Question 11 Mark
Identify the functional groups in the following compounds:
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Identify the functional groups in the following compounds:
Question 21 Mark
Give the IUPAC names of the following compounds: 
Question 31 Mark
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2$.
Answer
The structure of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2$ is:
The resonating structures of nitro benzene are represented as:
View full question & answer→The structure of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2$ is:
The resonating structures of nitro benzene are represented as:
Question 41 Mark
Identify the functional groups in the following compounds:
Question 51 Mark
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{CH}_3\text{H}=\text{CHCH}_2$
Answer
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{CH}_3\text{H}=\text{CHCH}_2$
The resonating structures of the given compound are:
View full question & answer→$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{CH}_3\text{H}=\text{CHCH}_2$
The resonating structures of the given compound are:
Question 61 Mark
Give the IUPAC names of the following compounds:
Question 71 Mark
Give the IUPAC names of the following compounds:
$\mathrm{Cl}_2 \mathrm{CHCH}_2 \mathrm{OH} \text {. }$
$\mathrm{Cl}_2 \mathrm{CHCH}_2 \mathrm{OH} \text {. }$
Answer
View full question & answer→2,2-Dichloroethanol.
Question 81 Mark
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \\ \text{CH}_3\text{COOH}_3+\text{CN}\rightarrow (\text{CH}_3)_2\text{C(CN)(OH)}$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \\ \text{CH}_3\text{COOH}_3+\text{CN}\rightarrow (\text{CH}_3)_2\text{C(CN)(OH)}$
Answer
View full question & answer→Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \\ \text{CH}_3\text{COOH}_3+\text{CN}\rightarrow (\text{CH}_3)_2\text{C(CN)(OH)}$
Here, ${}^-\text{CN}$ acts as a nucleophiles as it is an electron-rich species, i.e., it is a nucleus-seeking species.
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \\ \text{CH}_3\text{COOH}_3+\text{CN}\rightarrow (\text{CH}_3)_2\text{C(CN)(OH)}$
Here, ${}^-\text{CN}$ acts as a nucleophiles as it is an electron-rich species, i.e., it is a nucleus-seeking species.
Question 91 Mark
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{C}_6\text{H}_6+ \text{CH}_3\text{ CO}\rightarrow \text{C}_6\text{H}_5\text{COCH}_3$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{C}_6\text{H}_6+ \text{CH}_3\text{ CO}\rightarrow \text{C}_6\text{H}_5\text{COCH}_3$
Answer
View full question & answer→Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons. $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{C}_6\text{H}_6+ \text{CH}_3\text{ CO}\rightarrow \text{C}_6\text{H}_5\text{COCH}_3$Here, $\ \ \ \ \ \ \ + \\ \text{CH}_3\text{CO}$ acts as an electrophile as it is an electron-deficient species.
MCQ 101 Mark
In the organic compound $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH}$, the pair of hydridised orbitals involved in the formation of: $\mathrm{C}_2$ $\mathrm{C}_3$ bond is:
- A$\mathrm{sp}-\mathrm{sp}_2$.
- B$\mathrm{sp}-\mathrm{sp}_3$.
- ✓$\mathrm{sp}_2-\mathrm{sp}_3$
- D$\mathrm{sp}_3-\mathrm{sp}_3$.
Answer
View full question & answer→Correct option: C.
$\mathrm{sp}_2-\mathrm{sp}_3$
The carbon atom in the given compound may be numbered as:
$ \ \ \ \ \ 1 \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ 6 \\ \text{H}_2\text{C}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}\equiv\text{CH}$
Thus, $\mathrm{C}^2-\mathrm{C}^3$ are bonded by $\mathrm{sp}^2-\mathrm{sp}^3$ hybrid orbitals.
$ \ \ \ \ \ 1 \ \ \ \ \ \ 2 \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ 5 \ \ \ \ \ \ \ \ \ 6 \\ \text{H}_2\text{C}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}\equiv\text{CH}$
Thus, $\mathrm{C}^2-\mathrm{C}^3$ are bonded by $\mathrm{sp}^2-\mathrm{sp}^3$ hybrid orbitals.
Question 111 Mark
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{HO}^{-} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O}$
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{HO}^{-} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O}$
Answer
View full question & answer→Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{HO}^{-} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O}$
Here, $\mathrm{HO}^{-}$acts as a nucleophiles as it is an electron-rich species, i.e., it is a nucleus-seeking species.
