Tag: chemistry

Questions Related to chemistry

In order to distinguish between $\displaystyle C _{2}H _{5}NH _{2}$ and $\displaystyle C _{6}H _{5}NH _{2}$, which of the following reagents is useful? 

  1. Hinsberg reagent

  2. $\beta$-Naphthol

  3. Benzene diazonium chloride

  4. None of the above


Correct Option: B

Pyridine will give azo dye test.

  1. True

  2. False


Correct Option: A
Explanation:

Pyridine is an aromatic amine and aromatic amine gives azo dye test which implies pyridine would also give azo-dye test.

p-chloroaniline and anilinium hydrochloride can be distinguished by :

  1. $Sandmeyer $ $reaction$

  2. $\displaystyle NaHCO _{3}$

  3. $\displaystyle AgNO _{3}$

  4. $carbylamines$ $ test$


Correct Option: B,C
Explanation:

p-chloroaniline and anilinium hydrochloride can be distinguished by by using $AgNO _3$ and $NaHCO _3$, With $AgNO _3 $ anilinium hydrochloride gives white precipitate of $AgCl$. 


$NaHCO _3$ is a weak acid  which solublize anilinium hydrochloride and $CO _2$ gas liberates but  p-chloroaniline do not react with them.

Mixture of $1^0, 2^0 $ and $3^0$ amines can be separated by :

  1. Hinsberg's method

  2. Hofmann's isocyanide test

  3. Fractional distillation

  4. $NaNO _2/ HCl$


Correct Option: A,B,C,D
Explanation:
All the given methods can be used for the separation of amines.

(AHinsberg reaction is a chemical test for the detection of amines. A primary amine will form a soluble sulfonamide salt which precipitates after addition of diluted hydrochloric acid. A secondary amine in the same reaction will directly form an insoluble sulfonamide. A tertiary amine will not react with the sulfonamide but is insoluble. After adding dilute acid this insoluble amine is converted to a soluble ammonium salt. 

(BIn Hofmann's isocyanide test the mixture of three amines is treated with diethyl oxalate. The primary amine forms a solid oxamide, a secondary amine gives a liquid oxamic ester while tertiary amine does not react.

(CIn fractional distillation primary amine is recovered when solid oxamide is heated with caustic potash solution and collected as distillate on distilling the reaction mixture. The liquid is subjected to fractional distillation when tertiary amine distils over. The remaining liquid is distilled with KOH to recover secondary amine.

(DNitrosation of Amines can also be used for separation as primary amine with $NaNO _2 + HCl$ forms $RN _2$, secondary amine forms $R _2N _2O$ and tertiary amine forms $R _3NO$ which decomposes into $NR _3$.

Propylamine and aniline can be distinguished by azo dye test.

  1. True

  2. False


Correct Option: A
Explanation:

As only aryl systems undergo azo dye test. Therefore, propylamine gives negative azo dye test.

Aniline gives positive azo dye test.

When chloroform reacts with ethylamine in presence of alc. $KOH$, the compound formed is:

  1. ethyl cyanide

  2. ethyl isocyanide

  3. formic acid

  4. phosgene


Correct Option: B
Explanation:

Reaction:
$C{ H } _{ 3 }N{ H } _{ 2 }+CHCl _3+KOH\rightarrow C{ H } _{ 3 }NC$

Aniline and benzylamine  cannot be distinguished by azo dye test

  1. True

  2. False


Correct Option: A
Explanation:

Aniline and benzylamine both are aryl systems as only highly activated benzene system gives positive azo dye test below room temperature. Therefore, aniline and benzylamine cannot be distinguished by azo dye test as both give positive azo dye test.

A compound $(X)$ with molecular formula $C _3H _9N$ react with $C _6H _5SO _2Cl$ to give a solid which is insoluble in alkali. $(X)$ is:

  1. $CH _3CH _2CH _2NH _2$

  2. $CH _3 - {\underset{CH _3}{\underset{|}{\overset{CH _3}{\overset{|}{N:}}}}}$

  3. $CH _3 - NH - CH _2CH _3$

  4. $CH _3 - {\underset{CH _3}{\underset{|}{CH}}} - NH _2$


Correct Option: C
Explanation:

Since the compound reacts with benzenesulphonyl chloride to give a product which is insoluble in alkali, it shows there is no $H$ attached to $N$ in the product . Hence, the compound $X$ is a secondary amine.
$CH _3 - NH - C-2H _5 + C _6H _5SO _2Cl \rightarrow \underset{N - Ethyl - N - methylbenzene sulphonamide}{CH _3 - {\underset{C _2H _5}{\underset{|}{N}}} - SO _2C _6H _5}$

During preparation of arene diazonium salts, the excess of nitrous acid, if any, is destroyed by adding?

  1. $aq. \, NaOH$

  2. $aq. \, Na _2CO _3$

  3. $aq. \, NH _2CONH _2$

  4. $aq. \, KI$


Correct Option: C
Explanation:

Urea decomposes $HNO _2$ to give $N _2, \, CO _2 $ and $ H _2O$
$NH _2CONH _2+\, 2HONO \rightarrow 2N _2 \uparrow + \, CO _2 \uparrow +\, 3H _2O$

A colourless liquid $A(b.p:184^\circ:C)$ is sparingly soluble in warm water to which it gives feebly alkaline reaction on treating with $NaNO _{2}:and:dil:HCl$ in the cold solution, it yields a solution which reacts with an alkaline solution of $\beta-$napthol to give an orange yellow precipitate. Compound $A$ is :

  1. $C _{6}H _{5}N _{2}Cl$

  2. $C _{6}H _{5}NHNH _{2}$

  3. $n-C _{4}H _{9}NH _{2}$

  4. $C _{6}H _{5}NH _{2}$


Correct Option: D
Explanation:

Aniline is the colourless liquid A having boiling point $184 ^oC$.
On treatment with $NaNO _{2}:and:dil:HCl$ at low temperature, it forms benzne diazonum salt.
The diazonium salt reacts with napthol in alkaline medium to form azo dye which is n orange yellow precipitate.