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Aldehydes, Ketones, and Carboxylic Acids — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryAldehydes, Ketones, and Carboxylic Acids4 Marks
Question
Read the passage given below and answer the following questions :
When an aldehyde with no a-hydrogen reacts with concentrated aqueous NaOH, half the aldehyde is converted to carboxylic acid salt and other half is converted to an alcohol. In other words, half of the reactant is oxidized
and other half is reduced. This reaction is known as Cannizzaro reaction

The following questions are multiple choice questions. Choose the most appropriate answer:
  1. A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives:
  1. Benzyl alcohol and sodium formate.
  2. Sodium benzoate and methyl alcohol.
  3. Sodium benzoate and sodium formate.
  4. Benzyl alcohol and methyl alcohol.
  1. Which of the following compounds will undergo Cannizzaro reaction?
  1. CH3CHO
  2. CH3COCH3
  3. C6H5CHO
  4. C6H5CH2CHO
  1. Trichloroacetaldehyde is subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate ion and another compound. The other compounds is:
  1. 2, 2, 2-trichloroethanol
  2. Trichloromethanol
  3. 2, 2, 2-trichloropropanol
  4. Chloroform
  1. Which of the following reaction will not result in the formation of carbon-carbon bonds?
  1. Cannizzaro reaction
  2. Wurtz reaction
  3. Reimer- Tiemann reaction
  4. Friedel - Crafts acylation
✓

Answer

  1. (a) Benzyl alcohol and sodium formate.

Explanation:

It is an example of cross Cannizzaro reaction where aromatic aldehyde gets reduced to alcohol and aliphatic aldehyde gets oxidised to its sodium salt (both aldehydes must not contain any αα-hydrogen).

CBSE 12th Chemistry Aldehydes , Ketones and Carboxylic Acids Case Study Questions With Solution 2021

  1. (c) C6H5CHO
  1. (a) 2, 2, 2-trichloroethanol

Explanation:

The Cannizzaro product of given reaction yields 2, 2, 2-trichloroethanol.

CBSE 12th Chemistry Aldehydes , Ketones and Carboxylic Acids Case Study Questions With Solution 2021

  1. (a) Cannizzaro reaction

Explanation:

C - C bond is not formed in Cannizzaro reaction while other reactions result in the formation of C - C bond.

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Similar questions

Read the passage given below and answer the following questions:

The order of reactivity towards SN1 reaction depends upon the stability of carbocation in the first step. Greater the stability of the carbocation, greater will be its ease of formation from alkyl halide and hence faster will be the rate of the reaction. As we know, 3º carbocation is most stable, therefore, the tert-alkyl that halides will undergo SN1 reaction very fast. For example, it has been observed that the reaction (CH3)3CBr with OH- ion to give 2-methyl-2-propanol is about I million times as fast as the corresponding reaction of the methyl bromide to give methanol.

The primary alkyl halides always react predominantly by SN2 mechanism. On the other hand, the tertiary alkyl halides react predominantly by SN1 mechanism. Secondary alkyl halides may react by either mechanism or by both the mechanisms without much preference depending upon the nature of the nucleophile and solvent.

In these questions (Q. No. i-tv), a statement of assertion followed by a statement of reason is given. Choose tile correct answer out of tile following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Low concentration of nudeophile favours SN1 mechanism.

Reason: 2º alkyl halides are less reactive than 1º towards SN1 reactions.

  1. Assertion: Polar solvent slows down SN2 reactions.

Reason: CH3-Br is less reactive than CH3Cl.

  1. Assertion: Benzyl bromide when kept in acetone- water it produces benzyl alcohol.

Reason: The reaction follows SN2 mechanism.

  1. Assertion: Rate of hydrolysis of methyl chloride to methanol is higher in DMF than in water.

Reason: Hydrolysis of methyl chloride follows second order kinetics.

  1. Assertion: SN1 reaction is carried out in the presence of a polar protic solvent.

Reason: A polar protic solvent increases the stability of carbocation due to solvation.

