MCQ Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 6

Which functional group is present in a molecule of

${CH}_{3}OH$ ?

0%
Ketone
0%
Aldehyde
0%
Ester
0%
Alcohol
0%
Halide

Explanation

The functional group present in the molecule of methanol

$\displaystyle CH_3OH$ is hydroxyl group (

$\displaystyle -OH$ ) group. Methanol is an alcohol. Hence, the functional group is alcoholic

$\displaystyle -OH$ group.
Note: In general, we can represent alcohol as R-OH.

Which of the following factors can increase the amount of

$E_1$ product in the following reaction?

0%
Increasing polarity of solvent
0%
Using a strong base
0%
Rising reaction temperature
0%
Using a Lewis acid catalyst

Carbon dioxide formed after reaction with cupric oxide is tested by :

0%
caustic soda
0%
quick lime
0%
lime water
0%
baking soda

Explanation

The given organic compound is mixed with dry copper oxide (

$CuO$ ) and heated in a hard glass tube. The products of the reaction are passed over (white) anhydrous copper sulphate and then bubbled through lime water. If copper sulphate turns blue due to the formation of

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.

In the detection test of carbon and hydrogen with cupric oxide, carbon is oxidized to :

0%
carbon monoxide
0%
carbon dioxide
0%
carbonic acid
0%
copper sulphate

Explanation

The given organic compound is mixed with dry copper oxide (

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.So, the carbon is oxidized to

$C{ O }_{ 2 }$ .

In the detection test of carbon and hydrogen with cupric oxide, hydrogen is oxidized to __________.

0%
water
0%
methane
0%
hydride
0%
lime water

Explanation

The given organic compound is mixed with dry copper oxide (

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.

$2H+CuO \rightarrow X+Y$

The compound

$X$ and

$Y$ are:

0%
$H_2O, Cu$
0%
$H_2, CuO$
0%
both $A$ and $B$
0%
none of the above

Explanation

Compounds X and Y can be identified from the reaction below:

$2H+CuO\longrightarrow { H }_{ 2 }O+Cu$

The IUPAC name of the given compound is :

0%
$3$ -Phenyl- $4$ -methylhexene
0%
$3$ -Phenyl- $4$ -methylhexane
0%
$3$ -Phenyl- $4$ -methylhex- $2$ -ene
0%
$4$ -methyl- $3$ -phenylhexene

Explanation

The IUPAC name of the given compound is

$\text{4-methyl-3-phenylhexene}$

Hence, the correct option is

$D$

1431570_577099_ans_9c04d97110d04f05b96e8857d27740ca.png

IUPAC name of given compound is :

0%
Cyclohexanol
0%
Cyclohexanone
0%
Cyclohexanal
0%
None of these

The given diagram is an apparatus for carbon detection. Fill in the blank of the labelled diagram.

0%
Cupric oxide
0%
Anhydrous copper sulphate
0%
Copper precipitate
0%
Calcium carbonate

Explanation

The given organic compound is mixed with dry copper oxide ( $CuO$ ) and heated in a hard glass tube. The products of the reaction are passed over (white) anhydrous copper sulphate and then bubbled through lime water. If copper sulphate turns blue due to the formation of $CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by $C{ O }_{ 2 }$ , then the compound contains carbon.

Carbon and hydrogen are detected by heating the organic compound with ________.

0%
magnesium oxide
0%
cupric oxide
0%
zinc oxide
0%
sulphur dioxide

Explanation

The detection of various elements present in an organic compound is called qualitative analysis. Carbon and hydrogen are present in almost all the organic compounds. Carbon and hydrogen are detected by heating the organic compound with cupric oxide ( $CuO$ ) strongly, where carbon is oxidized to carbon dioxide and hydrogen to water. Carbon dioxide is tested by lime water test, whereas water is tested by anhydrous copper sulphate test.

Water formed is tested by :

0%
Copper nitrate test
0%
Anhydrous copper sulphate test
0%
Aqueous copper sulphate test
0%
Copper

Explanation

The given organic compound is mixed with dry copper oxide (

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.

