Question
Convert:
- Ethylene to Nitrobenzene.
- Write short note on Markovnikov's Rule.
✓
Answer
- Markovnikov's Rule: When a polar compound is a unsymmetrical alkene or alkyne, negative part goes to least substituted carbon e.g., $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}=\text{CH}_2+\text{HCl}\rightarrow\text{CH}_3-\text{CH}-\text{CH}_3$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The $K_{s p}$ values of two slightly soluble salts $A B$ and $\mathrm{PQ}_2$ are each equal to $4.0 \times 10^{-18}$. Which salt is more soluble?
→-
Account for the following:
-
Boron halides do not dimerise like $BH_3$.
-
Carbon shows catenation remarkably.
-
$PbCl_4$ is a good oxidising agent.
-
Complete the following reactions:
- $\text{Na}_{2}\text{B}_{4}\text{O}_{7} + 2\text{HCl} + \text{5H}_{2}\text{O} \xrightarrow{\ \ \ \ \ \ \ \ }$
- $\text{B}_{2}\text{H}_{6} + 6\text{NH}_{3} \xrightarrow[\ \ \ \ \ \ ]{\Delta}$
Explain the following terms:
- Photochemical smog.
- Acid rain.
- Eutrophication.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/ have the same energy lists:
→- $\text{n}=4,\text{l}=2,\text{m}_\text{l}=-2,\text{m}_{\text{s}}=-\frac{1}{2}$
- $\text{n}=3,\text{l}=2,\text{m}_\text{l}=1,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=4,\text{l}=1,\text{m}_\text{l}=0,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=3,\text{l}=2,\text{m}_\text{l}=-2,\text{m}_{\text{s}}=-\frac{1}{2}$
- $\text{n}=3,\text{l}=1,\text{m}_\text{l}=-1,\text{m}_{\text{s}}=+\frac{1}{2}$
- $\text{n}=4,\text{l}=1,\text{m}_\text{l}=0,\text{m}_{\text{s}}=+\frac{1}{2}$
Following data is given for the reaction: $\text{CaCO}_3\text{(s)}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{CaO(s)}+\text{CO}_2\text{(g)}$
$\Delta_\text{f}\text{H}^\ominus[\text{CaO}\text{(s)}]=-635.1\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CO}_2\text{(g)}]=-393.5\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CaCO}_3\text{(s)}]=-1206.9\text{KJ mol}^{-1}$
Predict the effect of temperature on the equilibrium constant of the above reaction.
→$\Delta_\text{f}\text{H}^\ominus[\text{CaO}\text{(s)}]=-635.1\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CO}_2\text{(g)}]=-393.5\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CaCO}_3\text{(s)}]=-1206.9\text{KJ mol}^{-1}$
Predict the effect of temperature on the equilibrium constant of the above reaction.
Use the information and data given below to answer the questions (a) to (c):
- Stronger intermolecular forces result in higher boiling point.
- Strength of London forces increases with the number of electrons in the molecule.
- Boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively.
- Which type of intermolecular forces are present in the molecules HF, HCl, HBr and HI?
- Looking at the trend of boiling points of HCl, HBr and HI, explain out of dipole interaction and London interaction, which one is predominant here.
- Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?
Write important chemical reaction of ethene.
If the speed of light is $3.0 \times 10^8 \mathrm{~ms}^{-1}$, calculate the distance covered by light in 2.00 ns .
→Calculate the number of electrons which will together weigh one gram.
The magnitude of surface tension of liquid depends on the attractive forces between the molecules. Arrange the following in increasing order of surface tension:
water, alcohol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)$ and hexane $\left.\left[\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4 \mathrm{CH}_3\right)\right]$.
→water, alcohol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)$ and hexane $\left.\left[\mathrm{CH}_3\left(\mathrm{CH}_2\right)_4 \mathrm{CH}_3\right)\right]$.