NEET Practice Questions (MCQs) with Answers & Solutions

Halogens are the most reactive elements. High reactivity is due to their low dissociation energy. Further among halogens reactivity decreases as we move down the group. 

In case of different nucleophiles, but present in the same group in the periodic table, then larger is the atomic mass, higher is the nucleophilicity. Hence the decreasing order of nucleophilicity of the halide ions is I- > Br- > Cl- > F- 

When the Lassaigne's extract is boiled with concentrated HNO3, it decomposes Na2S and NaCN (if formed during the test) to prevent their interference in the detection of halogens. This step ensures that the precipitation of AgCl is due to the presence of halogens in the given organic compound.

Ether Boiling point = 34.5° C

Benzene boiling point = 80°C 

Both can be separated by Distillation.

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