Question:medium

The half-life period of a radioactive element is $\mathrm{1.5\times10^{10}}$ years. Calculate the time in which the activity of the element is reduced to 75% of its original value. [Given : log 2 = 0·30, log 3 = 0·48, log 4 = 0·60]

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First order: t = (2.303/k) log(100/75), with k = 0.693/t½ and log(4/3)=0.12.
Updated On: Jun 16, 2026
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Solution and Explanation

(a)
To turn aniline into sulphanilic acid, treat it with hot concentrated sulphuric acid. The acid first protonates the amine and the system rearranges, finally planting a $\mathrm{-SO_3H}$ group at the para position. The product is the inner salt sulphanilic acid:
\[ C_6H_5NH_2 + H_2SO_4 \rightarrow p\text{-}H_2N\text{-}C_6H_4\text{-}SO_3H \]

(b)
The $\mathrm{-NH_2}$ group is a powerful ring activator that strongly directs incoming groups to the ortho and para spots. So when aniline is shaken with bromine water, no catalyst is needed and bromine goes into all three of those positions at once, giving 2,4,6-tribromoaniline:
\[ C_6H_5NH_2 + 3Br_2 \rightarrow 2,4,6\text{-tribromoaniline} + 3HBr \]

(c)
To make acetanilide, cap the amino group with an acetyl group using acetic anhydride. The lone pair on nitrogen attacks the anhydride, replacing one N to H bond with an N to $\mathrm{COCH_3}$ link and releasing acetic acid:
\[ C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH \]
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