by sideways overlapping of half filled p -orbitals

by the overlapping of atomic orbitals on the axis of nuclei

by mutual sharing of p - electrons

by overlapping of s-orbitals with p-orbital

MCQ Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 2

Organic compounds are non-conductors of electricity .

State True or false.

0%
True
0%
False

${ C }_{ 6 }{ H }_{ 5 }-O-{ C }_{ 2 }{ H }_{ 5 }$ IUPAC name of the compound is:

0%
ethoxybenzene
0%
phenoxyethane
0%
benzoxyehane
0%
none of these

The correct IUPAC name of:

0%
3-Cyclohexanol Propyne
0%
3-[3-Hydroxy Cyclohexyl] Propyne
0%
3-Propynyl Cyclohexanol
0%
3-(2-propynyl)Cyclohexanol

Explanation

The IUPAC name of the given compound is

$3-\left( \text{2-propynyl} \right)\text{Cyclohexanol}$ .

$\left( \text{2-propynyl} \right)$ is for the propynyl group attached with a triple bond at

$C-2$ of propynyl group and the cyclohexanol is for the ring of six carbon atoms with an alcohol group at

$C-3$ of ring structure.

Which of the following is correct IUPAC name of compound?

$CH_{3}-CH_{2}-CH=CH_{}-CH_{3}$

0%
Pent-2-ene
0%
Prop-1-en-2-yne
0%
1-Methylpropane
0%
Pen-4-ene

Explanation

$\overset {5}{C}H_{3}-\overset {4}{C}H_{2}-\overset {3}{C}H=\overset {2}{C}H-\overset {1}{C}H_{3}$

The IUPAC name of the given compound is Pent-2-ene.

The correct IUPAC name of

$H-\overset{\overset{O}{||}}C-CHO$ is:

0%
Formyl methanal
0%
1,2-Ethanedione
0%
2-Oxoethanal
0%
Ethanedial

Explanation

$O$

$H-\overset {||}{C}-\overset {||}{C}-H$

The given compound contain two aldehydes group. The common name of the compound is Glyoxal.

IUPAC name: Ethanedial

Which of the following statements is incorrect regarding the naming of complex compounds?

0%
$NH^{2-}$ , will be named as 'amido'
0%
'en' will be named as 'ethylenediamine'
0%
$C_{2}O_{2}S^{2-}_{2}$ will be named as 'dithio oxalato'
0%
${CH_{3}}^-$ will be named as 'methyl'

Explanation

$O$

$NH_2-\overset {||}{C}-$ named as amido.

$NH^{2-}$ will be named as amido is incorrect regarding the naming of complex.

IUPAC name of

$CH_3-O- \underset { CH_ 3 }{ \underset{| }{CH}}-CH_3$ is _____.

0%
3-Methoxy propane
0%
2-Methyl methoxy ethane
0%
1-Methoxy-1-methyl ethane
0%
Methoxy isopropane

IUPAC name of

$CH_{3}-O-C_{2}H_{5}$ is-

0%
Ethoxymethane
0%
1-methoxyethane
0%
Methylethyl ether
0%
Ethylmethyl ether

Explanation

The IUPAC name of

$CH_3-O-CH_2-CH_3$

An ether's IUPAC name should always follow this rule,

It is always name as small Alkoxy big alkane

Small-substituent here is

$CH_3$

Big-chain here is

$C_2H_5$

therefore it is named as

$1-$ methoxyethane.

Write the IUPAC name of

0%
$5,5-dimethyl-5,5-diethylpentane$
0%
$4,4-dimethyl-4,4-diethylpentane$
0%
$3-ethyl-4,4-diethylpentane$
0%
$1,1-dimethyl-2,2-diethylpentane$

The name of the following compound is :

${ CH }_{ 3 }{ CH }_{ 2 }({ CH }_{ 3 }){ CHNH }_{ 2 }$

0%
1-methyl-1-aminopropane
0%
2-aminobutane
0%
1-aminobutane
0%
1-amino-1-methylpropane

Explanation

The given compound is :

$CH_{3}CH_{2}(CH_{3})CHNH_{2}$

Its IUPAC name is :

1-amino-1-methylpropane

Hence, option D is correct.

The IUPAC name of

${\text{C}}{{\text{H}}_{\text{3}}} - {\text{CH}}\,{\text{ = }}\,{\text{CH}}\, - {\text{C}} \equiv \,{\text{CH}}$ is :

0%
2-Penten-4-yne
0%
1-Penten-3-ene
0%
Pent-3-en-1-yne
0%
None of these

IUPAC name of

$CH_3-C \equiv C-HC-(CH_3)_2$

0%
4-methyl pent-2-yne
0%
1,1- dimethyl-2-butyne
0%
2-methyl-4-pentyne
0%
4, 4-dimethyl-2-butyne.

