Cyclohexene
Cyclohexadiene
Cyclogexatriene
None of these

Solution

Solution

Solution

Compounds I and II are formed in nearly equal amounts
Compounds I is formed in larger proportions
Compounds II is formed in larger proportions
Relative amount of II cannot be predicted

Solution

Bromination of alkanes in presence of UV light takes place through free radical mechanism. Bromine substitutes that hydrogen, abstraction of which leads to formation of stablest free radical. (Reactivity Selectivity Principle). So in isobutane, bromine will come at second carbon, since after abstraction of hydrogen it will form 3° free radical giving II as the major product.

The reaction of propene with HOCl proceeds via the addition of

H⁺ in the first step
Cl⁺ in the first step
OH^- in the first step
Cl⁺ and OH^- in a single step

Combustion of which of the following compound (in the presence of excess of oxygen) does not result in the change in hybrid state of Catom –

CH₄
CH₂ = CH₂
CH₃ – CH₃
HC = CH

Solution

Combustion of hydrocarbons change them to CO₂ gas in which C is sp-hybridised. Therefore, HC = CH,in which carbon is in sp-hybrid state will not result in the change in hybrid state of C-atom.

A compound (X) (C₅H₈) reacts with ammonical AgNO₃ to give a white precipitate,and on oxidation with hot alkaline KMnO₄ gives the acid, (CH₃)₂CHCOOH, therefore X is –

CH₂= CH –CH = CH –CH₃
CH₃– CH = CH – CH₂ – CH₃
(CH₃)₂CH – C = CH
(CH₃)₂C = C = CH₂

Solution

Compound X reacts with ammonical AgNO₃ solution,so it must be a terminal alkyne.
Formation of (CH₃)₂CH COOH on oxidation of X with hot alkaline KMnO₄ further confirms that X is(CH₃)₂ CHC = CH.

Solution

Reaction (b) does not involve the formation of Cl⁺, the electrophile.

The conversion of ClCH=CHCl to Cl₂CH–CHCl₂ can be carried out with

Cl₂
Cl₂/hν
Cl₂/AlCl₃
Cl₂/aq. NaOH

The alkene C₆H₁₀ producing OHC(CH₂)₄CHO on ozonolysis is

Hexene-1
Hexene-3
Cyclohexene
1-Methylcyclohexene

Solution

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete