Introduction To Organic Chemistry

Step 1: Elements with atomic numbers beyond 100 were initially assigned systematic names based on their digits.
Step 2: The element with atomic number 111 was initially called Unununnium (Uuu), following the IUPAC temporary naming convention.
Step 3: Later, IUPAC officially named element 111 as Roentgenium (Rg) in honor of Wilhelm Roentgen, the discoverer of X-rays.
Step 4: Therefore, the correct answer is (D).

Step 1: Conformational isomerism occurs due to rotation around single bonds (sigma bonds), allowing the molecule to adopt different spatial orientations without breaking any bonds.
Step 2: Ethene ( ) contains a double bond between the two carbon atoms, which restricts rotation and thus prevents any conformational changes.
Step 3: In contrast, molecules like ethane, n-butane, 2,3-dimethylbutane, and cyclohexane all contain single sigma bonds that allow for such rotations.
Step 4: Therefore, ethene does not exhibit conformational isomerism due to the rigidity imposed by its double bond.

Step 1: In reactions, the reactivity of alkyl halides decreases with increasing steric hindrance around the carbon atom bearing the leaving group (bromine in this case).
Step 2: (n-butyl bromide) is the least hindered and thus most reactive in reactions due to having a primary carbon.
Step 3: (isobutyl bromide) and (sec-butyl bromide) are more hindered than n-butyl bromide. Isobutyl bromide, being less hindered than sec-butyl bromide, is more reactive.
Step 4: (tert-butyl bromide) is the most hindered with a tertiary carbon, making it the least reactive in reactions.
Step 5: Thus, the correct order from most reactive to least reactive for reactions is .

Step 1: The acidity of phenols is significantly influenced by substituents on the aromatic ring.
Step 2: Nitro groups are strong electron-withdrawing groups due to their -I (inductive) and -R (resonance) effects, which stabilize the phenoxide ion formed upon deprotonation.
Step 3: In para positions, the nitro group exerts a stronger electron-withdrawing effect compared to the ortho position due to better resonance stabilization of the negative charge on the oxygen atom of the phenoxide ion.
Step 4: Therefore, p-nitrophenol is more acidic than o-nitrophenol, cresols, and phenol, making it the most acidic compound among the given options.

Step 1: The Hell-Volhard-Zelinsky reaction specifically involves the halogenation of the alpha-carbon of carboxylic acids.
Step 2: This reaction uses a halogen (Br ) and red phosphorus, which generates phosphorus tribromide (PBr ), acting as a catalyst.
Step 3: Phosphorus tribromide converts the carboxylic acid to an acyl bromide, which subsequently undergoes alpha-bromination.
Step 4: The product, 2-bromopropanoic acid, confirms the pathway and mechanism of the Hell-Volhard-Zelinsky reaction.

Step 1: Halogens, including chlorine, are generally deactivating because of their strong electronegativity, which tends to withdraw electron density from the benzene ring through the inductive effect.
Step 2: Despite being deactivating, halogens are ortho-para directors because they can donate electron density back to the ring through resonance (mesomeric effect).
Step 3: Chlorine, therefore, is an ortho-para director but is overall deactivating due to its electron-withdrawing inductive effect being stronger than its electron-donating resonance effect.

Step 1: The Gatterman reaction is a well-known method for introducing a chlorine atom into an aromatic ring.
Step 2: This reaction involves the use of benzene diazonium chloride, which, under the presence of copper(I) chloride (CuCl) and HCl, forms chlorobenzene.
Step 3: The reaction proceeds through the replacement of the diazonium group ( ) by a chlorine atom.
Step 4: Therefore, benzene diazonium chloride is specifically converted to chlorobenzene in the Gatterman reaction.

Step 1: Aniline ( ) is less basic than ammonia ( ) due to the electron-withdrawing nature of the phenyl group via resonance.
Step 2: Benzylamine ( ) is more basic than aniline because the benzyl group is less electron-withdrawing compared to a direct phenyl attachment.
Step 3: Ethylamine ( ) is more basic than both benzylamine and ammonia due to the electron-donating effect of the ethyl group, enhancing the electron density on the nitrogen atom.
Step 4: The correct order of increasing basicity, considering the electronic effects, is therefore aniline, ammonia, benzylamine, and ethylamine.

Step 1: Animal starch is a common term used to refer to the primary form of stored carbohydrate in animals.
Step 2: Glycogen is a polysaccharide that serves as a form of energy storage in fungi and animals. It is highly branched and compact, making it ideal for quick energy release.
Step 3: Lactose, cellulose, amylase, and maltose have different roles: lactose is a sugar found in milk; cellulose is a structural component in plants; amylase is an enzyme that breaks down starch; and maltose, a disaccharide, is made from two glucose units.
Step 4: Therefore, the correct answer is glycogen, which matches the description of animal starch.

This statement is incorrect because electromeric effects are stronger and more immediate compared to inductive effects, meaning the electromeric effect will dominate when they oppose each other.

Ozonolysis is the reaction of alkenes with ozone to produce ozonides. It can be used to identify both the number of double bonds and the position of the double bonds in unsymmetrical alkenes.
Thus, the statement in option (D) is incorrect because ozonolysis can indeed be used to detect the position of the double bonds.

Step 1: The reduction of phenol (X) by zinc dust gives benzene (Y).
Step 2: Benzene (Y) reacts with acetyl chloride (CH COCl) in presence of AlCl to give acetophenone (Z).

Starch is a polysaccharide made up of glucose units and consists of two main components: amylose and amylopectin. Amylose is a linear polymer of glucose, while amylopectin has a branched structure. On average, amylose makes up about 15-20% of starch, while the remaining 80-85% is amylopectin.
Thus, the correct answer is , corresponding to option (E).

The Reimer-Tiemann reaction involves the electrophilic substitution of phenol with chloroform to form substituted benzal chloride.

Step 1: The MnO ion has four oxygen atoms surrounding the central Mn atom.
Step 2: These four oxygen atoms arrange themselves in a tetrahedral geometry to minimize repulsion.
Step 3: Therefore, the structure of MnO ion is tetrahedral.

Step 1: When benzene diazonium fluoroborate is treated with sodium nitrite and copper, a substitution reaction takes place.
Step 2: This reaction leads to the formation of nitrobenzene by the electrophilic aromatic substitution of the diazonium group with a nitro group.