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Assertion: In the E2 elimination, beta-H...

Assertion: In the E2 elimination, `beta-H` and leaving group should be antiperiplanar.
Reason: In the E2 elimination, base always abstracts unhindered `beta-H`.

A

If both assertionn and reason are true, and reason is the true explanation of the assertion.

B

If both assertion and reason are true, but reason is not the true explanation of the assertion.

C

If assertion is true, but reason is false.

D

If both assertion and Reason are false.

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The correct Answer is:
To solve the question, we need to analyze both the assertion and the reason provided. ### Step 1: Understand the Assertion The assertion states that in the E2 elimination mechanism, the beta-hydrogen (β-H) and the leaving group must be in an antiperiplanar arrangement. **Explanation:** - In E2 elimination reactions, the transition state is best stabilized when the β-H and the leaving group are positioned 180 degrees apart. This arrangement allows for optimal overlap of orbitals during the elimination process, leading to the formation of a double bond. ### Step 2: Understand the Reason The reason provided states that in E2 elimination, the base always abstracts unhindered β-H. **Explanation:** - While it is true that the base abstracts β-H, the assertion that it must be unhindered is not entirely accurate. The key factor for E2 reactions is the antiperiplanar arrangement rather than the steric hindrance of the β-H. The β-H can be hindered or unhindered as long as it is in the correct orientation relative to the leaving group. ### Step 3: Conclusion Based on the analysis: - The assertion is **true**: β-H and the leaving group must be antiperiplanar in E2 elimination. - The reason is **false**: The base does not always abstract unhindered β-H; it is the orientation that matters. ### Final Answer The correct conclusion is that the assertion is true, but the reason is false. Therefore, the answer is option number 3: Assertion is true, but the reason is false. ---
To solve the question, we need to analyze both the assertion and the reason provided. ### Step 1: Understand the Assertion The assertion states that in the E2 elimination mechanism, the beta-hydrogen (β-H) and the leaving group must be in an antiperiplanar arrangement. **Explanation:** - In E2 elimination reactions, the transition state is best stabilized when the β-H and the leaving group are positioned 180 degrees apart. This arrangement allows for optimal overlap of orbitals during the elimination process, leading to the formation of a double bond. ...
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Elimination E2

Elimination examples OF E2

The removal of two atoms or groups one generally hydrogen (H^(+)) and the other a leaving group (L^(-)) resulting in the formation of unsaturated compound is known as elimination reaction. In E_(1) (elimination) reactions the C-L bond is broken heterolytically (in step 1) to form a carbocation (" as in "S_(N^(1)) reaction) in which (L^(-)) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta- carbon atom by a base (nucleophile) to form an alkene. E_(1) reaction is favoured in compounds in which the leaving group is at secondar (2^(@)) or tertiary (3^(@)) Position. In E_(2) (elimination) reactions two sigma bonds are broken and a pi- bond is formed simultaneously. E_(2) reactions occur in one step through a transition state. E_(2) reactions are most common in haloalkanes (particulary 1^(@) ) and better the leaving group higher is the E_(2) reaction. In E_(2) reactions, both the leaving groups should be antiplaner. E_(1) cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step R_(2)CH-underset(X)underset(|)C-R_(2) '+:B^(-) "(base)" to R_(2)C=CR_(2) +H-B+X^(-) this reaction is an example of:

The removal of two atoms or groups one generally hydrogen (H^(+)) and the other a leaving group (L^(-)) resulting in the formation of unsaturated compound is known as elimination reaction. In E_(1) (elimination) reactions the C-L bond is broken heterolytically (in step 1) to form a carbocation (as in S_(N^(1)) reaction) in which (L^(-)) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta- carbon atom by a base (nucleophile) to form an alkene. E_(1) reaction is favoured in compounds in which the leaving group is at secondar (2^(@)) or tertiary (3^(@)) Position. In E_(2) (elimination) reactions two sigma bonds are broken and a pi- bond is formed simultaneously. E_(2) reactions occur in one step through a transition state. E_(2) reactions are most common in haloalkanes (particulary 1^(@) ) and better the leaving group higher is the E_(2) reaction. In E_(2) reactions, both the leaving groups should be antiplaner. E_(1) cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step Neopentyl bromide undergoes dehydrohalogenation to give alkene even though it has no beta- hydrogen.This is due to :

The removal of two atoms or groups one generally hydrogen (H^(+)) and the other a leaving group (L^(-)) resulting in the formation of unsaturated compound is known as elimination reaction. In E_(1) (elimination) reactions the C-L bond is broken heterolytically (in step 1) to form a carbocation (" as in "S_(N^(1)) reaction) in which (L^(-)) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta- carbon atom by a base (nucleophile) to form an alkene. E_(1) reaction is favoured in compounds in which the leaving group is at secondar (2^(@)) or tertiary (3^(@)) Position. In E_(2) (elimination) reactions two sigma bonds are broken and a pi- bond is formed simultaneously. E_(2) reactions occur in one step through a transition state. E_(2) reactions are most common in haloalkanes (particulary 1^(@) ) and better the leaving group higher is the E_(2) reaction. In E_(2) reactions, both the leaving groups should be antiplaner. E_(1) cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step 2-Bromonbutane on heating with concentrated solution of alcoholic KOH gives major product as :

The removal of two atoms or groups one generally hydrogen (H^(+)) and the other a leaving group (L^(-)) resulting in the formation of unsaturated compound is known as elimination reaction. In E_(1) (elimination) reactions the C-L bond is broken heterolytically (in step 1) to form a carbocation (" as in "S_(N^(1)) reaction) in which (L^(-)) is lost (rate determining step). The carbocation (in step 2) loses a proton from the beta- carbon atom by a base (nucleophile) to form an alkene. E_(1) reaction is favoured in compounds in which the leaving group is at secondar (2^(@)) or tertiary (3^(@)) Position. In E_(2) (elimination) reactions two sigma bonds are broken and a pi- bond is formed simultaneously. E_(2) reactions occur in one step through a transition state. E_(2) reactions are most common in haloalkanes (particulary 1^(@) ) and better the leaving group higher is the E_(2) reaction. In E_(2) reactions, both the leaving groups should be antiplaner. E_(1) cb (Elimination unimolecular conjugate base) reaction involves the removal of proton by a conjugate base (step 1) to produce carbanion which loses a leaving group to form an alkene (step 2) and is a slow step Isopropyl chloride on heating with concentrated solution of ethanolic KOH gives mainly:

|| E2 and E1 elimination reaction ||