A hydrocarbon A on chlorination gives B which on heating with alcoholic potassium hydroxide changes into another hydrocarbon C. The latter decolourises Baeyer’s reagent and on ozonolysis forms formaldehyde only. A is

Ethane
Butane
Methane
Ethene

Solution

In the presence of platinum catalyst, hydrocarbon A adds hydrogen to form n-hexane. When hydrogen bromide is added to A instead of hydrogen, only a single bromo compound is formed. Which of the following is A?

CH₃ — CH₂ — CH = CH — CH₂ — CH₃
CH₃ — CH₂ — CH₂ — CH = CH — CH₃
CH₃ — CH = CH — CH₂ — CH₂ — CH₃
CH₂ = CH — CH₂ — CH₂ — CH₂ — CH₃

Solution

Which is least reactive towards addition of HCl ?

CH₃CH₂CH=CH₂
CH₂ = CH–CH= CH₂
CH₃CH₂C = CCH₃
CH₃CH=CHCHO

Solution

Presence of electron-withdrawing substituent (e.g. C =O, –COOH, –COOR, –CN, etc.) deactivates the carbon-carbon double bond towards electrophilic addition.

Which is the most suitable reagent among the following to distinguish compound (3) from rest of the compounds ?
1.CH₃ – C = C- CH₃
2.CH₃ – CH₂ – CH₂ – CH₃
3.CH₃ – CH₂C = CH
4.CH₃ – CH = CH₂

Bromine in carbon tetrachloride
Bromine in acetic acid
Alk KMnO₄
Ammonical silver nitrate

Solution

Br₂ in CCl₄(a), Br₂ in CH₃COOH(b) and alk. KMnO₄(c) will react with all unsaturated compounds, i.e., 1,3 and 4 while ammonical AgNO₃ (d) reacts only with terminal alkynes, i.e., 3 and hence 3 can be distinguished from 1, 2 and 4 by.ammonical AgNO₃(d).

The probability factor for the replacement of hydrogen atom during chlorination/bromination is

more important in chlorination
more important in bromination
equally important in chlorination/bromination
not definite

Solution

More is the reactivity of a free radical, lesser is its selectivity for different type of H atoms (1º, 2º, or3º)and hence the probability factor will be more in such reagent. Since Cl^• is more reactive, it is less selective and hence it will be more influenced by the probability factor.

Acetylenic hydrogens are acidic because

Sigma electron density of C – H bond in acetylene is nearer to carbon, which has 50% s-character
Acetylene has only open hydrogen in each carbon
Acetylene contains least number of hydrogens among the possible hydrocarbons having two carbons
Acetylene belongs to the class of alkynes with molecular formula, C_nH_2n-2

Solution

The acidity of acetylene or 1--alkyne can be explained on the basis of molecular orbital concept according to which formation of C—H bond in acetylene involves sp-hybridised carbon atom. Now since s electrons are closer to the nucleus than p electrons, the electrons present in a bond having more s character will be correspondingly more closer to the nucleus.

Thus owing to high s character of the C—H bond in alkynes (s = 50%), the electrons constituting this bond are more strongly held by the carbon nucleus i.e., the acetylenic carbon atom or the sp orbital acts as more electronegative species than the sp² and sp³ with the result the hydrogen present on such a carbon atom(=C—H) can be easily removed as a proton.

Solution

Br is less reactive and more selective and so the most stable free radical (3°) will be the major product.

Birch's reduction
Clemensen's reduction
Catalytic reduction
All of these

Solution

The given reaction is Birch reduction.

The nodal plane in the ?-bond of ethene is located in

the molecular plane
a plane parallel to the molecular plane
a plane perpendicular to the molecular plane which bisects the carbon - carbon ?-bond at right angle
a plane perpendicular to the molecular plane which contains the carbon - carbon ?-bond

Solution

The ? bond is formed by the sideways overlapping of two p-orbitals of the two carbon atoms.The molecular plane does not have any ? electron density as the p-orbitals are perpendicular to the plane containing the ethene molecule.The nodal plane in the ?-bond of ethene is located in the molecular plane.

an optically active compound
an optically inactive compound
a racemic mixture
a diastereomeric mixture

Solution

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