Neopentyl bromide: why SN2 is very slow and solvolysis gives rearranged products
By Vimal Swarnkar (VKC) · UPSC Chemistry · Civil Services, Combined Geo-Scientist and IFoS
A primary bromide that refuses to do SN2, and an SN1 that gives a skeleton the starting material never had. Neopentyl bromide breaks two rules students learn in Class 12, and that is exactly why UPSC likes it. In this post I will show you how often this idea comes, which book to read, how I plan the answer, and how it should look in your booklet.
Neopentyl bromide, (CH3)3C–CH2Br, is a primary alkyl halide, yet it reacts very slowly by SN2. On solvolysis in aqueous ethanol it gives rearranged products such as 2-methylbut-2-ene and 2-methylbutan-2-ol. Explain with mechanisms.
SN2 is slow because the nucleophile must attack from the back, at 180° to the C–Br bond, and the three β-CH3 groups of the tert-butyl group fill that side. The five-coordinate transition state is crowded, the activation energy is high, and the rate drops to about 10−5 of ethyl bromide (4.2 × 10−6 with ethanol in Hughes’ data). Solvolysis gives rearranged products because a free 1° carbocation is too unstable to form. As Br− leaves, a CH3 moves with its bond pair to the CH2 carbon, a 1,2-methyl shift (Wagner–Meerwein), giving the 3° tert-pentyl cation. Water, ethanol or loss of H+ then give 2-methylbutan-2-ol, the ethyl tert-pentyl ether and 2-methylbut-2-ene (Saytzeff, major alkene). No product keeps the neopentyl skeleton.
Watch the video
Day 03 of my Daily Answer Writing series on YouTube: I write this answer step by step, both mechanisms included.
A word from me
Most students learn one line in school: “primary halides go SN2, tertiary halides go SN1.” Then they meet neopentyl bromide and the line fails twice. It is primary, but SN2 hardly happens. It cannot form a stable cation, but in a polar solvent it still reacts, and the product has a different carbon skeleton. If you write only “steric hindrance” and stop, you get two or three marks out of ten. The examiner wants to see the crowded transition state, the curly arrows of the methyl shift, and every product named.
So I ask my students to do one thing. Take one book, read the neopentyl example in it, and then close the book and draw the methyl-shift arrows from memory three times. Once you can draw that one picture, pinacol–pinacolone, the camphene rearrangement and every “why is the product rearranged?” question become the same question. If you get stuck, write to me on Telegram or at [email protected]. I would rather you ask than lose these marks.
In 11 of the last 15 papers
From 2021 to 2025, nucleophilic substitution (SN1, SN2, neighbouring groups) and the carbocation rearrangements that go with it came in 11 of 15 papers across Civil Services (Paper II), IFoS (Paper II) and Combined Geo-Scientist (Paper III): every year in Civil Services, every year in IFoS, and once in Geo-Scientist (2022). That is about 34 questions in five years. The neopentyl idea itself, a crowded SN2 or a solvolysis that rearranges through a 1,2-shift, came 4 times directly: IFoS 2022 Q1(g) (2-iodo-3,3-dimethylbutane with AgNO3, a near twin), Civil Services 2022 Q3(c)(ii) (SN2 order and steric hindrance, 10 marks), Civil Services 2024 Q4(a)(ii) (SN1, SN2, E1 or E2 in hot ethanol) and IFoS 2024 Q2(b)(iii) (a 1,2-hydride shift during solvolysis). The other 30 times it came indirectly, as pinacol–pinacolone, Wagner–Meerwein in bornane and camphene, ring expansion, neighbouring-group participation and solvent effects.
Syllabus and books
In Civil Services it is Paper II, Unit 2, Reaction mechanisms, parts (ii), (iii) and (vi). IFoS asks it in Paper II and the Geo-Scientist exam in Mains Paper III, organic section. The Civil Services syllabus reads:
“Reactive intermediates: generation, geometry, stability and reactions of carbonium ions … Substitution reactions: SN1, SN2 and SNi mechanisms; neighbouring group participation … Reactions and rearrangements: Pinacol-pinacolone … and Wagner-Meerwein rearrangements.”
