What are ambident nucleophiles? Give two examples of it.

Ambident Nucleophile- An anionic nucleophile, which has two nucleophilic centers, or two negative sites is known as an ambident nucleophile. This negative charge is delocalized due to resonance. Example – Cyanide and Thiocyanate are examples of ambident nucleophiles.

Give reason: Racemic mixture is optically inactive.

Racemic mixture contains two enantiomers (d and l forms) in equal proportions. The rotation caused by the molecules of one enantiomer is cancelled by the rotation caused by the molecules of other enantiomer. Therefore, it has zero optical rotation and hence, it is optically inactive.

How will you convert the following? Give chemical equations only. Aniline to chlorobenzene

The conversion of aniline to chlorobenzene actually consists of 2 steps. The first step is the reaction of aniline with NaNO 2 / HCl . This results in the formation of a diazonium salt, benzene diazonium chloride. This reaction occurs in the temperature range of 273 K - 278 K . The next step is treating the diazonium salt with CuCl . This process is commonly known as the Sandmeyer reaction. This reaction is an example of a nucleophilic aromatic substitution.

How will you convert the following? Give chemical equations only. Benzene to biphenyl

Benzene is converted to biphenyl by first reacting benzene with bromine in presence of iron to form bromobenzene. Bromo benzene is then reacted with sodium in presence of dry ether to form biphenyl and sodium bromide. C 6 H 6 + Br 2 → Fe C 6 H 5 − Br → Na / Dry ether C 6 H 5 − C 6 H 5 + NaBr

How will you convert the following? Give chemical equations only. Ethane to bromoethene

Ethane is converted to bromoethene by first reacting ethane with bromine in presence of U.V light to form bromoethane. Then bromoethane is reacted with alcoholic potassium hydroxide to form ethene. Ethene is then treated with bromine to form dibromoethane and on reacting with alcoholic potassium hydroxide forms bromoethene. CH 3 − CH 3 + Br 2 → uv light CH 3 − CH 2 − Br → alc . KOH CH 2 = CH 2 → Br 2 Br − CH 2 − CH 2 − Br → alc . KOH CH 2 = CHBr

Give reasons: The presence of the nitro group - NO 2 at 0/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.

The presence of an electron-withdrawing group at ortho- and para-positions increases the reactivity of haloarenes. The effect is pronounced when a nitro group is introduced at ortho and para- positions. This is often because the group -positions withdraw the electrons from the benzene formula which facilitates the attack of the nucleophile. The charge within the carbanion formed at o/p positions with reference to halogen atom is stabilized by the presence of nitro groups and resonance respectively.

Give reasons: Haloalkanes easily dissolve in organic solvents.

Haloalkanes easily tend to dissolve in organic solvents because the new intermolecular attractions between haloalkanes and solvent molecules have much the same strength as the ones being broken in the separate haloalkane and solvent molecules.

Give reasons: C — Cl bond length in chlorobenzene is shorter than C — Cl bond length in chloromethane.

C — Cl bond length in chlorobenzene is shorter than C — Cl bond length in chloromethane because in chlorobenzene each carbon atom is sp 2 hybridised due to resonance there is a partial double bond character, so the bond length is short.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Haloalkanes and Haloarenes>Assam Board-2019>Q 6

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Give reason:

n-Butyl bromide has higher boiling point than t-Butyl bromide.

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