind two oxygen atoms. If the question says non-chelating conditions, use the polar Felkin-Anh model.
Where do I check whether my exam covers this topic?
Stereoselective addition to carbonyls is part of organic stereochemistry. Read the current official syllabus for your exam to confirm the exact scope.
