Question 11 Mark
How would you convert the following compound into benzene?
Ethyne.
Ethyne.
Answer
View full question & answer→Benzene from Ethyne:
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Question 21 Mark
Write IUPAC names of the products obtained by the ozonolysis of the following compound:
3,4-Dimethylhept-3-ene.
3,4-Dimethylhept-3-ene.
Answer
View full question & answer→$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{H}_3\text{C}-\text{H}_2\text{C}-\text{H}_2\text{C}-\text{C}=\text{C}-\text{CH}_2-\text{CH}_3\xrightarrow{\text{ozonolysis}\ \ }\text{H}_3\text{C}-\text{CH}_2-\text{CH}_2-\text{C}=\text{O}+\text{CH}_3-\text{CH}_2-\text{C}=\text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Pentan-2-one}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Butan-2-one}$
Question 31 Mark
Write IUPAC names of the products obtained by the ozonolysis of the following compound:
Pent-2-ene.
Pent-2-ene.
Answer
View full question & answer→$\text{CH}_3-\text{CH}_2-\text{HC}=\text{CH}-\text{CH}_3\xrightarrow{\ \text{ozonolysis}\ \ }\text{CH}_3-\text{CH}_2-\text{CHO}+\text{CH}_3\text{CHO}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propanal}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethanal}$
Question 41 Mark
Write IUPAC names of the products obtained by the ozonolysis of the following compound:
1-Phenylbut-1-ene.
1-Phenylbut-1-ene.
Question 51 Mark
Write IUPAC names of the following compound:
Answer
IUPAC name: 2-Methyl phenol.
View full question & answer→IUPAC name: 2-Methyl phenol.Question 61 Mark
Write IUPAC names of the following compound:
Answer
can be written as:$\text{H}_2\stackrel{1}{\hbox{C}}=\stackrel{2}{\hbox{C}}\text{H}-\stackrel{3}{\hbox{C}}\text{H}=\stackrel{4}{\hbox{C}}\text{H}_2$
IUPAC name: 1, 3-Butadiene or Buta-1, 3-diene.
View full question & answer→can be written as:$\text{H}_2\stackrel{1}{\hbox{C}}=\stackrel{2}{\hbox{C}}\text{H}-\stackrel{3}{\hbox{C}}\text{H}=\stackrel{4}{\hbox{C}}\text{H}_2$IUPAC name: 1, 3-Butadiene or Buta-1, 3-diene.
Question 71 Mark
Write IUPAC names of the following compound:
Answer
IUPAC name: 4-Phenyl but-1-ene.
View full question & answer→IUPAC name: 4-Phenyl but-1-ene.Question 81 Mark
Write IUPAC names of the products obtained by the ozonolysis of the following compound:
2-Ethylbut-1-ene.
2-Ethylbut-1-ene.
Answer
View full question & answer→$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_2\text{H}_5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_2\text{H}_5\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{H}_3\text{C}-\text{H}_2\text{C}-\text{C}=\text{CH}_2\xrightarrow{\ \text{ozonolysis}\ }\text{CH}_2=\text{O}+\text{CH}_3-\text{CH}_2-\text{C}=\text{O}\\\\ \ \ \ \ \ \ \ \ \ \ \ \text{Methanal}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Pentan}-3-\text{one}$
Question 91 Mark
Write IUPAC names of the following compound:
$\text{CH}_3\text{CH}=\text{C}(\text{CH}_3)_2$
$\text{CH}_3\text{CH}=\text{C}(\text{CH}_3)_2$
Answer
View full question & answer→$\text{H}_3\stackrel{4}{\hbox{C}}-\stackrel{3}{\hbox{C}}\text{H}=\stackrel{{2}}{\hbox{C}}-\stackrel{1}{\hbox{C}}\text{H}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
IUPAC name: 2-Methylbut-2-ene.
IUPAC name: 2-Methylbut-2-ene.
Question 101 Mark
Write IUPAC names of the following compound:
$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}-\text{CH}_2-\text{CH}=\text{CH}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_2\text{H}_5$
$\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}-\text{CH}_2-\text{CH}=\text{CH}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_2\text{H}_5$
Answer
View full question & answer→$\stackrel{10}{\hbox{C}}\text{H}_3-\stackrel{9}{\hbox{C}}\text{H}=\stackrel{8}{\hbox{C}}\text{H}-\stackrel{7}{\hbox{C}}\text{H}_2-\stackrel{6}{\hbox{C}}\text{H}=\stackrel{5}{\hbox{C}}\text{H}-\stackrel{4}{\hbox{C}}\text{H}-\stackrel{3}{\hbox{C}}\text{H}_2-\stackrel{2}{\hbox{C}}\text{H}-\stackrel{1}{\hbox{C}}\text{H}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_2\text{H}_5$
IUPAC name: 4-Ethyldeca-1, 5, 8-triene.
