Question

A bromoalkane 'X' reacts with magnesium in dry ether to form compound 'y'. The reaction of 'y' with methanal followed by hydrolysis yields an alcohol having molecular formula ${{C}_{4}}{{H}_{10}}O$. The compound 'X' is

Answer 1

Hint: An attempt to this question can be made by understanding the action of the reagents given in the question. Magnesium in the presence of dry ether is used to prepare the reacting molecule of Grignard's reagent. Determine the action of Grignard's reagent on carbonyl compounds. Based on this you can find the initial reacting molecule using the reaction mechanisms for the given named reactions.

Complete solution:
> Grignard's reagent is made by the addition of magnesium metal to alkyl halides. The reagent then reacts with a carbonyl compound. The anionic part of reagent attacks the carbon containing the carbonyl group and attaches to it, increasing the length of the chain.
> The product compound has 4 carbon atoms and the reagent reacts with methanal. So the initial reactant molecule will consist of only 3 carbon atoms.
> We will assume the compound 'X' to be Bromopropane and then carry out the reactions to see if our assumption is correct.
> Bromopropane on reaction with Mg/dry ether gives the following product.
$C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\,\xrightarrow{Mg/dry\,ether}\,\,\,C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}^{-}{{(MgBr)}^{+}}$
> Mg atoms are inserted between the C-X bonds converting the covalent compound into ionic compounds.
This is the possible structure for compound 'Y'.
> Now the compound 'Y' reacts with methanal. As mentioned above, the propyl anion attacks the carbon atom having a carbonyl group. The reaction is given below.
$C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}MgBr\,\xrightarrow{HCHO}\,\,\,C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}{{O}^{-}}\,M{{g}^{2+}}B{{r}^{-}}$
> Upon hydrolysis we obtain,
$C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}{{O}^{-}}\,M{{g}^{2+}}B{{r}^{-}}\,\xrightarrow{{{H}_{2}}O}\,\,\,C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\,+\,MgBr$
> The product obtained in condensed form is ${{C}_{4}}{{H}_{10}}O$. Thus, our assumption is correct and the initial reacting molecule 'X' is bromopropane.

So, the correct answer is “Option B”.

Note: Dry ether is used for the preparation of Grignard's reagent. This is because it acts as a solvent as well as stabilises the reaction by acting as an aprotic solvent.