Aldehydes, Ketones and Carboxylic Acids - Class 12 Chemistry - Chapter 8 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Aldehydes, Ketones and Carboxylic Acids | NCERT | Chemistry | Class 12
The hydrolysis of optically active 2-bromobutane with aqueous NaOH results in the formation of:
Option A) $(+)$butan – $2$ – ol
Option B) $(-)$butan – $2$ – ol
Option C) $(\pm)$butan – $1$ – ol
Option D) $(\pm)$butan – $2$ – ol
The hydrolysis of 2-bromobutane using aqueous NaOH is a typical example of a nucleophilic substitution reaction, specifically SN2 (bimolecular nucleophilic substitution) mechanism. In this mechanism, the nucleophile (OH⁻ from NaOH) attacks the electrophilic carbon (the one attached to Br) from the opposite side of the bromine atom. This leads to the inversion of configuration at the carbon center.
Since 2-bromobutane is optically active, it suggests it has a chiral center; thus, the product formed will also possess a new configuration opposite to that of the starting material. This change is commonly referred to as "Walden inversion."
However, it's also important to consider that under usual conditions hydrolysis reactions can sometimes proceed via SN1 mechanism, especially in a more polar solvent or with a less strong nucleophile where a carbocation intermediate formation favors racemization (formation of a mixture of enantiomers). Given the strong nucleophilic conditions described (NaOH in water), an SN2 reaction is more likely.
Therefore, the hydrolysis of optically active 2-bromobutane usually results in:
2-butanol (butan-2-ol) with inversion of configuration.
Because the original 2-bromobutane was optically active but the stereochemistry is inverted, we get the enantiomer of the original stereoisomer. Therefore, the resultant 2-butanol must be:
optically active
have the opposite configuration of the original
Since the stereochemistry of the original 2-bromobutane isn't specified in terms of being R or S, the exact configuration of the butan-2-ol (either $(+)$ or $(-)$) cannot be determined definitively from the information provided.
The best answer with the given options assuming no partial racemization would be:
Option D) $(\pm)$butan – $2$ – ol
Conjugate base of $\mathrm{HCl}$ is -
A) $H^{\oplus}$
B) $Cl$
C) $\mathrm{H}_{2}\mathrm{Cl}$
D) None of these
The conjugate base of an acid is formed by removing a proton ($H^+$) from it. For the acid $\mathrm{HCl}$, the reaction can be represented as:
$$ \mathrm{HCl} \rightarrow H^+ + Cl^- $$
Here, $\mathrm{HCl}$ loses a proton, resulting in the formation of $Cl^-$. Therefore, the conjugate base of $\mathrm{HCl}$ is $Cl^-$.
Thus, the correct answer is:
Option (B) $Cl^-$
Some pairs of acids are given below. Select the pair in which the second acid is stronger than the first.
A $\mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{2} \mathrm{FCOOH}$
B $\mathrm{CHF}_{2} \mathrm{COOH}$ and $\mathrm{CH}_{2} \mathrm{ClCOOH}$
C $\mathrm{CH}_{2} \mathrm{ClCOOH}$ and $\mathrm{CH}_{2} \mathrm{BrCOOH}$
D $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHFCOOH}$ and $\mathrm{CH}_{3} \mathrm{CHFCH}_{2} \mathrm{COOH}$
To determine the stronger acid between two given in any pair, we need to assess the stability of their conjugate bases. A more stable conjugate base implies a stronger parent acid. The inductive effect (-I effect) is key here. It refers to the electron-withdrawing ability of substituents which can stabilize the negative charge on a conjugate base.
Option A: $\mathrm{CH}{3}\mathrm{COOH}$ and $\mathrm{CH}{2}\mathrm{FCOOH}$
Fluorine exhibits a stronger -I effect than hydrogen. This enhancement in electron-withdrawing ability due to fluorine in $\mathrm{CH}{2}\mathrm{FCOOH}$ leads to greater stabilization of its conjugate base compared to that of $\mathrm{CH}{3}\mathrm{COOH}$. Thus, $\mathrm{CH}_{2}\mathrm{FCOOH}$ is a stronger acid.
Option B: $\mathrm{CHF}{2}\mathrm{COOH}$ and $\mathrm{CH}{2}\mathrm{ClCOOH}$
Despite chlorine having a good -I effect, fluorine's effect is much stronger, and here it is present in two moieties. Therefore, the first acid here, $\mathrm{CHF}_{2}\mathrm{COOH}$, actually has a more stable conjugate base, making it the stronger acid.
Option C: $\mathrm{CH}{2}\mathrm{ClCOOH}$ and $\mathrm{CH}{2}\mathrm{BrCOOH}$
Chlorine's -I effect surpasses that of bromine, making the conjugate base of the first acid ($\mathrm{CH}{2}\mathrm{ClCOOH}$) more stable than that of the second. This means $\mathrm{CH}{2}\mathrm{ClCOOH}$ is a stronger acid.
Option D: $\mathrm{CH}{3} \mathrm{CH}{2} \mathrm{CHFCOOH}$ and $\mathrm{CH}_{3} \mathrm{CHFCH}_2 \mathrm{COOH}$
Fluorine in the first acid is closer to the carboxyl group, hence its -I effect is more strongly felt here than in the second acid. Consequently, the conjugate base of the first acid is more stable, rendering it a stronger acid.
From the analysis, only Option A fits the criteria where the second acid ($\mathrm{CH}{2}\mathrm{FCOOH}$) is stronger than the first ($\mathrm{CH}{3}\mathrm{COOH}$). Here, the proximity and strength of the inductive effect play a pivotal role in making $\mathrm{CH}_{2}\mathrm{FCOOH}$ more acidic.
Propene, $\mathrm{CH_{3}-CH=CH_{2}}$ can be converted to 1-propanol by oxidation. Which set of reagents among the following is ideal to make this conversion happen?
A Alkaline $\mathrm{KMnO_{4}}$
B $\mathrm{B}{2} \mathrm{H}{6}$ and alkaline $\mathrm{H}{2} \mathrm{O}{2}$
C $\mathrm{O}_{3} / \mathrm{Zn}$ dust
D $\mathrm{OsO}_{4}$
The ideal reagents for converting propene ($\mathrm{CH_{3}-CH=CH_{2}}$) to 1-propanol via oxidation are $\mathrm{B}_{2} \mathrm{H}_{6}$ (diborane) followed by treatment with alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$ (hydrogen peroxide under basic conditions). This approach uses an anti-Markovnikov hydroboration-oxidation reaction. Therefore, the correct answer is:
B) $\mathrm{B}_{2} \mathrm{H}_{6}$ and alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$
Column 1: Reaction Column 2: Type of reactions Column 3: Lab test for reactants \begin{tabular}{|c|c|c|} \hline (I) & $\stackrel{\oplus}{\mathrm{H}^{\oplus}}$ & (i) Acid-base reaction, (P) Tollen's test \ \hline (II) & $\xrightarrow{\mathrm{PhMgBr}}$ & (ii) Nucleophilic addition reaction, (Q) Carbylamine test \ \hline (III) & $\frac{\mathrm{NaBD}{4}}{\mathrm{H}{2} \mathrm{O}}$ & (iii) Nucleophilic substitution reaction, (R) Lucas test \ \hline (IV) & $\xrightarrow[\mathrm{CH}{3} \mathrm{MgBr}]{ }$ & (iv) Fisher esterification, (S) Neutral $\mathrm{FeCl}{3}$ test \ \hline \end{tabular}
The only correct combination that gives two different stereoisomeric product is:
A (II) (i) (S)
B (III) (iii) (P)
C (IV) (iii) (Q)
D (I) (iv) (R)
The correct option is D (I) (iv) (R).