Three phrases in it decide this question: SN1, SN2, carbonium ions (their stability) and Wagner-Meerwein.
You do not need all of these books. Any one of them is enough for this question. I have listed three so that you can use whichever you already have.
- Book 1: Clayden, Greeves and Warren, Organic Chemistry (2nd edition), Chapter 15 “Nucleophilic substitution at saturated carbon”, with the summary table that marks ‘neopentyl’ as bad for both SN1 and SN2 (page 347), and Chapter 36 “Participation, rearrangement and fragmentation”, where neopentyl iodide with AgNO3 in water gives the rearranged alcohol and the methyl shift is drawn with its cyclic transition state. If you have this book, this is the one I would use.
- Book 2: Morrison and Boyd, Organic Chemistry, Section 14.15 “The SN1 reaction: rearrangement”: neopentyl bromide with ethoxide gives neopentyl ethyl ether slowly by SN2, while with ethanol it gives almost entirely rearranged products by SN1. Problem 14.6 and the end-of-chapter problem on neopentyl alcohol (an 85 : 15 mixture of two alkenes) are good practice.
- Book 3: March’s Advanced Organic Chemistry, Table 9.2 (relative rates of RBr with ethanol: ethyl 1, neopentyl 4.2 × 10−6) and Section 10.G on branching at the α and β carbons. Read this only if you want the numbers behind the claim.
How to build the answer
Read the tree from left to right. Every leaf is a point to write, in that order. The highlighted words are the ones the examiner looks for.
Given information
- 1° at the α-carbon
- β-carbon quaternary (tert-butyl)
Assumptions
- SN2: one step, back-side attack
- SN1: ionisation is the RDS
- Cation stability 3° > 2° > 1°
- A 1,2-shift happens if it gives a more stable cation
Key law
- rate = k[RBr][Nu] against k[RBr]
SN2 part
- 5-coordinate TS sketch
- three β-CH3 shield the back face
- rate ≈ 10−5 of EtBr
Solvolysis part
- 1° cation avoided
- curly arrows: 1,2-CH3 shift as Br− leaves
- 3° carbocation
Products
- alcohol and ether (SN1)
- 2-methylbut-2-ene, Saytzeff (E1)
Check
- no product keeps the neopentyl skeleton
Reading the result
- sterics block SN2
- rearrangement rescues SN1
Principle
- β-branching acts like α-branching
Application
- isobornyl chloride → camphene
Final answer
- one line, double underlined
Two pictures that earn marks
What gains marks
- The opening line: 1° at the α-carbon, quaternary at the β-carbon. It tells the examiner you have seen the trap.
- A sketch of the SN2 transition state, with Nu and Br at 180° and the three β-CH3 groups on the side of attack.
- A number for the rate: about 105 times slower than ethyl bromide.
- Curly arrows for the 1,2-methyl shift, drawn as one step with Br− leaving, and the name Wagner–Meerwein.
- The 3° cation labelled, with the reason it is more stable: hyperconjugation and +I.
- Every product named, with its path: alcohol and ether (SN1), 2-methylbut-2-ene major and 2-methylbut-1-ene minor (E1, Saytzeff).
- One application from the syllabus: isobornyl chloride to camphene, or pinacol–pinacolone.
Match the length to the marks
I always tell my students to let the marks decide the length. For 5 marks, as in IFoS 2022 Q1(g), I would draw only the ionisation, the methyl shift and the major product, with two lines of reason, about half a page. For 10 marks, as in this Day 03 question, write both parts, the SN2 sketch and the solvolysis mechanism, the two pages below. For 15 marks, add the energy profile of the SN1 path, the evidence that the methyl group participates as Br− leaves (Book 3), and a second example such as pinacol–pinacolone.
How it looks in the booklet
This is the full 10-mark answer, in the order of the tree above. Page 1 sets up the two mechanisms and explains why SN2 is slow, and page 2 does the solvolysis and the conclusion.
