Organic Chemistry – Some Basic Principles and Techniques - Class 11 Chemistry - Chapter 8 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Organic Chemistry – Some Basic Principles and Techniques | NCERT | Chemistry | Class 11
What is the correct IUPAC name of the compound given below?
A. 4-(propoxycarbonyl)cyclohexanecarboxylic acid
B. 4-(butoxycarbonyl)cyclohexanecarboxylic acid
C. 6-(propoxyloxy)cyclohexanecarboxylic acid
D. 4-(butanoyloxy)cyclohexanecarboxylic acid
Answer: A. 4-(propoxycarbonyl)cyclohexanecarboxylic acid
Explanation:
In IUPAC nomenclature:
Suffix for carboxylic acids ($\mathrm{-COOH}$) is typically "-oic acid" or "carboxylic acid".
Prefix used is "carboxy-" if another functional group is present.
For Esters ($\mathrm{-COOR}$):
Suffix used is "-oate" or "carboxylate".
When the carbonyl group ($\mathrm{C=O}$) of an ester is directly attached to the parent chain, "alkoxy carbonyl" is used as a prefix.
Given that cyclohexane is the backbone and an ester group is linked as a substituent at the 4th position, the appropriate prefix is "propoxycarbonyl" because it denotes a $\mathrm{OCO}$ linkage wherein a propoxy group ($\mathrm{C3H7O}$) is attached.
Thus, the IUPAC name for the compound is:
4-(propoxycarbonyl)cyclohexanecarboxylic acid.
What does IUPAC mean?
IUPAC stands for International Union of Pure and Applied Chemists. It serves as an international governing body that primarily sets standards in chemistry. IUPAC's responsibilities include making recommendations for naming new chemical elements, as well as establishing various other chemistry-related standards, such as labeling groups on the periodic table.
Assuming general laboratory reagents and butadiene are available, which of the following sequences could accomplish the following synthesis?
A) 1: Dissolved metal reduction of butadiene; 2: Catalytic hydrogenation
B) 1: Kolbe's electrolysis; 2: Catalytic hydrogenation
C) 1: Stephen's aldehyde synthesis; 2: Catalytic hydrogenation
D) 1: Diels-Alder of butadiene and the substrate; 2: Catalytic hydrogenation
The correct option is D) where the steps are as follows:
- Diels-Alder reaction between butadiene and a suitable dienophile
- Catalytic hydrogenation
Explanation:
-
In the Diels-Alder reaction, butadiene acts as the diene and the other substrate serves as the dienophile, forming a cyclohexene derivative which maintains the relative stereochemistry. The groups attached to the newly formed ring (like Cl and COOH in this hypothetical) will remain trans to each other, crucial for preserving the required stereochemistry of the product.
-
The next step of catalytic hydrogenation reduces any double bonds present in the cyclohexene from the Diels-Alder product, completing the synthesis by providing a fully saturated compound as needed in the target molecule.
This sequence utilizes well-known and practical reactions suitable for synthesizing complex molecular structures from simpler compounds like butadiene, showcasing the elegance of organic synthesis techniques like these.
The points to be considered while writing the IUPAC names are:
Number of carbon atoms
Unsaturation
Valency of the heteroatom
Functional group
The correct options are:
A Number of carbon atoms
B Unsaturation
D Functional group
Important factors to consider when writing IUPAC names:
Number of Carbon Atoms: Identify the parent chain, which is the longest continuous chain in a compound. This determines the root word of the compound.
Unsaturation: Determine the presence or absence of unsaturation (double or triple bonds) in the carbon chain. This affects the primary suffix of the compound, indicating whether the compound is an alkane, alkene, or alkyne.
Functional Group: Identify the functional group present (e.g., alcohol, aldehyde, ketone) as it influences the secondary suffix in the name of the compound.
Heteroatoms: These are atoms other than carbon and hydrogen that are attached to one or more carbon atoms in an organic compound. However, the valency of these heteroatoms is not considered in the naming of the compound.
Thus, the correct points to consider are A, B, and D. The valency of heteroatoms is not a factor in determining the IUPAC name.