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{HO}^{-} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O}$
Here, $\mathrm{HO}^{-}$acts as a nucleophiles as it is an electron-rich species, i.e., it is a nucleus-seeking species.
Question 121 Mark
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. $\ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{C}_6\text{H}_5-\text{CH}_2$
Answer
$\ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{C}_6\text{H}_5-\text{CH}_2$
The resonating structures of the given compound are:
View full question & answer→$\ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{C}_6\text{H}_5-\text{CH}_2$
The resonating structures of the given compound are:
MCQ 131 Mark
The best and latest technique for isolation, purification and separation of organic compounds is :
- ACrystallisation.
- BDistillation.
- CSublimation.
- ✓Chromatography.
Answer
View full question & answer→Correct option: D.
Chromatography.
Chromatography is the most useful and the latest technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances.
Question 141 Mark
Give the IUPAC names of the following compounds:
Question 151 Mark
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}\mathrm{CHO}$.
Answer
View full question & answer→$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$
The resonating structures of the given compound are represented as:
The resonating structures of the given compound are represented as:
Question 161 Mark
Give the IUPAC names of the following compounds:
Question 171 Mark
Give the IUPAC names of the following compounds:
Question 181 Mark
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CHO}$.
Answer
The structure of $\mathrm{C}_6 \mathrm{H}_5\mathrm{CHO}$ is:
The resonating structures of benzaldehyde are represented as:
View full question & answer→The structure of $\mathrm{C}_6 \mathrm{H}_5\mathrm{CHO}$ is:
The resonating structures of benzaldehyde are represented as:
MCQ 191 Mark
Which of the following carbocation is most stable?
- A$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ (\text{CH}_3)_3\text{C}.\text{CH}_2$
- ✓$\ \ \ \ \ \ \ \ \ \ \ + \\ (\text{CH}_3)_3\text{C}$
- C$\ \ \ \ \ \ \ \ \ \ \ \ \ \ + \\ \text{CH}_3\text{CH}_2\text{CH}_2$
- D$\ \ \ \ \ \ \ + \\ \text{CH}_3 \text{CH CH}_2\text{CH}3$
Answer
View full question & answer→Correct option: B.
$\ \ \ \ \ \ \ \ \ \ \ + \\ (\text{CH}_3)_3\text{C}$
$\ \ \ \ \ \ \ \ \ \ \ + \\ (\text{CH}_3)_3\text{C}$ is a tertiary carbocation. A tertiary carbocation is the most stable carbocation due to the electron releasing effect of three methyl groups. An increased $+\ I$ effect by three methyl groups stabilizes the positive charge on the carbocation.
Question 201 Mark
Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. $\mathrm{C}_6 \mathrm{H}_5\mathrm{OH}$.
Answer
The structure of $\mathrm{C}_6 \mathrm{H}_5\mathrm{OH}$ is:
The resonating structures of phenol are represented as:
View full question & answer→The structure of $\mathrm{C}_6 \mathrm{H}_5\mathrm{OH}$ is:
The resonating structures of phenol are represented as:
MCQ 211 Mark
The reaction : $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{KI}$ is classified as :
- AElectrophilic substitution.
- ✓Nucleophilic substitution.
- CElimination.
- DAddition.
Answer
View full question & answer→Correct option: B.
Nucleophilic substitution.
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{Kl}$
In the given reaction, the $\mathrm{I}^{-}$from the alkyl iodide is replaced by the $\mathrm{OH}^{-}$ ion. Thus, it is substitution reaction.
The substitution is brought about by the $\mathrm{OH}^{-}$ ion which is a nucleophile.
$\because$ The reaction is a nucleophilic substitution reaction.
In the given reaction, the $\mathrm{I}^{-}$from the alkyl iodide is replaced by the $\mathrm{OH}^{-}$ ion. Thus, it is substitution reaction.
The substitution is brought about by the $\mathrm{OH}^{-}$ ion which is a nucleophile.
$\because$ The reaction is a nucleophilic substitution reaction.
MCQ 221 Mark
In the Lassaigne's test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of :
- A$\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$.
- ✓$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$.
- C$\mathrm{Fe}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]$.
- D$\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_4$.
Answer
View full question & answer→Correct option: B.
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$.
