Test: Elimination Reaction - Chemistry MCQ

Rate of formation of II and IV similar because H & D are approximately of same size.
C–H & C–D bond breaking is the part of RDS hence rate in I is slower than III
It undergo E2 not via E1 because carbocation is not stable.

For 5-Bromocycloheptene > 4-Bromocycloheptene, more stable product leads to faster rate of reaction. Product formed by E2 of 4-Bromocycloheptene is more stable than 5-Bromocycloheptene. So, rate of reaction 4-Bromocycloheptene is fater than 5-Bromocycloheptene. So option 5-Bromocycloheptene > 4-Bromocycloheptene is incorrect.

For 2-Bromo- I -phenylbutane > 3-Bromo- I-phenylbutane, it is true rate of reaction for 2-Bromo- I -phenylbutane is more than 3-Bromo- I-phenylbutane as product formed in former is more stable due to resonance.

For 3-Bromocyclohexene > Bromocyclohexane, it is true rate of reaction for 3-Bromocyclohexeneis more than Bromocyclohexane as product formed in former is more stable.

For 3-Bromo-2-methylpentane > 2-Bromo-4-methylpentane, it is true rate of reaction for 3-Bromo-2-methylpentane more stable than 2-Bromo-4-methylpentane as product formed in former is more stable.

Here correct is statement 1 and 4 are true. Due to steric hinderence tert-Butyl alcohol anion will attack less substituted C-H bond.
Ethanol anion will attack more substituted C-H bond to form more stable alkene.

Compounds a, b and c will follow Addition- Elimination mechanism.

Compound will follow elimination -addition mechanism, as shown in below reaction. NH^– anion first attack on the ring which lead to formation of benzyne ring and further addition of -NH₂ will take place.

In E1 elimination RDS is the formation of carbocation from halide hence C – H and C – D are not the part of RDS there K_H/K_D is approx. 1.

Here three times Elimination(E2) occurs and three pie bond will be formed and three water molecule will be removed as shown below.

E2 is a concerted reaction in which bonds break and new bonds form at the same time in a single step. Order of reactivity of alkyl halides towards E2 dehydrohalogenation is found to be 3^o > 2^o > 1^o. In E2 reaction both hydrogen and leaving group should be antiperiplanar and that’s why In E2 elimination different stereoisomer (diastereomer) converts into different stereo product.

All of the above molecules will form product by hoffmann rule that is less substituted C-H bond will be attacked because more stable product will be formed via this in all cases.
I.

II.

When leaving group are F & NH2 type then they give hoffmann product.

III.

IV.

Due to inductive effect of halogen it stabilize the negative charge on adjacent to carbon atom, as X will withdraw the electron towards itself.

Iodine is a better leaving group than chlorine, so attach will be at iodine site.

In most unimolecular reactions of alkyl halide SN1 reaction is favoured over E1 reaction. E1 mechanism is favoured as compared to S_N1 mechanism by branching at carbon increases, as more steric hinderence at the attacking site will leads to E1 mechanism.

Exocyclic alkene are more stable than endo beacuse of but due to “steric hindrance” the less substituted alkene is formed.

From the following isomers of C₅H₁₁Br, except IX all are capable of E2 elimination.

Compound d will give E₂ reaction, CL is a better leaving group so formation of carboanion and removal of leaving group occurs simultaneously. But E₁CB reaction shown by poor leaving group like in compounds a, b and c.