The molecular mass of an organic compound with a mass ratio of C, H, and N being 9:1:3.5 is 108 grams per mole ($ \text{g mol}^{-1} $). Its molecular formula is:
A. $ \text{C}_6\text{H}_8\text{N}_2 $
B. $ \text{C}_7\text{H}_{10}\text{N} $
C. $ \text{C}_5\text{H}_6\text{N}_3 $
D. $ \text{C}_4\text{H}_{18}\text{N}_3 $
To find the molecular formula of an organic compound with a given molecular mass of 108 g/mol and a mass ratio of C, H, and N of 9:1:3.5, we proceed as follows:
Determine Moles of Each Element:
The given mass ratio can be expressed as:
Carbon (C): $ \frac{9}{12} \text{ mol} $
Hydrogen (H): $ \frac{1}{1} \text{ mol} $
Nitrogen (N): $ \frac{3.5}{14} \text{ mol} $
Simplify to Smallest Whole Number Ratio:
Convert these values to a simpler ratio: $$ \frac{9/12}{9/12} : \frac{1/1}{9/12} : \frac{3.5/14}{9/12} $$ This simplifies to: $$ \frac{3}{4} : 1 : \frac{1}{4} = 3 : 4 : 1 $$
Construct Empirical Formula:
Hence, the empirical formula of the compound is $ \mathbf{C_3H_4N} $.
Calculate the Molecular Formula:
The molar mass of the empirical formula $ \mathbf{C_3H_4N} $ is: $$ 3 \times 12 + 4 \times 1 + 1 \times 14 = 54 \ \text{g/mol} $$
Given that the molecular mass is 108 g/mol, we find the multiplicand ( n ) by: $$ n = \frac{108}{54} = 2 $$
Determine the Final Molecular Formula:
Therefore, the molecular formula is: $$ \left(C_3H_4N\right)_2 = \mathbf{C_6H_8N_2} $$
Final Answer: The molecular formula is A) $C(_6)H(_8)N(_2)$.
What is a homologous series? Explain with an example.
A homologous series refers to a class of organic compounds that share the same general molecular formula and exhibit similar chemical properties. Successive members of this series differ by a $\mathrm{CH}_{2}$ group.
Properties of Homologous Series
Consistent Increment: Each member varies from the next by a $\mathrm{CH}_{2}$ group and a molecular mass of $14 : \mathrm{amu}$ (atomic mass unit).
Uniform Elements and Functional Groups: All members contain the same elements and have the same functional groups.
Common Molecular Formula: The general molecular formulas for different types of hydrocarbons are:
Alkanes: $\mathrm{C}_{n} \mathrm{H}_{2n+2}$
Alkenes: $\mathrm{C}_{n} \mathrm{H}_{2n}$
Alkynes: $\mathrm{C}_{n} \mathrm{H}_{2n-2}$
Gradation in Physical Properties: There is a regular progression in physical properties (such as boiling point, melting point) with increasing molecular mass.
Similar Chemical Properties: Members exhibit similar chemical properties due to the presence of the same functional groups.
Uniform Preparation Method: All members can be synthesized using the same general method.
Example:
For alkanes, the series starts from methane ($\text{CH}_4$) and proceeds as ethane ($C_2H_6$), propane (C$_3$H$_8$), and so on, each differing by a $\mathrm{CH}_{2}$ unit.
To ensure resonance, it is essential that the included atoms in the resonant structure are as much planar as possible in the surface or almost flat. If this condition is not fully satisfied, the included atoms cannot be parallel to each other and as a result, planarization cannot occur. The presence of a large group of atoms on the atoms obstructs the planarity of the resonant structures. We recognize this phenomenon as a three-dimensional hindrance to resonance. The three-dimensional hindrance of resonance has a very significant impact on the following:
Physical properties
Acidity and basicity
Reactivity of organic compounds
Examine the overview of the two structures and choose the correct statement -
Structure I has longer carbon-nitrogen bonds than structure II.
Structure I has shorter carbon-nitrogen bonds than structure II.
The carbon-nitrogen bond length is the same in the two structures.
Comparison cannot be made.
To ensure resonance, it is essential that the included atoms in the resonant structure are as planar as possible, meaning they should lie on the surface or be almost flat. If this condition is not fully satisfied, the included atoms cannot be parallel to each other, and as a result, planarization cannot occur. The presence of large groups of atoms can obstruct the planarity of the resonant structures. This phenomenon is recognized as three-dimensional hindrance to resonance.
The three-dimensional hindrance of resonance significantly impacts the following:
Physical properties
Acidity and basicity
Reactivity of organic compounds
Now, considering the provided structures:
Structure (I)
Structure (II), which has bulky groups attached.
