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Practice each question, then open the answer or worked solution.
Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Ans:
The structure of 2-methylbutane can be stated as (skeleton);
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On the basis of the above structure, various alkenes that will give 2-methylbutane on hydrogenation are given as;
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Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, ${{E}^{+}}$
a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
b) Toluene, $p-C{{H}_{3}}-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$ , $p-{{O}_{2}}N-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$ .
Ans:
Electrophiles are reagents that participate in a reaction by accepting an electron pair in order to bond to the corresponding nucleophiles.
The higher the electron density on a benzene ring, the more reactive is the compound towards an electrophile.
Here,
The presence of an EWG deactivates the aromatic ring by decreasing the electron density. Now, as the $N{{O}_{2}}^{-}$ group is more EWG that $C{{l}^{-}}$ group.
Thus, the decreasing order of EWG is given as;
Chlorobenzene > p-nitrochlorobenzene > 2,4-dinitrochlorobenzene
Ans:
Here, $C{{H}_{3}}^{-}$ is an EDG and $N{{O}_{2}}^{-}$is an EWG.
Thus, toluene will have the maximum electron density and is most easily attacked by ${{E}^{+}}$ .
The number of $N{{O}_{2}}^{-}$substituents define the order as;
Toluene > $p-C{{H}_{3}}-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$ > $p-{{O}_{2}}N-{{C}_{6}}{{H}_{4}}-N{{O}_{2}}$
Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?
Ans:
The ease of nitration depends on the presence of electron density on the compound to form nitrates. Nitration reactions are examples of electrophilic substitution reactions where an electron-rich species is attacked by a nitronium ion ($N{{O}_{2}}^{+}$ ).
Now as we know, $C{{H}_{3}}^{-}$group is electron donating and $N{{O}_{2}}^{-}$ is electron withdrawing. Therefore, toluene will have the maximum electron density among the 3 compounds followed by benzene. On the other hand, m– Dinitrobenzene will have the least electron density. Thus, it will undergo nitration with difficulty. Hence, the increasing order of nitration is as;
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Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Ans:
The ethylation reaction of benzene involves the addition of an ethyl group on the benzene ring. This reaction is called Friedel-Craft alkylation reaction and takes place in the presence of a Lewis acid.
Any Lewis acid like anhydrous $FeC{{l}_{3}},SnC{{l}_{4}},B{{F}_{3}}$ etc. can be used during the ethylation of benzene.
Why is the Wurtz reaction not preferred for the preparation of alkanes containing an odd number of carbon atoms? Illustrate your answer by taking one example.
Ans:
Wurtz reaction is limited for the synthesis of symmetrical alkanes (alkanes with an even number of carbon atoms). In the reaction, two similar alkyl halides are taken as reactants and an alkane, containing double the number of carbon atoms, are formed.
Example:
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This reaction cannot be used for the preparation of unsymmetrical alkanes because if two dissimilar alkyl halides are taken as the reactants, then a mixture of alkanes is obtained as the products.
Example:
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The boiling points of the above alkanes are very close. Hence, it becomes difficult to separate them.