In the Lassaigne's test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron $\text{(II)}$ sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron $\text{(II)}$ sulphate and forms sodium hexacyanoferrate $\text{(II)}.$ Then, on heating with sulphuric acid, some iron $\text{(II)}$ gets oxidised to form iron $\text{(III)}$ hexacyanoferrate $\text{(II)},$ which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as
$6 \mathrm{CN}^{-}+\mathrm{Fe}^{2+} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
$3\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}+4 \mathrm{Fe}^{3+} \xrightarrow{3 \mathrm{H}_2 \mathrm{O}} \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH}_2 \mathrm{O}$
Hence, the Prussian blue colour is due to the formation of $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$.
$6 \mathrm{CN}^{-}+\mathrm{Fe}^{2+} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
$3\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}+4 \mathrm{Fe}^{3+} \xrightarrow{3 \mathrm{H}_2 \mathrm{O}} \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot \mathrm{xH}_2 \mathrm{O}$
Hence, the Prussian blue colour is due to the formation of $\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$.
Question 231 Mark
Arrange the following in increasing order of $\mathrm{C}-\mathrm{C}$ bond length: $\mathrm{C}_2 \mathrm{H}_6, \mathrm{C}_2 \mathrm{H}_4, \mathrm{C}_2 \mathrm{H}_2$.
Answer
View full question & answer→$\mathrm{C}_2 \mathrm{H}_2(120 \mathrm{pm})<\mathrm{C}_2 \mathrm{H}_4(134 \mathrm{pm})<\mathrm{C}_2 \mathrm{H}_6(154 \mathrm{pm})$.
Question 241 Mark
How will you separate a mixture of iodine and sodium chloride?
Answer
View full question & answer→Sublimation.
Question 251 Mark
What is the suitable adsorbent in the process of column chromatography?
Answer
View full question & answer→$\mathrm{Al}_2, \mathrm{O}_3$, (alumina) is most suitable adsorbent for column chromatography.
Question 261 Mark
Which method is used to extract a organic compound in aqueous solution?
Answer
View full question & answer→Differential extraction is used to extract organic compound by adding organic solvent in which organic compound is more soluble. Organic layer is separated and on evaporation we get organic compound.
Question 271 Mark
Give an example of metamerism.
Answer
View full question & answer→Diethyl ether $\left(\mathrm{C}_2 \mathrm{H}_3 \mathrm{OC}_3 \mathrm{H}_7\right)$ and methyl propyl ether $\left(\mathrm{CH}_3 \mathrm{OC}_3 \mathrm{H}_7\right)$ are related to each other by metamerism.
Question 281 Mark
In which of the following compounds will Kjeldahl's method be suitable for nitrogen estimation? Give reason. 
Answer
View full question & answer→It is suitable only for Benzamide (IV) because it can form ammonia on reaction with $\mathrm{H}_2 \mathrm{SO}_4$.
Question 291 Mark
Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:
$\mathrm{R}-\mathrm{CONH}_2$
$\mathrm{R}-\mathrm{CONH}_2$
Answer
View full question & answer→$\mathrm{R}-\mathrm{CONH}_2$: it shows resonance due to the presence of lone pair on N atom and $\pi$ electrons on C =0 bond,Hence amide can be represented by three resonating structures.
Question 301 Mark
What is the basic principle of chromatography?
Answer
View full question & answer→Chromatography is based on the principle of differential adsorption due to difference in solubility. The substance which is less adsorbed will move faster and will get separated.
Question 311 Mark
What are primary and secondary suffixes as applied to IUPAC nomenclature?
Answer
View full question & answer→The primary suffix indicates whether the carbon chain is saturated or unsaturated while the secondary suffix indicates the functional group present in the molecule.
Question 321 Mark
Using curved arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.
- $\text{CH}_3-\text{SCH}_3$
- $\text{CH}_3-\text{CN}$
- $\text{CH}_3-\text{Cu}$
Answer
View full question & answer→- $\text{CH}_3-\text{SCH}_3\leftrightarrow\text{C}^-\text{H}_3+\stackrel{+}{\text{SCH}_3}$
- $\text{CH}_3-\text{CN}\leftrightarrow\stackrel{+ \ \ \ \ \ }{\text{CH}_3}+\text{CN}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{Carbocation}}$
- $\text{CH}_3-\text{Cu}\leftrightarrow\text{C}^-\text{H}_3+\text{Cu}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{Carbonion}}$
Question 331 Mark
Name the process of separating benzoic acid and naphthalene.