Due to the bulky groups in structure II, there will be greater three-dimensional hindrance, leading to less effective resonance. Consequently, the carbon-nitrogen bonds in structure II are expected to be longer because resonance is less effective in shortening the bond length.
Therefore, the correct statement is:
Structure I has shorter carbon-nitrogen bonds than structure II.
Ortho-nitrophenol, compared to p- and m-nitrophenol in water, has a lower value because -
o-nitrophenol shows intramolecular H-bonding
o-nitrophenol shows intermolecular H-bonding
The melting point of o-nitrophenol is lower than m- and p-isomers
o-nitrophenol is more volatile in vapor phase than m- and p-isomers
Ortho-nitrophenol exhibits intramolecular hydrogen bonding compared to para- and meta-nitrophenol, which significantly influences its properties in water. This form of hydrogen bonding occurs within the same molecule, creating a more stable internal structure and decreasing the likelihood of intermolecular hydrogen bonding interactions with water molecules.
Therefore, due to intramolecular H-bonding:
o-nitrophenol is less able to form hydrogen bonds with water molecules, leading to a lower solubility in water.
It is more volatile as it has fewer interactions with other molecules in both vapor and liquid phases compared to its meta and para isomers.
It has a lower melting point than m-nitrophenol and p-nitrophenol because the intramolecular hydrogen bonding reduces the effective intermolecular force.
Thus, the correct answer is:
o-nitrophenol shows intramolecular H-bonding.
Resonance refers to the intramolecular delocalization of unbound electrons without any change in the positions of atoms. This delocalization can occur in conjugated systems involving carbon atoms as well as other atoms besides carbon. The delocalization makes the system more stable. A system's stability increases with the number of resonant structures. A conjugated structure is least stable when there is a positive charge on a more electronegative atom and similar charges on neighboring atoms. The decreasing order of stability for the following resonant structures would be-
I > II > III
II > III > I
III > II > I
I > III > II
Resonance is a process whereby there is intramolecular delocalization of unbound electrons without any change in the positions of atoms. This delocalization can occur in conjugated systems involving carbon atoms as well as other atoms.
From the perspective of stability, delocalization makes the entire system more stable. When a center has more resonant structures, the entire structure becomes more stable.
Priority order:
A conjugated structure is the least stable when a more electronegative atom has a positive charge and neighboring atoms have similar charges.
For the given example of conjugated structures, the order of stability is as follows: $$ \text{I} > \text{II} > \text{III} $$
Structure I is the most stable.
Next is structure II.
And structure III is the least stable.
Therefore, the correct answer is:
I > II > III
The Huckel's rule dictates that a compound is aromatic if it follows the formula 4n+2 for the number of outer electrons. What is the value of n?
To determine the value of ( n ) for the given compound to be aromatic according to Huckel's rule, we need to count the number of π-electrons in the molecule.
In the image, we observe that the molecule is composed of three fused benzene rings. Each benzene ring has 6 π-electrons.
Total π-electrons for the entire molecule:
$$ 6 + 6 + 6 = 18 $$
According to Huckel's rule, aromatic compounds must have $ 4n + 2 $ π-electrons, where $ n $ is a non-negative integer.
Setting up the equation: $$ 4n + 2 = 18 $$
Solving for $ n $:
$$ 4n + 2 = 18 $$
$$ 4n = 16 $$
$$ n = 4 $$
Thus, the value of ( n ) for this compound to be aromatic is: $ n = 4 $
Given pairs of compounds, how many pairs are there in which
(i) the reaction is more reactive and
(ii) as compared to $\ce{AgNo3}$?
For determining the number of pairs in which the reaction is more reactive compared to silver nitrate ($\text{AgNO}_3$), we must evaluate the structure and reactivity of each pair of compounds. Here is the detailed analysis:
Pair (A):
(i) Bromo cyclohexane (Alkyl bromide): Moderately reactive.
(ii) Bromo cyclohexene (Allylic bromide): More reactive due to the double bond adjacent to the bromine.
Pair (B):
(i) Bromo cyclopentane (Alkyl bromide): Moderately reactive.
(ii) Bromo cyclopentene (Allylic bromide): More reactive than (i) due to the double bond adjacent to the bromine.