IUPAC name: 4-Ethyldeca-1, 5, 8-triene.
Question 111 Mark
Write IUPAC names of the following compound:
$\text{CH}_2=\text{CH}-\text{C}\equiv\text{C}-\text{CH}_3$
$\text{CH}_2=\text{CH}-\text{C}\equiv\text{C}-\text{CH}_3$
Answer
View full question & answer→$\stackrel{1}{\hbox{C}}\text{H}_2=\stackrel{2}{\hbox{C}}\text{H}-\stackrel{3}{\hbox{C}}\equiv\stackrel{4}{\hbox{C}}-\stackrel{5}{\hbox{C}}\text{H}_3$
IUPAC name: Pen-1-ene-3-yne.
IUPAC name: Pen-1-ene-3-yne.
Question 121 Mark
Question 131 Mark
What is decarboxylation? Give an example.
Answer
View full question & answer→The process of removing carbon dioxide from sodium salt of acid with the help of soda lime is called decarboxylation, e.g.
$\text{CH}_3\text{COONa}+\text{NaOH(CaO)}\xrightarrow{\text{heat}}\text{CH}_4+\text{Na}_2\text{CO}_3\\ _\text{Sodium ethanaste}\ \ \ \ \ \ \ \ \ _\text{Soda lime}\ \ \ \ \ \ \ \ \ \ \ \ \ \ _\text{Methane}\ \ _\text{Sodium carbonate}$
$\text{CH}_3\text{COONa}+\text{NaOH(CaO)}\xrightarrow{\text{heat}}\text{CH}_4+\text{Na}_2\text{CO}_3\\ _\text{Sodium ethanaste}\ \ \ \ \ \ \ \ \ _\text{Soda lime}\ \ \ \ \ \ \ \ \ \ \ \ \ \ _\text{Methane}\ \ _\text{Sodium carbonate}$
Question 141 Mark
Explain why alkynes are less reactive than alkenes toward addition of $\text{Br}_2$.
Answer
View full question & answer→The three-membered ring bromonium ion formed from the alkyne (A) has a full double bond causing it to be more strained and less stable than the one from the alkene (B),
Also, the carbon's of (A) that are part of the bromonium ion have more s-character than (B), further making (A) less stable than (B).
Also, the carbon's of (A) that are part of the bromonium ion have more s-character than (B), further making (A) less stable than (B).
Question 151 Mark
Question 161 Mark
Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.
Answer
View full question & answer→Nitro group is an electron withdrawing group (-R and -I effects). It deactivates the ring by decreasing nucleophilicity for further substitution.
Question 171 Mark
Why iodination of benzene is carried out in the presence of nitric acid or iodic acid?
Answer
View full question & answer→The iodination of benzene is usually brought about by refluxing benzene with iodine and conc. $\mathrm{HNO}_3$ or $\mathrm{HIO}_3$.
$\mathrm{HNO}_3$ or $\mathrm{HIO}_3$ oxidises HI to $\mathrm{I}_2$ and plevents the backward reaction to occur.
$\mathrm{HNO}_3$ or $\mathrm{HIO}_3$ oxidises HI to $\mathrm{I}_2$ and plevents the backward reaction to occur.
Question 181 Mark
Explain the reason for extra ordinary stability of benzene in spite of presence of three double bonds in it.
Answer
View full question & answer→It is due to resonance, $6\pi-$electrons are delocalised.
Question 191 Mark
Give the structure of the alkene $\left(\mathrm{C}_4 \mathrm{H}_8\right)$ which adds on HBr in the presence and in absence of peroxide to give the same product, $\mathrm{C}_4 \mathrm{H}_9 \mathrm{Br}$.
Answer
But-2-ene is symmetric alkene, therefore, it gives 2-Bromobutane in presence as well as in absence of peroxide.
View full question & answer→But-2-ene is symmetric alkene, therefore, it gives 2-Bromobutane in presence as well as in absence of peroxide.
Question 201 Mark
Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?
Answer
Thus, but-2-ene is capable of showing geometrical isomerism.
View full question & answer→Thus, but-2-ene is capable of showing geometrical isomerism.
Question 211 Mark
Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.
