Question
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
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Answer
Benzene is a planar molecule having delocalized electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles.
Therefore, it undergoes electrophilic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophilic substitutions with difficulty.
Therefore, it undergoes electrophilic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophilic substitutions with difficulty.
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Note: Consider structures I to VII and answer the questions:
→- $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}$
- $\text{CH}_3-\text{CH}_2-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
- $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
- $\text{CH}_3-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
- $\text{CH}_3-\text{CH}_2-\text{O}-\text{CH}_2-\text{CH}_3$
- $\text{CH}_3-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3$
- $\text{CH}_3-\text{O}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Answer the following questions:
- B, Al, Ga arrange in decreasing order of atomic radii.
- C, F, O first electron gain enthalpy in increasing order.
- Al forms amphoteric oxide, why?
On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$
What will be the effect of addition of argon to the above reaction mixture at constant volume?
→$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$
What will be the effect of addition of argon to the above reaction mixture at constant volume?
Write the anode reaction, the cathode reaction and the net cell reaction in the following cells. Which electrode would be positive terminal in each cell?
→- $\text{Zn}\text{(s)}|\text{Zn}^{2+}||\text{Br}_2|\text{Br}^-|\text{Pt}\text{(s)}$
- $\text{Cr}\text{(s)}|\text{Cr}^{3+}||\text{I}_2|\text{I}^-|\text{Pt}\text{(s)}$
- $\text{Pt}\text{(s)}|\text{H}_2\text{(g)}||\text{H}^+\text{(aq)}||\text{Cu}^{2+}|\text{Cu}\text{(s)}$
A white solid 'P', on strong heating, decomposes to give reddish brown gas 'R' and a residue 'Q' and a colourless gas. The residue 'R' is yellow when hot and white when cold. The solution of white solid 'P' in water gives white precipitate 'S' with conc. NaOH. The white form colourless solution. Identify 'P', 'Q', 'R', 'S' and write the chemical reactions involved.
- State Hess's Law of constant heat summation. How does it follow from the first law of thermodynamics.
- Determine $\Delta_\text{r}\text{H}^\circ,$ at 298K for reaction:
You are given:
- $\text{C(graphite)}+\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ \text{CO}_2(\text{g});$ $\Delta_\text{r}\text{H}^\circ=-393.51\text{kJ/ mol}$
- $\text{H}_2\text{(g)}+\frac{1}{2}\text{O}_2(\text{g})\overrightarrow{ \ \ \ \ \ \ \ }\ \text{H}_2\text{O(l)};$ $\Delta_\text{r}\text{H}^\circ=-285.8\text{kJ/ mol}$
- $\text{CO}_2(\text{g})+2\text{H}_2\text{O(l)}\overrightarrow{\ \ \ \ \ \ \ \ \ }\ \text{CH}_4(\text{g})+2\text{O}_2(\text{g});$ $\Delta_\text{r}\text{H}=+890.3\text{kJ/ mol}$
Match the intermediates given in Column I with their probable structure in Column II.
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Column I
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Column II
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(i)
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Free radical
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(a)
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Trigonal planar
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(ii)
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Carbocation
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(b)
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Pyramidal
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(iii)
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Carbanion
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(c)
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Linear
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Consider the reactions: $\mathrm{O}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+2 \mathrm{O}_2(\mathrm{~g})$ Why it is more appropriate to write these reactions as:
$\mathrm{O}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
Also suggest a technique to investigate the path of the above redox reactions.
→$\mathrm{O}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
Also suggest a technique to investigate the path of the above redox reactions.
Which of the following Lewis structure of $\text{CO}_2$ molecule is least significant resonating form?
→Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?