Chemistry

Organic Chemistry Fundamentals

287 Questions

Organic chemistry fundamentals cover the structure, properties, and reactions of carbon containing compounds. The questions explore functional groups, isomerism, homologous series, and basic reaction mechanisms. This forms the basis of general chemistry across multiple examination formats.

Functional groupsIsomerism typesHomologous seriesCycloalkanesOrganic monomers

Organic Chemistry Fundamentals Questions

Multiple choice chemistry chemical reactions of organic compounds alkanoic acids nomenclature of carboxyl group nomenclature of carboxylic acids

The correct structural formula of butanoic acid is

  1. $H-\overset{H}{\overset{|}{\underset{H}{\underset{|}{C}}}}-\overset{H}{\overset{|}{C}}=\overset{H}{\overset{|}{C}}-\overset{O}{\overset{||}{C}}-OH$
  2. $H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{O}{\overset{||}{C}}}}-OH$
  3. $H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-OH$
  4. $H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\overset{O}{\overset{||}{C}}-OH$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Butanoic acid is a carboxylic acid with the structural formula CH3CH2CH2-COOH. where $-COOH$ is the functional group.

$H-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\underset{H}{\underset{|}{\overset{H}{\overset{|}{C}}}}-\overset{O}{\overset{||}{C}}-OH$

So, the correct representation of butanoic acid is $D$
Multiple choice chemistry chemical reactions of organic compounds alkanoic acids nomenclature of carboxyl group nomenclature of carboxylic acids

The functional group present in carboxylic acids is ?

  1. -COOH

  2. -CO-

  3. -OH

  4. -CHO
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Carboxylic acids are represented by -COOH- functional group. It ionizes to give carboxylate ions and hydrogen ions. This group present in a compound makes it an acid and if present with other groups in the compound it is always present at position no.1 and carbon is included in the carbon chain atoms.

Multiple choice chemistry organic nitrogen compounds nitroarenes- preparation and properties nitro compounds types of organic reactions

Nitro benzene is used _________.

  1. as a solvent

  2. In the manufacture of aniline

  3. In cheap scent

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

It is a solvent because it can dissolve a large number of polar substances (polar aprotic solvent)
It can be used to manufacture aniline because it can be reduced by stephen's reduction to anniline
It is also used as a scent because it has pleasant odour

Multiple choice chemistry hydrocarbons introduction to alkenes - ethyne acetylene alkynes

$H-C\equiv C-H+NaNH _{2}\longrightarrow A\overset{2 \ mole \ CH _3I}{\longrightarrow}B$


Then B is :

  1. 1-Butyne

  2. 2-Butyne

  3. 2-Pentyne

  4. Propyne
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reaction of process:
$CH\equiv CH+2NaNH _2\rightarrow NaC\equiv CNa\xrightarrow {2 mole CH _3I} CH _3CH\equiv CHCH _3$


Option B is correct.

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

Which of the following is not a property of diazonium salts?

  1. Diazonium salts are colourless crystalline solids.

  2. Being ionic in nature they are soluble in water.

  3. Most of these salts explode when dried.

  4. The aqueous solutions of these salts are poor conductors of electricity.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Salt solution such as sodium chloride ($NaCl$) conducts an electric current because it has ions in it that have the freedom to move about in solution. These ions are produced when sodium chloride dissolves in pure water to produce sodium ($Na^+$) and chloride ions ($Cl^–$).

Diazonium salts ($ArN _2^+X$) aqueous solutions are neutral to litmus and conduct electricity due to the presence of ions.
In an aqueous solution (that is, dissolved in water), the ions and electrons are much more mobile, thereby allowing electricity to flow through the solution.

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Which one of the following sets of compound correctly illustrates the law of reciprocal proportions?

  1. $P _2O _3, PH _3, H _2O$
  2. $P _2O _5, PH _3, H _2O$
  3. $N _2O _5, NH _3, H _2O$
  4. $N _2O, NH _3, H _2O$
  5. $NO _2, NH _3, H _2O$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $PH _3$, the ratio by weight of $P:H=31:3$

For $H _2O, O:H= 16:2=8:1$
Keeping the weight of $H(=1)$ fixed, $P:O=\cfrac {31}{3}:\cfrac {8}{1}=31:24\longrightarrow (1)$
In $P _2O _5, P:O$ is $(2 \times 31):(5 \times 16)$
$=62:80$ or $31:40 \longrightarrow (2)$
Keeping the weight of $P(=31)$ fixed in equation (1) & (2), the ratio of oxygen is $24:40$ or $3:5$ which is a simple ratio.