Answer
View full question & answer→Alkenes and arenes both are unsaturated and electron rich. Olefins or alkenes undergo addition reaction to give more stable saturated product; in this reaction hybridization changes from $\text{sp}^2$ to $\text{sp}^3$. Arenes are stabilised by resonance w delocalization of $\pi$-electrons. On addition reaction to the double bond of arene, we get a product which is not resonance stabilised. Thus, arenes prefer to undergo substitution reaction while alkenes prefer to undergo addition reaction.
$\text{CH}_2=\text{CH}_2+\text{H}_2\xrightarrow{\text{Ni/573K}}\text{CH}_3-\text{CH}_3\$\text{less stable})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{more stable})$
$\text{CH}_2=\text{CH}_2+\text{H}_2\xrightarrow{\text{Ni/573K}}\text{CH}_3-\text{CH}_3\$\text{less stable})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{more stable})$
Question 221 Mark
Convert 1-bromopropane to 2-bromopropane.
Answer
View full question & answer→$\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\xrightarrow{\text{KOH(alc)}}\text{CH}_3-\text{CH}=\text{CH}_2\\ \ \ \ _{1-\text{bromopropane}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{Propene}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\xrightarrow[\text{Markownikoff addition}]{\text{HBr}}\text{CH}_3-\text{CH}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{2-\text{bromopropane}}$
Question 231 Mark
Why is dipole moment of trans-1, 2-dichloroethene zero?
Answer
It is because individual dipoles are equal and opposite such that net dipole moment is zero.
View full question & answer→It is because individual dipoles are equal and opposite such that net dipole moment is zero.
Question 241 Mark
Write an equation of the reaction of propyne with water in the presence of $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HgSO}_4$. Show the intermediate.
Answer
View full question & answer→$\text{CH}_3-\text{C}\equiv\text{CH}+\text{H}_2\text{O}\xrightarrow{\text{HgSO}_4/\text{H}_2\text{SO}_4}$$\begin{bmatrix}\\\\\text{CH}-\text{C}=\text{CH}_2\\ |\\\ \ \text{OH}\\ _\text{(Enol) unstable}\end{bmatrix}{\xrightarrow{\text{Tautomerism}}\text{CH}_3-\text{C}-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \|\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{Acetone}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{(\text{a ketone stable})}}$
The Markownikoff's addition of $\mathrm{H}_2\mathrm{O}$ gives an unstable enol form that rearranges to the more stable ketone.
The Markownikoff's addition of $\mathrm{H}_2\mathrm{O}$ gives an unstable enol form that rearranges to the more stable ketone.
Question 251 Mark
Convert methane to ethane.
Answer
View full question & answer→$\ \text{CH}_4+\text{Cl}_2\xrightarrow{\text{sunlight}}\text{CH}_3\text{Cl}+2\text{Na}+\text{ClCH}_3\\_\text{Methane}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _\text{Chloromethane}\ _\text{Sodium}\ \ _\text{Chloromethane}\\\xrightarrow[\text{Ether}]{\text{Dry}}\text{CH}_3-\text{CH}_3+2\text{NaCl}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _\text{Ethane}$
Question 261 Mark
Draw the Newman's projection formula of the staggered form of 1, 2-dichloroethane.
Question 271 Mark
How will you separate propene from propyne?
Answer
View full question & answer→On passing the mixture through ammoniacal $\mathrm{AgNO}_3$ solution or ammoniacal $\mathrm{CuCl}_2$ solution when propyne reacts while propene passes over.
Question 281 Mark
How will you distinguish the following pair of organic compounds by giving chemical tests? Propane and propene.
Answer
View full question & answer→Propane and propene can be distinguished by bromine water test and alkaline $\mathrm{KMnO}_4$ test.
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{Br}_2 / \mathrm{H}_2 \mathrm{O} \rightarrow \text { No change }$
$\text{CH}_3-\text{CH}=\text{CH}_2+\text{Br}_2/\text{H}_2\text{O}\overrightarrow{\ \ \ \ \ }\ \text{CH}_3-\text{CH}-\text{CH}_2\\ \ \ \ \ \ \ \ \ \ _\text{Propene} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _\text{Brown}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br} \ \ \ \ \ \ \ \text{Br}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{(\text{Colourless})}$
Pink colour of alkaline $\mathrm{KMnO}_4$ turns colourless in the presence of unsaturated compound.
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{Br}_2 / \mathrm{H}_2 \mathrm{O} \rightarrow \text { No change }$
$\text{CH}_3-\text{CH}=\text{CH}_2+\text{Br}_2/\text{H}_2\text{O}\overrightarrow{\ \ \ \ \ }\ \text{CH}_3-\text{CH}-\text{CH}_2\\ \ \ \ \ \ \ \ \ \ _\text{Propene} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _\text{Brown}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br} \ \ \ \ \ \ \ \text{Br}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{(\text{Colourless})}$
Pink colour of alkaline $\mathrm{KMnO}_4$ turns colourless in the presence of unsaturated compound.