Answer
View full question & answer→Fractional crystallisation using benzene as a solvent because both are soluble in benzene but to different extent. Naphthalene is more soluble, whereas benzoic acid is less soluble which will crystallise out first and after separating benzoic acid on further crystallization naphthalene will be obtained.
Question 341 Mark
What is the state of hybridisation of carbon atoms in $\text{CH}_3-\text{CH}=\text{CH}_2?$
Answer
View full question & answer→$\text{CH}_3-\text{CH}=\text{CH}_2\\\ \ ^{\text{sp}^3}\ \ \ \ \ \ \ \ ^{\text{sp}^2}\ \ \ \ \ \ \ ^{\text{sp}^2}$
Question 351 Mark
Write the $\text{IUPAC}$ name of Following?
- $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ _6\ \ \ \ \ \ \ \ \ \ _5|\ \ \ \ \ \ _4\ \ \ \ \ \ \ \ \ _3\ \ \ \ \ \ \ \ \ _2\ \ \ \ \ \ \ \ \ \ \ _1\\\text{CH}_3-\text{C}-\text{CH}=\text{CH}-\text{CH}_2-\text{COOH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Answer
View full question & answer→- $1-$Methyl ethyl cyclohexane.
- $5, 5-$dimethyl hex$-3-$enoic acid.
Question 361 Mark
If we get blood red colouration on adding $\mathrm{FeSO}_4$ and dil. $\mathrm{H}_2 \mathrm{SO}_4$ to the sodium extract, what do you infer about the elements present in organic compound?
Answer
View full question & answer→The organic compound contains nitrogen and sulphur.
$\because\text{Na}+\text{S}+\text{C}+\text{N}\xrightarrow{\ \ \ \ \ \ \ }\text{NaSCN}$
$\text{Fe}^{3+}\text{NaSCN}\xrightarrow{\ \ \ \ \ }\text{Fe}(\text{SCN})_3+3\text{Na}^+\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{Ferric thiocyanate}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{(\text{Blood red colouration)}}$
$\because\text{Na}+\text{S}+\text{C}+\text{N}\xrightarrow{\ \ \ \ \ \ \ }\text{NaSCN}$
$\text{Fe}^{3+}\text{NaSCN}\xrightarrow{\ \ \ \ \ }\text{Fe}(\text{SCN})_3+3\text{Na}^+\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{Ferric thiocyanate}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{(\text{Blood red colouration)}}$
Question 371 Mark
How many $\sigma\text{ and }\pi$ bonds are present in
Answer
View full question & answer→There are $9\sigma\text{ and }9\pi$ bonds.
Question 381 Mark
$\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}$ is more stable than $(\text{CH}_3)_2\stackrel{{+}}{\hbox{C}}\text{H}.$ Explain why.
Answer
View full question & answer→It is due to more +1 effect (positive inductive effect) of three alkyl groups which are electron releasing groups, make it more stable than $2^{\circ}$ carbocation which has two electron releasing alkyl groups. Secondly, $\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{+}$ is most stable due to hyperconjugation.
Question 391 Mark
How many $\sigma$ and $\pi$ bonds gre there in,
Answer
View full question & answer→$\ \ \ \ \ \ \ \ \text{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ _1|\ \ \ \ \ \ _2\ \ \ \ \ \ _3\ \ \ \ \ _4||\ \ \ \ \ _5\\\text{H}-\text{C}-\text{C}\equiv\text{C}-\text{C}-\text{C}\equiv\text{N}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}$
There are $9\sigma$ bonds. $5\pi$ bonds [All single bonds are $\sigma$ bonds, double bonds have $1\sigma$ and $1\pi,$ Triple bonds have $1\sigma$ and $2\pi$ bonds].
There are $9\sigma$ bonds. $5\pi$ bonds [All single bonds are $\sigma$ bonds, double bonds have $1\sigma$ and $1\pi,$ Triple bonds have $1\sigma$ and $2\pi$ bonds].
