Aniline is diazotized and the diazonium salt hydrolysed to yield phenol which is brominated to produce $C_{6} H_{2} (B r_{3}) O H .$ Calculate the mass of the final product obtained from 9.3 g of aniline if the yield in the two steps is 45% and 70% (Atomic mass of Br = 80) ?

1.04 g

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10.43 g

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14.89 g

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23.17 g

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Solution

The correct option is B
10.43 g

$C_{6} H_{5} N H_{2} 45 % y i e l d - ---- \to C_{6} H_{5} O H 70 % y i e l d - ---- \to C_{6} H_{2} (B r_{3}) O H$
Mol of $C_{6} H_{5} O H$ formed $= \frac{9.3}{9.3} \times \frac{45}{100} = 0.045$
Mol of $C_{6} H_{2} (B r_{3}) O H = 0.045 \times \frac{70}{100} = 0.0315$
Mass of $C_{6} H_{2} (B r_{3}) O H$ formed $= 0.0315 \times 331 = 10.43 g$

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Aniline is diazotized and the diazonium salt hydrolysed to yield phenol which is brominated to produce $C_{6} H_{2} (B r_{3}) O H .$ Calculate the mass of the final product obtained from 9.3 g of aniline if the yield in the two steps is 45% and 70% (Atomic mass of Br = 80) ?

Q. Aniline reacts with ___ to yield ___ as the final product.

Q. When aniline is treated with benzene diazonium chloride in acidic medium, the final product obtained is :

Regarding benzene diazonium fluoroborate, correct statements are

A.Prepared from aniline by diazotization followed by treating the salt with HBF₄

B.Fluorobenzene is obtained by direct fluorination of benzene with F₂ gas

C.Unstable at room temperature and soluble in water

D.Stable at room temperature and insoluble in water

Q. Iodobenzene is prepared from aniline

(C_{6} H_{5} N H_{2})

in a two step process as shown here:

C_{6} H_{5} N H_{2} + H N O_{2} + H C l \to C_{6} H_{5} N_{2}^{+} C l^{-} + 2 H_{2} O

C_{6} H_{5} N_{2}^{+} C l^{-} + K l \to C_{6} H_{5} I + N_{2} + K C l

In an actual preparation, 9.30 g of aniline was converted to 12.32 g of iodobenzene. The percentage yield of iodobenzene is:

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Aniline is diazotized and the diazonium salt hydrolysed to yield phenol which is brominated to produce C6H2(Br3)OH. Calculate the mass of the final product obtained from 9.3 g of aniline if the yield in the two steps is 45% and 70% (Atomic mass of Br = 80) ?

The correct option is B 10.43 g C6H5NH245% yield−−−−−→C6H5OH70%yield−−−−−→C6H2(Br3)OH Mol of C6H5OH formed =9.39.3×45100=0.045 Mol of C6H2(Br3)OH=0.045×70100=0.0315 Mass of C6H2(Br3)OH formed =0.0315×331=10.43g

Aniline is diazotized and the diazonium salt hydrolysed to yield phenol which is brominated to produce $C_{6} H_{2} (B r_{3}) O H .$ Calculate the mass of the final product obtained from 9.3 g of aniline if the yield in the two steps is 45% and 70% (Atomic mass of Br = 80) ?