Question 291 Mark
Which of the two trans-but-2-ene or trans-pent-2-ene is non-polar?
Answer
View full question & answer→In trans-but-2-ene, the dipole moments of the two $\mathrm{C}-\mathrm{CH}_3$ bonds are equal and opposite. Therefore, they cancel out each other. Hence, trans but-2-ene is non-polar.
Question 301 Mark
Explain why n-pentane has higher boiling point than neo-pentane.
Answer
View full question & answer→n-Pentane has more surface area than neo-pentane, therefore, has more van der Waals' forces of attraction hence has higher boiling point.
Question 311 Mark
Why do the $\mathrm{C}-\mathrm{C}$ bonds rather than $\mathrm{C}-\mathrm{H}$ bonds break during cracking of alkanes?
Answer
View full question & answer→Since the bond dissociation energy of $\mathrm{C}-\mathrm{C}$ bonds $\left(348 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ is lower than bond dissociation energy of $\mathrm{C}-\mathrm{H}$ bonds ( $414 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ), therefore, during cracking of alkanes, $\mathrm{C}-\mathrm{C}$ bonds break more easily than C-H bonds.
Question 321 Mark
What happens when 2-bromobutane is being treated with KOH (alcoholic)?
Answer
View full question & answer→$\text{CH}_3-\text{CH}-\text{CH}_2-\text{CH}_3+\text{KOH(alc.)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Br}\\ \ \ \ _{2-\text{bromobutane}}\\\overrightarrow{\ \ \ \ \ \ \ }\text{ CH}_3-\text{CH}=\text{CH}-\text{CH}_3+\text{KBr}+\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{But-2-ene}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{But-2-ene will be formed.}}$
Question 331 Mark
Which alkyne would you start with to Prepare $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CO}-\text{CH}_3?$
Answer
View full question & answer→$\text{HC}\equiv\text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_3+\text{H}_2\text{O}\xrightarrow[\text{HgSO}_4]{\text{H}_2\text{SO}_4}\\ \ \ \ \ \ \ \ \ \ _\text{Pent-1-yne}$$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{H}_2\text{C}=\text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\rightleftharpoons\text{CH}_3-\text{C}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{\text{Pentan-2-one}}$
Question 341 Mark
Explain why p-xylene has higher melting point than the corresponding ortho or meta-isomers.
Answer
View full question & answer→The para-isomer has a more symmetrical structure, allowing it to fit better into the crystal lattice than the other or meta-isomer.
Question 351 Mark
Identify the structure of A and B;
$\text{CH}_3-\text{CH}_2\text{COOH}\xrightarrow{\text{KOH}}\text{A}\xrightarrow{\text{Kolbe's electrolysis}}\text{B}$
$\text{CH}_3-\text{CH}_2\text{COOH}\xrightarrow{\text{KOH}}\text{CH}_3\text{CH}_2\text{COO}^-\text{K}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\\\xrightarrow{\text{Kolbe's electrolysis}}\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{B}$
$\text{CH}_3-\text{CH}_2\text{COOH}\xrightarrow{\text{KOH}}\text{A}\xrightarrow{\text{Kolbe's electrolysis}}\text{B}$
$\text{CH}_3-\text{CH}_2\text{COOH}\xrightarrow{\text{KOH}}\text{CH}_3\text{CH}_2\text{COO}^-\text{K}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\\\xrightarrow{\text{Kolbe's electrolysis}}\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{B}$
Answer
View full question & answer→$\text{A}\overrightarrow{\ \ \ \ }\text{CH}_3-\text{CH}_2\text{COO}^--\text{K}^+;$$\text{B}\overrightarrow{\ \ \ \ \ \ }\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Butane}$
Question 361 Mark
Despite their -I effect, halogens are o- and p-directing in haloarenes. Explain.
Answer
View full question & answer→In case of aryl halides, halogens are little deactivating because of their strong -I effect. Therefore, overall electron density on the benzene ring decreases. In other words, halogens are deactivating due to -I effect. However, because of the +R-effect, i.e., participation of lone pairs of electrons on the halogen atom with the π-electrons of the benzene ring, the electron density increases more at o- and p-positions than at m-positions.
As a result, halogens are o-, p-directing. The combined result of +R-effect and -I-effect of halogens is that halogens are deactivating but o, p-directing.
