Animated Solution for Chemistry - Organic Chemistry: [where, $\text{Et}={\text{C}}_2{\text{H}}_5$, ${}^t\text{Bu}=({\text{CH}}_3{)}_3\text{C}\text{—}$] Consider the above reaction sequence, product A and product B formed respectively are

The problem presents us with a two-step organic synthesis starting from 1-bromoacetaldehyde ($Br-C{H}_2-CHO$). We are tasked with identifying two major products, A and B, formed sequentially.

The first step involves reacting the aldehyde with excess ethanol ($EtOH$) in the presence of dry $HCl$ gas. The second step takes the resulting Product A and treats it with potassium tert-butoxide (${}^{t}Bu{O}^{-}{K}^{+}$), a well-known strong and bulky base.

When an aldehyde is treated with an alcohol in an acidic medium, it undergoes a nucleophilic addition reaction. The dry $HCl$ acts as a catalyst, protonating the carbonyl oxygen to make the carbonyl carbon highly electrophilic.

Because ethanol is present in excess, the reaction doesn't stop at the hemiacetal stage. Instead, it proceeds all the way to form an acetal. The carbonyl oxygen is completely replaced by two ethoxy ($-OEt$) groups.

Applying this to our reactant, the $CHO$ group transforms into a $CH(OEt{)}_2$ group. The bromine atom on the adjacent carbon remains untouched because it does not react under these specific conditions. Thus, Product A is 1-bromo-2,2-diethoxyethane.

Next, Product A is subjected to potassium tert-butoxide. This reagent is a classic example of a sterically hindered, strong base. Due to its bulky nature, it is a poor nucleophile but an excellent proton acceptor, making it the perfect reagent to drive an E2 elimination reaction.

For an E2 elimination to occur, the base must abstract a proton from the $\beta$-carbon (the carbon adjacent to the one holding the leaving group). In Product A, the leaving group is the bromide ion ($B{r}^{-}$) on the $\alpha$-carbon. The adjacent carbon, which holds the two ethoxy groups, is the $\beta$-carbon.

The proton attached to this $\beta$-carbon is surprisingly acidic. This enhanced acidity is due to the strong $-I$ (inductive) effect of the two highly electronegative oxygen atoms pulling electron density away from the carbon.

The bulky tert-butoxide ion abstracts this acidic $\beta$-proton. Simultaneously, the electrons from the $C-H$ bond cascade down to form a carbon-carbon double bond, and the bromide ion is expelled.

This concerted mechanism yields Product B, which is 1,1-diethoxyethene. Comparing our derived structures for A and B with the given options, we find that Option (a) is the perfect match.