Question
Read the passage given below and answer the following questions:
$RCONH_2$ is converted into $RNH_2$ by means of Hoffmann bromamide degradation. During the reaction amide is treated with $Br_2$ and alkali to get amine. This reaction is used to descend the series in which carbon atom is removed as carbonate ion $(\text{CO}^{2-}_3)$ Hoffmann bromide degradation reaction can be written as:

The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Hoffmann bromamide degradation is used for the preparation of
  1. Primary amines.
  2. Secondary amines.
  3. Tertiary amines.
  4. Secondary aromatic amines.
  1. Which is the rate determining step in Hoffmann bromamide degradation?
  1. Formation of (i)
  2. Formation of (ii)
  3. Formation of (iii)
  4. Formation of (iv).
  1. Which of the following are used for the conversion of (i) to (ii)?
  1. $KBr$
  2. $KBr + CH_3ONa$
  3. $KBr + KOH$
  4. $Br_2 + KOH$
  1. Identify Bin the following reaction.
$\text{R}-\text{C}\equiv\text{N}\xrightarrow[\text{(Partially hydrolysis) }]{\text{Cone. HCI}}\text{A}\xrightarrow{\frac{\text{Br}_2}{\text{KOH}}}\text{B}$
  1. $RCONH_2$
  2. $RNH_2$
  3. $RNHBr$
  4. $R = N = C = O$
  1. What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hoffmann bromamide degradation?

Answer

  1. (a) Primary amines.
  1. (d) Formation of (iv).
Explanation:
The rate determining step is probably loss of $Br^-$ to form is ocyanate as this is the slowest step.
  1. (d) $Br_2 + KOH$
Explanation:
  1. (b) $RNH_2$
Explanation:
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{R}-\text{C}\equiv\text{N}\xrightarrow[\text{(partially hydrolysis)}]{\text{Cone. HCI}}\text{R}-\text{C}-\text{NH}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \xrightarrow{\frac{\text{br}_2}{\text{KHO}}}\text{R}-\text{NH}_2$
  1. (b) In ethylamine there is no resonance while in acetamide the lone pair of electrons on N-atom is delocalised and is less available for protonation.
Explanation:
Since, the overall reaction is intermolecular, hence there will be no effect on product fonnation.