Pair (C):
(i) Bromo cyclohexane (Alkyl bromide): Moderately reactive.
(ii) Bromo cyclohexene (Allylic bromide): More reactive due to resonance stabilization from the double bond.
Pair (D):
(i) Bromo cycloheptene (Alkyl bromide): Moderately reactive.
(ii) Bromo cyclopentene (Allylic bromide): More reactive because of the double bond adjacent to bromine.
Pair (E):
(i) Bicyclo[2.2.1] hept-2-yl chloride: Less reactive.
(ii) Bicyclo[2.2.1] hept-2-yl chloride (with a double bond): More reactive due to the strained ring and double bond.
Pair (F):
(i) Chlorocyclopentane: Moderately reactive.
(ii) Chlorocyclohexane: Slightly less reactive than chlorocyclopentane due to steric factors.
Pair (G):
(i) Chlorocyclopropane: More reactive due to ring strain.
(ii) Cyclobutane: Less reactive since it lacks halogen substitution.
Pair (H):
(i) Chloromethylcyclopropane: More reactive due to the strained cyclopropane ring.
(ii) Cyclobutylchloride: Less reactive due to larger ring and lack of additional strain.
From the pairs analyzed, the most reactive compounds compared to $\text{AgNO}_3$ because of either resonance from a double bond or ring strain are found in pairs (A)i, (B)ii, (C)ii, (D)ii, and (E)ii.
Final Answer: 5 pairs are more reactive compared to $\text{AgNO}_3$.
Compounds:
(a) $\ce{CH3CO2H}$
(b) $\ce{MeOCH2CO2H}$
(c) $\ce{CF3CO2H}$
(d)
The correct order of increasing acid strength is -
d < a < c < b
d < a < b < c
a < d < c < b
b < d < a < c
To determine acidity, we can analyze the electron-withdrawing effects of the substituents on the compounds, particularly on the carbon next to the -COOH group.
(a) $\ce{CH3CO2H}$:
This has only a CH_3 group, which has a slight +I (inductive) effect. This pushes electrons towards the COOH group, slightly decreasing acidity.
(b) $\ce{MeOCH2CO2H}$:
This contains the MeO group, which exhibits a +M (mesomeric) effect but also a -I (inductive) effect. This can increase acidity slightly.
(c) $\ce{CF3CO2H}$:
This has the very strong electron-withdrawing $\ce{CF3}$ group, which has a strong -I effect, significantly increasing acidity.
(d) The compound (as shown in the image):
This has two Me groups, which exhibit a +I effect. These groups push electrons towards the COOH group, reducing acidity.
Based on these factors, the correct order of increasing acidity is:
d < a < b < c
Therefore, the correct answer is: B (d < a < b < c)
The order of basicity in the following compounds is as follows:
IV > III > II > I
II > III > I > IV
I > III > II > IV
III > I > II > IV
To understand which option is correct, we need to analyze the basicity of the given compounds. The basicity of a compound is determined by the ability of its lone pair of electrons to accept protons (H$^+$).
Piperidine (I): This is a six-membered ring with one nitrogen atom. The lone pair on the nitrogen in piperidine is freely available, making it a strong base.
Pyridine (II): This is an aromatic six-membered ring with one nitrogen atom. The lone pair of electrons on nitrogen is in an sp2 hybrid orbital, making it less available for protonation due to delocalization into the ring, reducing its basicity.
Morpholine (III): This is a six-membered ring with one nitrogen and one oxygen atom. The presence of an electron-withdrawing oxygen reduces the electron density on nitrogen, slightly decreasing its basicity compared to piperidine.
Pyrrole (IV): This is a five-membered aromatic ring with one nitrogen atom. The lone pair on nitrogen is involved in the aromatic sextet, making it much less available to accept protons and, thus, the least basic.
Based on this assessment:
Compound I (Piperidine) will be the most basic because the lone pair on nitrogen is fully available for protonation.
Compound III (Morpholine) will be second because the oxygen reduces the electron density only slightly.
Compound II (Pyridine) will be third due to the delocalization of the lone pair in the aromatic ring.
Compound IV (Pyrrole) will be the least basic as the lone pair on nitrogen is involved in maintaining the aromaticity.
Thus, the order of basicity is: I > III > II > IV, which matches option C: I > III > II > IV.