What is the IUPAC name of the given compound

${ CH }_{ 3 }-\underset { \underset { OH }{ | } }{ CH } -{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 2 }-\overset { \underset { | }{ Br } }{ \underset { \overset { | }{ Br } }{ C } } -{ CH }_{ 3 }$

0%
6, 6-Dibromoheptan-2-ol
0%
2, 2-dibromoheptan-6-ol
0%
6, 6-Dibromoheptan-2-al
0%
None of these

The IUPAc name of

0%
4,5-Diethylocta-1,4,7-triene
0%
4,5-Diethylocta-1,4,6-triene
0%
4,5-Dimethylocta-1,4,6-triene
0%
4,5-Diethylocta-1,4,6-triyne

The correct name of

$H_3C-CH_2-C\equiv C-CH=CH_2$ is

0%
hex-3-yn-5-ene
0%
hex-5-en-3-yne
0%
hex-3-yn-1-ene
0%
hex-1-en-3-yne

IUPAC nomenclature of

$H_3C-\overset{\,\,CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH=\overset{\,\,\,\,CH_3}{\overset{|}{C}}-CH_3$

0%
2,4,4 - trimethylpent -2-ene
0%
2,4,4 - trimethylpent -3-ene
0%
2,2,4 - trimethylpent -3-ene
0%
2,2,4 - trimethylpent -2-ene

The IUPAC name of following compound is

$H_3C-\overset{\overset{\,\,\,\,\,\,\,C_2H_5}{|}}CH-\underset{\underset{\,\,\,\,\,CHO}{|}}CH-CH_3$

0%
3-ethyl-2-methylbutanal
0%
2,3 dimethylpentanal
0%
2-ethyl 3-methylbutanal
0%
2-ethyl-3-methylbutan-3-al

The IUPAC name of the compound is

0%
2-Ethyl-ethyl acetate
0%
Ethyl-3-methyl butanoate
0%
Ethyl-2-methyl butanoate
0%
2-methyl butanoic acid

The name of the given according to IUPAC nomenclature system is:

0%
2,3-dibromo-1,4-dichlorobut-2-ene
0%
1,4-dibromo-2,3-dichlorobut-2-ene
0%
Dichlorobromobutene
0%
Dichlorobromobutane

Which of the functional group from following contains the compound of minimum 3 carbon atoms?

0%
-COOH
0%
-CHO
0%
> C=O
0%
-C -O

The IUPAC name of the given is:

0%
2-bromo-3-methylbut-1-ene
0%
4-bromo-3-methylpent-2-ene
0%
2-bromo-3-methylpent-3-ene
0%
4-bromo-3-methylpent-1-ene

In which of the following, carbon principal group can not be taken in parent chain ?

Which group is always taken as a substituent in the IUPAC system of nomenclature?

0%
$-NO_{2}$
0%
$-C\equiv N$
0%
0%
$-NH_{2}$

Which is not the characteristic of

$\pi-$ bond?

0%
$\pi$ bond is formed when a sigma bond already exists
0%
$\pi$ bonds are formed from hybrid orbitals
0%
$\pi$ bond may be formed by the overlapping of p orbitals
0%
$\pi$ bond results from lateral overlap of atomic orbitals

Explanation

$\pi$ bond is formed between two atoms if a sigma bond already exists between them.

$\pi$ bonds are formed only by pure orbitals (

$p-$ orbitals), not from hybrid orbitals. They formed by lateral or sideways overlapping.

The number of sigma and pi bonds in a molecule of cyanogen are:

0%
$4,3$
0%
$3,4$
0%
$5,2$
0%
$3,5$

Explanation

Cyanogen,

$N\equiv C-C\equiv N$ has three sigma and four 'pi' bonds.

Pi bond is formed:

0%
by the overlapping of atomic orbitals on the axis of nuclei
0%
by mutual sharing of p - electrons
0%
by sideways overlapping of half filled p -orbitals
0%
by overlapping of s-orbitals with p-orbital

Explanation

If a bond between two atoms is broken when one atom is rotated around the bond axis, that bond is called a pi bond. Pi bonds are formed by the sideways overlap of parallel half filled $p-$ orbitals on adjacent atoms as shown in the figure. They are not formed from hybrid orbitals.

Thus, option (C) is correct.

Main source of most of the organic compounds is :

0%
coal tar
0%
petroleum
0%
both A & B
0%
ammonia

Which of the following has

$p \pi - d \pi$ bonding ?