Question 401 Mark
Complete the following:$(\text{CH}_3\text{COO})_2\text{Pb}+\text{Na}_2\text{S}\xrightarrow{\ \ \ \ \ \ }$
Answer
View full question & answer→$(\text{CH}_3\text{COO})_2\text{Pb}+\text{Na}_2\text{S}\xrightarrow{\ \ \ \ \ \ }\text{PbS}+2\text{CH}_3\text{COONa}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{(\text{black ppt.)}}$
Question 411 Mark
Write IUPAC name of the following organic compound:
$\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}=\text{CH}_2$
$\text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}=\text{CH}_2$
Answer
View full question & answer→$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ _4\ \ \ \ \ \ \ \ \ \ _3|\ \ \ \ \ \ _2\ \ \ \ \ \ \ \ \ _1\\\text{CH}_3-\text{C}-\text{CH}=\text{CH}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ _{3,3\text{Dimethylbut-1-ene}}$
Question 421 Mark
Arrange the following in decreasing order of -I effect:
$-\text{Br},-\text{Cl},-\text{F}$
$-\text{Br},-\text{Cl},-\text{F}$
Answer
View full question & answer→$\text{F}>\text{Cl}>\text{Br}.$
Question 431 Mark
Name the stationary and mobile phases in paper chromatography.
Answer
View full question & answer→Liquid on paper is stationary phase and the solution of substance to be separated is mobile phase in paper chromatography.
Question 441 Mark
Draw all position isomers of an alcohol with molecular formula, $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$.
Answer
View full question & answer→$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\\ \ \ \ \ \ \ _{\text{Propan-1-ol}}$ and $\text{CH}_3\text{CH(OH)CH}_3\\ \ \ \ \ \ \ \ \ _{\text{Propan-2-ol}}$
Question 451 Mark
Identify the tertiary (3°) and quarternary (4°) carbon in,
Answer
Tertiary 'C' is carbon attached to three more carbon atoms. 4° carbon is attached to four carbon atoms.
View full question & answer→Tertiary 'C' is carbon attached to three more carbon atoms. 4° carbon is attached to four carbon atoms.
Question 461 Mark
Write the correct stability order for the following carbocations.
$(\text{CH}_3)_3\stackrel{+ }{\text{CCH}}_2,(\text{CH}_3)_3\stackrel{+}{\text{C}},\text{CH}_3\text{CH}_2\stackrel{+ \ \ \ \ }{\text{CH}}_2$
$(\text{CH}_3)_3\stackrel{+ }{\text{CCH}}_2,(\text{CH}_3)_3\stackrel{+}{\text{C}},\text{CH}_3\text{CH}_2\stackrel{+ \ \ \ \ }{\text{CH}}_2$
Answer
View full question & answer→The order of stability of carbocation is 3° > 2° >1°.
$(\text{CH}_3)_3\stackrel{+}{\text{C}}>(\text{CH}_3)_3\stackrel{+ }{\text{CCH}}_2>\text{CH}_3\text{CH}_2\stackrel{+ \ \ \ \ }{\text{CH}}_2$
3° carbocation, i.e.$(\text{CH}_3)_3\stackrel{+}{\text{C}}$ is the most stable carbocation.
$(\text{CH}_3)_3\stackrel{+}{\text{C}}>(\text{CH}_3)_3\stackrel{+ }{\text{CCH}}_2>\text{CH}_3\text{CH}_2\stackrel{+ \ \ \ \ }{\text{CH}}_2$
3° carbocation, i.e.$(\text{CH}_3)_3\stackrel{+}{\text{C}}$ is the most stable carbocation.
Question 471 Mark
Lassaigne's test is not shown by diazonium salts, though they contain nitrogen. Why?
Answer
View full question & answer→Diazonium salts $(\text{C}_{6}\text{H}_{5}\text{N}^{+}_{2}\text{X}^{-})$ readily lose $\mathrm{N}_2$ on heating before reacting with fused sodium metal. Therefore, these do not give positive Lassaigne's test for nitrogen.
Question 481 Mark
Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:
$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{NH}_2$
$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{NH}_2$
Answer
View full question & answer→$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{NH}_2$: As the lone pair of electrons on N atom, is not conjugated with the $\pi$ electrons of double bond.
Hence, it does not show resonance.
Hence, it does not show resonance.
Question 491 Mark
How will you separate a mixture of o-nitro phenol and p-nitrophenol?
Answer
View full question & answer→A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation, o-nitrophenol being less volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.
Question 501 Mark
How are liquids with high boiling point or liquids which decompose at or below their normal boiling points, purified?
Answer
View full question & answer→They are purified by distillation under reduced pressure so that their boiling point is reduced and they do not decompose but get purified at lower boiling point.