Chemistry MCQs for NEET — Practice Questions with Answers

10 ml of $1M\; H2SO4$ corresponds to 10 mmol.

 Friedel-Craft's alkylation : When benzene reacts with alkyl halide in presence of anhy. AlCl₃, toluene is obtained. It is called Friedel-Craft's alkylation. Friedel-Craft's reaction is type of electrophilic substitution.

  C₆H₆ + CH₃Cl $\stackrel{\mathrm{Anhy}.{\mathrm{AlCl}}_3}{\to }$C₆H₅CH₃ + HCl

                                      Toluene

$3{\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2\stackrel{{\mathrm{B}}_2{\mathrm{H}}_6}{\to }{\left({\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2\right)}_3\mathrm{B}\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2/{\mathrm{OH}}^{-}}{\to }\underset{\mathrm{Propan}-1-\mathrm{ol}}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{OH}}$

Here, Half mole of (B₂H₆) diborane reacts with propane by Markownikoffs addition it gives tripropyl borane called hydroboration. In presence of H₂O₂ in basic medium tripropyl borane gives alcohol. Remember that product is Anti-Markownikoff's rule that is 1-propanol. Reaction is called hydroboration oxidation.

In the Friedel-Crafts alkylation of benzene to produce toluene, the reactants are benzene (C6H6) and methyl chloride (CH3Cl) in the presence of anhydrous aluminum chloride (AlCl3) as a Lewis acid catalyst. The methyl group from methyl chloride gets attached to the benzene ring, forming toluene.

Acidic character of alkynes can be explained on the basis of sp-hybridisation state of the carbon atom in alkynes. We know that an electron in s orbital is more tightly held than in a p orbital because s-electrons are more closer to the nucleus. In sp-hybridisation, s-character is more (50%) as compared to ${\mathrm{sp}}^2$-(33%) or${\mathrm{sp}}^3$-(25%) hybrid orbitals. Due to very large s-character, the electrons in sp-hybrid orbitals are held tightly by the nucleus and are quite electronegative. Consequently, the electron pair of $\mathrm{H}-\mathrm{C}\equiv$ bond gets displaced more towards the carbon atom and help in the release of ${\mathrm{H}}^{+}$ ions.

For  example ethyne react with sodium metal and release the ${\mathrm{H}}^{+}$ ions.

           $\mathrm{CH}\equiv \mathrm{CH}$ + Na$\to \mathrm{HC}\equiv {\mathrm{C}}^{-}{\mathrm{Na}}^{+}$

(a) Sodamide is strong base, attracts the more acidic hydrogen and gives sodium salt

$\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$ + ${\mathrm{NaNH}}_2$ $\to$$\mathrm{H}-\mathrm{C}\equiv {\mathrm{C}}^{-}{\mathrm{Na}}^{+}$ + $\frac{1}{2}{\mathrm{H}}_2$

                                            Sodium ethynide

$$\begin{gathered} {\mathrm{Cl}}_2+{\mathrm{H}}_2\mathrm{O}\to \underset{\mathrm{hypochlorous} \mathrm{Acid}}{\mathrm{HOCl}+\mathrm{HCl}} \\ {\mathrm{CH}}_2={\mathrm{CH}}_2\stackrel{\mathrm{HOCl}}{\to }\underset{\mathrm{OH}}{\underset{|}{{\mathrm{CH}}_2}}-\underset{\mathrm{Cl}}{\underset{|}{{\mathrm{CH}}_2}} \left(\mathrm{Ethylene} \mathrm{chlorohydrin}\right) \end{gathered}$$