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Read the passage given below and answer the following questions:

Both alcohols and phenols are acidic in nature, but phenols are more acidic than alcohols. Acidic strength of alcohols mainly depends upon the inductive effect. Acidic strength of phenols depends upon a combination of both inductive effect and resonance effects of the substituent and its position on the benzene ring. Electron withdrawing groups increases the acidic strength of phenols whereas electron donating groups decreases the acidic strength of phenols. Phenol is a weaker acid than carboxylic acid.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Phenols are highly acidic as compare to alcohols due to:
  1. The higher molecular mass of phenols.
  2. The stronger hydrogen bonds in phenols.
  3. Alkoxide ion is a strong conjugate base.
  4. Phenoxide ion is resonance stabilised.
  1. The correct order of acidic strength among the following is:

  1. (III) > (IV) > (II) > (I)
  2. (IV) > (III) > (I) > (II)
  3. (IV) > (III) > (II) > (I)
  4. (I) > (II) > (IV) > (III)
  1. The correct decreasing order of pKa value is:

  1. II > IV > I > III
  2. IV > II > III > I
  3. II I> II > IV > I
  4. IV > I > II > III
  1. The compound that does not liberate CO2, on treatment with aqueous sodium bicarbonate solution is:
  1. Benzoic acid.
  2. Benzenesulphonic acid.
  3. Salicylic acid.
  4. Carbolic acid.
  1. Most acidic amongst the following is:

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Number of molecules which must collide simultaneously to give product is called molecularity. It is equal to sum of coefficients of reactants present in stoichiometric chemical equation.

For reaction, m1A + m2B → Product

Molecularity = [m1 + m2]

ln complex reaction each step has its own molecularity which is equal to the sum of coefficients of reactants present in a particular step. Molecularity is a theoretical property. Its value is any whole number. Number of concentration terms on which rate of reaction depends is called order of reaction or sum of powers of concentration terms present in the rate equation is called order of reaction.

If rate equation ofreaction is: Rate $=\text{k}\cdot\text{C}^{\text{m}_1}_\text{A}\cdot\text{C}^{\text{m}_2}_\text{B}$

Then order of reaction = m1 + m2.

ln simple reaction, order and molecularity are same.

ln complex reaction, order of slowest step is the order ofover all reaction. This step is known as rate determining step. Order is an experimental property. Its value may be zero, fractional or negative.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Higher order(> 3) reactions are rare due to:
  1. Shifting of equilibrium towards reactants due to elastic collisions.
  2. Loss of active species on collision.
  3. Low probability of simultaneous collision of all the reacting species.
  4. Increase in entropy and activation energy as more molecules are involved.
  1. The molecularity of the reaction:

$6\text{FeSO}_4+3\text{H}_2\text{SO}_4+\text{KClO}_3\rightarrow\text{KCl}+3\text{Fe}_2(\text{SO}_4)_3+3\text{H}_2\text{O}$ is:

  1. 6
  2. 10
  3. 3
  4. 7
  1. Which of the following statements is false in the following?
  1. Order of a reaction may be even zero.
  2. Molecularity of a reaction is always a whole number.
  3. Molecularity and order always have same values for a reaction.
  4. Order of a reaction depends upon the mechanism of the reaction.
  1. The rate of reaction, A + 2B → products, is given by the following equation:

$-\frac{\text{d}[\text{A}]}{\text{dt}}=\text{k}[\text{A}][\text{B}]^2$

If B is present in large excess, the order of the reaction is:

  1. Zero
  2. First
  3. Second
  4. Third
  1. The rate of the reaction, A + B + C → products, is given by $\text{r}=\frac{\text{d}[\text{A}]}{\text{dt}}=\text{k}[\text{A}]^\frac{1}{2}[\text{B}]^\frac{1}{3}[\text{C}]^\frac{1}{4}.$ The order of the reaction is:

  1. $\frac{1}{3}$​​​​​​​

  2. $\frac{1}{4}$

  3. $\frac{1}{2}$

  4. $\frac{13}{12}$

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Read the passage given below and answer the following questions:

Aldehydes and ketones having acetyl group $\left(\begin{array}{c}\ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \text{CH}_3-\text{C}-\end{array}\right)$ are oxidised by sodium hypohalate (NaOX) or halogen and alkali (X2 + OH-) to corresponding sodium salt having one carbon atoms less than the carbonyl compound and give a haloform.