Fischer esterification is the reaction type in option (I) (iv) which can yield optically active products, implying the possible formation of two different stereoisomers. This reaction involves the transformation of a carboxylic acid and an alcohol under acidic conditions, typically resulting in the formation of an ester and water.
Overall, the scenario provided in Option D aligns correctly with the criteria of yielding two stereoisomeric products due to the nature of Fischer esterification.
According to Cahn-Ingold-Prelog sequence rules, the correct order of priority of the given groups is:
(A) $-\mathrm{COOH} > -\mathrm{CH}_{2}\mathrm{OH} > -\mathrm{OH} > -\mathrm{CHO}$
(B) $-\mathrm{COOH} > -\mathrm{CHO} > -\mathrm{CH}_{2}\mathrm{OH} > -\mathrm{OH}$
(C) $-\mathrm{OH} > -\mathrm{CH}_{2}\mathrm{OH} > -\mathrm{CHO} > -\mathrm{COOH}$
(D) $-\mathrm{OH} > -\mathrm{COOH} > -\mathrm{CHO} > -\mathrm{CH}_{2}\mathrm{OH}$
Solution: The correct choice is (D): $$ -\mathrm{OH} > -\mathrm{COOH} > -\mathrm{CHO} > -\mathrm{CH}_2 \mathrm{OH} $$
According to the Cahn-Ingold-Prelog sequence rules, priority is assigned based on the atomic number of atoms directly attached to the chirality center. If the atoms have similar atomic numbers, multiple bonding (i.e., double or triple bonds) influences the priority order.
In this comparison:
- Hydroxyl (-OH) directly bonds to oxygen (atomic number 8), giving it the highest priority.
- Carboxyl (-COOH) and formyl (-CHO) groups also attach through an oxygen, but the presence of additional oxygen and/or double bonds in carboxyl groups grants it higher priority over formyl.
- Hydroxymethyl (-CH_2OH) has oxygen too but is ranked lowest as it attaches through a carbon (atomic number 6) first rather than the oxygen.
Thus, the sequence is rightly: $$ -\mathrm{OH} > -\mathrm{COOH} > -\mathrm{CHO} > -\mathrm{CH}_2 \mathrm{OH} $$
In nucleophilic addition reactions, the reactivity of carbonyl compounds follows the order -
(A) $\mathrm{H}{2}\mathrm{C}=\mathrm{O} > \mathrm{R}{2}\mathrm{C}=\mathrm{O} < \mathrm{Ar}_{2}\mathrm{C}=\mathrm{O} > \mathrm{RCHO} > \mathrm{ArCHO}$
(B) $\mathrm{H}{2}\mathrm{C}=\mathrm{O} > \mathrm{RCHO} > \mathrm{ArCHO} > \mathrm{R}{2}\mathrm{CO} > \mathrm{Ar}_{2}\mathrm{C}=\mathrm{O}$
(C) $\mathrm{Ar}{2}\mathrm{C}=\mathrm{O} > \mathrm{R}{2}\mathrm{C}=\mathrm{O} > \mathrm{ArCHO} > \mathrm{RCHO} > \mathrm{H}_{2}\mathrm{C}=\mathrm{O}$
(D) $\mathrm{ArCHO} > \mathrm{Ar}{2}\mathrm{C}=\mathrm{O} > \mathrm{RCHO} > \mathrm{R}{2}\mathrm{C}=\mathrm{O} > \mathrm{H}_{2}\mathrm{C}=\mathrm{O}$
Solution
The correct answer is Option B: $$ \mathrm{H}_2\mathrm{C}=\mathrm{O} > \mathrm{RCHO} > \mathrm{ArCHO} > \mathrm{R}_2\mathrm{CO} > \mathrm{Ar}_2\mathrm{C}=\mathrm{O} $$
This order can be justified considering that the reactivity in nucleophilic addition reactions decreases with increasing steric hindrance near the carbonyl carbon. Formaldehyde ($\mathrm{H}_2\mathrm{C}=\mathrm{O}$), lacking any alkyl or aryl groups, is the least hindered and thus the most reactive. As we move toward dialkyl ketones ($\mathrm{R}_2\mathrm{CO}$) and diaryl ketones ($\mathrm{Ar}_2\mathrm{C}=\mathrm{O}$), the presence of bulky groups around the carbonyl carbon impairs the approach of the nucleophile, thereby decreasing the reactivity.
Butanal and 2-methyl propanal are examples of which type of isomerism?
A) Chain isomerism
B) Tautomerism
C) Position isomerism
D) Functional isomerism
The correct answer is A) Chain isomerism.
Chain isomerism occurs when compounds have the same molecular formula but different arrangements of the carbon chain. Butanal and 2-methylpropanal both have the molecular formula $C_4H_8O$, yet they differ in their carbon backbone structures. Butanal has a straight chain, while 2-methylpropanal includes a branched chain. Hence, they are examples of chain isomerism.
(P) (Q) $(P)$ and $(Q)$ undergo acidic hydrolysis. Products formed by them can be differentiated by:
A) 2,4-DNP
B) Lucas reagent C) $\mathrm{NaHSO}_{3}$
D) Fehling solution
The correct answer is D) Fehling solution.
The structures of compounds $(P)$ and $(Q)$ can be hypothesized based on their behavior with different reagents. Since Fehling solution is mentioned as a differentiating agent, it suggests that the products of acidic hydrolysis of either $(P)$ or $(Q)$ involve an aldehyde. Fehling solution specifically tests for the presence of aldehydes, which reduce the Cu(II) ion in the reagent from its blue color to a red precipitate of Cu(I) oxide. Therefore, the product of $(Q)$ upon acidic hydrolysis must be an aldehyde as it reacts with Fehling solution to yield a color change from blue to brick red.
Which one is more soluble in water amongst the following?
A) Acetone
B) Butanone
C) Pentanone
D) Hexanone
The most soluble compound in water among the given options is A) Acetone.
Solubility in water generally decreases as the length of the hydrocarbon chain in a compound increases because longer hydrocarbon chains lead to reduced polarity. Acetone, having the shortest chain, exhibits the highest solubility due to its relatively higher polarity compared to the longer ketones listed.
How many NADH molecules are produced during the formation of acetyl CoA from pyruvic acid?
A) 4
B) 2
C) 1
D) 0
The correct answer is Option C: 1.
In the metabolic pathway where pyruvic acid is converted into acetyl CoA, 1 molecule of NADH is produced. This reaction takes place in the mitochondria and is catalyzed by the enzyme complex pyruvate dehydrogenase. The overall process is crucial for linking glycolysis to the citric acid cycle.
"What is an alkyne?"
Solution
Alkynes are a fundamental group of hydrocarbons. They exhibit a characteristic general formula of $$ \mathrm{C}n\mathrm{H}{2n-2} $$ where n represents the number of carbon atoms in the molecule. This formula designates alkynes as unsaturated hydrocarbons because, unlike alkanes, they contain at least one triple bond between carbon atoms. These triple bonds include both sigma and pi bonds, which involve different types of overlapping atomic orbitals. An example of an alkyne is ethyne (commonly known as acetylene), with the chemical structure $$ \mathrm{C}_2\mathrm{H}_2 \quad \text{or} \quad \mathrm{CH} \equiv \mathrm{CH} $$.
Potassium, when heated strongly in oxygen, forms:
(A) $\mathrm{KO}_{2}$
(B) $\mathrm{K}_{2} \mathrm{O}$
(C) $\mathrm{K}{2} \mathrm{O}{2}$
Solution
The correct option is (A) $\mathrm{KO}_{2}$.
When potassium is heated strongly in the presence of oxygen, it predominantly forms superoxide ($\mathrm{KO}_{2}$). This compound has a distinctive attribute compared to other typical oxides or peroxides formed by different metals.