Arrange the following sets of intermediates in decreasing order of their stability:
(i) The stability order of carbocations:
Cycloheptatrienyl cation (tropylium ion) (c)
Triphenylmethyl cation (a)
tert-Butyl cation (d)
Benzyl cation (a)
Phenylmethyl cation (c)
The tropylium ion is exceptionally stable due to its aromaticity contributing to its stability. The triphenylmethyl cation is stabilized by resonance involving three phenyl groups. The other cations follow in decreasing order of stability based on the inductive effect and hyperconjugation.
(ii) The stability order of radicals:
Benzyl radical (c)
Cycloheptatrienyl radical (d)
Allyl radical (a)
Vinyl radical (b)
Benzylic and cycloheptatrienyl radicals are greatly stabilized by resonance, making them the most stable. Allyl radical is also stabilized by resonance but to a lesser extent. The vinyl radical is the least stable due to the sp² hybridization of carbon.
(iii) The stability order of radicals:
Benzyl radical (a)
Benzyl radical with chlorine in para position (c)
Benzyl radical with chlorine in ortho position (b)
Substituent effects and the ability to delocalize the unpaired electron via resonance typically determine radical stability. Here, (a) and (c) are both benzyl radicals; however, the chlorine substituent's inductive effects reduce stability slightly more in the ortho position than in the para position.
(iv) The stability order of free carbocations (positive charge carbon atoms):
Cyclopentyl cation (c)
Cyclohexyl cation (b)
Cyclopropyl cation (a)
Larger ring sizes usually correlate with greater carbocation stability due to reduced angle strain and increased hyperconjugation and delocalization potential.
(v) The stability order of radical intermediates:
Cyclopentadienyl radical (c)
Cycloheptatrienyl radical (b)
Cyclopropenyl radical (a)
Similar to carbocations, the resonance stabilization in these species considerably influences their stability, with larger rings often providing more stable structures.
(vi) The stability order of small ring cations:
Cyclopentadienyl cation (c)
Cycloheptatrienyl cation (b)
Cyclopropyl cation (a)
Resonance and aromatic stabilization are dominant in contributing to the cation's stability. Cyclopentadienyl and cycloheptatrienyl cations are stabilized through aromaticity whereas cyclopropyl cation does not have such stabilization mechanisms.
Note: For radicals and carbocations, resonance stabilization is a key factor. Aromaticity in cyclic structures greatly enhances the stability of cationic and radical intermediates.
Select aromatic species in each pair of species.
Identifying Aromatic Species
An aromatic compound must follow Huckel's rule, which states that a molecule must have $\left(4n + 2\right)\pi$ electrons (where n is a non-negative integer) to be aromatic. Additionally, the molecule must be cyclic, planar, and fully conjugated.
(i) Pair of Species
(a)
Structure: Contains a 3-membered ring fused to a 5-membered ring.
Electrons Count: 6 π electrons (1 double bond and a lone pair forming a sextet).
Planarity and Conjugation: Cyclic and planar, fully conjugated.
Conclusion: Aromatic due to 6 π electrons.
(b)
Structure: Thiophene-like sulfur-containing 4-membered ring.
Electrons Count: 4 π electrons.
Planarity and Conjugation: Cyclic and planar, fully conjugated.
Conclusion: Non-aromatic due to 4 π electrons (not fitting Huckel's rule).
(ii) Pair of Species
(a)
Structure: Benzene derivative.
Electrons Count: 6 π electrons (3 double bonds within a 6-membered ring).
Planarity and Conjugation: Cyclic and planar, fully conjugated.
Conclusion: Aromatic due to 6 π electrons.
(b)
Structure: Cycloheptatrienyl cation ($\text{C}_7\text{H}_7^+$).
Electrons Count: 6 π electrons (due to the positive charge delocalizing the lone pair).
Planarity and Conjugation: Cyclic and planar, fully conjugated.
Conclusion: Aromatic due to 6 π electrons.
(iii) Pair of Species
(a)
Structure: Cyclopropenyl cation ($\text{C}_3\text{H}_3^+$).
Electrons Count: 2 π electrons.
Planarity and Conjugation: Cyclic and planar, fully conjugated.
Conclusion: Aromatic due to 2 π electrons.
(b)
Structure: Cyclobutadienyl cation ($\text{C}_4\text{H}_4$).
Electrons Count: 4 π electrons.
Planarity and Conjugation: Cyclic and planar, fully conjugated.