$\ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\text{CH}_3\xrightarrow[\text{orX}_2+\text{NaOH}]{\text{NaOX}}\\ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}-\stackrel{-}{\hbox{ O}}\stackrel{+}{\hbox{Na}}+\text{CHX}_3(\text{X = Cl, Br, I})$

Sodium hypoiodite (NaOl) when treated with compounds containing CH3CO-group gives yellow precipitate of iodoform. Haloform reaction does not affect a carbon-carbon double bond present in the compound.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which of the following compounds will give positive iodoform test?
  1. Isopropyl alcohol.
  2. Propionaldehyde.
  3. Ethylphenyl ketone.
  4. Benzyl alcohol.
  1. Which of the following compounds is not formed in iodoform reaction of acetone?
  1. CH3COCH2l
  2. lCH2COCH2l
  3. CH3COCHl2
  4. CH3COCl3
  1. For the given set of reactions,

starting compound A corresponds to:

  1. In the following reaction sequence, the correct structures of E, F and G are:

(* implies 13C labelled carbon)

  1. An organic compound 'A' has the molecular formula C3H6O. It undergoes iodoform test. When saturated with HCl it gives 'B' of molecular formula C9H14O. 'A' and 'B' respectively are:
  1. propanal and mesityl oxide.
  2. Propanone and mesityl oxide.
  3. propanone and 2,6-dimethyl-2,5-hepta-dien-4-one.
  4. propanone and propionaldehyde.
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Read the passage given below and answer the following questions:
To explain bonding in coordination compounds various theories were proposed. One of the important theory was valence bond theory. According to that, the central metal ion in the complex makes available a number of empty orbitals for the formation of coordination bonds with suitable ligands. The appropriate atomic orbitals of the metal hybridise to give a set of equivalent orbitals of definite geometry.
The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e., (n - 1)d or outer d-orbitals i.e., nd. For example, Co3+ forms both inner orbital and outer orbital complexes, with ammonia it forms [Co(NH3)6]3+ and with fluorine it forms [CoF6]3- complex ion.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following is not true for [CoF6]3-?
  1. It is paramagnetic.
  2. It has coordination number of 6.
  3. It is outer orbital complex.
  4. It involves d2sp3 hybridisation.
  1. [Cr(H2O)6]Cl3 (at. no. of Cr = 24) has a magnetic moment of 3.83B.M. The correct distribution of 3d-electrons in the central metal of the complex is:
  1. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_\text{yz}$
  2. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}$
  3. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{zy}},3\text{d}^1_{\text{z}^2}$
  4. $3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_{\text{z}^2},3\text{d}^1_\text{xz}$
  1. Which of the following is true for [Co(NH3)6]3+?
  1. It is an octahedral, di magnetic and outer orbital complex.
  2. It is an octahedral, paramagnetic and outer orbital complex.
  3. It is an octahedral, paramagnetic and inner orbital complex.
  4. It is an octahedral, di magnetic and inner orbital complex.
  1. The paramagnetism of [CoF6]3- is due to.
  1. 3 electrons.
  2. 4 electrons.
  3. 2 electrons.
  4. 1 electron.
  1. Which of the following is an inner orbital or low spin complex?
  1. [Ni(H2O)6]3+
  2. [FeF6]3-
  3. [Co(CN)6]3-
  4. [NiCl4]2-
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Read the passage given below and answer the following questions:
A compound (X) containing C, H and O is unreactive towards sodium. It also does not react with Schiff s reagent. On refluxing with an excess of hydroiodic acid, (X) yields only one organic product ( Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorus and iodine. The compound (Z) on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The compound (X) is an:
  1. Acid.
  2. Aldehyde.
  3. Alcohol.
  4. Ether.
  1. The IUPAC name of the acid formed is:
  1. Methanoic acid.
  2. Ethanoic acid.
  3. Propanoic acid.
  4. Butanoic acid.
  1. Compound (Y) is:
  1. Ethyl iodide.