0%
$NO^{-}_{3}$
0%
$SO^{2-}_{3}$
0%
$BO^{3-}_{3}$
0%
$CO^{2-}_{3}$

Explanation

$N, B, C$ do not have d-orbitals. Only

$SO_3^{2-}$ has

$p\pi - d \pi$ bonding.

Which one of the following is an incorrect statement?

0%
A pi bond is formed when a sigma bond already exists.
0%
A pi bond may be formed by the overlapping of p or d orbitals.
0%
A pi bond is formed by the overlapping of hybrid orbitals.
0%
A pi bond is formed by the lateral overlapping of atomic orbitals.

Which of the following is gemdihalide?

0%
$CH_{3} - CH(Br) - CH_{2}Br$
0%
$CH_{3} -CHBr_{2}$
0%
$CH_{3}- CH(Br) -CH_{2} -CH_{2}Br$
0%
$BrCH_{2} CH_{2}Br$

Explanation

Gem dihalides have halogens attached on the same carbon atom.

$-\overset{|}{\underset{|}{C}}-\overset{X}{\overset{|}{\underset{X}{\underset{|}{C}}}}-$

Option B is correct.

Generally pi bond is formed from the lateral overlap of:

0%
$s-s$ orbitals
0%
$p-p$ orbitals
0%
$d-d$ orbitals
0%
both $B$ and $C$

Explanation

$\pi$ bond is formed by lateral (side ways) overlap of either

$p-p$ ,

$d-d$ or

$d-p$ orbitals.

Hence, option

$D$ is correct.

IUPAC name of the above compound is :

0%
Bromo methyl benzene
0%
Phenyl methyl bromide
0%
Benzyl bromide
0%
Benzal bromide

The first organic compound was synthesized in laboratory by:

0%
Wohler
0%
Kolbe
0%
Berzelius
0%
Neil Barthlot

The best solvent for the crystallization of organic solid is in which:

0%
organic compound is highly soluble
0%
compound is soluble on warming
0%
compound is soluble at $90^{0}$
0%
compound is insoluble

$CH_3\! -\! C\equiv CH$

The IUPAC name of the compound is:

0%
propane
0%
propene
0%
propyne
0%
1-methyl-propane

Explanation

It contains 3 carbon chain and the functional group is alkyne so the name is

$H$

$H\!\! -\! \overset { |}{ \underset { | }{ C} }\! -\! C\equiv C\! - \! H$

$H$

prop

$+$ yne

$=$ propyne

So, the correct option is

$C$ .

In the detection of hydrogen from organic
the compound, the compound is heated with CuO
and water produced is tested with__________.

0%
lime water
0%
anhydrous $CuSO_{4}$
0%
potash solution
0%
calcium chloride solution

Explanation

$C$ and

$H$ are oxidized to

$CO_{2}$ and

$H_{2} O$ when heated with

$CuO$ .

$CO_{2}$ turns lime water milky and

$H_{2}O$ turns anhydrous

$CuSO_{4}$ to hydrated

$CuSO_{4}$ which is blue in carbon.

$H+CuO\overset{\Delta }{\rightarrow}H_{2}O+Cu$

$CuSO_{4}+5H_{2}o\rightarrow CuSO_{4},5H_{2}O$
White Blue

In sodium extract nitrogen is present as:

0%
sodamide
0%
sodium cyanide
0%
sodium nitrite
0%
sodium nitrate

Explanation

In sodium extract nitrogen is present as sodium cyanide.

$Na+C+N \overset{\Delta }{\rightarrow} NaCN$

In the detection of carbon, lime water turns
milky due to the formation of:

0%
$CaO$
0%
$Ca(OH)_{2}$
0%
$CaCO_{3}$
0%
$Ca(HCO_{3})_{2}$

Explanation

Carbon dioxide passes through limewater, giving a milky solution. This is due to the insoluble suspension of calcium carbonate formed:

$Ca(OH)2(aq) + CO2(g)\rightarrow CaCO3(s) + H2O(l)$

Sodium extract gives blood red colour when treated with

$FeCl_{3}$ . Formation of blood red colour confirms the presence of:

0%
only nitrogen
0%
only sulphur
0%
only halogens
0%
both nitrogen and sulphur

Explanation

$Na+N+C+S \rightarrow Na SCN$
Sodium thiocyanate

$SCN^- +Fe^{3+}\rightarrow [Fe (SCN)]^{2+}$
Ferric thiocyanate (Blood-red colour)

It confirms the presence of both

$N$ and

$S$ .

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

0%
N
0%
S
0%
Cl
0%
H

Which of the following statements is not applicable to Lassaigne's test?

