Chemistry

Chemical Reactions and Equations

684 Questions

Chemical reactions and equations form a foundational chemistry topic where students identify reaction products, balance chemical formulas, and classify reaction types. It covers critical mechanisms like Markovnikov's rule, precipitation, and endothermic or exothermic processes. These questions are highly common in general science sections of state and central government competitive exams.

Endothermic reactionsPrecipitation reactionsDehydrohalogenation of alkyl halidesChemical equation balancingReaction product identification

Chemical Reactions and Equations Questions

Multiple choice allotropes of carbon chemistry occurrence of carbon compounds in nature polymers

${ C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 }\xrightarrow { \quad \quad Conc.{ H } _{ 2 }S{ O } _{ 4 }\quad \quad  } 12C+11{ H } _{ 2 }O$
Which of the following is obtained by the above reaction?

  1. Animal charcoal

  2. Sugar charcoal

  3. Coke

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

Sugar when treated with con. $H _2SO _4$ it gives carbon as sugar charcoal and water.

${ C } _{ 12 }{ H } _{ 22 }{ O } _{ 11 }\xrightarrow { \quad \quad Conc.{ H } _{ 2 }S{ O } _{ 4 }\quad \quad  } 12C+11{ H } _{ 2 }O$

Multiple choice methane study of methane carbon- an important element hydrocarbons chemistry

Methane reacts with chlorine in the presence of sunlight. If this reaction is allowed to continue, the end product formed will be :

  1. carbon tetrachloride

  2. chloroform

  3. ethylene chloride

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

The free-radical chlorination of methane is a chain reaction that continues until all hydrogen atoms are replaced by chlorine, resulting in carbon tetrachloride (CCl4).

Multiple choice methane study of methane carbon- an important element hydrocarbons chemistry

Chlorination of methane does not occur in dark because:

  1. methane can form free radicals in presence of sunlight only
  2. to get chlorine free radicals from $Cl _2$ molecules energy is required. It cannot happen in dark.

  3. substitution reaction can take place only in sunlight and not in dark

  4. termination step cannot take place in dark. It requires sunlight.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Chlorination of methane is a free radical substitution. To get chlorine free radical, UV light is required to initiate the substitution reaction, which is provided by sunlight.
$Cl _2\xrightarrow [  ]{ hv } Cl^{\bullet}+Cl^{\bullet}$
Multiple choice methane study of methane carbon- an important element hydrocarbons chemistry

Reaction of methane with halogens under appropriate conditions is called as :

  1. Halogenation

  2. Free radical halogenation

  3. Methylation

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

Methane reacts with halogens given appropriate conditions as follows:

${X} _{2} + UV \rightarrow  2 X \cdot$

$X \cdot + C{H} _{4} \rightarrow HX + C{H} _{3} \cdot$

$C{H} _{3} \cdot + {X} _{2} \rightarrow C{H} _{3}X + X \cdot$

where $X$ is a halogen: fluorine $\left(F\right)$, chlorine $\left(Cl\right)$, bromine $\left(Br\right)$, or iodine $\left(I\right)$. This mechanism for this process is called free radical halogenation.

Multiple choice acyl chlorides carboxylic acids and their derivatives functional derivatives of carboxylic acids carbonyl compounds chemistry

With which of the following can ethyl ethanoate not undergo a substitution reaction?

  1. $CH _3CO _2Na$
  2. aqueous $NaOH$
  3. $CH _3OH$, $H^+$
  4. aqueous $NH _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Ethyl ethanoate is an ester. Sodium acetate (CH3CO2Na) is the salt of the carboxylic acid component of the ester and does not act as a nucleophile to substitute the ethoxy group in a standard substitution reaction.

Multiple choice chemistry alcohols, esters and carboxylic acids reduction of carboxylic acids chemical properties chemical properties of carboxylic acids

How  - $COOH$ can be converted into $CH _3$ group?

  1. $\;Na\;\&\;alcohol$
  2. $\;Zn+HCl$
  3. $\;LiAlH _4$
  4. $\;Hl\;\&\;red\;phosphorous$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$-COOH$ can be converted into $CH _3$ group by heating with a mixture of HI and red phosphorus.
$CH _3COOH  \xrightarrow {\;Hl\;&\;red\;phosphorous} CH _3-CH _3$

Multiple choice chemistry alcohols, esters and carboxylic acids reduction of carboxylic acids chemical properties chemical properties of carboxylic acids

Which of the following reaction is not the oxidation?

  1. $CH _{3} - CHO \rightarrow CH _{3}COOH$
  2. $C _{2}H _{5}OH \rightarrow CH _{3} - CHO$
  3. $C _{2}H _{5}OH \rightarrow CH _{3} - COOH$
  4. $CH _{3}COOH \rightarrow C _{2}H _{5}OH$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Oxidation involves the loss of electrons or an increase in oxidation state (often adding oxygen or removing hydrogen). Option D shows the reduction of acetic acid to ethanol (adding hydrogen), which is the opposite of oxidation.

