When an SN2 reaction takes place at a chiral carbon, what happens to the configuration at that carbon?

It becomes planar in the product

Haloalkanes & Haloarenes

10 NEET-UG practice questions on Haloalkanes & Haloarenes, part of Chemistry. The first 10 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Easy
Ethylmagnesium bromide reacts at once with water. Name the organic product.
1. AEthane
2. BButane
3. CEthene
4. DEthanol
Show answerHide answer
Answer: A · Ethane
The carbon bound to magnesium is strongly carbanion-like and takes a proton from water: CH₃CH₂MgBr + H₂O → CH₃CH₃ + Mg(OH)Br.
- B: This is a coupling product, not formed by reaction with water.
- C: This would require elimination, which does not occur here.
- D: This assumes OH is transferred to carbon, but carbon picks up H instead.
Remember this
Grignard reagents are destroyed by any source of acidic hydrogen, giving the hydrocarbon, so they must be kept dry.
Question 2 · Easy
When an SN2 reaction takes place at a chiral carbon, what happens to the configuration at that carbon?
1. AIt is fully racemised
2. BIt becomes planar in the product
3. CIt is retained
4. DIt is inverted
Show answerHide answer
Answer: D · It is inverted
The nucleophile attacks from the side opposite the leaving group, in a single step, so the other three groups flip over like an umbrella.
- A: Racemisation is the feature of SN1, which has a planar intermediate.
- B: The product carbon is still sp³; only the transition state has the planar arrangement.
- C: Retention would need attack from the same side as the leaving group.
Remember this
SN2 means backside attack and inversion; SN1 goes through a planar carbocation and leads to racemisation.
Question 3 · Hard
Follow this sequence. 1-Bromopropane + Mg, dry ether → P P + propanone, then H₃O⁺ → Q Q + conc. H₂SO₄, heat → R (major) What is R?
1. A2-Methylpent-1-ene
2. B2-Methylpent-2-ene
3. C4-Methylpent-2-ene
4. DHex-3-ene
Show answerHide answer
Answer: B · 2-Methylpent-2-ene
P is CH₃CH₂CH₂MgBr. It adds to the C=O of propanone, giving the tertiary alcohol Q, 2-methylpentan-2-ol. Acid dehydration then follows Saytzeff's rule, losing H from the chain CH₂ next to the carbocation centre to give (CH₃)₂C=CHCH₂CH₃.
- A: This is the less substituted alkene, the minor product.
- C: This moves the methyl branch and the double bond away from the tertiary carbon.
- D: This has a straight chain, ignoring the branch set by the propanone carbon.
Remember this
A Grignard reagent and a ketone give a tertiary alcohol, which dehydrates easily to the more substituted alkene.
Question 4 · Medium
Arrange these bromides in increasing order of reactivity by the SN1 mechanism. A. (CH₃)₃CBr B. CH₃CH₂Br C. (CH₃)₂CHBr D. CH₃Br
1. AD, B, C, A
2. BA, C, B, D
3. CD, C, B, A
4. DB, D, C, A
Show answerHide answer
Answer: A · D, B, C, A
The rate-determining step of SN1 forms a carbocation, so the rate follows carbocation stability: methyl < 1° < 2° < 3°.
- B: This is the SN2 order, which is controlled by crowding.
- C: This places the secondary bromide below the primary one, against carbocation stability.
- D: This places ethyl below methyl, though a primary cation is more stable than methyl.
Remember this
SN1 reactivity is 3° > 2° > 1° > CH₃X, exactly the reverse of the steric order for SN2.
Question 5 · Medium
On heating 2-bromo-2-methylbutane with alcoholic KOH, which alkene is formed as the major product?
1. A2-Methylbut-1-ene
2. B2-Methylbutan-2-ol
3. C2-Methylbut-2-ene
4. D3-Methylbut-1-ene
Show answerHide answer
Answer: C · 2-Methylbut-2-ene
Alcoholic KOH causes β-elimination. By Saytzeff's rule the more substituted alkene is preferred, so H is lost from the CH₂ group to give (CH₃)₂C=CHCH₃.
- A: This is the less substituted alkene, formed in the smaller amount.
- B: This is the substitution product expected with aqueous alkali, and it is not an alkene.
- D: This has its double bond away from the carbon that held bromine, which simple elimination cannot give.
Remember this
In dehydrohalogenation the major alkene carries the greater number of alkyl groups on the double bond.
Question 6 · Medium
Counting enantiomers separately, how many isomers in all have the molecular formula C₄H₉Br?
1. A4
2. B3
3. C5
4. D6
Answer and explanation in LearnFree account
Question 7 · Medium
Assertion A: Chlorobenzene is much less reactive than chloroethane towards nucleophilic substitution. Reason R: The C-Cl bond in chlorobenzene is longer than that in chloroethane. Choose the correct answer.
1. AA is correct but R is not correct
2. BBoth A and R are correct but R is not the correct explanation of A
3. CBoth A and R are correct and R is the correct explanation of A
4. DA is not correct but R is correct
Answer and explanation in LearnFree account
Question 8 · Hard
Follow the sequence shown, starting from bromobenzene. What is Y?
Figure: MedExams
1. APhenylethyne
2. BEthylbenzene
3. C1-Phenylethanol
4. DEthenylbenzene
Answer and explanation in LearnFree account
Question 9 · Medium
The enantiomerically pure bromide drawn is converted to an alcohol by a purely SN2 pathway. Name the product P.
Figure: MedExams
1. A(S)-Butan-2-ol
2. B(±)-Butan-2-ol
3. C(R)-Butan-2-ol
4. DButan-1-ol
Answer and explanation in LearnFree account
Question 10 · Easy
Refer to bromides X and Y in the figure. Assertion A: Towards OH⁻ under SN2 conditions, X reacts faster than Y. Reason R: The carbocation formed from Y is more stable than the one formed from X. Choose the correct answer.
Figure: MedExams
1. AA is correct but R is not correct
2. BBoth A and R are correct and R is the correct explanation of A
3. CA is not correct but R is correct
4. DBoth A and R are correct but R is not the correct explanation of A
Answer and explanation in LearnFree account

Haloalkanes & Haloarenes

Ethylmagnesium bromide reacts at once with water. Name the organic product.

When an SN2 reaction takes place at a chiral carbon, what happens to the configuration at that carbon?

Follow this sequence. 1-Bromopropane + Mg, dry ether → P P + propanone, then H₃O⁺ → Q Q + conc. H₂SO₄, heat → R (major) What is R?

Arrange these bromides in increasing order of reactivity by the SN1 mechanism. A. (CH₃)₃CBr B. CH₃CH₂Br C. (CH₃)₂CHBr D. CH₃Br

On heating 2-bromo-2-methylbutane with alcoholic KOH, which alkene is formed as the major product?

Counting enantiomers separately, how many isomers in all have the molecular formula C₄H₉Br?

Assertion A: Chlorobenzene is much less reactive than chloroethane towards nucleophilic substitution. Reason R: The C-Cl bond in chlorobenzene is longer than that in chloroethane. Choose the correct answer.

Follow the sequence shown, starting from bromobenzene. What is Y?

The enantiomerically pure bromide drawn is converted to an alcohol by a purely SN2 pathway. Name the product P.

Refer to bromides X and Y in the figure. Assertion A: Towards OH⁻ under SN2 conditions, X reacts faster than Y. Reason R: The carbocation formed from Y is more stable than the one formed from X. Choose the correct answer.