HALOALKANES AND HALOARENES CLASS 12 NCERT SOLUTIONS PDF

Thus, C atom of chlorobenzene has less tendency to release electrons to Cl than carbon atom of cyclohexylchloride. As a result, C — Cl bond in chlorobenzene is less polar than in cyclohexylchloride. Further, due to delocalization of lone pairs of electrons of the Cl atom over the benzene ring, C-Cl bond in chlorobenzene acquires some double bond character while the C — Cl in cyclohexy! In other words, C-Cl bond in chlorobenzene is shorter than in cyclohexyl chloride. Since dipole moment is a product of charge and distance, therefore, chlorobenzene has lower dipole moment than cyclohexylchloride due to lower magnitude of negative charge on the Cl atom and shorter C-Cl distance. The molecules of H2O are hold together by H-bonds.

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Draw the structures of major monohalo products in each of the following reactions : Solution: The major haloderivatives formed in the given reactions are Question 6. Arrange each set of compounds in order of increasing boiling points. Solution: The boiling points of organic compounds depend on the strength of the intermolecular forces in them. These forces are : a van der Waals forces and b dipole-dipole interactions These forces are dependent on the i molecular mass and ii surface area of the molecules i As the molecular mass of the compound increases, the boiling point also increases.

Branched compounds are more compact and therefore have less surface area as compared to their straight chain counterparts and therefore lower boiling point. Explain your answer. There are 5 atoms simultaneously bonded to it. Thus, for such a transition state to be possible, there should be minimum steric hindrance.

In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction? This intermediate is formed by the cleavage of the C — X bond.

More stable is the resultant carbocation faster is the SN1 reaction. Solution: A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight.

Identify the hydrocarbon. The hydrocarbon with molecular formula C5H10 can either a cycloalkane or an alkene. Since the compound does not react with Cl2 in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since the cycloalkane reacts with Cl2 in the presence of bright sunlight to give a single monochloro compound, C5H9Cl, therefore, all the ten hydrogen atoms of the cycloalkanes must be equivalent.

Thus, the cycloalkane is cyclopentane.

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