Q.)Neopentyl bromide is a 1° alkyl halide, yet it reacts very slowly by SN2; on solvolysis in aqueous ethanol it gives rearranged products. Explain with mechanisms. [Day 03; cf. IFoS 2022], 10M
Ans.Given: neopentyl bromide, (CH3)3C–CH2Br (1-bromo-2,2-dimethylpropane). The C–Br carbon is 1° (α-carbon), but the next carbon is quaternary (β-carbon), carrying three CH3 groups.
For substitution at a saturated carbon, assume that
- i) SN2 is one step: the nucleophile attacks from the back side, at 180° to C–Br, through a 5-coordinate TS.
- ii) SN1 is two steps: ionisation is the RDS, giving a planar carbocation.
- iii) carbocation stability is 3° > 2° > 1° (hyperconjugation, +I).
- iv) a 1,2-shift of H or CH3 happens whenever it gives a more stable cation.
(a) Why SN2 is very slow. In the TS of (i) the α-carbon is half-bonded to both Nu and Br. The three β-CH3 groups sit exactly on the side the nucleophile must enter, so they shield the back face: the TS is crowded, its energy is high, and k2 falls sharply. Here β-branching acts like α-branching.
| R–Br (all 1° except Me) | CH3 | Et | n-Pr | i-Bu | neopentyl |
|---|---|---|---|---|---|
| relative k2 (EtOH) | 17.6 | 1 | 0.28 | 0.030 | 4.2 × 10−6 |
Even I− in acetone, the best SN2 combination, hardly reacts with it. With EtO− the SN2 product, neopentyl ethyl ether, forms only slowly and without rearrangement.
All other symbols have their usual meaning.
(b) Solvolysis in aqueous ethanol: SN1 / E1 with rearrangement. A polar protic solvent favours ionisation (ii), but a free 1° carbocation is too unstable. So, as Br− leaves, a CH3 group moves with its bond pair to the CH2 carbon: a 1,2-methyl shift (Wagner–Meerwein rearrangement). The charge lands on the old quaternary carbon, now a 3° carbocation.
The 3° cation, (CH3)2C+CH2CH3, then reacts in three ways:
| Step | Product | Path |
|---|---|---|
| + H2O, − H+ | (CH3)2C(OH)CH2CH3, 2-methylbutan-2-ol | SN1 |
| + EtOH, − H+ | (CH3)2C(OEt)CH2CH3, 2-ethoxy-2-methylbutane | SN1 |
| − H+ from CH2 | (CH3)2C=CHCH3, 2-methylbut-2-ene (major) | E1, Saytzeff |
| − H+ from CH3 | CH2=C(CH3)CH2CH3, 2-methylbut-1-ene (minor) | E1 |
Check: every product has the branched C5 skeleton of tert-pentyl, and no neopentyl alcohol is formed ✓. The trisubstituted alkene is the more stable one, so it is the major alkene ✓.
Reading the result:
- i) SN2 is blocked by steric hindrance from the β-carbon, although the halide is primary.
- ii) SN1 is possible only because the methyl shift bypasses the 1° cation; that is why every product is rearranged.
- iii) With a strong nucleophile (EtO−) the slow SN2 gives the unrearranged ether; in a weakly nucleophilic, ionising solvent the rearranged products take over.
Principle: branching at the β-carbon slows SN2 as much as branching at the α-carbon, and a carbocation rearranges by a 1,2-shift whenever a more stable cation can form.
Application: the same Wagner–Meerwein shift turns isobornyl chloride into camphene in terpene chemistry, and drives the pinacol–pinacolone rearrangement.
So, neopentyl bromide: SN2 very slow (steric, ≈ 105 × slower than EtBr); solvolysis goes by SN1 / E1 through a 1,2-CH3 shift and gives only rearranged 3° products
The same question in other exams
IFoS 2022, Paper II, Q1(g), 5 marks: major product of 2-iodo-3,3-dimethylbutane with AgNO3 in hot ethanol, with mechanism. Ag+ pulls I− off as AgI, giving the 2° cation CH3CH+C(CH3)3. A CH3 moves across to give the 3° cation (CH3)2CH–C+(CH3)2, and loss of H+ gives 2,3-dimethylbut-2-ene. Same tert-butyl group, same methyl shift.
