Test: Alkene Preparation - NEET MCQ

With ethanolic KOH, we have E₂ elimination.

So, we have 3 products in all

Note that  H₂SO₄, ­H3PO₄ and KHSO₄ are dehydrating agents with rearrangement possible. While P₂O­₅, P­₄O₁₀, ThO₂, POCl₃ and Al₂O₃ are dehydrating agents without any possibility of rearrangement. 

The correct answer is option B

As there is no hydrogen at adjacent position for E₂ elimination.

Both are the same alkene. So only one alkene is formed.

The optically active C6H1 2 hydrocarbon is 3-Methylpent-1-ene, having one chiral carbon shown in fig A.
On catalytic hydrogenation , the compound obtained is shown in fig B, and the molecule does not have chiral carbon.
The reaction is C6H1 2 → C6H1 4 and this reaction takes place inthe presence of H2 and Pd.

Option a) Diazene(N₂H₂) is a hydrogenating agent. So, there will be no reaction.
Option b) Al₂O₃+CrO₃ acts as a dehydrogenation catalyst and so an alkene is formed. (Here 1-propene is formed)
Option c) This reaction is Wolff Kishner reaction. Here, acetone would be converted to alkane.
Option d) Zn/CH₃COOH substitutes Cl with H and an alkane is formed.

Alkenes do not undergo mercuration, indeed they undergo oxymercuration , a process in which an alkene is converted into an alcohol.

A Lindlar catalyst is a heterogeneous catalyst that consists of palladium deposited on calcium carbonate or barium sulphate which is then poisoned with various forms of lead or sulphur. It is used for the hydrogenation of alkynes to cis alkenes (i.e. without further reduction into alkanes) When the triple bond is (2-butyne) hydrogenated over the Lindlar’s catalyst i.e.Pd/CaCO₃
 it gives predominantly cis alkene(2-butene). 

Zn reacts with 2,3 - dibromobutane to give alkene. As trans 2-butene is more stable than cis 2-butene(owing tosterric hinderancce), former is major product.

The correct answer is Option D.
The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes get stronger with increase in the size of the molecules. In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane.

 2 is the right answer.

Alcoholic KOH causes dehydrohalogenation reaction i.e. elimination reaction.

1. methyl 1 pentene

2. 3 methyl 2 butene

3. 2 ethyl 1 butene

Hence 3 is the correct answer.

The correct answer is 5.
C=C-C-C-C-C-C , no geometrical isomerism so 1
C-C=C-C-C-C-C 2 , geometrical cis trans
C-C-C=C-C-C-C , cis trans again so 2
∴ 2+2+1=5