Eliminating Alcohols: Why Do (A) and (B) Give Different Products?

In summary, E2 elimination using alc.KOH is a chemical reaction that involves treating an alkyl halide with an alcoholic solution of potassium hydroxide to form an alkene and a salt. The conditions required for this reaction include a strong base, an alcoholic solvent, and elevated temperatures. The mechanism of E2 elimination using alc.KOH is a one-step process that forms a carbanion intermediate before eliminating a halide ion. The major products of this reaction are an alkene and a salt. The advantages of E2 elimination using alc.KOH include high selectivity, mild reaction conditions, and the use of readily available reagents, making it suitable for industrial production.
  • #1
baldbrain
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Homework Statement
I just have a problem with (A) & (B)
IMG_20180719_173721.JPG

The attempt at a solution
Analysing all possible products;
(A) can give 1-pentene (minor) + 2-pentene (major). But you also have the E-Z diastereomers of 2-pentene. So that's total 3 possible products.
(B) will give only the diastereomers E & Z 2-pentene. So 2 possible products
Hence, (A)→(s) & (B)→(r)
But they've given (A)→(r) & (B)→(s)
How can it be?
 

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  • #2
Books are not infallible.
 
  • #3
Are you sure? No hitch? We aren't missing anything?
 
  • #4
Nope. S, R, Q, P.
 
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1. What is E2 elimination using alc.KOH?

E2 elimination using alc.KOH is a chemical reaction in which an alkyl halide is treated with an alcoholic solution of potassium hydroxide (KOH) to form an alkene and a salt. This reaction is known as an elimination reaction because it eliminates a molecule of HX (hydrogen halide) to form the alkene.

2. What are the conditions required for E2 elimination using alc.KOH?

The conditions required for E2 elimination using alc.KOH include a strong base (such as KOH) in an alcoholic solvent (such as ethanol or methanol) and a suitable alkyl halide. The reaction is typically carried out at elevated temperatures (>100°C) to facilitate the elimination process.

3. What is the mechanism of E2 elimination using alc.KOH?

The mechanism of E2 elimination using alc.KOH involves a one-step concerted process in which the base abstracts a proton from the beta carbon (adjacent to the halide) to form a carbanion intermediate. This intermediate then eliminates the halide ion to form the alkene.

4. What are the major products of E2 elimination using alc.KOH?

The major products of E2 elimination using alc.KOH are an alkene and a salt (typically potassium halide). The alkene is formed from the elimination of a molecule of HX (hydrogen halide) from the alkyl halide, while the salt is formed from the reaction of the base (KOH) with the eliminated halide ion.

5. What are the advantages of E2 elimination using alc.KOH?

E2 elimination using alc.KOH has several advantages, including high selectivity for the formation of the desired alkene product, mild reaction conditions, and the use of relatively inexpensive and readily available reagents. Additionally, the reaction can be easily scaled up for industrial production of alkenes.

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