$C+2CuO \rightarrow X+Y$

The compound

$X$ and

$Y$ are :

0%
$CO_2, \ 2Cu$
0%
$CO, \ Cu_2O$
0%
both $A$ and $B$
0%
none of the above

Explanation

The reaction takes place as follows:

$C+2CuO\longrightarrow C{ O }_{ 2 }+2Cu$

Carbon present in the compounds is oxidised to carbon dioxide which:

0%
turns blue litmus red
0%
turns lime water milky
0%
turns solution to green
0%
has no effect on solution

Explanation

The given organic compound is mixed with dry copper oxide (

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapour) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.

Amongst the following elements present in an organic compound, the one which has no direct test for detection is :

0%
$Cl$
0%
$N$
0%
$S$
0%
$O$

In Lassaigne's test, the organic compound is fused with a piece of sodium metal in order to:

0%
increase the ionisation of the compound
0%
decrease the melting point of the compound
0%
increase the reactivity of the compound
0%
convert the covalent compound into a mixture of ionic compounds

Water turns anhydrous copper sulphate solution in the carbon-hydrogen detection experiment :

0%
brown
0%
red
0%
blue
0%
green

Explanation

The given organic compound is mixed with dry copper oxide (

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.

Among the following, which one is a wrong statement?

0%
${ SeF }_{ 4 }$ and ${ CH }_{ 4 }$ have same shape.
0%
${ PH }_{ 5 }$ and ${ BiCl }_{ 5 }$ do not exist.
0%
$\rho \pi -d\pi$ bonds are present in ${ SO }_{ 2 }$ .
0%
${ I }_{ 3 }^{ + }$ has bent geometry.

Which of the following elements in an organic compound cannot be detected by Lassaigne's test?

0%
Nitrogen
0%
Sulphur
0%
Chlorine
0%
Hydrogen

The I.U.P.A.C. name of acetylene is :

0%
Propane
0%
Propyne
0%
Ethene
0%
Ethyne

Explanation

In acteylene 2 carbon atoms are present hence general name is ethane and triple bond is present so the suffix to be added is 'yne'. Hence the I.U.P.A.C name is ethyne.

$H-C\equiv C-H$

The _________ present in the compound is oxidised to water which condenses on the cooler parts of the test-tube or in the bulb of delivery tube :

0%
Hydrogen
0%
Copper
0%
Carbon soot
0%
None of the above

Explanation

The given organic compound is mixed with dry copper oxide (

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.

Which one of the following processes will produce hard water ?

0%
Saturation of water with $CaCO_3$
0%
Saturation of water with $MgCO_3$
0%
Saturation of water with $MgSO_4$
0%
none of the above

The presence of carbon in an organic compound is detected by heating it with :

0%
sodium metal to convert it to $NaCN$
0%
$CaO$ to convert it into $CO$ which burns with a blue flame
0%
$CuO$ to convert it into $CO_2$ which turns lime water milky
0%
$Cu$ wire to give a bluish-green flame

Explanation

The given organic compound is mixed with dry copper oxide (

$CuS{ O }_{ 4 }.5{ H }_{ 2 }O$ (by water vapor) then the compound contains hydrogen. If lime water is turned milky by

$C{ O }_{ 2 }$ , then the compound contains carbon.

Identify the functional group present in acetic acid:

0%
$O-\underset{O}{\underset{||}{C}}-OH$
0%
$O-\overset{C}{\overset{||}{\underset{O}{\underset{||}{C}}}}-OH$
0%
$-\underset{O}{\underset{||}{C}}-OH$
0%
$H-\overset{H}{\overset{|}{\underset{H}{\underset{|}{C}}}}=OH$

Explanation

The functional group present in acetic acid is carboxylic group

$\displaystyle -\underset{ \displaystyle O}{\underset{||}{C}}-OH$ .

Acetic acid can be represented as

$\displaystyle CH_3-\underset{ \displaystyle O}{\underset{||}{C}}-OH$

Hence, the correct option is

$C$

The atom or the group of atoms by which the characteristic reactions of organic compounds are determined, that atom or group of atoms is called ____________.