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Which one of the following sets of compounds correctly illustrate the law of reciprocal proportions.

  1. $P _{2}O _{3}, PH _{3}, H _{2}O$
  2. $P _{2}O _{5}, PH _{3}, H _{2}O$
  3. $N _{2}O _{5}, NH _{3}, H _{2}O$
  4. $N _{2}O, NH _{3}, H _{2}O$
  5. $NO _{2}, NH _{3}, H _{2}O$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Law of Reciprocal proportions:-

When two different elements combine with the same mass of a $3^{rd}$ element, the ratio of masses in which they do so must be same or multiple of the mass ratio in which they combine with each other.
$A)$ $P _2O _3,PH,H _2O$
The ratio of phosphorous combining with oxygen in $P _2O _3$ is 
$P:O=62:48=31:24$
The ratio of Phosphorous combining with hydrogen in $PH _3$ is
$P:H=31:3$
The ratio of hydrogen combining with oxygen in $H _2O$ is 
$H:O=2:16=1:8$       $\longrightarrow 1$
Now, the ratio of hydrogen & oxygen combining with $P$
$\Rightarrow 3:24=1:8$     $\longrightarrow 2$
From $1$ & $2$,  $1:8$ is same in both cases
$\therefore$ Law of reciprocal proportions is illustrated correctly.

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Which of the following sets of compounds correctly illustrate the law of reciprocal proportion ?

  1. $N _2O,NH _3,SO _2$
  2. $P _2O _5,PH _3,H _2O$
  3. $NO _2,NH _3,SO _3$

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

The law of reciprocal proportions is one of the basic laws of stoichiometry. It relates the proportions in which elements combine across a number of different elements. A simple statement of the law is:

 Element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A.

Hydrogen combines with Oxygen to form ${ H } _{ 2 }O$

$2:16\\ 1:8$

Phosphorous combines with Oxygen to form ${ P } _{ 2 }O _{ 5 }$

$62:80\\ 31:40\\ $

Phosphorous combines with Hydrogen to form $P{ H } _{ 5 }$

$31:5\\ \cfrac { 31 }{ 40 } \times \cfrac { 8 }{ 1 } =\cfrac { 31 }{ 5 } $

Multiple choice chemistry the language of chemistry percent composition percentage composition and empirical formula empirical formula, percentage composition and molecular formula

Which of the following compound represents an alkane?

  1. ${C} _{5}{H} _{8}$
  2. ${C} _{7}{H} _{16}$
  3. ${C} _{8}{H} _{6}$
  4. ${C} _{9}{H} _{10}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Alkanes have general molecular formula is $C _nH _{2n+2}$


$\therefore C _7H _{16}$ is Alkane.

Hence, the correct option is $\text{B}$

Multiple choice polyhalogen compounds haloalkanes and haloarenes chemistry

0.0852 g of an organic halide (A) when dissolved in 2.0 g of camphor, the melting point of the mixture was found to be $167^{\circ}$C. Compound (A) when heated with sodium gives a gas (B). 280 mL of gas (B) at STP weighs 0.375g. What would be 'A' in the whole process? $K _f$ for camphor = 40, m.pt. of camphor = $179^{\circ}$C.

  1. $C _2H _5Br$
  2. $CH _3I$
  3. $(CH _3) _2CHI$
  4. $C _3H _7Br$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\delta$=179-167=12, w=0.0852g, W=2g,Kf= 40 
Moleculer weight of (A)
=$\dfrac{1000\times{Kf}\times{w} }{\delta{T}\times {W}}$
=$\dfrac{1000\times{40}\times{0.0852} }{\delta{12}\times {2}}$
=142
(A) undergoes wortz reaction to form (B) i.e
$(A)\rightarrow{Na} (B)+NaX$
(B) is an alkane say $C _nH _2n+2$
$\therefore$ 280 mL of (B) weighs 0.375 g at NTP
$\therefore$ 22400 mL of (B) weighs 
=$\dfrac{0.375\times{22400}}{280}$
=30 g at NTP
M.wt of (B)=30, 12n+2n+2=30,n=2
Thus (B) is ethane and therefore (A) is $CH _3X$
The m.wt of $CH _3X$=142
At.wt of X=127
$\therefore$ is iodine 
Therefore alkyl halide is $CH _3I$.
This reaction is 
2$CH _3I\xrightarrow[Na]{ether} C _2h _6$
(A)                         (B)