\[{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{2}}MgBr\xrightarrow{{{C}_{2}}{{H}_{5}}OH}{{\left( C{{H}_{3}} \right)}_{2}}CHC{{H}_{3}}\]

So $$F$$ can have three possible structures

$$t - BuOK$$   is a bulky strong base and causes elimination (dehydrohlogenation) reaction
The given reaction proceeds via biomolecular elimination.

In (i) and (iv) due to the presence of Lucas reagent $$\left( {HCl + anh.\,ZnC{l_2}} \right)$$     alcohols give alkyl halides while in (iii) alkyl halide is formed due to $${S_N}1$$  reaction.

In $${S_N}2$$  mechanism transition state is pentavelent. For bulky alkyl group it will have sterical hinderance and smaller alkyl group will favour the $${S_N}2$$  mechanism. So the decreasing order of reactivity of alkyl halides is $$RC{H_2}X > {R_2}CX > {R_3}CX$$

Due to steric hindrance tertiary alkyl halide do not react by $${S_N}2$$  mechanism they react by $${S_N}1$$  mechanism. $${S_N}2$$  mechanisam is followed in case of primary and secondary alkyl halides
The order is
$$C{H_3} - X > C{H_3} - C{H_2}X > {\left( {C{H_3}} \right)_2}CH.X > {\left( {C{H_3}} \right)_3}C - X$$

Electrophilic addition reaction takes place via more stable carbocation.

When chlorine gas is reacted with propene at high temperature $$\left( {{{400}^ \circ }C} \right),$$   then substitution occurs in place of addition reaction. Hence, allyl chloride is formed

$$F, Cl,$$  $$Br$$  and $$I$$ are the elements of $$VII$$ $$A$$ group. In a group atomic radii increases from top to bottom and the bond dissociation energy decreases as
$$R - F > R - Cl > R - Br > R - I$$
So, during dehydrohalogenation $$R-I$$  bond breaks more easily than $$R-I$$  bond. Hence, order of reactivity will be
$$R - I > R - Br > R - Cl > R - F$$