0%
hydrocarbon
0%
functional group
0%
atom
0%
compound

Explanation

Functional groups are specific substituents or moieties within molecules that are responsible for the characteristic chemical reactions of those molecules.

Examples:

$-OH,-COOH,-COOR$ etc.

Hence, the atom or the group of atoms by which the characteristic reactions of organic compounds are determined, that atom or group of atoms is called functional group.

Identify the correct decreasing order of priority of the functional groups from the following:

0%
$-CONH_2, -C=O, -HC=O, -CN$
0%
$-CONH_2, -CN, -HC=O, -C=O$
0%
$-CN, -CONH_2, -C=O, -HC=O$
0%
$-CN, -CONH_2, -HC=O, -C=O$

Explanation

$Amides \ > \ Nitriles \ > \ Aldehydes \ > \ ketones$

Option B is correct.

An organic compound is a clear liquid having a molecular formula

${C}_{4}{H}_{8}O$ . It has an open-chain structure. Without any carbon-carbon double bond. The compound can be:

a) An alcohol
b) An ester
c) An aldehyde
d) A ketone

0%
(a) and (b)
0%
(c) and (d)
0%
(b) and (d)
0%
(d) and (a)

Explanation

An organic compound is a clear liquid having a molecular formula C

$_4$ H

$_8$ O. It has an open chain structure. Without any carbon-carbon double bond. The compound can be either an aldehyde or a ketone.

How many delocalised

$\pi$ - electrons are there in the following compound?

0%
1
0%
2
0%
4
0%
6

Explanation

In the given compound. 4

$\pi$ -electrons of double bond and 1 lone pair on

$N-$ atom leads to delocalization of

$\pi$ -electrons.

The IUPAC name of compound

$CH_2 = CH -CH(CH_{3})_2$ is:

0%
1, 1-dimethyl-2-propane
0%
2-vinyl propene
0%
3-methyl-1-butene
0%
2-vinyl propane

Explanation

The IUPAC name of

$\displaystyle CH_2 = CHCH (CH_{3})_2$ is

$\displaystyle 3-methyl-1-butene$ . The numbering is started from the end which gives the lowest locants to doubly bonded C atoms.
Hence, the option (C) is the correct answer.

The functional group present in a molecule having the formula

${C}_{12}{O}_{9}$ is

0%
Carboxylic acid
0%
Anhydride
0%
Aldehyde
0%
Alcohol

Among the given compounds 2, 3-dimethylhexane is:

0%
i
0%
ii
0%
iii
0%
iv

Explanation

$\displaystyle 2, 3-dimethylhexane$ is compound

$\displaystyle ii$ . The parent chain contains 6 C atoms. Hence it is derivative of hexane. The numbering is started from the end which gives lowest locants to the substituents.

Which substance has the lowest boiling point?

0%
$CH_3 CH_2 CH_2 CH_2 OCH_3$
0%
$CH_3 CH_2 OCH_2 CH_3$
0%
$CH_3 CH_2 CH_2 CH_3$
0%
$CH_3 CH_2 C = OCH_3$

Explanation

$\displaystyle CH_3 CH_2 CH_2 CH_3$

$\displaystyle (n-butane)$ has the lowest boiling point as it has lowest molecular weight and is non polar. It has weakest intramolecular forces of attraction.

All others are polar comp[ounds and so has dipole-dipole inbteraction.

Hence, the correct option is

$C$

$O=C=O$

How many pi bonds does given compound have?

0%
2
0%
3
0%
4
0%
5

The IUPAC name for the hydrocarbon represented by:

0%
3, 3-diethylpentane
0%
2-ethylpentane
0%
tetraethyl carbon
0%
neo-nonane

Passing

${H}_{2}S$ gas into a mixture of

${ Mn }^{ 2+ },{ Ni }^{ 2+ },{ Cu }^{ 2+ },{ Hg }^{ 2+ }$ ions in acidified aqueous solution precipitates:

0%
$CuS$ and $HgS$
0%
$MnS$ and $CuS$
0%
$MnS$ and $NiS$
0%
$NiS$ and $HgS$

Explanation

${Cu}^{2+}$ and

${Hg}^{2+}$ lie in IInd group of qualitative analysis

${H}_{2}S$ in acid medium is reagent for it.