The compound $\mathrm{CH}{3} \mathrm{COOC}{2} \mathrm{H}_{5}$ belongs to the family of carboxylic acids.
A) True
B) False
The correct answer is B) False.
The compound $$\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5$$ is classified as an ester, which is characterized by the functional group $$-COO$$. In contrast, compounds belonging to the carboxylic acid family exhibit the functional group $$-COOH$$. Therefore, the provided compound does not fit within the carboxylic acids category.
Among the following compounds, the most acidic is:
A. p-nitrophenol
B. p-hydroxybenzoic acid
C. o-hydroxybenzoic acid
D. p-toluic acid
The correct answer is C. o-hydroxybenzoic acid.
o-Hydroxybenzoic acid is the most acidic compound among the options given. This enhanced acidity arises from the intramolecular hydrogen bonding in its conjugate base. The close proximity of the hydroxyl (-OH) and carboxyl (-COOH) groups allows for this hydrogen bonding, which significantly stabilizes the anion formed after deprotonation. This stabilization effect is stronger than the electron-donating resonance effect of the hydroxyl group, leading to a more acidic nature compared to the other options.
A carbon compound 'P' on heating with excess concentrated $\mathrm{H}{2} \mathrm{SO}{4}$ forms another carbon compound 'Q' which on addition of hydrogen in the presence of a nickel catalyst forms a saturated carbon compound 'R'. One molecule of 'R' on combustion forms two molecules of carbon dioxide and three molecules of water. Identify P, Q, and R, and write chemical equations for the reactions involved.
Solution:
Starting Compound (P) Identification:
- When P is heated with excess $\mathrm{H}{2}\mathrm{SO}{4}$, it results in the formation of another compound Q. Given that $\mathrm{H}{2}\mathrm{SO}{4}$ is often used as a dehydrating agent, it suggests that P might be an alcohol.
- Further, since Q readily adds hydrogen to form a saturated hydrocarbon R, which on complete combustion forms equal amounts of $\mathrm{CO}_2$ and water, P is likely an ethyl group bearing compound. Hence, P is ethyl alcohol or ethanol ($\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$).
Intermediate Compound (Q) Identification:
- The reaction of $\mathrm{C}2\mathrm{H}5\mathrm{OH}$ with $\mathrm{H}{2}\mathrm{SO}{4}$ mainly involves the elimination of a water molecule, a dehydration reaction, leading to the formation of an alkene.
- Thus, Q is ethene ($\mathrm{C}_2\mathrm{H}_4$).
Final Product (R) Identification:
- Q (ethene) upon hydrogenation, with the presence of a nickel catalyst, leads to a saturated hydrocarbon.
- Given the number of carbons in ethene remains unchanged and merely gains hydrogens, R is ethane ($\mathrm{C}_2\mathrm{H}_6$).
Reaction Equations:
- Dehydration of Ethanol to Ethene: $$ \mathrm{C}_2\mathrm{H}_5\mathrm{OH} \rightarrow \mathrm{C}_2\mathrm{H}_4 + \mathrm{H}_2\mathrm{O} $$
- Hydrogenation of Ethene to Ethane: $$ \mathrm{C}_2\mathrm{H}_4 + \mathrm{H}_2 \rightarrow \mathrm{C}_2\mathrm{H}_6 $$ (catalyzed by nickel)
- Combustion of Ethane: $$ 2\mathrm{C}_2\mathrm{H}_6 + 7\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} $$
In this process, we can trace the transformation of ethanol as it is dehydrated to ethene, then hydrogenated to ethane, which fully oxidizes upon combustion to yield carbon dioxide and water.
A four-carbon dicarboxylic acid that is an intermediate of C4 pathway, CAM pathway, and Krebs cycle is:
A) Aspartic acid
B) Oxaloacetate
C) Malic acid
D) Succinic acid
The correct answer is B) Oxaloacetate.
Oxaloacetate is a key intermediate in multiple metabolic pathways, including the C4 pathway, CAM pathway, and the Krebs cycle. This four-carbon dicarboxylic acid plays a crucial role in each of these processes, transitioning and interconnecting various metabolic functions essential for cellular energy production and carbon fixation.
In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
(a) Assertion and reason both are correct statements and reason explains the assertion. (b) Both assertion and reason are wrong statements. (c) Assertion is a correct statement and reason is a wrong statement. (d) Assertion is a wrong statement and reason is a correct statement. (e) Assertion and reason both are correct statements, but the reason does not explain the assertion.
Assertion (A): Deoxyribose, $\mathrm{C}{5} \mathrm{H}{10} \mathrm{O}{4}$ is not a carbohydrate. Reason (R): Carbohydrates are hydrates of carbon, so compounds which follow $\mathrm{C}{\mathrm{x}}\left(\mathrm{H}{2} \mathrm{O}\right){y}$ formula are carbohydrates.
Solution
Correct Answer: (c) Assertion is a correct statement and reason is a wrong statement.
Assertion Explanation:The assertion states that Deoxyribose, with the molecular formula $\mathrm{C}{5} \mathrm{H}{10} \mathrm{O}_{4}$, is not a carbohydrate. This statement is correct. Deoxyribose is indeed a carbohydrate, specifically a monosaccharide called 2-deoxyribose, which plays an essential role in the structure of DNA. However, it is specifically differentiated from typical carbohydrates due to the absence of an oxygen atom in the second position of the ring, which is notable hence the prefix "deoxy".
Reason Explanation:The reason provided claims that carbohydrates are hydrates of carbon that follow the generalized formula $\mathrm{C}{\mathrm{x}}(\mathrm{H}{2}\mathrm{O})_{\mathrm{y}}$. This statement is incorrect as it overly simplifies the classification of carbohydrates. Not all compounds fitting this stoichiometry are carbohydrates. Other structural features, such as the presence of a carbonyl group and multiple hydroxyl groups arranged on a carbon backbone, classify a compound as a carbohydrate. The given molecular formula for carbohydrates does not guarantee the presence of such features.
Therefore, while the assertion is true (Deoxyribose is a carbohydrate), the reason given is not a correct statement regarding the classification of carbohydrates.
The combination of which of the following compounds will give an ester?
A) Methanone and propanal
B) Methanal and propanol
C) Methanol and propanoic acid
D) Methanone and formic acid
Correct Answer: C) Methanol and Propanoic Acid
Esters are organic compounds typically formed by the reaction between a carboxylic acid and an alcohol. In this scenario, methanol (CH3OH) reacts with propanoic acid (CH3CH2COOH) to yield an ester, specifically methyl propanoate (CH3CH2COOCH3), through a reaction mechanism that could involve a catalyst like concentrated sulfuric acid. The byproduct of this reaction is water (H2O). The chemical reaction can be represented as: $$ \text{CH}_{3}\text{OH}(\text{aq}) + \text{CH}_{3}\text{CH}_{2}\text{COOH}(\text{aq}) \rightarrow \text{CH}_{3}\text{CH}_{2}\text{COOCH}_{3}(\text{aq}) + \text{H}_{2}\text{O} $$
$$ \mathrm{CaC}{2} + \mathrm{H}{2}\mathrm{O} \rightarrow \mathrm{A} \xrightarrow{\mathrm{H}{2}\mathrm{SO}{4} / \mathrm{HgSO}_{4}} \mathrm{~B} . $$
Identify $A$ and $B$ in the given reaction.