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Read the passage given below and answer the following questions:
Ethers are readily cleaved by HI or HBr at 373K to form an alcohol and an alkyl halide.
$\text{R}-\text{O}-\text{R}+\text{HX}\xrightarrow{373\text{K}}\text{R}-\text{X}+\text{R}-\text{OH}$
$\text{R}-\text{OH}+\text{HX}\xrightarrow{373\text{K}}\text{R}-\text{X}+\text{H}_2\text{O}$
Mixed ether, containing primary or secondary alkyl group, when heated with hydrogen halide, the lower alkyl group forms halide and higher will form an alcohol. Tertiary alkyl ether when heated with hydrogen halide gives tertiary alkyl halide.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
  1. $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{O}- \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 $
  2. $\text{CH}_3-\text{CH}-\text{CH}_2-\text{O}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
  3. $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{O}-\text{CH}_3$
  4. $\text{CH}_3-\text{CH}_2-\text{CH}-\text{O}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
  1. When $CH_2 = CH - O - CH_2 - CH_3$ reacts with one mole of HI, one of the products formed is:
  1. Ethane.
  2. Ethanol.
  3. Iodoethene.
  4. Ethanal.
  1. $(CH_3)_3COCH_3$ and $CH_3OC_2H_5$​​​​​​​ are treated with hydroiodic acid. The fragments obtained after reactions are respectively:
  1. $(CH_3)_3CI + CH_3OH; CH_3I + C_2H_5OH$
  2. $(CH_3)_3CI + CH_3OH; CH_3OH + C_2H_5I$
  3. $(CH_3)_3COH + CH_3I; CH_3OH + C_2H_5I$
  4. $CH_3I + (CH_3)_3COH; CH_3I + C_2H_5OH$
  1. Which of the following ether is unlikely to be cleaved by hot cone. HBr?
Read the passage given below and answer the following questions:
The f-block elements are those in which the differentiating electron enters the $(n -2)f$ orbital. There are two series of F-block elements corresponding to filling of $4f $ and $5f-$orbitals. The series of $4f-$orbitals is called lanthanides. Lanthanides show different oxidation states depending upon stability of $f^0, f^7$ and $F^{14}$ configurations, though the most conunon oxidation states is $+3$. There is a regular decrease in size oflanthanides ions with increase in atomic number which is known as lanthanide contraction.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The atomic numbers of three lanthanide elements $X, Y$ and $Z$ are $65, 68$ and $70$ respectively, their $Ln^{3+}$ electronic configuration is:
  1. $4f^8, 4f^{11}, 4f^{13}$
  2. $4f^{11}, 4f^8, 4f^{13}$
  3. $4f^0, 4f^2, 4f^{11}$
  4. $4f^3, 4f^7, 4f^9$
  1. Lanthanide contraction is observed in:
  1. $Gd$
  2. $At$
  3. $Xe$
  4. $Te$
  1. Which of the following is not the configuration oflanthanoid?
  1. $[Xe]4f^{10}6s^2$
  2. $[Xe]4f^15d^16s^2$
  3. $[Xe]4d^{14}5d^{10}6s^2$
  4. $[Xe]4f^75d^16s^2$
  1. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
  1. Cerium $(Z = 58)$
  2. Europium $(Z = 63)$
  3. Lanthanum $(Z = 57)$
  4. Gadolinium $(Z = 64)$
  1. Identify the incorrect statement among the following.
  1. Lanthanoid contraction is the accumulation of successive shrinkages.
  2. The different radii of $Zr$ and $Hf$ due to consequence of the lanthanoid contraction.
  3. Shielding power of $4f$ electrons is quite weak.
  4. There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.
Read the passage given below and answer the following questions: The order of reactivity towards $S_N1$ 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, $3^\circ$ carbocation is most stable, therefore, the tert-alkyl that halides will undergo $S_N1$ reaction very fast. For example, it has been observed that the reaction $(CH_3)_3CBr$ 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 $S_N2$ mechanism. On the other hand, the tertiary alkyl halides react predominantly by $S_N1$ 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 $S_N1$ mechanism.
Reason: $2^\circ$ alkyl halides are less reactive than $1^\circ$ towards $S_N1$ reactions.
  1. Assertion: Polar solvent slows down $S_N2$ reactions.
Reason: $CH_3-Br$ is less reactive than $CH_3Cl.$
  1. Assertion: Benzyl bromide when kept in acetone- water it produces benzyl alcohol.
Reason: The reaction follows $S_N2$ 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: $S_N1$ 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.
For the reaction : $2\text{NO}_\text{(g)}+\text{Cl}_{2\text{(g)}}\rightarrow2\text{NOCl}_\text{(g)},$ the following data were collected. All the measurements were taken at $263K.$
Experiment No.
Initial [NO] (M)
Initial $[Cl_2]$ (M)
Initial rate of disapp. of $Cl_2 $ (M/ min)
$1.$
$0.15$
$0.15$
$0.60$
$2.$
$0.15$
$0.30$
$1.20$
$3.$
$0.30$
$0.15$
$2.40$
$4.$
$0.25$
$0.25$
$?$
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The molecularity of the reaction is:
  1. $1$
  2. $2$
  3. $3$
  4. $4$
  1. The expression for rate law is:
  1. $\text{r}=\text{k}[\text{NO}][\text{Cl}_2]$
  2. $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]$
  3. $\text{r}=\text{k}[\text{NO}][\text{Cl}_2]^2$
  4. $\text{r}=\text{k}[\text{NO}]^2[\text{Cl}_2]^2$
  1. The overall order of the reaction is:
  1. $2$
  2. $0$
  3. $1$
  4. $3$
  1. The value of rate constant is:
  1. $150.32\ M^{-2} \min^{-1}$
  2. $200.08\ M^{-1} \min^{-1}$
  3. $177.77\ M^{-2} \min^{-1}$
  4. $155.75\ M^{-1} \min^{-1}$
  1. The initial rate of disappearance of $Cl_2$ in experiment $4$ is:
  1. $1.75M\ \min^{-1}$
  2. $3.23M\ \min^{-1}$
  3. $2.25M\ \min^{-1}$
  4. $2.77M\ \min^{-1}$
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 $\ce{(NaOX)}$ or halogen and alkali $(X_2 + OH^-)$ 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 $\ce{(NaOl)}$ when treated with compounds containing $\ce{CH_3CO}-$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. $\ce{CH_3COCH_2l}$
  2. $\ce{ICH_2COCH_2l}$
  3. $\ce{CH_3COCHl_2}$
  4. $\ce{CH_3COCl_3}$
  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 $^{13}C$ labelled carbon)
  1. An organic compound $'A\ '$ has the molecular formula $\ce{C_3H_6O}$. It undergoes iodoform test. When saturated with $\ce{HCl}$ it gives $'B\ '$ of molecular formula $\ce{C_9H_{14}O}. 'A\ '$ 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.
Read the passage given below and answer the following questions:
Due to intermolecular hydrogen bonding, the boiling points of alcohols and phenols are much higher than those of corresponding haloalkanes, haloarenes, aliphatic and aromatic hydrocarbons. Among isomeric alcohols, the boiling points follow the order: primary > secondary > tertiary. Boiling points of ethers are much lower than those of isomeric alcohols. The solubility of alcohols in water decreases as the molecular mass of alcohols increases. Amongst isomeric alcohols solubility increases with branching. The solubility of phenols in water is much lower than that of alcohols. Lower ethers such as dimethyl ether and ethyl methyl ether are soluble in water, but the solubility decreases as the molecular mass increases.
In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the 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: Alcohols have higher boiling points than ethers of comparable molecular masses.
Reason: Alcohols and ethers are isomeric in nature.
  1. Assertion: The solubility of phenols in water is much lower than that of alcohols.
Reason: Phenols do not form H-bonds with water.
  1. Assertion: Among n-butane, ethoxyethane, 1-propanol and 2-propanol, the increasing order of boiling points is, 1-butanol < 1-propanol < ethoxyethane < n-butane.
Reason: Boiling point increases with increase in molecular mass.
  1. Assertion: Dimethyl ether and diethylether are soluble in water.
Reason: As the molecular mass increases, solubility of ethers in water decreases.
  1. Assertion: Butan-2-ol has higher boiling point than 2-methylpropan-2-ol.
Reason: Amongst isomeric alcohols, the boiling points decreases with branching.
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 $pK_a$ 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 $CO_2$, 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:
Read the passage given below and answer the following questions:
A mixture of two aromatic compounds $(A)$ and $(B)$ was separated by dissolving in chloroform followed by extraction with aqueous $KOH$ solution. The organic layer containing compound $(A)$, when heated with alcoholic solution of KOH produce $C_7H_5N (C)$ associated with unpleasant odour.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. What is A?
  1. $C_6H_5NH_2$
  2. $C_6H_5CH_3$
  3. $C_6H_5CH_3$
  4. None of these.
  1. The reaction of $(A)$ with alcoholic solution of $KOH$ to produce $(C)$ of unpleasant odour is called:
  1. Sandmeyer reaction.
  2. Carbylamine reaction.
  3. Ullmann reaction.
  4. Reimer-Tiemann reaction..
  1. The alkaline aqueous layer $(B)$ when heated with chloroform and then acidified give a mixture of isomeric compounds of molecular formula $C_7H_6O_2. (B)$ is:
  1. $C_6H_5CHO$
  2. $C_6H_5COOH$
  3. $C_6H_5CH_3$
  4. $C_6H_5OH$
  1. In the chemical reaction, $CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow (A)+ (B) + 3H_2O,$
  1. $C_2H_5NC$ and $KCl$
  2. $C_2H_5CN$ and $KCl$
  3. $CH_3CH_2CONH_2$ and $KCl$
  4. $C_2H_5NC$ and $K_2CO_3$
  1. Direct nitration of an aromatic compound (A) is not feasible because:
  1. The reaction cannot be stopped at the mononitration stage.
  2. A mixture of o, m and p-nitroaniline is always obtained.
  3. Nitric acid oxidises most of the aromatic compound to give oxidation products along with only a small amount of nitrated products.
  4. All of the above.
Read the passage given below and answer the following questions:
The addition reaction of enol or enolate to the carbonyl functional group of aldehyde or ketone is known as aldol addition. The $\beta-$ hydroxyaldehyde or $\beta-$hydroxyketone so obtained undergo dehydration in second step to produce a conjugated enone. The first part of reaction is an addition reaction and the second part is an elimination reaction. Carbonyl compound having $\propto-$hydrogen undergoes aldol condensation reaction.