$HCl\rightarrow {H}^{+}+{Cl}^{-}$

${H}_{2}S\leftrightharpoons 2{H}^{+}+{S}^{2-}$

Which of the following is a 3-methylbutyl group?

0%
$CH_3 CH_2 CH_2 CH_2-$
0%
$( CH_3 CH_2 )_2 CH-$
0%
$(CH_3)_3 CCH_2-$
0%
$(CH_3)_2 CHCH_2CH_2-$

Explanation

When a hydrogen is removed from butane, butyl group is obtained.
Hence, the structure of 3-methylbutyl group is obtained, when one hydrogen atom is removed from 3-methylbutane.

i.e.

$CH_3 - \underset { \overset { | }{ CH_3 } }{ CH }{CH_2 CH_3}$ after removing one hydrogen

$(CH_3)_2 CHCH_2CH_2-$

3-methylbutane 3-methylbutyl

The molecular formulae for phosgene and tear gas are .......... and ......... respectively.

0%
$COCl_2$ and $CCl_3 NO_2$
0%
$SOCl_2$ and $CCl_3 NO_2$
0%
$COCl_2$ and $CCI_3NO_2$
0%
$SOCl_2$ and $CCI_3 NO_2$

Pi bond is formed between the two atoms when their is a:

0%
donation of electron pair from one atom to another.
0%
axial overlapping of atomic orbitals
0%
transfer of one or more than one electron from one atom to another
0%
lateral overlapping of atomic orbitals

Among the xylenes, which is thermodynamically most stable?

0%
0%
0%
0%
All are equally stable

How do we differentiate between

${ Fe }^{ 3+ }$ and

${ Cr }^{ 3+ }$ in group III?

0%
By taking excess of ${ NH }_{ 4 }OH$
0%
By increasing ${ NH }_{ 4 }^{ + }$ ion cocentration
0%
By decreasing ${ OH }^{ - }$ ion concentration
0%
Both (b) and (c)

What is the IUPAC name of the following compound ?

0%
1, 2 - dimethyl cyclopropyl cyclohexane
0%
1, 2 - dimethyl spiro [2, 5] octane
0%
7, 8 - dimethyl spiro [5, 2] octane
0%
1, 2 - dimethyl bicyclo [5, 2] octane

In presence of

$HCl(aq.);{H}_{2}S$ gas results the precipitation of group-2 cations but not of group-4 cations during qualitative analysis. It is due to:

0%
Lower concentration of ${ S }^{ 2- }$
0%
Higher concentration of ${ S }^{ 2- }$
0%
Lower concentration of ${ H }^{ + }$
0%
Higher concentration of ${ H }^{ + }$

Explanation

Due to common ion effect, the ionisation of

${H}_{2}S$ is suppressed and at low concentration of

${S}^{2-}$ ions, the second group sulphides are precipitated.

How many

${1}^{o},{2}^{o}$ and

${3}^{o}$

$H$ atoms are present in toluene?

0%
$3,0,5$
0%
$3,5,0$
0%
$4,3,0$
0%
$3,6,0$

$HCHO$ and

$HCOOH$ are detected by

0%
$NaH{ CO }_{ 3 }$
0%
$Cu{ SO }_{ 4 }/NaOH$
0%
$Ag{ NO }_{ 3 }/{ NH }_{ 4 }OH$
0%
All of these

Explanation

Both

$HCHO$ and

$HCOOH$ react with

$Ag{ NO }_{ 3 }/{ NH }_{ 4 }OH\quad$

$Cu{ SO }_{ 4 }/NaOH\quad$ solution. But

$HCOOH$ gives effervescences of

${CO}_{2}$ when reacts with

$NaH{ CO }_{ 3 }$ , whereas

$HCHO$ does not give any effervescence of

${CO}_{2}$ when reacts with

$NaH{ CO }_{ 3 }$ .