A $\mathrm{C}{2}\mathrm{H}{2}$ and $\mathrm{CH}_{3}$
B $\mathrm{CH}_{4}$ and $\mathrm{HCOOH}$
C $\mathrm{C}{2}\mathrm{H}{4}$ and $\mathrm{CH}_{3}\mathrm{COOH}$
D $\mathrm{C}{2}\mathrm{H}{2}$ and $\mathrm{CH}_{3}\mathrm{COOH}$
Solution
The correct option is D $\mathrm{C}{2}\mathrm{H}{2}$ and $\mathrm{CH}_{3}\mathrm{COOH}$
-
First step (Formation of $A$):
The reaction between calcium carbide ($\mathrm{CaC}{2}$) and water ($\mathrm{H}{2}O$) produces acetylene ($\mathrm{C}{2}\mathrm{H}{2}$) and calcium hydroxide ($\mathrm{Ca}(\mathrm{OH})2$). This is known as the Wohler reaction: $$ \mathrm{CaC}{2} + 2 \mathrm{H}{2}O \rightarrow \mathrm{C}{2}\mathrm{H}_{2} + \mathrm{Ca}(\mathrm{OH})_2 $$
Therefore, compound $A$ is acetylene ($\mathrm{C}{2}\mathrm{H}{2}$).
-
Second step (Formation of $B$):
The acetylene then undergoes a hydration reaction in the presence of sulfuric acid ($\mathrm{H}{2}\mathrm{SO}{4}$) and mercuric sulfate ($\mathrm{HgSO}{4}$) to form acetic acid ($\mathrm{CH}{3}\mathrm{COOH}$): $$ \mathrm{C}{2}\mathrm{H}{2} \xrightarrow{\mathrm{H}{2}\mathrm{SO}{4} / \mathrm{HgSO}{4}} \mathrm{CH}{3}\mathrm{COOH} $$
So, compound $B$ is acetic acid ($\mathrm{CH}_{3}\mathrm{COOH}$).
The final correct answer is based on these product identifications and is Option D.
$\mathrm{CO}_{2}$ on reaction with ethyl magnesium bromide gives
A) Ethane
B) Propanoic acid
C) Acetic acid
D) None of these
The Grignard synthesis of carboxylic acids involves reacting $\mathrm{CO}_2$ (carbon dioxide) with a Grignard reagent, which in this case is ethyl magnesium bromide ($\mathrm{C_2H_5MgBr}$). The general reaction can be described using the following equation:
$$ \mathrm{R-MgBr + CO}_2 \rightarrow \mathrm{R-COOH} $$
where $\mathrm{R}$ represents an alkyl group. Applying this to the specific case provided:
-
Ethyl magnesium bromide reacts with carbon dioxide: $$ \mathrm{C}_2\mathrm{H}_5\mathrm{MgBr} + \mathrm{CO}_2 \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{COO}^- \mathrm{Mg}^{2+}\mathrm{Br}^- $$
-
On acidic workup, the reaction further processes to provide a carboxylic acid: $$ \mathrm{C}_2\mathrm{H}_5\mathrm{COO}^- \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{COOH} $$
Thus, the product formed is acetic acid ($\mathrm{CH_3COOH}$), noting that when $\mathrm{C}_2\mathrm{H}_5$ is bonded to the carbonyl group $(\mathrm{COOH})$, the compound is acetic acid extended by one methylene group, hence forming propanoic acid ($\mathrm{C}_2\mathrm{H}_5\mathrm{COOH}$).
Therefore, the answer is Option B: Propanoic acid.
Oxidative decarboxylation of pyruvic acid results in the formation of: I. Acetyl Co-A II. CO2 III. ATP IV. NADH + H+
Select the correct answer using the codes given below:
A. Only I B. I and II C. I, II, and III D. I, II, and IV E. III and IV
The correct answer is D. I, II, and IV.
In the mitochondrial matrix, pyruvic acid undergoes a reaction known as oxidative decarboxylation when it is in the presence of oxygen. This process is catalyzed by the enzyme pyruvic acid dehydrogenase. During this reaction, several products are formed:
- Acetyl CoA - This is a crucial molecule that enters the citric acid cycle.
- CO2 (Carbon Dioxide) - This is released as a by-product.
- NADH + H+ - These are important reducing agents used in the electron transport chain to generate ATP through oxidative phosphorylation.
However, ATP is not directly produced in this reaction, making options that include ATP (like option C) incorrect. Therefore, the formation only involves Acetyl Co-A, CO2, and NADH + H+, corresponding to answer D.
$2.4 \mathrm{~kg}$ of carbon is made to react with $1.35 \mathrm{~kg}$ of aluminium to form $\mathrm{Al}_{4} \mathrm{C}_{3}$. The maximum amount of aluminium carbide formed is:
A) $5.44 \mathrm{~kg}$
B) $3.75 \mathrm{~kg}$
C) $1.05 \mathrm{~kg}$
D) $1.80 \mathrm{~kg}$
Correct Option: D) $1.80 , \mathrm{kg}$
Balanced Chemical Equation:$$ 4 \mathrm{Al} + 3 \mathrm{C} \rightarrow \mathrm{Al}_4 \mathrm{C}_3 $$
Molar mass of Carbon (C): $12 , \mathrm{g/mol}$
Total moles of Carbon: $$ \frac{2400 , \mathrm{g}}{12 , \mathrm{g/mol}} = 200 , \mathrm{mol} $$
Molar mass of Aluminium (Al): $27 , \mathrm{g/mol}$
Total moles of Aluminium: $$ \frac{1350 , \mathrm{g}}{27 , \mathrm{g/mol}} = 50 , \mathrm{mol} $$
Stoichiometry of reaction: 3 moles of C react with 4 moles of Al.
Required moles of Aluminium to react with 200 moles of Carbon: $$ \left(\frac{4}{3} \times 200 , \mathrm{mol}\right) = 266.67 , \mathrm{mol} $$
Since only 50 moles of Aluminium are available, Aluminium is the limiting reagent.
Moles of $\mathrm{Al}_4 \mathrm{C}_3$ formed: $$ \frac{1}{4} \times 50 , \mathrm{mol} = 12.5 , \mathrm{mol} $$
Molar mass of $\mathrm{Al}_4 \mathrm{C}_3$: $$ (4 \times 27 + 12 \times 3) , \mathrm{g/mol} = 144 , \mathrm{g/mol} $$
Mass of $\mathrm{Al}_4 \mathrm{C}_3$ produced: $$ 144 , \mathrm{g/mol} \times 12.5 , \mathrm{mol} = 1800 , \mathrm{g} = 1.8 , \mathrm{kg} $$
Thus, the maximum amount of aluminium carbide, $\mathrm{Al}_4 \mathrm{C}_3$, that can be formed is $1.8 , \mathrm{kg}$, corresponding to option D.
Which of the following can exhibit cis-trans isomerism?
(A) $\mathrm{HC} \equiv \mathrm{CH}$ (B) $\mathrm{ClCH}=\mathrm{CHCl}$
(C) $\mathrm{CH}{3}\mathrm{CHCl}.\mathrm{COOH}$ (D) $\mathrm{ClCH}{2}-\mathrm{CH}_{2}\mathrm{Cl}$
The correct option that can exhibit cis-trans isomerism is (B) $\mathrm{ClCH}=\mathrm{CHCl}$.
Cis-trans isomerism, also known as geometric isomerism, occurs in compounds that contain a double bond with two different substituents on each carbon atom of the double bond. For a molecule to display cis-trans isomerism, each carbon atom in the double bond must have two different groups attached to it.
-
Option (A) $\mathrm{HC} \equiv \mathrm{CH}$ has a triple bond between the carbon atoms and thus does not allow for geometric isomerism.
-
Option (B) $\mathrm{ClCH}=\mathrm{CHCl}$ has a double bond with chlorine atoms on each carbon along with hydrogen atoms. This arrangement fulfills the requirement for cis-trans isomerism, where the substituents differ on each carbon.
-
Option (C) $\mathrm{CH}_{3}\mathrm{CHCl}.\mathrm{COOH}$ does not contain the necessary double bond structure for cis-trans isomerism—it consists of a single bond throughout.
