You have been given four bottles marked $A, B, C$ and $D$ each containing one of the organic compounds given below:
$I: C_6H_5CH_2NH_2$ $II: C_6H_5CH_2COOH$ $III: C_6H_5CH_2CH_3$ $IV: C_6H_5CH_2CH(NH_2)COOH$
The following observations were made:
$(i)$ The compound in bottle $A$ did not dissolve in either $1 \ N \ NaOH$ or $1 \ N \ HCl$.
$(ii)$ The compound in bottle $B$ dissolved in $1 \ N \ NaOH$ but not in $1 \ N \ HCl$.
$(iii)$ The compound in bottle $C$ dissolved in both $1 \ N \ NaOH$ and $1 \ N \ HCl$.
$(iv)$ The compound in bottle $D$ did not dissolve in $1 \ N \ NaOH$ but dissolved in $1 \ N \ HCl$.
$(a)$ Indicate the compounds in: bottle $A = \dots$,bottle $B = \dots$,bottle $C = \dots$ and bottle $D = \dots$.
$(b)$ The compound with the highest solubility in distilled water is .......
| $I: C_6H_5CH_2NH_2$ | $II: C_6H_5CH_2COOH$ |
| $III: C_6H_5CH_2CH_3$ | $IV: C_6H_5CH_2CH(NH_2)COOH$ |
The following observations were made:
$(i)$ The compound in bottle $A$ did not dissolve in either $1 \ N \ NaOH$ or $1 \ N \ HCl$.
$(ii)$ The compound in bottle $B$ dissolved in $1 \ N \ NaOH$ but not in $1 \ N \ HCl$.
$(iii)$ The compound in bottle $C$ dissolved in both $1 \ N \ NaOH$ and $1 \ N \ HCl$.
$(iv)$ The compound in bottle $D$ did not dissolve in $1 \ N \ NaOH$ but dissolved in $1 \ N \ HCl$.
$(a)$ Indicate the compounds in: bottle $A = \dots$,bottle $B = \dots$,bottle $C = \dots$ and bottle $D = \dots$.
$(b)$ The compound with the highest solubility in distilled water is .......
(D) The solubility of organic compounds in $1 \ N \ NaOH$ (a base) and $1 \ N \ HCl$ (an acid) depends on their acidic or basic nature.
$(i)$ Compound $A$ is neutral as it does not react with either acid or base. Thus,$A = III$ $(C_6H_5CH_2CH_3)$.
$(ii)$ Compound $B$ is acidic as it dissolves in $NaOH$ but not $HCl$. Thus,$B = II$ $(C_6H_5CH_2COOH)$.
$(iii)$ Compound $C$ is amphoteric as it dissolves in both $NaOH$ and $HCl$. Thus,$C = IV$ $(C_6H_5CH_2CH(NH_2)COOH)$.
$(iv)$ Compound $D$ is basic as it dissolves in $HCl$ but not $NaOH$. Thus,$D = I$ $(C_6H_5CH_2NH_2)$.
$(b)$ The compound with the highest solubility in distilled water is $IV$ $(C_6H_5CH_2CH(NH_2)COOH)$ because it exists as a zwitterion $(C_6H_5CH_2CH(NH_3^+)COO^-)$ in water,which is highly polar.
$(i)$ Compound $A$ is neutral as it does not react with either acid or base. Thus,$A = III$ $(C_6H_5CH_2CH_3)$.
$(ii)$ Compound $B$ is acidic as it dissolves in $NaOH$ but not $HCl$. Thus,$B = II$ $(C_6H_5CH_2COOH)$.
$(iii)$ Compound $C$ is amphoteric as it dissolves in both $NaOH$ and $HCl$. Thus,$C = IV$ $(C_6H_5CH_2CH(NH_2)COOH)$.
$(iv)$ Compound $D$ is basic as it dissolves in $HCl$ but not $NaOH$. Thus,$D = I$ $(C_6H_5CH_2NH_2)$.
$(b)$ The compound with the highest solubility in distilled water is $IV$ $(C_6H_5CH_2CH(NH_2)COOH)$ because it exists as a zwitterion $(C_6H_5CH_2CH(NH_3^+)COO^-)$ in water,which is highly polar.
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Given below are two statements:
Statement $I$: The dipole moment of $R-CN$ is greater than $R-NC$ and $R-NC$ can undergo hydrolysis under acidic medium to produce $R-NH_2$ and $HCOOH$.
Statement $II$: $R-CN$ hydrolyses under acidic medium to produce a compound which on treatment with $SOCl_2$ followed by the addition of $NH_3$ gives another compound $(x)$. This compound $(x)$ on treatment with $NaOCl/NaOH$ gives a product, that on treatment with $CHCl_3/KOH/\Delta$ produces $R-NC$.
In the light of the above statements, choose the correct answer from the options given below:
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In the light of the above statements, choose the correct answer from the options given below:
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