NEET Practice Questions (MCQs) with Answers & Solutions

The Balz-Schiemann reaction involves the conversion of an aryl amine to an aryl fluoride via a diazonium salt intermediate. The reaction is represented as follows: $C_6H_5N_2Cl^- + HBF_4 ightarrow C_6H_5N_2^+BF_4^- ightarrow[ riangle] C_6H_5F$. This reaction is useful for introducing a fluorine atom into an aromatic ring.

The compound $CH_2=CH-CN$ can be named in different ways:

1. Acrylonitrile, which is the common name.
2. Vinyl cyanide, which indicates the presence of a vinyl group ($CH_2=CH-$) and a cyanide group ($-CN$).
3. Prop-2-enenitrile, which follows the IUPAC nomenclature rules, indicating a three-carbon chain with a double bond starting from the second carbon and a nitrile group ($-CN$). Therefore, all the given names are correct.

For the molecular formula $C_5H_{12}O$, there are 8 possible isomeric alcohols. These include different structural isomers with variations in the position of the hydroxyl group ($-OH$) and branching of the carbon chain. The isomers can be primary, secondary, or tertiary alcohols, each with a unique structure. Therefore, the correct answer is 8.

The conversion of $CH_3CH_2CH_2OH$ (1-propanol) to $CH_3CH_2CH_2COOH$ (butanoic acid) involves the following steps: first, reaction with $PBr_3$ to replace the hydroxyl group with a bromine atom to form $CH_3CH_2CH_2Br$ (1-bromopropane). Next, this compound reacts with $KCN$ to form $CH_3CH_2CH_2CN$ (butyronitrile). Finally, hydrolysis of the nitrile with $H_3O^+$ yields the carboxylic acid $CH_3CH_2CH_2COOH$.

In the dehydration of alcohols to alkenes using concentrated $H_2SO_4$, the initiation step is the protonation of the alcohol molecule. This step involves the alcohol accepting a proton from the $H_2SO_4$, forming an oxonium ion, which makes the -OH group a better leaving group.