Identify \(\mathrm{Z}\) in the following series: \(\mathrm{CH}_{2}=\mathrm{CH}_{2}\) \(\mathrm{HBr} \rightarrow \mathrm{X} \stackrel{\text { Hydrolysis }}{\longrightarrow} \mathrm{Y} \stackrel{\mathrm{I}_{2}, \text { Excess }}{\longrightarrow} \mathrm{Z}\) (a) \(\mathrm{CHI}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CHO}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)

When it comes to understanding electrophilic addition reactions in organic chemistry, Markovnikov's rule is a fundamental principle that provides insights into the orientation of the addition.

Simply stated, Markovnikov's rule predicts that in the addition of a protic acid, like HBr, to an unsymmetrical alkene, the hydrogen (H) from the acid will attach to the carbon with the greater number of hydrogen atoms, while the more electronegative part (in this case, bromide ion, Br-) will go to the carbon with more alkyl substituents. This outcomes in the more stabilized carbocation, hence leading to the most stable product.

In our exercise, the reactant is ethene (CH_2=CH_2), which is symmetrical. Even so, following Markovnikov's rule, the H from HBr attaches to one of the carbon atoms, and the Br attaches to the other, resulting in 2-bromoehtane (CH_3-CH_2-Br). Understanding how Markovnikov's rule influences the outcomes of reactions can greatly improve the student's ability to predict products in organic synthesis.

Hydrolysis of alkyl halides is a reaction where the halide ion is replaced by a hydroxide ion, converting the alkyl halide into an alcohol.

The reaction typically occurs by adding water or aqueous hydroxide, which attacks the carbon bonded to the halide. This reaction forms a carbon-oxygen bond while eliminating the halide ion. The process is a key transformation in organic synthesis as it extends the functional versatility of alkyl chains.

In the given exercise, the hydrolysis step involves transforming 2-bromoehtane (CH_3-CH_2-Br) to ethanol (CH_3-CH_2-OH). For NEET preparations, students should note that hydrolysis not only increases the molecule's polarity but also its functional potential, allowing further reactions to take place.

The iodoform test is an important organic chemistry reaction specific to identifying methyl ketones or secondary alcohols that can be oxidized to methyl ketones.

When a compound that has a methyl group attached to a carbonyl group (as in methyl ketones) or is a suitable alcohol, it is treated with iodine in the presence of a base. The methyl group is converted into iodoform (CHI3), which is characterized by its yellow precipitate. This test serves as a qualitative method for the presence of such structures in a molecule.

In the context of the exercise, ethanol undergoes vigorous conditions in the presence of excess iodine to yield iodoform (CHI3), demonstrating the practical application of the iodoform test for confirming the presence of a certain functional group within the molecule.

Halogenation reactions are crucial transformations in organic chemistry where a halogen replaces one or more hydrogen atoms in an organic compound.

There are various types of halogenation including fluorination, chlorination, bromination, and iodination, each involving the addition or substitution of a halogen for another substituent. The specific reaction conditions can lead to different products, and the presence of a catalyst or UV light may be required to initiate the reaction.

In our exercise, the conversion of ethanol to iodoethane or iodoform depends on the amount of iodine and the reaction conditions. The excess iodine indicated points towards the formation of iodoform through the halogenation of the methyl group in ethanol. Understanding how the quantity of reactants and condition specifics affect the halogenation outcome is vital for accurately predicting reaction products.