Conclusion: Non-aromatic due to 4 π electrons (not fitting Huckel's rule).
Species Selected as Aromatic:
(i) (a): 6-membered ring with 6 π electrons.
(ii) (a): Benzene with 6 π electrons.
(ii) (b): Cycloheptatrienyl cation with 6 π electrons.
(iii) (a): Cyclopropenyl cation with 2 π electrons.
The following products will be produced:
The given chemical reactions are:
(a) The hydroboration-oxidation of a cyclooctene: $$ \text{Cyclooctene} \xrightarrow{\text{BH}_3 \cdot \text{THF}} \xrightarrow{\text{H}_2\text{O}_2, \text{OH}^-} \text{Cyclooctanol} $$
(b) The hydroboration-deuteration of a molecule containing a double bond with carboxyl groups at the ends: $$ \text{(COOH)HC=CH(COOH)} \xrightarrow{1. \text{BD}_3, \text{THF}} \xrightarrow{2. \text{D}_2\text{O}, \text{D}_3\text{O}^+} \text{(COOH)DCH-CHD(COOH)} $$
Detailed Explanation:
(a) The hydroboration-oxidation reaction of cyclooctene:
Hydroboration:
Reacting cyclooctene with borane ($\text{BH}_3 \cdot \text{THF}$) leads to the formation of an intermediate organoborane. This process involves the addition of a $\text{BH}_2$ group to one carbon of the double bond and a hydrogen atom to the other carbon. This occurs in a syn-addition manner, i.e., both the $\text{BH}_2$ group and the hydrogen add to the same side of the double bond.
Oxidation:
The organoborane intermediate is then treated with hydrogen peroxide ($\text{H}_2\text{O}_2$) in a basic medium (usually sodium hydroxide, $\text{NaOH}$). This leads to the replacement of the $\text{B}$ group with a hydroxyl group ($\text{-OH}$), producing cyclooctanol.
(b) Hydroboration followed by deuteration:
Hydroboration:
The molecule with the double bond and carboxyl groups undergoes hydroboration with deuterated borane ($\text{BD}_3$). The $\text{BD}_2$ group and a deuterium atom ($\text{D}$) add to the double bond in a syn-addition manner.
Deuteration:
The intermediate organoborane is then treated with deuterium oxide ($\text{D}_2\text{O}$) and deuterium acid ($\text{D}_3\text{O}^+$), replacing the boron with deuterium atoms. This leads to the final product where both hydrogens initially present in the double bond are replaced by deuterium atoms.
The products will be made from the dehydration of the following.
(a) The compound shown is 1,1-dimethylcyclohexanol. Upon dehydration, it will form the following product:
$$ \text{CH}_3 - \underset{\displaystyle |}{\underset{|}{C}} - \text{OH (1,1-dimethylcyclohexanol)} \rightarrow \text{(CH}_3\text{)}_2\text{C = CH}_2 $$
(b) The compound is cyclohexanol. Dehydration of cyclohexanol will produce cyclohexene:
$$ \text{Cyclohexanol} (\text{C}_6\text{H}_{11}\text{OH}) \rightarrow \text{Cyclohexene} (\text{C}_6\text{H}_{10}) $$
(c) The compound is cyclopentylmethanol. Upon dehydration, it will yield methylenecyclopentane:
$$ \text{Cyclopentylmethanol} (\text{C}_6\text{H}_{11}\text{OH}) \rightarrow \text{(CH}_3\text{-CH}_2\text{-C}_5\text{H}_6\text{)} $$
The main product of the following reaction is:
1)
2)
3)
4)
The main product of the given reaction is Option 1.
Explanation:
The reaction involves the cyclization of the given ketone compound.
The starting material is a cyclohexanone derivative.
Upon reaction, intramolecular aldol condensation occurs where two carbonyl groups react intramolecularly.
This leads to the formation of a new carbon-carbon bond, resulting in the formation of a product with an additional ring structure, specifically forming a bicyclic ketone.
Thus, the main product formed is represented by Option 1.
