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

Though polar, alkyl halides cannot form hydrogen bonds with water hence they are insoluble in water.

$${\left( {C{H_3}} \right)_3}CCl > {\left( {C{H_3}} \right)_2}CHC{H_2}Cl$$        is not correct is not correct $${\left( {C{H_3}} \right)_3}CCl$$    is smaller than $${\left( {C{H_3}} \right)_2}CHC{H_2}Cl.$$     It is due to large surface area and shorter alkyl chain.

Nuclear substitution will not take place.

For a given alkyl group, the order of reactivity is \[\xleftarrow[\text{increasing bond energy}]{R-I>R-Br>R-Cl>R-F}\]     decreasing halogen reactivity.
This order depends on the carbon-halogen bond energy; the carbon-fluorine bond energy is maximum and thus fluorides are least reactive while carbon-iodine bond energy is minimum hence iodides are most reactive.

Bond strength follows the order
$$C - F > C - Cl > C - Br > C - I$$

The reactivity of different alkyl halides towards $${S_N}2$$  reaction decreases in the order :
methyl halides > $${1^ \circ }$$ halides > $${2^ \circ }$$ halides > $${3^ \circ }$$ halides.
Since, 2-bromo-2-methylbutane is a tertiary bromide hence it is least reactive among the given.

$$\Delta T = 179 - 167 = 12,w = 0.0852\,g,$$       $$w = 2\,g,{K_f} = 40,$$    Molecular weight of $$(A)$$
$$\eqalign{ & = \frac{{1000 \times {K_f} \times w}}{{\Delta T \times W}} \cr & = \frac{{1000 \times 40 \times 0.0852}}{{12 \times 2}} \cr & = 142 \cr} $$
$$(A)$$  undergoes Wurtz reaction to form $$(B)$$  i.e.,
\[\left( A \right)\xrightarrow{Na}\left( B \right)+NaX\]
$$(B)$$  is an alkane say $${C_n}{H_{2n + 2}}$$
$$280\,mL$$   of $$(B)$$  weighs $$0.375\,g$$   at $$NTP$$
$$\therefore \,22400\,mL$$    of $$(B)$$  weighs $$ = \frac{{0.375 \times 22400}}{{280}} = $$     $$30\,g$$  at $$NTP$$
$$\therefore \,M.wt.\,\,{\text{of}}\,\,\left( B \right) = 30,$$     $$12n + 2n + 2 = 30,\,n = 2$$
Thus $$(B)$$  is ethane and therefore $$(A)$$  is $$C{H_3}X.$$
The $$m.wt.$$  of $$C{H_3}X = 142$$
$$At.\,wt.$$   of $$X = 127$$   $$\therefore \,X$$  is iodine
Therefore. alkyl halide is $$C{H_3}I.$$   This reaction is
\[\underset{\left( A \right)}{\mathop{2C{{H}_{3}}I}}\,\xrightarrow[\text{either}]{Na}\underset{\left( B \right)}{\mathop{{{C}_{2}}{{H}_{6}}}}\,\]

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At bridge head position $${S_N}1$$  and $${S_N}2$$  do not takes place.

Nuclear substituted halogen has double bond character due to resonance.