As a result, halogens are o-, p-directing. The combined result of +R-effect and -I-effect of halogens is that halogens are deactivating but o, p-directing.
Question 371 Mark
Arrange the three isomers of pentane in increasing order of their boiling points.
Answer
View full question & answer→2, 2-Dimethylpropane < 2-Methylbutane < Pentane because surface area increases, van der Waal's forces of attraction increases, hence boiling point decreases.
Question 381 Mark
Name the reagent to convert
Answer
View full question & answer→Li/ liq $\mathrm{NH}_3$ is used to prepare trans-2-butene to But-2-yne.
Question 391 Mark
Give reason whether 
is aromatic or not?
is aromatic or not?Answer
View full question & answer→It is aromatic because:
- It is planar.
- It has $6\pi-$electrons or $(4\text{n}+2)\pi$ electrons.
- $6\pi$ electrons are delocalised.
Question 401 Mark
Give major product only.
Question 411 Mark
$\text{n-hexane}\xrightarrow[\text{V}_2\text{O}_5]{770\text{K, 10-20atm}}?$
View full question & answer→Question 421 Mark
Is pyrolle
aromatic compound or not? Give reason.Answer
View full question & answer→It is aromatic because it has 6 electrons ($2\pi$ bonds + llone pair) which are delocalised and it is planar molecule.
Question 431 Mark
Name the rule which decide stability of alkene. State the rule.
Answer
View full question & answer→Saytzeff Rule: It states ‘More alkyl substituted alkenes are more stable’.
Question 441 Mark
Give IUPAC name of acetylene.
Answer
View full question & answer→Ethyne.
Question 451 Mark
Write IUPAC names of the following compound:
Answer
IUPAC name: 4-Phenyl but-1-ene.
View full question & answer→IUPAC name: 4-Phenyl but-1-ene.Question 461 Mark
Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile.Give reason.
Answer
View full question & answer→The methoxy group $\left(-\mathrm{OCH}_3\right)$ is electron releasing group. It increases the electron density in beniene nucleus due to resonance effect (+R-effect). Hence, it makes anisole more reactive than benzene towards the electrophile.+
In case of alkyl halides, the electron density increases at ortho and para positions due to +R effect. However, the halogen atom also withdraws electrons from the ring because of its -1 effect. Since the -1 effect is stronger than the +R effect, the halogens are moderately deactivating. Thus, overall electron density on benzene ring decreases, which makes further substitution difficult. $-\mathrm{NO}_2$ group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong-R-effect and strong-I-effect. Hence, it makes nitrobenzene less reactive. Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order:
In case of alkyl halides, the electron density increases at ortho and para positions due to +R effect. However, the halogen atom also withdraws electrons from the ring because of its -1 effect. Since the -1 effect is stronger than the +R effect, the halogens are moderately deactivating. Thus, overall electron density on benzene ring decreases, which makes further substitution difficult. $-\mathrm{NO}_2$ group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong-R-effect and strong-I-effect. Hence, it makes nitrobenzene less reactive. Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order:
Question 471 Mark
Arrange the following alkenes in the decreasing order of stability.
- $\ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ |\\\text{CH}_3\text{C}=\text{CHCH}_3$
- $ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3\text{CHCH}=\text{CH}_2$
- $ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_2=\text{CCH}_2\text{CH}_3$
Answer
View full question & answer→Stability of an alkene increases with increase in the number of electron donating substituent at the double bonded carbon atom. Thus, $\text{(i) > (iii) > (ii)}$.
Question 481 Mark
What is Lindlar's catalyst? Give its use.
Answer
View full question & answer→Lindlar's catalyst is $\mathrm{Pd} / \mathrm{BaSO}_4$. It is used to convert alkynes to alkenes with the help of H , e.g., But-2yne, on reduction with $\mathrm{Pd}-\mathrm{BaSO}_4$ forms cis-2-butene.
Question 491 Mark
Who discovered benzene?
Answer
View full question & answer→Michael Faraday discovered benzene in 1825.
MCQ 501 Mark
In which of the following pair both angle strain and torsional strain is present?
- ✓Cyclopropane, Cyclobutane
- BCyclopropane, Cyclohexane
- CCyclobutane, Cyclohexane
- DCyclohexane, Cyclopentane
Answer
View full question & answer→Correct option: A.
Cyclopropane, Cyclobutane
$3-$ Membered and $4-$ membered cycloalkane rings are the least stable containing angle strain $($due to large deviation from tetrahedral angle$)$ as well as torsional strain $($due to electronic repulsions between the bonds$),$ lying close to each other.