Multiple choice chemistry alcohols, esters and carboxylic acids reduction of carboxylic acids chemical properties chemical properties of carboxylic acids

$CH _{3}COOH \xrightarrow[]{LiAlH _{4}}(A)\xrightarrow[H _{2}]{Ni} B$ In this reaction A and B respectively are :

  1. $CH _{3}OH and CH _{4}$
  2. $C _{2}H _{5}OH and C _{2}H _{6}$
  3. $CH _{3}CHO and C _{2}H _{5} OC _{2}H _{5}$
  4. $C _{2}H _{5}OH and CH _{3}OCH _{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

B
$LiAlH _{4}$ reduces ethanoic acid to ethanol and then $H _{2}$ / Ni reduces it further to alkane.

Multiple choice chemistry alcohols, esters and carboxylic acids reduction of carboxylic acids chemical properties chemical properties of carboxylic acids

Which optically active compound on reduction with $\displaystyle LiAlH _{4}$ will give optically inactive compound?

  1. $CH _3-CH(OCH _3)-COOH$
  2. $CH _3-CH _2-CH(OH)-COOH$
  3. $CH _3-CH _2-CH(CH _2OH)-COOH$
  4. $CH _3-CH(OH)-CH _2-COOH$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Reduction of CH3-CH2-CH(CH2OH)-COOH with LiAlH4 yields CH3-CH2-CH(CH2OH)-CH2OH. This molecule has a chiral center at the carbon attached to the ethyl group, but the two CH2OH groups are not identical in a way that would make it achiral. Wait, actually, the reduction of the carboxylic acid to a primary alcohol creates a molecule where the two groups attached to the chiral center might become identical. Let's re-evaluate: CH3-CH2-CH(CH2OH)-COOH reduces to CH3-CH2-CH(CH2OH)-CH2OH. The chiral carbon is attached to H, ethyl, CH2OH, and CH2OH. Since two groups (CH2OH) are identical, it becomes achiral.

Multiple choice chemistry alcohols, esters and carboxylic acids reduction of carboxylic acids chemical properties chemical properties of carboxylic acids

The organic product formed in the reaction. $C _{6}H _{5}COOH \xrightarrow[(II)H _{3}O]{(I)LiAIH _{4}}$?

  1. $C _{6}H _{5}CH _{2}OH$
  2. $C _{6}H _{5}OHCH _{4}$
  3. $C _{6}H _{5}CH _{3} and CH _{3} OH$
  4. $C _{6}H _{5}CH _{3} and CH _{4} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

using LAH (lithium aluminium hydride), the COOH group is reduced to  $ CH _{2} OH$  group

Multiple choice chemistry alcohols, esters and carboxylic acids reduction of carboxylic acids chemical properties chemical properties of carboxylic acids

Consider the following reaction $CH _3-(CH _2) _{14}-COOH \overset{LiAlH _4}{\longrightarrow} \,X \overset{HCl}{\longrightarrow}$
Palmitic acid
$ Y \overset{1.\, Mg/Et _2O}{\underset{2.\, Oxirane}{\longrightarrow}} Z \overset{KMnO _4}{\longrightarrow} W$
The correct identity of the compounds is/are :

  1. $X \,is \,CH _3-(CH _2) _{14}-CH _2OH$
  2. $Y\, is \, CH _3-(CH _2) _{14}-CH _2Cl$
  3. $Z \,is\, CH _3-(CH _2) _{16}-CH _2OH$
  4. $W \, is\, CH _3-(CH _2) _{16}-CHO$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Reaction:
$CH _3-(CH _2) _{14}-COOH \overset{LiAlH _4}{\longrightarrow} \,CH _3-(CH _2) _{14}-CH _2OH \overset{HCl}{\longrightarrow} CH _3-(CH _2) _{14}-CH _2Cl$
$ CH _3-(CH _2) _{14}-CH _2Cl \overset{1.\, Mg/Et _2O}{\underset{2.\, Oxirane}{\longrightarrow}} CH _3-(CH _2) _{16}-CH _2OH \overset{KMnO _4}{\longrightarrow} CH _3-(CH _2) _{16}-COOH$

Multiple choice reactions of haloarenes haloalkanes and haloarenes chemistry

The order of reactivities of methyl halides in the formation of Grignard reagent is :

  1. $CH _3I> CH _3Br> CH _3Cl$
  2. $CH _3Cl> CH _3Br> CH _3I$
  3. $CH _3Br> CH _3Cl> CH _3I$
  4. $CH _3Br> CH _3I> CH _3Cl$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Grignard reagents are highly reactive and react with any source of a proton to give hydrocarbons. Water, alcohol, amines are sufficiently acidic to convert them to corresponding hydrocarbons. Halide reactivity increases in the order $Cl < Br < I$

Therefore A is the correct option.