0%
The extra elements (N, S and halogens) are tested as their inorganic salts.
0%
During fusion of the organic compound with sodium, $Na_{2} S$ , NaCN and NaX are formed.
0%
It can even be used for the detection of hydrogen in the organic compound.
0%
It is possible to differentiate between the different halogens.

Explanation

Lassaignes test

$Na+C+N\ \overset{a}{\rightarrow}NaCN$

$2Na+S\rightarrow Na_{_{2}}S$

$Na+X\ \rightarrow Na_{2}S$
Option (a) is correct
when s is also present

$Na+C+N+S\rightarrow NaCNS$

$3CNS^{-}+Fe^{^{3}+}\rightarrow Fe(CNS)_{3}$
Blood red

$NaSCN+2Na\rightarrow NaCN+Na_{2}S$
option (b) is correct
for testing halogens

$AsNo_{3}+Na x\rightarrow Ag\times \downarrow + NaNO_3$
when X=Cl Agcl (white ppt) soluble in

$NH_{4}OH$
X=Br AgBr (pate-yellow ppt) partially soluble in

$NH_{4}OH$
X=I AsI (Dark-yellow ppt) is soluble in

$NH_{4}OH$
option (D) is correct.

It cant be used for the detection of hydrogen in the organic compound.

So the answer is the option(C).

Neo-heptyl alcohol is correctly represented as :

Explanation

Hint: In a trivial system, The prefix "neo" is used when all but two carbons form a continuous chain, and these two carbons are part of a terminal tert-butyl group

Explanation:

In the trivial system of naming organic compounds, the names of compounds are dependent on the source of the organic compounds or their properties.

If the prefix is iso, then there will be one methyl group on the secondary carbon of the parent chain. And if the prefix is neo, then there will be two methyl groups on the secondary carbon of the parent chain.

Also, heptyl is there that means there are a total of $7 \ C$ present and the alcohol is primary.

Hence, Neo-heptyl alcohol is correctly represented as:

Correct Option: $C$

Which of the following is called ethanoic acid :

0%
$HCOOH$
0%
$CH_{3}COOH$
0%
$CH_{3}CH_{2}COOH$
0%
$CH_{3}CH_{2}CH_{2}COOH$

Explanation

Methanoic acid :

$H COOH$
ethanoic acid :

$CH_3 COOH$

The IUPAC name of the above shown compound is :

0%
But-2-ene-2,3-diol
0%
Pent-2-ene-2,3-diol
0%
2-methylbut-2-ene-2,3-diol
0%
Hex-2-ene-2,3-diol

Explanation

Here,

$-OH$ is the main functional group.

The number will start from the lower

$-OH$ group as shown in the structure.

The name is Pent-2-ene-2,3-diol.

Option B is correct.

Which of the following is branched chain alkane?

0%
n-pentane
0%
pyridine
0%
iso-butane
0%
n-hexane

The incorrect IUPAC name is:

0%
0%
0%
$CH_3CCCH(CH_3)_2$ 4-methyl-2-pentyne
0%

Correct IUPAC Name of given compound :

0%
3-Ethenyl-2-methylhexa-2,5-dienoicacid
0%
2-Methyl-3-ethenylhexa-2,5-dien-1-oic acid
0%
3-Ethenyl-2-methyhexa-2,5-dien-1-carboxylicacid
0%
2-Methyl-3-eth(-2-enyl)hexa-2,5-dienoicacid

Common name of

$CH_{3}CH_{2}COOH$ :

0%
Propionic acid
0%
Butyric acid
0%
Vinegar
0%
Methanoic acid

Explanation

$^{3}CH_{3}^{2}CH_{2}^{1}COOH \Rightarrow$ Propionic acid

$\Rightarrow$ 3 corbons, so common name is
common name

$\Rightarrow$ propionic acid, IUAC

$\Rightarrow$ Propanoic acid

The IUPAC name of the compound

$CH_3CH=CHC\equiv CH$ is:

0%
pent - 3 - en - 1 - yne
0%
pent - 2 - en - 4 - yne
0%
pent - 1 - yn - 3 - ene
0%
pent - 4 - yn - 2 ene

$CH_{2}OHCHOHCH_{2}OH$ is glycerol.
It doesn't contain ___________ alcoholic group.

0%
$1^{0}$
0%
$2^{0}$
0%
$3^{0}$
0%
both $1^{0}$ & $2^{0}$

Explanation

$3^{0}$ -OH group

$\begin{matrix} 1^{ 0 } & CH_ 2 & - &OH \\ & |& & \\2^0 & CH&-&OH\\&|&&\\3^0&CH_2&-&OH \end{matrix}$

CBSE Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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