-
Option (D) $\mathrm{ClCH}{2}-\mathrm{CH}{2}\mathrm{Cl}$ also lacks a double bond and thus, cannot exhibit cis-trans isomerism.
Therefore, option (B) is the molecule that can demonstrate cis-trans isomerism with chlorine and hydrogen positioned differently around the double bond.
A new carbon-carbon bond formation is possible in [IIT-JEE 1998].
A. Cannizzaro reaction
B. Friedel-Crafts alkylation
C. Clemmensen reduction
D. Reimer-Tiemann reaction
The correct response is D. Reimer-Tiemann reaction.
In both Friedel-Crafts alkylation and Reimer-Tiemann reaction, new carbon-carbon (C-C) bonds are formed. In the case of Friedel-Crafts alkylation, the mechanism leads to the formation of a new C-C bond between the carbon atom of the benzene ring and an alkyl group:
Similarly, in the Reimer-Tiemann reaction, a new C-C bond is formed between the carbon of the benzene ring and a formyl (-CHO) group:
Thus, for new carbon-carbon bond formation in the provided options, the Reimer-Tiemann reaction is a definitive choice.
Which is the wrong IUPAC name of chlorocyclohexane-dione?
A) 3-Chloro-1,4-cyclohexanedione
B) 2-Chloro-1,3-cyclohexanedione
C) 4-Chloro-1,3-cyclohexanedione
D) 5-Chloro-1,3-cyclohexanedione
The correct answer is A) 3-Chloro-1,4-cyclohexanedione.
To determine the numbering, consider the following points:
- The cyclohexane ring is the parent chain.
- Halogens such as chlorine (Cl) and groups like ketones (>$\text{C=O}$) are substituents on this ring.
- There is no special priority order for halogens vs. alkyl groups in the numbering scheme. Halogens are treated like alkyl substituents.
- For the ketones, the functional group suffix is "one," reflecting their presence as ketonic groups.
Given these rules, the proper positioning for the functional groups based on IUPAC nomenclature would be 2-Chloro-1,4-cyclohexanedione. Thus, the chloro group should be at the 2-position, not at the 3-position as denoted in option A. Therefore, option A's name is incorrect, making it the correct choice for this question.
In which of the following reactions, oxygen is an oxidant?
A $2 \text{ F}_{2} + \text{ O}_{2} \rightarrow 2 \text{ F}_{2} \text{O}$ B $\text{ C} + \text{ O}_{2} \rightarrow \text{ CO}_{2}$ C $2 \text{ C} + \text{ O}_{2} \rightarrow 2 \text{ CO}$ D $2 \text{ N}_{2} + \text{ O}_{2} \rightarrow 2 \text{ N}_{2} \text{O}$
To determine in which reactions oxygen acts as an oxidant, we need to understand the role of an oxidant. An oxidant or oxidizing agent is a substance that has the ability to oxidize other substances by undergoing a reduction itself, which means it gains electrons or its oxidation state decreases.
Let's go through each reaction:
Reaction A
$$ 2 \text{F}_2 + \text{O}_2 \rightarrow 2 \text{F}_2\text{O} $$
Fluorine ($\text{F}_2$) starts with an oxidation state of 0.
Oxygen ($\text{O}_2$) starts with an oxidation state of 0.
In $2 \text{F}_2\text{O}$, fluorine has an oxidation state of -1, and oxygen has an oxidation state of +2.
The oxidation state of oxygen increases from 0 to +2, indicating oxygen is being oxidized, not reduced. Hence, oxygen is not an oxidant in this reaction.
Reaction B
$$ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 $$
Carbon (C) starts with an oxidation state of 0.
Oxygen ($\text{O}_2$) starts with an oxidation state of 0.
In $\text{CO}_2$, carbon has an oxidation state of +4, and each oxygen atom has an oxidation state of -2.
The oxidation state of oxygen decreases from 0 to -2, showing that oxygen is being reduced. Therefore, oxygen acts as an oxidant in this reaction.
Reaction C
$$ 2 \text{C} + \text{O}_2 \rightarrow 2 \text{CO} $$
Carbon (C) starts with an oxidation state of 0.
Oxygen ($\text{O}_2$) starts with an oxidation state of 0.
In $\text{CO}$, carbon has an oxidation state of +2, and oxygen has an oxidation state of -2.
Again, the oxidation state of oxygen decreases from 0 to -2, indicating oxygen is being reduced. Thus, oxygen is an oxidant in this reaction as well.
Reaction D
$$ 2 \text{N}_2 + \text{O}_2 \rightarrow 2 \text{N}_2\text{O} $$
Nitrogen ($\text{N}_2$) starts with an oxidation state of 0.
Oxygen ($\text{O}_2$) starts with an oxidation state of 0.
In $2 \text{N}_2\text{O}$, nitrogen has an oxidation state of +1, and oxygen has an oxidation state of -2.
The oxidation state of oxygen decreases from 0 to -2, indicating a reduction of oxygen. Therefore, oxygen is an oxidant in this reaction.
Conclusion
In reactions B, C, and D, oxygen undergoes reduction (decrease in oxidation state), meaning it acts as an oxidant.
So, the correct answer is:
$$ \boxed{B, C, D} $$
Consider the following reaction: $\mathrm{CO(g)} + 2\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3}\mathrm{OH}(g)$.
Given: $\Delta_{f}H^{\circ}(\mathrm{CH}_{3}\mathrm{OH}, g) = -201 \mathrm{kJ/mol}$, $\Delta_{f}H^{\circ}(\mathrm{CO}, g) = -114 \mathrm{kJ/mol}$, $S^{\circ}(\mathrm{CH}_{3}\mathrm{OH}, g) = 240 \mathrm{J/K-mol}$, $S^{\circ}(\mathrm{H}_{2}, g) = 29\mathrm{J/K-mol}$, $S^{\circ}(\mathrm{CO}, g) = 198\mathrm{J/mol-K}$, $C_{p, m}^{\circ}(\mathrm{H}_{2}) = 28.8\mathrm{J/mol-K}$, $C_{p, m}^{\circ}(\mathrm{CO}) = 29.4\mathrm{J/mol-K}$, $C_{p, m}^{\circ} (\mathrm{CH}_{3}\mathrm{OH}) = 44\mathrm{J/mol-K}$, and $\ln(\frac{320}{300}) = 0.06$, all data at $300\mathrm{K}$.
$\Delta_{r}G^{\circ}$ at $320\mathrm{K}$ is :
A $155.18\mathrm{J/mol-K}$,
B $150.02\mathrm{J/mol-K}$,
C $172\mathrm{J/mol-K}$,
None of these.
To solve for $\Delta_r G^{\circ}$ at $320\ \mathrm{K}$, we need to use the thermodynamic data provided and the following equations:
$\Delta S^\circ_\text{reaction} = S^\circ_\text{products} - S^\circ_\text{reactants}$
$\Delta H^\circ_\text{reaction} = H^\circ_\text{products} - H^\circ_\text{reactants}$
$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$
Correcting $\Delta H^\circ_\text{reaction}$ and $\Delta S^\circ_\text{reaction}$ from 300 K to 320 K.
Let's begin by calculating $\Delta S^\circ_\text{reaction}$ and $\Delta H^\circ_\text{reaction}$ at 300 K.
