Haloalkanes and Haloarenes

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{OCH}}_2 \stackrel{{\mathrm{H}}^{+}}{\to } {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\stackrel{+}{\mathrm{H}}}_2 +{\mathrm{CH}}_3\mathrm{OH}$

                        $$\begin{gathered} \downarrow {\mathrm{I}}^{-} \\ {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{I} \end{gathered}$$

This can be explained on the basis of S_N1 mechanism, the carbonium ion produced being benzylium ion, since this type is more stable than alkylium ion.

Compounds containing CH₃CO group give the iodoform test. The reaction is

CH₃CO-CH₃ + 3I₂ +NaOH $\to$CHI₃ + HCOONa + H₂O

                                           iodoform (yellow ppt)

Hence, only acetone gives the iodoform test.

${\mathrm{CH}}_3\stackrel{+}{\mathrm{N}}{\mathrm{H}}_3\stackrel{-}{\mathrm{Cl}}$ is a salt which can be completely ionized in aqueous solution.

CH₃O^- is stronger nucleophilic reagent because methyl group increase the electron density on oxygen, since it show +I effect.

In the above reaction, alkyl halide is 1°. It always gives substitution reaction by S_N2/E-2 mechanism.

E.g. with (CH₃)₃C-O^-. It gives mainly elimination reaction and involves carbocation intermediate, i.e.primary carbocation. It rearranges itself to form tertiary carbocation. It is mainly because of the stability of a charged system which is increased by dispersal of the charge. Stability of carbocation is 3° > 2° > 1°. Rearrangement can be done in 2 ways- 

Therefore,

2-bromobutane on reaction with sodium ethoxide in ethanol gives 2-butene as a major product.

This is according to Saytzeffs rule i.e. the predominant product is the most substituted alkene. 2-butene is more stable than 1-butene due to presence of large number of hyperconjugating structures in 2-butene.

Primary halide undergoes alkaline hydrolysis through SN² mechanism .

Hydra - acids of halogens in increasing order of acidity :

HF < HCl < HBr < HI

As the strength of hydrogen bonding decreases in the order of :

HF > HCl > HBr > HI

CH₃CH₂C(Br)=CH-Cl

2 - bromo - 1 - chloro butene - 1

According to Saytzeff's rule, the major product will be the one that contains more number of substituents around the double bond. Therefore, the correct answer is CH₃CH = CHCH₃.