Find out the final product C

Question

Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).

Answer 1

The given reaction involves a series of steps leading to the final product C. Let's analyze each step to identify the compounds formed at each stage:

Initial Reaction: The compound given is \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br}. It reacts with alcoholic KOH.
This is an elimination reaction where hydrogen halide (HBr) is removed, resulting in the formation of an alkene. Thus, \text{A} = \text{Propene (CH}_3\text{-CH}=\text{CH}_2\text{)}.
First Intermediate Reaction: The alkene propene reacts with HBr following Markovnikov's rule, adding the Br to the more substituted carbon atom, giving the compound 2-bromopropane (B).
\text{B} = \text{CH}_3\text{-CHBr-CH}_3.
Final Reaction: 2-bromopropane reacts with alcoholic KOH. This is another elimination reaction similar to the initial step, resulting in the formation of another alkene. However, since the solvent is alcoholic, it favors substitution, converting the bromide to an alcohol. The product formed is propan-2-ol.
\text{C} = \text{Propan-2-ol (CH}_3\text{-CHOH-CH}_3\text{)}.

Thus, the final product C is Propan-2-ol. This matches the given correct answer option.

Answer 2

The given reaction involves a series of steps leading to the final product C. Let's analyze each step to identify the compounds formed at each stage:

Initial Reaction: The compound given is \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br}. It reacts with alcoholic KOH.
This is an elimination reaction where hydrogen halide (HBr) is removed, resulting in the formation of an alkene. Thus, \text{A} = \text{Propene (CH}_3\text{-CH}=\text{CH}_2\text{)}.
First Intermediate Reaction: The alkene propene reacts with HBr following Markovnikov's rule, adding the Br to the more substituted carbon atom, giving the compound 2-bromopropane (B).
\text{B} = \text{CH}_3\text{-CHBr-CH}_3.
Final Reaction: 2-bromopropane reacts with alcoholic KOH. This is another elimination reaction similar to the initial step, resulting in the formation of another alkene. However, since the solvent is alcoholic, it favors substitution, converting the bromide to an alcohol. The product formed is propan-2-ol.
\text{C} = \text{Propan-2-ol (CH}_3\text{-CHOH-CH}_3\text{)}.

Thus, the final product C is Propan-2-ol. This matches the given correct answer option.

The Correct Option is B