Alkynes

The problem involves a sequence of reactions that transform compound 'A' into compound 'B'. Let's analyze the steps to determine the structure and chirality of compound 'B'.

Step 1: Formation of Grignard Reagent

The first step involves the reaction of compound 'A' with magnesium (Mg) in the presence of diethyl ether (Et₂O). This is a typical method to form a Grignard reagent. The bromine atom in compound 'A' is replaced by a magnesium-bromide (MgBr) unit:

Step 2: Reaction with D₂O

In the second step, the Grignard reagent reacts with D₂O (deuterium oxide). Grignard reagents are highly reactive with protic solvents like water and deuterium oxide. The deuterium from D₂O replaces the magnesium-bromide group, resulting in the formation of a deuterated compound.

The overall reaction can be summarized as the introduction of a deuterium (D) atom in place of the bromine atom:

Conclusion: Structure and Chirality of Compound 'B'

Compound 'B' is a deuterated isobutane. The introduction of deuterium does not create a new chiral center; therefore, the compound remains achiral.

Out of the provided options, this analysis aligns with the option that shows isobutane with D in place of the Br group, having no chiral center.

Step 1: Reaction with and Dilute HCl (Formation of A)
The starting compound is butan-2-one. The reaction with followed by dilute HCl results in the formation of a cyanohydrin. The CN ion from NaCN attacks the carbonyl carbon, and the oxygen picks up a proton. The major product is 2-hydroxy-2-methylbutanenitrile.
Step 2: Reduction with (Formation of B)
Lithium aluminum hydride is a strong reducing agent. It reduces the nitrile group (CN) to an amine group . The product is 1-amino-2-methylbutan-2-ol.
Step 3: Hydrolysis with and Heat (Formation of C)
The amine group in is hydrolyzed with and heat ( ) into a carboxylic group.
The major product is 2-hydroxy-2-methylbutanoic acid.
Conclusion
The structures of , , and correspond to option (4).

To solve this question about the properties of ethyne, we need to analyze each statement given in the options:

1. The C–C bonds in ethyne is shorter than that in ethene: This is a true statement. In ethyne (acetylene, C₂H₂), the carbon-carbon bond is a triple bond, which is shorter than the double bond in ethene (C₂H₄).
2. Both carbons are sp hybridised: This is also true. In ethyne, each carbon atom is sp hybridized, forming a linear molecule with an angle of 180 degrees between bonds.
3. Ethyne is linear: This statement is true. Due to sp hybridization, ethyne has a linear geometry.
4. The carbon-carbon bonds in ethyne is weaker than that in ethene: This is an incorrect statement. The carbon-carbon triple bond in ethyne is stronger than the carbon-carbon double bond in ethene due to the presence of more overlapping orbitals in a triple bond.

Based on the analysis, the incorrect statement is: The carbon-carbon bonds in ethyne is weaker than that in ethene.