Question
$SF_6$ is known but $SCl_6$ is not. Why?
✓
Answer
Due to small size, six fluorine atoms can be accommodated around sulphur atom while chlorine atoms being larger in size are difficult to accommodate.
The other reason is that the fluorine being highly electronegative and oxidising in nature is capable of unpairing the paired orbitals of the values shell of sulphur atom and thereby showing the highest, oxidation state of $+6$ while chlorine is not able to do this.
The other reason is that the fluorine being highly electronegative and oxidising in nature is capable of unpairing the paired orbitals of the values shell of sulphur atom and thereby showing the highest, oxidation state of $+6$ while chlorine is not able to do this.
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How would you bring about the following conversions?
i. Propanal to butanone
ii. Benzaldehyde to benzophenone
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i. Propanal to butanone
ii. Benzaldehyde to benzophenone
iii. Benzoyl chloride to benzonitrile
Explain the following items:
Biocatalysts.
Biocatalysts.
Match the example given in Column I with the name of the reaction in Column II.
| Column I (Example) | Column II (Reaction) | ||
| (i) | $\ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ || \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{Cl}+\text{H}_2\xrightarrow{\text{Pd-C/BaSO}_4}\text{CH}_3-\text{C}-\text{H}$ | (a) | Friedel Crafts acylation |
| (ii) |  |
(b) | HVZ reaction |
| (iii) |  |
(c) | Aldol condensation |
| (iv) | $\text{R}-\text{CH}_2-\text{COOH}\xrightarrow{\text{Br}_2/\text{Red}\ \text{P}}\text{R}-\text{CH}-\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{Br}$ | (d) | Cannizaro’s reaction |
| (v) | $\text{CH}_3-\text{CN}\xrightarrow[\text{(ii)}\ \text{H}_2\text{O}/\text{H}^+]{\text{(i)}\ \text{SnCl}_2/\ \text{HCl}}\text{CH}_3\text{CHO}$ | (e) | Rosenmund’s reduction |
| (vi) | $2\text{CH}_3\text{CHO}\xrightarrow{\text{NaOH}}\text{CH}_3-\text{CH}=\text{CH}=\text{CHCHO}$ | (f) | Stephen’s reaction |
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A certain reaction is $50 \%$ complete in $20$ minutes at $300$ K and the same reaction is again $50 \%$ complete in 5 minutes at 350 K . Calculate the activation energy if it is a first order reaction.
$\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \log 4=0.602\right]$
→$\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \log 4=0.602\right]$
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