Step 1: Calculate $\Delta S^\circ_\text{reaction}$ at 300 K
$$\Delta S^\circ_\text{reaction} = S^\circ(\text{CH}_3\text{OH}(g)) - \left[S^\circ(\text{CO}(g)) + 2 \times S^\circ(\text{H}_2(g))\right]$$
Given:
$S^\circ(\text{CH}_3\text{OH}(g)) = 240\ \mathrm{J/K \cdot mol}$
$S^\circ(\text{CO}(g)) = 198\ \mathrm{J/K \cdot mol}$
$S^\circ(\text{H}_2(g)) = 29\ \mathrm{J/K \cdot mol}$
Substitute these values: $$\Delta S^\circ_\text{reaction} = 240 - (198 + 2 \times 29)$$ $$= 240 - 256$$ $$= -16\ \mathrm{J/K \cdot mol}$$
Step 2: Calculate $\Delta H^\circ_\text{reaction}$ at 300 K
$$\Delta H^\circ_\text{reaction} = H^\circ_f(\text{CH}_3\text{OH}(g)) - \left[H^\circ_f(\text{CO}(g)) + 2 \times H^\circ_f(\text{H}_2(g))\right]$$
Given:
$H^\circ_f(\text{CH}_3\text{OH}(g)) = -201\ \mathrm{kJ/mol}$
$H^\circ_f(\text{CO}(g)) = -114\ \mathrm{kJ/mol}$
$H^\circ_f(\text{H}_2(g)) = 0\ \mathrm{kJ/mol}$ (Standard enthalpy of formation for $\mathrm{H}_2(g)$ is zero)
Substitute these values: $$\Delta H^\circ_\text{reaction} = -201 - (-114 + 2 \times 0)$$ $$= -201 + 114$$ $$= -87\ \mathrm{kJ/mol}$$
Step 3: Correct $\Delta H^\circ_\text{reaction}$ and $\Delta S^\circ_\text{reaction}$ for 320 K
First, we need the heat capacity values to correct $\Delta H^\circ$ and $\Delta S^\circ$:
$C_{p, m}^{\circ}(\text{CH}_3\text{OH}) = 44\ \mathrm{J/mol \cdot K}$
$C_{p, m}^{\circ}(\text{CO}) = 29.4\ \mathrm{J/mol \cdot K}$
$C_{p, m}^{\circ} (\text{H}_2) = 28.8\ \mathrm{J/mol \cdot K}$
Correcting $\Delta S^\circ_\text{reaction}$: $$\Delta C_p = C_{p, m}^{\circ}(\text{CH}3\text{OH}) - [C{p, m}^{\circ}(\text{CO}) + 2 \times C_{p, m}^{\circ}(\text{H}_2)]$$ $$= 44 - (29.4 + 2 \times 28.8)$$ $$= 44 - 87$$ $$= -43\ \mathrm{J/mol \cdot K}$$
Now correct $\Delta S^\circ_\text{reaction}$ to 320 K: $$\Delta S^\circ_\text{reaction}(320\ \mathrm{K}) = \Delta S^\circ_\text{reaction}(300\ \mathrm{K}) + \Delta C_p \ln\left(\frac{320}{300}\right)$$ $$= -16 + (-43 \times 0.06)$$ $$= -16 - 2.58$$ $$= -18.58\ \mathrm{J/K \cdot mol}$$
For $\Delta H^\circ$ at 320 K: $$\Delta H^\circ_\text{reaction}(320\ \mathrm{K}) = \Delta H^\circ_\text{reaction}(300\ \mathrm{K}) + \Delta C_p (T_2 - T_1)$$ $$= -87 + (-43 \times 20/1000)$$ $$= -87 - 0.86$$ $$= -87.86\ \mathrm{kJ/mol}$$
Step 4: Calculate $\Delta_r G^{\circ}$ at 320 K
Use the Gibbs free energy equation: $$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T \Delta_r S^{\circ}$$ $$= -87.86\ \mathrm{kJ/mol} - 320 \times (-0.01858)\ \mathrm{kJ/mol}$$ (Note: Convert $ \Delta_r S^\circ$ to $\mathrm{kJ/K \cdot mol}$)
Convert $S$ to kJ: $$= -87.86 + 320 \times 0.01858$$ $$= -87.86 + 5.9456$$ $$= -81.9144\ \mathrm{kJ/mol}$$
Therefore, the value of $\Delta_r G^{\circ}$ at $320\ \mathrm{K}$ is (-81.91\ \mathrm{kJ/mol}).
So, the correct answer is None of these.
Based on the following equilibrium reactions:
$\mathrm{HNO}_{2} + \mathrm{HF} \rightarrow \mathrm{H}_{2}\mathrm{F}^{+} + \mathrm{NO}_{2}^{-}$
$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{HF} \rightarrow \mathrm{F} + \mathrm{CH}_{3}\mathrm{COOH}_{2}^{+}$
$\mathrm{H}_{2}\mathrm{O} + \mathrm{CH}_{3}\mathrm{COOH} \rightarrow \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{CH}_{3}\mathrm{COO}^{-}$
The correct order(s) regarding acid strength are:
A) $\mathrm{H}_{2}\mathrm{O} < \mathrm{CH}_{3}\mathrm{COOH}$
B) $\mathrm{HF} < \mathrm{HNO}_{2}$
C) $\ce{H2O} < \ce{HF}$
D) $\ce{CH2COOH} < \ce{HNO2}$
To determine the order of acid strength based on the given equilibrium reactions, let’s analyze each reaction and the options provided.
Equilibrium Reactions:
$\mathrm{HNO}_{2} + \mathrm{HF} \rightarrow \mathrm{H}_{2}\mathrm{F}^{+} + \mathrm{NO}_{2}^{-}$
$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{HF} \rightarrow \mathrm{F} + \mathrm{CH}_{3}\mathrm{COOH}_{2}^{+}$
$\mathrm{H}_{2}\mathrm{O} + \mathrm{CH}_{3}\mathrm{COOH} \rightarrow \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{CH}_{3}\mathrm{COO}^{-}$
Analyzing Acid Strength:
For $\mathrm{HNO}_2$ and $\mathrm{HF}$:
In the reaction $\mathrm{HNO}_{2} + \mathrm{HF} \rightarrow \mathrm{H}_{2}\mathrm{F}^{+} + \mathrm{NO}_{2}^{-}$, $\mathrm{HNO}_2$ donates a proton to $\mathrm{HF}$. This indicates that $\mathrm{HNO}_2$ is a stronger acid compared to $\mathrm{HF}$.
Thus, $\mathbf{\mathrm{HNO}_2 > \mathrm{HF}}$.
For $\mathrm{CH}_3\mathrm{COOH}$ and $\mathrm{HF}$:
When $\mathrm{CH}_3\mathrm{COOH}$ reacts with $\mathrm{HF}$ resulting in $\mathrm{F}^{-}$ and $\mathrm{CH}_3\mathrm{COOH}_{2}^{+}$, $\mathrm{HF}$ retains its proton, indicating $\mathrm{HF}$ is a stronger acid than $\mathrm{CH}_3\mathrm{COOH}$.
Thus, $\mathbf{\mathrm{HF} > \mathrm{CH}_3\mathrm{COOH}}$.
For $\mathrm{H}_2\mathrm{O}$ and $\mathrm{CH}_3\mathrm{COOH}$:
In the reaction $\mathrm{H}_2\mathrm{O} + \mathrm{CH}_3\mathrm{COOH} \rightarrow \mathrm{H}_3\mathrm{O}^{+} + \mathrm{CH}_3\mathrm{COO}^{-}$, $\mathrm{H}_2\mathrm{O}$ accepts a proton from $\mathrm{CH}_3\mathrm{COOH}$, indicating $\mathrm{CH}_3\mathrm{COOH}$ is a stronger acid.
Thus, $\mathbf{\mathrm{CH}_3\mathrm{COOH} > \mathrm{H}_2\mathrm{O}}$.
Options Analysis:
Considering the acid strength evaluated:
Option A: $\mathrm{H}_2\mathrm{O} < \mathrm{CH}_3\mathrm{COOH}$ is correct.
