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Multiple choice chemistry hydrocarbons introduction to alkenes - ethyne acetylene alkynes

Which of the following reaction will give acetylene as a major product?

  1. $2[CH _{3}] \underset {\triangle}{\xrightarrow {Ag}}$
  2. $H _{3}C - CH _{2} - Br\underset {KOH}{\xrightarrow {alc.}}$
  3. $\underset {\underset {Br}{|}}{CH _{2}} - \underset {\underset {Br}{|}}{CH _{2}} \underset {\triangle}{\xrightarrow {Zn}}$

  4. All of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Only $2[CH _3] \xrightarrow[\Delta]{Ag} HC \equiv CH$ give acetylene

$B)$ $H _3C - CH _2 - Br \xrightarrow[KOH]{ak} CH _2 = CH _2$
$C)$ $\underset{Br}{\underset{|}{CH _2}} - \underset{Br}{\underset{|}{CH _2}} \rightarrow H _2C = CH _2$

Multiple choice chemistry hydrocarbons introduction to alkenes - ethyne acetylene alkynes

How many grams of hydrogen is required to saturate one mole of acetylene?

  1. 3 g

  2. 6 g

  3. 10 g

  4. 4 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The hydrogenation of acetylene can be written as:
$C _{2}H _{2} + 2H _{2} \rightarrow C _{2}H _{6}$
As 2 moles of hydrogen are required to saturate one mole of acetylene and 1 mole = 2g of $H _{2}$, therefore 4g of hydrogen are required to saturate one mole of acetylene .

Multiple choice chemistry hydrocarbons introduction to alkenes - ethyne acetylene alkynes

Ethene is obtained by dehydrobromination of:

  1. CHBr = CHBr
  2. $CH _3CHBr _2$
  3. $CH _3 - CH _2Br$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Dehydrobromination is an organic reaction in which alkyl halide when boiled with alc.alkali gives corresponding alkene called as beta-elimination reaction.

Multiple choice distinction between pairs of amines organic compounds containing nitrogen organic compounds with functional group containing nitrogen chemistry

Aniline and N-methylaniline can be distinguished by using the:
(A) Hoffman Mustard oil reaction
(B)  Hinsberg reagent
(C)  Carbylamine test
(D)   Mulliken Test

  1. B and C

  2. C only

  3. A, B and C

  4. A, B, C and D
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Aniline is a primary amine, while N-methylaniline is a secondary amine. The Carbylamine test (C) is specific to primary amines, and the Hinsberg reagent (B) reacts differently with primary and secondary amines, making both useful for distinction.

Multiple choice distinction between pairs of amines organic compounds containing nitrogen organic compounds with functional group containing nitrogen chemistry

Nitration of acetanilide followed by hydrolysis gives

  1. Ortho nitro aniline

  2. Para nitroaniline

  3. Ortho and para nitroaniline

  4. Ortho nitro anilinium ion
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Acetanilide is nitrated to form p-nitroacetanilide because the acetamido group is bulky, directing substitution to the para position. Subsequent hydrolysis of the amide group yields para-nitroaniline.

Multiple choice distinction between pairs of amines organic compounds containing nitrogen organic compounds with functional group containing nitrogen chemistry

The reaction between primary amine, chloroform and few drops of alcoholic KOH is known as: 

  1. Hofmann's reaction

  2. Reimer-Tiemann's reaction

  3. Carbylamine reaction

  4. Kolbe's reaction
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Carbylamine reaction:
$C{ H } _{ 3 }C{ H } _{ 2 }N{ H } _{ 2 }+CH{ Cl } _{ 3 }+3KOH\longrightarrow C{ H } _{ 3 }C{ H } _{ 2 }NC+3KCl+3{ H } _{ 2 }O$
It forms isocyanide.
Multiple choice distinction between pairs of amines organic compounds containing nitrogen organic compounds with functional group containing nitrogen chemistry

Aniline and diphenylamine may be distinguished by:

  1. lassaignes test

  2. schiffs test

  3. carbyl amine reaction

  4. solubility test
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Aniline and diphenylamine may be distinguished by carbylamine reaction.
Aniline is a primary amine and diphenylamine is a secondary amine.
carbylamine reaction is given by primary amines. It is not given by secondary or tertiary amines.
In Hoffmann's carbylamine test (also known as Isocyanide test), aliphatic or aromatic primary amine is heated with chloroform  in presence of alcoholic potassium hydroxide to give foul smelling alkyl/aryl  isocyanide or carbylamine.
$\underset {} {C _6H _5-NH _2} + CHCl _3  + 3KOH  \xrightarrow {heat} C _6H _5-NC  + 3KCl + 3H _2O$

Multiple choice distinction between pairs of amines organic compounds containing nitrogen organic compounds with functional group containing nitrogen chemistry

Which of the following is an example of azo dye?

  1. Orange-1

  2. Malachite green

  3. Indigo

  4. Martius yellow
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Azo dye are a large class of synthetic organic dyes, that contain nitrogen as the azo group or $−N=N−$ as a part of their molecular structures. Orange-1 is an example of azo dye.

Multiple choice distinction between pairs of amines organic compounds containing nitrogen organic compounds with functional group containing nitrogen chemistry

Primary, secondary, tertiary amines can be separated by the following except: 

  1. Fractional distillation

  2. Fractional method using diethyl oxalate
  3. Hinsberg's method using $\displaystyle { C } _{ 6 }{ H } _{ 5 }{ SO } _{ 2 }Cl$

  4. Selective crystallisation
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Separation of $1^o,\, 2^o$ and $3^o$ amines:

$1^o,\, 2^o$ and $3^o$ amines + $R _4NX \xrightarrow{distillation}$ Mixture of $1^o,\, 2^o,\, 3^o$ amine
Mixture  of $1^o,\, 2^o,\, 3^o$ amine can be separated by following methods.
(i) Fractional distillation: The mixture of amines may be fractional distillation because their boiling points are quite different. It is used in industry.

(ii) Hinsberg method: In this method mixture of amines is seperated by using benzene sulphonyl chloride (Hinsberg's reagent).
$C _6H _5SO _2Cl+1^o\,amine \to Product \xrightarrow{KOH} dissolve$
$C _6H _5SO _2Cl+2^o\,amine \to Product \xrightarrow{KOH} insoluble$
$3^o$ amine does not react with benzene sulphonyl chloride.

(iii) Hofman method: In this method mixture of amines is separated by using ethyl oxalate.
$1^o$ amine + ethyl oxalate $\to$ solid product
$2^o$ amine + ethyl oxalate $\to$ liquid product
$3^o$ amine + ethyl oxalate $\to$ No reaction