Electrophilic Aromatic Substitution (Haloalkanes and Haloarenes) MCQ Questions & Answers in Organic Chemistry | Chemistry

(ii) and (iii) carbon atoms are asymmetric.

$$A$$  is 1, 4-addition product due to conjugation $$B$$  is obtained by further addition of $$HBr.$$  $$\left( {C{H_3}CHBr.C{H_2}.C{H_2}Br} \right).B$$       gives $$C$$  by cyclisation If $$B$$  reacts by $${S_N}2$$  mechanism, $$Br$$  on $${1^ \circ }$$ carbon must be replaced by $$O{H^ - }$$  to give $$C{H_3}CHBr - C{H_2} - C{H_2}OH.$$

Let $$iso$$  - butane be $$a$$ $$g$$ and $$iso$$  - butene $$b$$ $$g. $$
Thus, $$a + b = 10\,g$$
\[\underset{\begin{align} & \,\,\,\,\,\,\,\,| \\ & \,\,\,\,\,\,C{{H}_{3}} \\ & \,\,\text{iso-butene} \\ & \,\,\,\,\,\,\,\,\text{(56)} \\ \end{align}}{\mathop{C{{H}_{3}}-C=C{{H}_{2}}}}\,\]    \[\underset{\left( 80\times 2 \right)}{\mathop{+\,B{{r}_{2}}\to }}\,C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,Br \\ | \end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{2}}Br\]
\[C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\,\, \\ C{{H}_{3}}\,\, \\ \text{iso-butane} \end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{3}}+B{{r}_{2}}\to \]      No addition reaction
Now $$160\,g$$  of $$B{r_2}$$  reacts with $$56\,g$$  of $$iso$$  - butene
$$\therefore \,20\,g$$   of $$B{r_2}$$  reacts with $$\frac{{56 \times 20}}{{160}} = 7\,g$$    of $$iso$$  - butene
Thus, amount of $$iso$$  - butane $$(a)$$  is $$10 - 7 = 3 \,g$$
\[\underset{\left( \text{Mol}\text{. mass = 56} \right)}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\, \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{-C=}}\,C{{H}_{2}}+{{H}_{2}}\to }}\,\]     \[\underset{\left( \text{Mol}\text{. mass = 58} \right)}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\, \\ C{{H}_{3}}\, \end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{3}}}}\,\]
Now $$56\,g$$  of $$iso$$  - butene gives $$58\,g$$  of $$iso$$  - butane
$$\therefore \,7\,g$$   of $$iso$$  - butene gives $$\frac{{58 \times 7}}{{56}} = 7.25\,g$$     of $$iso$$  - butane
Total amount of $$iso$$  - butane available for $$10 \,g$$  mixture $$ = \left( {7.25 + 3} \right) = 10.25\,g$$
\[\underset{\left( \text{Mol}\text{. mass = 58} \right)}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\, \\ C{{H}_{3}}\,\, \end{smallmatrix}}{\mathop{-CH-}}\,}}\,C{{H}_{3}}+B{{r}_{2}}\to \]      \[\underset{\begin{smallmatrix} \text{tert-butyl bromide} \\ \left( \text{Mol}\text{. mass = 137} \right) \end{smallmatrix}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,Br \\ | \end{smallmatrix}}{\mathop{-C-}}}\,C{{H}_{3}}}}\,\]
Now $$58\,g$$  of iso-butane gives $$137\,g$$  of $$tert$$  - butyl bromide
$$\therefore \,10.25\,g$$   of $$iso$$  - butane gives $$\frac{{137 \times 10.25}}{{58}} = 24.21\,g$$      of $$tert$$  - butylbromide.

As $$C-I$$  bond is weakest, $${\left( {C{H_3}} \right)_3}C - I$$    will undergo $${S_N}1$$  reaction most readily.

heat favours elimination.

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\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}I\xrightarrow{alc.\,KOH}\]      \[C{{H}_{3}}-\underset{\left( X \right)}{\mathop{CH}}\,=C{{H}_{2}}\]
\[C{{H}_{3}}-\underset{\left( X \right)}{\mathop{CH}}\,=C{{H}_{2}}\xrightarrow{B{{r}_{2}}}\]     \[C{{H}_{3}}\underset{\begin{smallmatrix} |\,\,\,\,\,\, \\ Br\,\,\,\, \\ \left( Y \right)\,\,\,\, \end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{2}}Br\]

In compound (i), $$Br$$  is directly attached to chiral $$C$$ atom thus, will give racemic mixture on nucleophilic substitution $$\left( {{S_N}1} \right)$$  by $$O{H^ - }$$ ion.

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$${C_4}{H_7}Cl$$   is a monochloro derivative of $${C_4}{H_8}$$  which itself exists in three acyclic isomeric forms.
\[\begin{align} & \underset{\begin{smallmatrix} I \\ \text{(Its four }C'\text{s are different)} \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}CH}}\,=C{{H}_{2}} \\ & \underset{\begin{smallmatrix} II \\ \text{(It has 2 types of carbon)} \end{smallmatrix}}{\mathop{C{{H}_{3}}CH}}\,=CHC{{H}_{3}} \\ \end{align}\]

Grand total of acyclic isomers $$ = 6 + 4 + 2 = 12$$