Option B: $\mathrm{HF} < \mathrm{HNO}_2$ is correct.
Other Options: If provided, need not be analyzed as per provided data.
Conclusion:
Based on the given data and the evaluated acid strength, the final conclusions are:
Option A and Option B are correct.
Complete the following reaction:
$$ \mathrm{CH} \equiv \mathrm{CH} \xrightarrow[\mathrm{HgSO}_{4}+\mathrm{H}_{2}]{\stackrel{\mathrm{H}_{2} \mathrm{O}}{ } \mathrm{SO}_{4}} \stackrel{\text{tautomerism}}{\rightleftharpoons} $$
(A) $\mathrm{CH}_{3} \mathrm{CHO}$
(B) $\mathrm{CH}_{3} \mathrm{COOH}$
(C) $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{OH}$
(D) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$
The correct options are (A) $\mathrm{CH_3CHO}$ and (C) $\mathrm{CH_2 = CH - OH}$.
When water is added to ethyne ($ \mathrm{CH} \equiv \mathrm{CH} $) in the presence of $\mathrm{H_2SO_4}$ (sulfuric acid) and $\mathrm{HgSO_4}$ (mercury(II) sulfate), it results in the formation of acetaldehyde ($ \mathrm{CH_3CHO} $).
The reaction sequence is as follows: $$ \mathrm{CH} \equiv \mathrm{CH} \xrightarrow[\mathrm{HgSO}_{4}+\mathrm{H}_{2}O]{\mathrm{H_2SO_4}} \mathrm{CH_3CHO} \stackrel{\text{tautomerism}}{\rightleftharpoons} \mathrm{CH_2 = CH - OH} $$
$\mathrm{CH}_{3} \mathrm{COOH} \xrightarrow{\mathrm{LiALH}_{3}} (\mathrm{A})+\mathrm{CH}_{3} \mathrm{COOH} \xrightarrow{\mathrm{H}_{3} \mathrm{O}^{+}}(\mathrm{B})+\mathrm{H}_{2} \mathrm{O}$
In the above reaction ' $A$ ' and ' $B$ ' respectively are: $\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ $\mathrm{CH}_{3} \mathrm{CHO}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ $9 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{CHO}$ $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}$
The correct option is D $ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} $
Explanation:
Reduction Reaction:$$ \mathrm{CH}_{3} \mathrm{COOH} \xrightarrow{\mathrm{LiAlH}_{4}} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} $$ Here, acetic acid ($ \mathrm{CH}_{3} \mathrm{COOH} $) undergoes __complete reduction__ by lithium aluminium hydride ($ \mathrm{LiAlH}_{4} $) to form ethanol ($ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} $).
Esterification Reaction:$$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} + \mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} + \mathrm{H}_{2} \mathrm{O} $$ The ethanol ($ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} $) formed in the first step reacts with acetic acid ($ \mathrm{CH}_{3} \mathrm{COOH} $) to produce ethyl acetate ($ \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} $) and water ($ \mathrm{H}_{2} \mathrm{O} $) in an esterification reaction.
Thus, the compounds ' $A$ ' and ' $B$ ' are ethanol and ethyl acetate respectively.
In the following reaction sequence, $Y$ is:
$$ \mathrm{PhOCOCH}_{3} \xrightarrow[\Delta]{\mathrm{AlCl}_{3}}(X) \xrightarrow[\mathrm{NaOH}]{\mathrm{I}_{2}}(Y) $$
A $\mathrm{PhCOONa}$
B $\mathrm{PhCOOH}$
C.
D.
Let's analyze the reaction sequence step-by-step to identify compound $Y$.
The first step involves the reaction of $\mathrm{PhOCOCH}_3$ with $\mathrm{AlCl}_3$ under heat ($\Delta$). This step likely involves a Friedel-Crafts acylation to form an intermediate product $X$.
$$ \mathrm{PhOCOCH}_3 \xrightarrow[\Delta]{\mathrm{AlCl}_3} (X) $$
During this step, a phenyl ester ($\mathrm{PhOCOCH}_3$) undergoes acylation to yield acetophenone (C$_6$H$_5$$\mathrm{COCH}_3$) as $X$.
The second step involves the reaction of $X$ with sodium hydroxide ($\mathrm{NaOH}$) and iodine ($\mathrm{I}_2$). This step points towards a Haloform reaction, typically used for methyl ketones.
$$ (X) \xrightarrow[\mathrm{NaOH}]{\mathrm{I}_2} (Y) $$
The intermediate product $X$ (acetophenone) will react with $\mathrm{NaOH}$ and $\mathrm{I}_2$ to perform the iodoform test, where the methyl group is transformed, yielding a carboxylate salt.
$$ \text{C}_6\text{H}_5\text{COCH}_3 + 3\mathrm{I}_2 + 4\mathrm{NaOH} \rightarrow \text{C}_6\text{H}_5\text{COONa} + 3\text{NaI} + 3\mathrm{H}_2\text{O} $$
In this reaction:
The Iodoform reaction produces sodium benzoate ($\text{PhCOONa}$).
Therefore, the correct compound $Y$ is:
Option A: $\mathrm{PhCOONa}$
In summary, the important steps are:
Friedel-Crafts acylation to form acetophenone.
Haloform reaction transforming acetophenone into sodium benzoate.
Thus, the correct answer is $\mathbf{\text{A:} , \mathrm{PhCOONa}}$.
Which one of the following statements is not true?
A. Lactose contains $\alpha$-glycosidic linkage between $C_{1}$ of galactose and $C_{4}$ of glucose.
B. Lactose $\left(\mathrm{C}_{11} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ is a disaccharide and it contains 8 hydroxyl groups.
C. Lactose is a reducing sugar and it gives Fehling's test.
D. On acid hydrolysis, lactose gives one molecule of $D(+)$-glucose and one molecule of $D(+)$ galactose.
To determine which statement is not true about lactose, let's evaluate each option carefully.
Statement A: "Lactose contains $\alpha$-glycosidic linkage between $C_{1}$ of galactose and $C_{4}$ of glucose."
Incorrect: Lactose actually contains a $\beta$-1,4-glycosidic linkage rather than an $\alpha$-glycosidic linkage.
Statement B: "Lactose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ is a disaccharide and it contains 8 hydroxyl groups."
Correct: This property is accurately described. Lactose is indeed a disaccharide with the given molecular formula and has 8 hydroxyl groups.
Statement C: "Lactose is a reducing sugar and it gives Fehling's test."
Correct: Lactose is a reducing sugar, as it contains a free anomeric carbon which allows it to give a positive Fehling's test.
Statement D: "On acid hydrolysis, lactose gives one molecule of $D(+)$-glucose and one molecule of $D(+)$-galactose."
Correct: Acid hydrolysis of lactose yields one molecule each of $D(+)$-glucose and $D(+)$-galactose.
Final Answer: Since the linkage specified in Statement A is incorrect, the statement A is not true.
Write down all possible methods to synthesize the following alcohols using Grignard reagents.
Methods for Preparing Alcohols Using Grignard Reagent
The Grignard reagent can be used to prepare alcohols from various compounds. Based on the desired structure of the alcohol, appropriate Grignard reagents and carbonyl compounds are selected to complete the reactions.