$HCOOH+NaH{ CO }_{ 3 }\longrightarrow HCOONa+{ H }_{ 2 }O+{ CO }_{ 2 }\uparrow \quad$

The IUPAC name of dimethyl ether is :

0%
ethoxymethane
0%
methoxymethane
0%
methoxyethane
0%
ethoxyethane

Explanation

Dimethyl ether is

$CH_3OCH_3$ .

It's IUPAC name is methoxymethane.

Hence, option

$B$ is correct.

IUPAC name of above compound is:

0%
5-ethenyleyclopenta-1,3-diene
0%
3-ethenylcyclopenta-1,4-diene
0%
1-ethenylcyclopenta-2,4-diene
0%
2-ethenylcyclopenta-1,3-diene

Explanation

The IUPAC has suggested few rules to be followed during the naming of a compound.

The naming should be in alphabetical order. From

$C_1$ , ethyne group is hanging and the rest is a ring of

$5-C$ .

So, the name goes as 1-ethenylcyclopenta-2,4-diene.

The IUPAC name of the following structure

$CH_3CH=C(CH_3)C\equiv CCH_3$ is:

0%
3-methylhex-2-en-4-yne
0%
3-methylhex-2-yn-4-ene
0%
4-methylhex-4-yn-4-ene
0%
all are correct

Explanation

$\mathbf{Given }$ : The structural formula of the compound is given

$\mathbf{Solution}$ :

1 2 3 4 5 6

since double and triple bond are on the same distance from both side so we will start numbering from double bond first as "double bond has has higher priority than triple bond" .

a methyl group is attached with 3rd carbon so the answer will be

3-methylhex-2-en-4-yne

$\mathbf{hence ,}$ option A is correct.

IUPAC name of the given is

0%
3-hydroxymethylbenzoate
0%
4-hydroxy ethylbenzoate
0%
2-hydroxyethylbenzoate
0%
2-hydroxymethylbenzoate

Explanation

Is a senior group as compared to

$-OH$

$CH_3-CH_2-\underset{CH_2}{\underset{||}{C}-}COOH$
The correct IUPAC name of the above compound is:

0%
$2$ -methyl butanoic acid
0%
$2$ -Ethylprop- $2$ -enoic acid
0%
$2$ -carboxy- $1$ -butene
0%
none of these

Explanation

According to IUPAC rules, the priority order is

Functional groups ${-COOH}$
Double bond at 2nd position
No. of carbons in a chain ${3}$

So the name of the compound is 2-ethylprop-2-enoic acid.

Which of the following is least stable?

0%
$CH_{3} - \overset {(+)}{CH} - CH_{3}$
0%
$CH_{3}CH_{2} -\overset{(+)} CH_{2}$
0%
$CH_{3} - \underset {\underset {CH_{3}}{|}}{\overset {(+)}{C}} - CH_{3}$
0%
$(CH_{3})_{3}C - \underset {\underset {C_{6}H_{5}}{|}}{\overset {(+)}{C}}-H$

Explanation

Stability of carbocation:

$CH_3^{\oplus} < \underset {CH_3CH_2^{\oplus}}{1^o carbocation} < \underset {CH_3-\underset {CH_3}{\underset {|}{CH^{\oplus}}}}{2^o carbocation} < \underset {CH_3-\underset {CH_3}{\underset {|}{C}}^{\oplus}-CH}{3^o} carbocation$

$CH_3-CH_2-CH_2^{\oplus}$ is

$1^o$ carbocation. Hence, it is least stable.

Which of the following can have vacant

$p-$ orbital?

0%
${ _{ }^{ \bullet }{ CH } }_{ 3 }$
0%
$\overset { \oplus }{ C } { H }_{ 3 }$
0%
$\upharpoonleft \downharpoonright { CH }_{ 2 }$
0%
Both (b) and (c)

Explanation

$CH^{-}_{3}\rightarrow$

$1s^{2}2s^{2} 2p^{4}\rightarrow$ vacant p- orbital not found.

$CH^{+}_{3}\rightarrow$

$1s^{2}2s^{2} 2p^{2}\rightarrow$ vacant p- orbital found.

$CH_{2}\rightarrow$

$1s^{2}2s^{2} 2p^{2}\rightarrow$ vacant p- orbital found.

CBSE Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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