Civil Services 2022, Paper II, Q3(c)(ii), 10 marks: arrange four tertiary bromides by SN2 reactivity and give the reason. The reason is the SN2 half of this post: back-side attack, a crowded five-coordinate TS, and a cage structure that cannot invert at all.
Civil Services 2024, Paper II, Q4(a)(ii), 5 marks: (CH3)3CBr with ethanol at 60 °C. A 3° bromide in a protic solvent: SN1 gives tert-butyl ethyl ether, and E1 gives 2-methylpropene, more of it as the temperature rises.
IFoS 2024, Paper II, Q2(b)(iii), 5 marks: CH3CH(Cl)CH(CH3)CH(CH3)CH3 in hot ethanol. Here the shift is of a hydride: the 2° cation becomes 3°, and loss of H+ gives 2,3-dimethylpent-2-ene.
IFoS 2022, Paper II, Q2(c), and Civil Services 2024, Paper II, Q4(c)(i): bicyclic (bornane) alcohols with acid. Same Wagner–Meerwein shift, this time of a ring bond, giving camphene-type products.
Every version is the same three questions. Can the nucleophile reach the back of the carbon? If the substrate ionises, is the cation 1° or 2°, with a more substituted carbon next door? If yes, move the H or CH3 (or ring bond) with its pair, and only then add the nucleophile or take off H+.
Questions students ask
Why does neopentyl bromide react so slowly by SN2 even though it is a primary halide?
In SN2 the nucleophile must attack the carbon from the side opposite to the C–Br bond. In neopentyl bromide, (CH3)3C–CH2Br, the next carbon carries three methyl groups, and they fill that side. The five-coordinate transition state becomes very crowded and its energy rises, so the rate is about 100,000 times lower than for ethyl bromide. This was the Day 03 question of VKC Daily Answer Writing (10 marks), built on IFoS 2022 Paper II Q1(g) and Civil Services 2022 Paper II Q3(c)(ii).
Why does solvolysis of neopentyl bromide give rearranged products?
A primary carbocation is too unstable to form on its own. As bromide leaves, one methyl group moves with its pair of electrons to the CH2 carbon, a 1,2-methyl shift (Wagner–Meerwein rearrangement), and the positive charge ends up on a tertiary carbon. Water, ethanol or loss of a proton then give 2-methylbutan-2-ol, 2-ethoxy-2-methylbutane and 2-methylbut-2-ene. No product keeps the neopentyl skeleton.
What is the major alkene from neopentyl bromide in aqueous ethanol?
2-Methylbut-2-ene, (CH3)2C=CHCH3. The tertiary cation can lose a proton from the CH2 group or from a CH3 group; losing it from CH2 gives the trisubstituted alkene, which is more stable (Saytzeff rule). 2-Methylbut-1-ene is the minor alkene. Heating neopentyl alcohol with acid gives the same two alkenes, in about an 85 : 15 ratio.
Does neopentyl bromide ever give an unrearranged product?
Yes, but only by SN2. With a strong nucleophile such as ethoxide, the slow SN2 reaction gives neopentyl ethyl ether without rearrangement, because SN2 has no carbocation. With ethanol alone, the reaction goes through ionisation and almost all the products are rearranged.
What is a Wagner–Meerwein rearrangement?
A 1,2-shift of a hydrogen, alkyl or aryl group, or a ring bond, with its bonding pair, to a neighbouring positive carbon, giving a more stable carbocation. The neopentyl-to-tert-pentyl shift is the simplest example; isobornyl chloride to camphene is the classic one in terpene chemistry. It is named in the Civil Services Paper II syllabus.
How often does UPSC ask SN1, SN2 and carbocation rearrangements?
From 2021 to 2025 the topic appeared in 11 of 15 organic chemistry papers across Civil Services Paper II, IFoS Paper II and Combined Geo-Scientist Paper III, about 34 questions in all: every year in Civil Services and IFoS, and in Geo-Scientist 2022. Four were direct versions of the neopentyl idea, such as IFoS 2022 Q1(g), 5 marks.
Read the complete answers
By VKC, your mentor. Integrated M.Sc. Chemistry, IIT Roorkee. CSIR NET Chemical Sciences, All India Rank 4. New posts are announced first on Telegram.