Question
Explain Crystal field splitting in tetrahedral complax.
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An organic compound $‘A’$ with molecular formula $C_8H_8O$ gives positive $\ce{DNP}$ and iodoform tests. It does not reduce Tollens’ or Fehling’s reagent and does not decolourise bromine water also. On oxidation with chromic acid $(H_2CrO_4),$ it gives a carboxylic acid $(B)$ with molecular formula $C_7H_6O_2$. Deduce the structures of $A$ and $B.$
→Explain the following:
Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the major product.
Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the major product.
Account for the following:
→- Electrophilic substitution in case of aromatic amines takes place more readily than benzene.
- $\ce{CH_3CONH_2}$ is a weaker base than $\ce{CH_3CH_2NH_{2}}.$
- Nitro compounds have higher boiling points than hydrocarbons having almost same molecular mass.
Arrange the following compounds in the increasing order of their property indicated:
$i.$ Acetaldehyde, Benzaldehyde, Acetophenone, Acetone $($Reactivity towards $HCN)$
$ii. \ce{(CH_3)_2CHCOOH, CH_3CH_2CH(Br)COOH, CH_3CH(Br)CH_2COOH} ($Acidic strength$)$
$iii. \ce{CH_3CH_2OH, CH_3CHO, CH_3COOH} ($Boiling point$)$
→$i.$ Acetaldehyde, Benzaldehyde, Acetophenone, Acetone $($Reactivity towards $HCN)$
$ii. \ce{(CH_3)_2CHCOOH, CH_3CH_2CH(Br)COOH, CH_3CH(Br)CH_2COOH} ($Acidic strength$)$
$iii. \ce{CH_3CH_2OH, CH_3CHO, CH_3COOH} ($Boiling point$)$
Calculate the emf of the following cell:
$ Mg ( s )\left| Mg ^{2+}(0.2 M ) \| Ag ^{+}\left(1 \times 10^{-3} M \right)\right| Ag ( s )$
$E ^0\left( Ag ^{+} / Ag \right)=0.80 V$
$E ^0\left( Mg ^{2+} / Mg \right)=-2.37 V$
→$ Mg ( s )\left| Mg ^{2+}(0.2 M ) \| Ag ^{+}\left(1 \times 10^{-3} M \right)\right| Ag ( s )$
$E ^0\left( Ag ^{+} / Ag \right)=0.80 V$
$E ^0\left( Mg ^{2+} / Mg \right)=-2.37 V$
Identify the compounds A, B and C in the following reaction:
$\text{CH}_3-\text{Be}\ \xrightarrow{\ \ \text{Mg/ Ether}}\ \text{(A)}\ \xrightarrow[\text{(ii)}\text{Water}]{\ \ \text{CO}_2}\ \text{(B)}\ \xrightarrow[\triangle]{\ \ \text{CH}_3\text{OH}/\text{H}^+}\ \text{(C)}$
→$\text{CH}_3-\text{Be}\ \xrightarrow{\ \ \text{Mg/ Ether}}\ \text{(A)}\ \xrightarrow[\text{(ii)}\text{Water}]{\ \ \text{CO}_2}\ \text{(B)}\ \xrightarrow[\triangle]{\ \ \text{CH}_3\text{OH}/\text{H}^+}\ \text{(C)}$
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH , alkenes are major products. Explain.
Given that $\Lambda^{0}_{\text{m}}(\text{HCl})=426\text{ S cm}^2\text{mol}^{-1},\Lambda^{0}_{\text{m}}(\text{NaCl})\\=126\text{ S cm}^2\text{mol}^{-1}$
$\Lambda^{0}_{\text{m}}(\text{CH}_3\text{COONa})=91\text{ S cm}^2\text{mol}^{-1}$
→$\Lambda^{0}_{\text{m}}(\text{CH}_3\text{COONa})=91\text{ S cm}^2\text{mol}^{-1}$
What are fuel cells? Explain the electrode reactions involved in the working of $H_2 - O_2$ fuel cell.
→Give reasons for the following:
→- Acetylation of aniline reduces its activation effect.
- $\ce{CH_3NH_2}$ is more basic than $\ce{C_6H_5NH_2}.$
- Although $-\ce{NH_2}$ is $o/p$ directing group, yet aniline on nitration gives a significant amount of $m-$nitroaniline.