  2. Methyl iodide.
  3. Propyl iodide.
  4. Mixture of (a) and (b).
  1. Compound (Z) is:
  1. Methanol.
  2. Ethanol.
  3. Propanol.
  4. Butanol.
  1. Compound (X) on treatment with excess of Cl2 in presence of tight gives:
  1. $\propto-$ Chlorodiethyl ether.
  2. $\propto,\propto'-$ Dichlorodiethyl ether.
  3. Perchlorodiethyl ether.
  4. None of these.
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Explain the D and L notation method of spatial arrangement with respect to glucose.
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Read the passage given below and answer the following questions:
Metal carbonyl is an example of coordination compounds in which carbon monoxide (CO) acts as ligand. These are also called homoleptic carbonyls. These compounds contain both $\sigma$ and $\pi$ character. Some carbonyls have metal-metal bonds. The reactivity of metal carbonyls is due to (i) the metal centre and (ii) the CO ligands. CO is capable of accepting an appreciable amount of electron density from the metal atom into their empty $\pi$ or $\pi-\text{orbital}.$ These types of ligands are called $\pi-\text{accepter}$ or $\pi-\text{acid}$ ligands. These interactions increases the $\Delta_0$ value.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. What is the oxidation state of metal in [Mn2(CO)10]?
  1. +1
  2. -1
  3. +2
  4. 0
  1. Among the following metal carbonyls, the C-O bond order is lowest in:
  1. [Mn(CO)6]+
  2. [Fe(CO)5]
  3. [Cr(CO)6]
  4. [V(CO)6]-
  1. Which of the following can be reduced easily?
  1. V(CO)6
  2. Mo(CO)6
  3. [Co(CO)4]-
  4. Fe(CO)5
  1. The oxidation state of cobalt in K[Co(CO)4] is:
  1. +1
  2. +3
  3. -1
  4. 0
  1. Structure of decacarbonyl manganese is:
  1. Trigonal bipyramidial
  2. Octahedral
  3. Tetrahedral
  4. Square pyramidal
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Explain the structure of nucleic acid compounds.
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Read the passage given below and answer the following questions:
Valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory considers the metal-ligand bond to be ionic arising from electrostatic interaction between the metal ion and the ligands. In coordination compounds, the interaction between the ligand and the metal ion causes the five d-orbitals to split-up. This is called crystal field splitting and the energy difference between the two sets of energy level is called crystal field splitting energy. The crystal field splitting energy $(\Delta_0)$ depends upon the nature of the ligand. The actual configuration of complexes is divided by the relative values of $\Delta_0$ and P (pairing energy).
If $\Delta_0<\text{P},$ then complex will be high spin.
If $\Delta_0>\text{P},$ then complex will be low spin
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following ligand has lowest $\Delta_0$ value?
  1. CN-
  2. CO
  3. F-
  4. NH3
  1. The crystal field splitting energy for octahedral $(\Delta_0)$ and tetrahedral $(\Delta_t)$ complex is related as:
  1. $\Delta_\text{t}=\frac{1}{2}\Delta_0$
  2. $\Delta_\text{t}=\frac{4}{9}\Delta_0$
  3. $\Delta_\text{t}=\frac{3}{5}\Delta_0$
  4. $\Delta_\text{t}=\frac{2}{5}\Delta_0$
  1. On the basis of crystal field theory, the electronic configuration of d4 in two situations : (i) t.0 > P and (ii) t.0
  (i) (ii)
(a) $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
(b) $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
(c) $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
(d) $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
  1. Using crystal field theory, calculate magnetic moment of central metal ion of [FeF6]4-.
  1. 1.79B.M.
  2. 2.83B.M.
  3. 3.85B.M.
  4. 4.9B.M.
  1. Electronic configuration of d-orbitals in [Ti(H2O)6]3+ ion in an octahedral crystal field is:
  1. $\text{t}^1_{2\text{g}}\text{e}^0_\text{g}$
  2. $\text{t}^2_{2\text{g}}\text{e}^0_\text{g}$
  3. $\text{t}^0_{2\text{g}}\text{e}^1_\text{g}$
  4. $\text{t}^1_{2\text{g}}\text{e}^1_\text{g}$
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