Analyse the mechanism of the following reaction:
Mechanism Analysis of the Reaction
Protonation Stage:The reaction begins with the protonation of cyclohexene by the acidic medium ($H_2SO_4$). $$ \text{Cyclohexene} + H^+ \rightarrow \text{Cyclohexyl cation (carbocation)} $$
Formation of the Carbocation:The double bond in the cyclohexene breaks, resulting in the formation of a cyclohexyl carbocation. $$ \text{Cyclohexene} \rightarrow \text{Cyclohexyl carbocation} $$
Nucleophilic Attack:Another molecule of cyclohexene acts as a nucleophile and attacks the positively charged carbon atom of the cyclohexyl carbocation, forming a dimeric carbocation intermediate. $$ \text{Cyclohexyl carbocation} + \text{Cyclohexene} \rightarrow \text{Dimeric carbocation intermediate} $$
Deprotonation:The final step involves the deprotonation of the dimeric carbocation intermediate, leading to the formation of bicyclohexyl (dicyclohexyl). $$ \text{Dimeric carbocation intermediate} \rightarrow \text{Bicyclohexyl} + H^+ $$
Overall Reaction:
The overall transformation is essentially the dimerization of cyclohexene under acidic conditions, resulting in the formation of a bicyclohexyl compound. $$ 2 \text{ Cyclohexene} \overset{H^+/ H_2SO_4}{\longrightarrow} \text{Bicyclohexyl} $$
Throughout this mechanism, the key intermediate is the cyclohexyl carbocation, which is formed through protonation and then participates in further steps to yield the final product.
The main products are:
1)
2)
3)
4)
Option B is identified as the major product.
Reason: In the given chemical reaction, a hydrogen ion (H⁺) is added to the organic compound. This is an addition reaction, where a proton attaches to the carbonyl group of the ketone. As a result, a carboxylic acid is formed in the compound.
This process can be represented by the following chemical equation:
$$ \text{Ketone} \overset{H^+}{\longrightarrow} \text{Carboxylic Acid} $$
Upon analyzing the entire mechanism and the structure of the product, it is clear that Option B is the major product.
What is A?
The molecule labeled as $\text{A}$ can be either $\text{CH}_2=\overset{+}{\text{N}}=\overset{-}{\text{N}}$ or $\text{CH}_2=\overset{O}{\overset{||}{\text{C}}}$ (a group that forms $\text{CH}_2$).
Analyse the mechanism of the following reaction:
The given reaction involves the transformation of cyclohexene into bicyclo[2.2.1]heptane using methylene iodide ($\text{CH}_2\text{I}_2$) in the presence of a zinc-copper couple (denoted by $\text{Zn(Cu)}$). Here's the step-by-step analysis of the mechanism:
Formation of the Carbene:
Methylene iodide ($\text{CH}_2\text{I}_2$) reacts with the zinc-copper couple to generate the carbene intermediate, diiodomethane ($: \text{CH}_2$).
Addition of the Carbene to the Alkene:
The carbene ($: \text{CH}_2$), which is highly reactive, adds to the double bond in cyclohexene. This forms a cyclopropane ring as the carbene inserts into the $\text{C}=\text{C}$ double bond of cyclohexene.
Formation of Bicyclo[2.2.1]heptane:
The overall result of this addition is the formation of bicyclo[2.2.1]heptane, a compound with a bicyclic structure, where a three-membered cyclopropane ring is fused with two six-membered rings.
Summarizing the steps:
$$ \text{Cyclohexene} + \text{CH}_2\text{I}_2 \xrightarrow{\text{Zn(Cu)}} \text{Bicyclo[2.2.1]heptane} $$
This reaction is a classic example of a cyclopropanation reaction, where a carbene intermediate adds across the double bond of an alkene, resulting in a three-membered ring structure.
A is equal to B?
No, A is not equal to B.
In this reaction, the organic compound undergoes a series of transformations:
The cyclohexane derivative (with attached triphenylphosphine) first transforms into compound A through specific chemical reactions.
Compound A is then subjected to ozonolysis to yield compound B, which is a ketone, specifically cyclohexanone.
Since A is an intermediate product that undergoes further reaction to become B, they are not the same compound.
In which of the following cases, Hoffmann product is the major product?
A.
B.
C.
D.
D:
Explanation:
Hoffman products, which are less substituted alkenes, can be formed under the following conditions:
Weak leaving groups such as F (fluorine) facilitate the formation of a carbanion transition state, leading to the formation of a less substituted alkene.
Hoffman elimination of quaternary ammonium salts results in a less substituted alkene.
When using a large bulky base, it abstracts the less hindered hydrogen from the substrate, resulting in the formation of a less substituted alkene.