To synthesize the specified alcohol, the following possible methods can be utilized:
Method 1:
Grignard Reagent: Cyclohexylmethylmagnesium bromide ($ C_6H_{11}CH_2MgBr $)
Carbonyl Compound: Cyclobutanone ($ C_4H_6O $)
Reaction:$$ C_4H_6O + C_6H_{11}CH_2MgBr \rightarrow \text{Product (after hydrolysis)} $$
Method 2:
Grignard Reagent: Biscyclohexylmagnesium bromide ($ C_{12}H_{22}MgBr_2 $)
Carbonyl Compound: Propanal ($ C_3H_6O $)
Reaction:$$ C_3H_6O + C_{12}H_{22}MgBr_2 \rightarrow \text{Product (after hydrolysis)} $$
Method 3:
Grignard Reagent: Ethylmagnesium Bromide ($ C_2H_5MgBr $)
Carbonyl Compound: Cyclohexanone with Cyclobutyl group ($ C_{10}H_{16}O $)
Reaction:$$ C_{10}H_{16}O + C_2H_5MgBr \rightarrow \text{Product (after hydrolysis)} $$
Through these procedures, the designated alcohol can be synthesized by the reactions between the Grignard reagent and the corresponding carbonyl compound.
Important Tip: In all methods, it is essential to consider the hydrolysis (decomposition) stage to obtain the alcohol in the correct form.
Which of the following solutions has the maximum pH?
Options:
A. $0.2 , M , \mathrm{HClO_4}$
B. $0.20 , M , \mathrm{CH_3COOH}$
C. $0.020 , M , \mathrm{HCl}$
D. $0.2 , M , \mathrm{NaCl}$
To determine the solution with the highest pH, we must first understand the nature of the dissolved substances:
HClO$_4$ is a strong acid.
CH$_3$COOH (acetic acid) is a weak acid.
HCl is a strong acid.
NaCl is a neutral salt.
The pH of a solution is a measure of its acidity or basicity. Given the nature of these substances:
Strong acids completely dissociate in water, resulting in a low pH.
For example, HClO$_4$ and HCl will both result in a low pH.
Weak acids only partially dissociate, resulting in a moderately low pH.
For example, CH$_3$COOH, as it partially dissociates, will have a higher pH than strong acids but still below neutral.
Neutral salts like NaCl do not affect the pH of the water significantly, so its solution will have a pH close to 7, which is higher than both strong and weak acids.
Thus, out of the given options:
A: 0.2 M HClO$_4$ and C: 0.020 M HCl will have a high concentration of H$^+$ ions, leading to very low pH.
B: 0.20 M CH$_3$COOH will have a slightly higher pH than option A and C, due to partial dissociation.
D: 0.2 M NaCl will have a pH of around 7 (neutral).
So, the highest pH is for:
Final Answer: D $0.2 , \mathrm{M} , \mathrm{NaCl}$
Statement: The volume of a gas is inversely proportional to the number of moles of the gas.
Reason: The ratio of gaseous reactants and gaseous products is according to their molar ratio.
If both the statement and the reason are true and the reason correctly explains the statement.
If both the statement and the reason are true but the reason does not correctly explain the statement.
If the statement is true but the reason is false.
If the statement is false but the reason is true.
The correct answer is D.
In the reaction:
$$ \mathrm{H}_{2} + \mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl} $$
the ratio of the gaseous reactants and products is in accordance with their molar ratio. Specifically, the volumes of $\mathrm{H}_{2}$, $\mathrm{Cl}_{2}$, and $\mathrm{HCl}$ are in the ratio $1:1:2$, matching the molar ratio. This indicates that the volume of a gas is directly related to the number of moles, rather than being inversely related.
Therefore, the assertion that the volume of a gas is inversely proportional to the number of moles of the gas is false.
However, the reason stating that the ratios of gaseous reactants and products correspond to their molar ratios is true.
Hence, the assertion is incorrect, but the reason is correct. So the final answer is:
D
Statement: The slope of the curve of the compressibility factor of hydrogen with pressure is positive at all pressures.
Reason: Even at low pressure, repulsive forces are effective in hydrogen gas.
If both the statement and the reason are true, then the reason provides a correct explanation for the statement.
If the statement is true but the reason is false, then the reason does not provide the correct explanation for the statement.
If both the statement and the reason are true but the reason, then no explanation is needed.
Correct Answer: A
The compressibility factor for $ \mathrm{H}_2 $ increases with increasing pressure. At $ 273 \mathrm{~K} $, $ Z > 1 $ indicates that it is harder to compress compared to an ideal gas. Here, repulsive forces become significant.
Final Answer: A
Statement: The heat absorbed during the isothermal expansion of an ideal gas against a vacuum is zero.
Reason: The volume enclosed by the molecule of an ideal gas is zero.
A. If both the statement and reason are true and the reason correctly explains the statement.
B. If both the statement and reason are true but the reason does not correctly explain the statement.
C. If the statement is true but the reason is false.
D. If both the statement and reason are false.
The correct answer is:$\mathbf{C}$
In an isothermal expansion of an ideal gas against a vacuum, the work done is zero because the expansion occurs without any external pressure.
The reason given, which is that the volume occupied by an ideal gas molecule is zero, is incorrect.
Therefore, the statement is true, but the reason is false.
Final Answer: C
According to the Brønsted-Lowry concept, when an acid donates a proton, it forms a conjugate base. The "strength factor" (1) / ("acid strength") of the conjugate base determines whether a strong acid can produce a weak acid in the reverse reaction. However, the reverse is not possible. Which reactions are possible. Y = Yes or N = No.
According to the Brønsted-Lowry concept, when an acid donates a proton, it forms a conjugate base. The strength factor (or acid strength) of the conjugate base determines whether a strong acid can produce a weak acid in the reverse reaction. However, the reverse is not possible. Based on this principle, the possibility of the reactions listed is as follows:
(i)$$\ce{RSO3H + Na -> RSO3Na + 1/2 H2(T)}$$ Possible: Yes (Y)
(ii)$$\ce{RCOOH + Na -> RCOO^-Na^+ + 1/2 H2(T)}$$ Possible: Yes (Y)
(iii)$$\ce{+ Na -> + 1/2 H2(T)}$$ Possible: Yes (Y)
(iv)$$\ce{R-OH + Na -> RO Na^+ + 1/2 H2(T)}$$ Possible: Yes (Y)
(v)$$\ce{ -> Na }$$ Possible: Not listed in given solution, so undetermined in this context.
(vi)$$\ce{CH \equiv CH + Na -> CH \equiv CNa^+ + 1/2 H2}$$ Possible: Not listed in given solution, so undetermined in this context.
(vii)$$\ce{CH \equiv CH + Na\ (Solvent\ as\ Liq.\ NH3) -> CH \equiv CNa^+ + NH3}$$ Possible: Yes (Y)
A sample of pure acetic acid undergoes 40% dimerization. Then the mole fraction of the dimer in the final mixture is:
0.16
0.75
0.4
0.25
The correct option is D, 0.25.
The dimerization reaction of acetic acid can be represented as follows: $$ 2 \mathrm{A} \rightarrow \mathrm{A}_{2} $$
Here, A represents acetic acid.
Assume the initial number of moles of acetic acid (A) is 1.
Given that 40% of acetic acid undergoes dimerization, the moles of acetic acid dimerized can be calculated as: $$ \text{Moles of A dimerized} = 0.4 \text{ (40\% of 1 mole)} $$
After the dimerization process, the remaining moles of acetic acid (A) are: $$ \text{Final moles of A} = 1 - 0.4 = 0.6 $$
In the dimerization reaction, 2 moles of A form 1 mole of $A_2$. Therefore, the moles of $A_2$ (dimer) formed are: $$ \text{Moles of } A_{2} = \frac{\text{Moles of A dimerized}}{2} = \frac{0.4}{2} = 0.2 $$
The total number of moles after the reaction are: $$ \text{Total moles} = \text{Moles of A} + \text{Moles of } A_{2} = 0.6 + 0.2 = 0.8 $$
Finally, the mole fraction of the dimer $A_2$ in the final mixture can be determined by: $$ \text{Mole fraction of } A_{2} = \frac{\text{Number of moles of } A_{2}}{\text{Total number of moles}} = \frac{0.2}{0.8} = 0.25 $$
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