The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Condensation reaction is the reverse of which of the following reaction?
  1. Lock and key hypothesis.
  2. Oxidation.
  3. Hydrolysis.
  4. Glycogen formation.
  1. Which of the following compounds would be the main product of an aldol condensation of acetaldehyde and acetone?
  1. $CH_3CH = CHCHO$
  2. $CH_3CH = CHCOCH_3$
  3. $(CH_3)_2C = CHCHO$
  4. $(CH_3)_2C = CHCOCH_3$
  1. Which combination of carbonyl compounds gives phenyl vinyl ketone by an aldol condensation?
  1. Acetophenone and Formaldehyde.
  2. Acetophenone and acetaldehyde.
  3. Benzaldehyde and acetaldehyde.
  4. Benzaldehyde and acetone.
  1. Which of the following will undergo aldol condensation?
  1. $HCHO$
  2. $CH_3CH_2OH$
  3. $C_6H_5CHO$
  4. $CH_3CH_2CHO$
  1. Which of the following does not undergo aldol condensation?
  1. $CH_3CHO$
  2. $CH_3CH_2CHO$
  3. $CH_3COCH_3$
  4. $C_3H_2CHO$
In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Lyophilic sols are prepared by directly mixing the substance with the dispersion medium.
Reason: Lyophobic sols can not be prepared by simply mixing the substance with the dispersion medium.