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IUPAC nomenclature Numerical - Organic Chemistry: Some Basic Principals & Techniques, Class 11 Video Lecture
FAQs on IUPAC nomenclature Numerical - Organic Chemistry: Some Basic Principals & Techniques, Class 11 Video Lecture
| 1. What are the basic principles of IUPAC nomenclature in organic chemistry? |  |
Ans. The basic principles of IUPAC nomenclature in organic chemistry include:
- The longest continuous carbon chain in the molecule is selected as the parent chain.
- The parent chain is numbered in such a way that the substituents, such as functional groups or branches, have the lowest possible locants.
- The names of substituents are written as prefixes and are arranged in alphabetical order.
- The type and position of functional groups are indicated by suffixes or prefixes.
- If there are multiple identical substituents, prefixes such as di-, tri-, tetra-, etc., are used to indicate their number.
| 2. What are the techniques used in IUPAC nomenclature in organic chemistry? | |
Ans. The techniques used in IUPAC nomenclature in organic chemistry are as follows:
- Identifying the parent chain: The longest continuous carbon chain in the molecule is identified as the parent chain.
- Numbering the parent chain: The parent chain is numbered in such a way that the substituents have the lowest possible locants.
- Naming substituents: Substituents, such as functional groups or branches, are named as prefixes and arranged in alphabetical order.
- Indicating functional groups: The type and position of functional groups are indicated by suffixes or prefixes.
- Using numerical prefixes: If there are multiple identical substituents, numerical prefixes such as di-, tri-, tetra-, etc., are used to indicate their number.
| 3. How do you select the parent chain in IUPAC nomenclature? | |
Ans. In IUPAC nomenclature, the parent chain is selected based on the longest continuous carbon chain in the molecule. The parent chain is usually the chain that contains the functional group with the highest priority according to the IUPAC priority rules. The chain may be straight or branched, and it should contain the maximum number of carbon atoms. If there are multiple chains of the same length, the chain with the maximum number of substituents or branches is selected as the parent chain.
| 4. How do you assign locants in IUPAC nomenclature? | |
Ans. Locants in IUPAC nomenclature are assigned to indicate the position of substituents or functional groups on the parent chain. The locants are assigned by numbering the parent chain in such a way that the substituents have the lowest possible locants. The numbering starts from the end that gives the lowest locant to the first substituent encountered. If there are multiple substituents, the numbering is continued in the direction that gives the lowest locants to all substituents. If there is a choice between two numbering systems, the one that gives the lowest locants to more substituents is selected.
| 5. How are functional groups indicated in IUPAC nomenclature? | |
Ans. Functional groups in IUPAC nomenclature are indicated by suffixes or prefixes. Suffixes are used when the functional group is the principal functional group in the molecule, while prefixes are used when there are other functional groups present. The suffixes are added to the end of the parent chain name and indicate the type of functional group present. For example, "ol" is used for alcohols, "one" is used for ketones, and "oic acid" is used for carboxylic acids. Prefixes, on the other hand, are added before the parent chain name and indicate the type and position of functional groups. For example, "hydroxy" is used for alcohols and "oxy" is used for ethers.
Text Transcript from Video
hello friends this video on organic
chemistry basic part 14 is what Q Bar
Exam feel calm no more fear from exam
this question
we should have following it represents
the correct IUPAC name the first one is
to diamond tread painted or to develop
eaten since it is dimethyl you should
have two numbers here this is wrong
right so this is correct to to dimethyl
pentane is correct so for a the answer
is to die tile in day let's take the
second one be two four seven time is
high octane and two five seven
try me trial opt-ins let's draw this
octane workup eight carbon three four
five six seven eight
right and this is number one two three
four five six seven eight
so it says that try me tied at second I
have a metal rod at fourth I have a
methyl group and in eighth seventh I
have type if you have second number the
same thing in the reverse way it will be
one two three four five six seven eight
the same thing I can write s257 triumph
accountant right but as I told the
Friday will be given to the lower number
so this is correct this is not correct
right because this is correct
this is not going because it is two five
seven this is two four seven for the
scene come back the next is to
chloroform methyl painting or 402 methyl
Benton again there is a conclusion here
because 2n 4 we can give any number
names so 2 will be 4 see this is correct
right because the told if you have the
conflict 2 plus 4 is equal to 6 in both
the cases so 2 will be for the lower
alphabetically-ordered group that is
close this is correct to Pluto is
correct and flu is wrong
similarly in this case you have all and
ein over and I'm so out of all lines are
of Alka alcohol and I'll kind I'll call
it a functional group so this would come
in the last this is wrong and this is
or me understand this so let me repeat
this this is dimethyl first one by
methane so it has to be two - it can't
be - this is correct in this case tri
methyl obtained we drizzle diagram and
we saw that both to 475 impossible so
two four seven we took because it has
lower number in this case we have two
one for commenting Lauro and methyl
since chloro comes first will give the
lower number to glow and in this case we
have a kind group and alcohol group this
is the primary function look this should
come in the last so this is correct so
we have to keep the condensed bond line
structure formula and also identify the
functional group for these three let's
start with the first so let's start - -
for triangle painting so we have paint
pains let us draw five carbon let's
number it one two three four five at
second position I have two methyl groups
at third position I have one with high
group that's all I have and let's other
thing I have to add carbon here this
carbon has only one bonds let me add
seeing would want this governor has one
two three four all complete this carbon
has only one bond so let me add skin one
bonds this carbon also has only one bond
so let me add three more bonds this
carbon has one two three what it we have
no hydrogen here this carbon has one
bonds let me add three more two CS three
this I have to add two more hydrogen and
this I can make it a ch3 so now this is
my formula I have to write condensed
formula right so let me write condensed
formula so it will be ch3 here one more
siesta como si señor 3 CS 3 actually
then I have one carbon here right and
then I have
ch3 and one carbon here I have this one
I have sorry one before this this is the
CSD is here here I have a reaction
correct so I have ch2 here right again I
have C H and two CS d this is my
condensed formula cut so I have cs3 here
one ch3 2 ch3 ch3 ch3 here good then I
have carbon then I have ch2 there then I
have C 1 hydrogen and two CS t let me
draw the bond length formula for this I
have five so let me first draw it is
painting so let me draw a 5 year 1 1 2 e
1 2 3 4 5 this pentane at first second I
had two methyl fail here 3 4 I have a
methyl some Hey right this is my one he
whoa
so I do I have methyl groups and for I
haven't there also if you see let me
number this ingredients that you know in
this chain matter
correct
so at the fall this carbon second Garmin
has if you see freedom diagnostic so I
have see a steep see there and then I
have one carbon right this is second
carbon act in this third carbonyls would
plugin and it is 15 things this is how
it is correct so let us draw the other
structure that is two hydroxy one two
three
propane a trick carboxylic acid let me
draw the bond line for my it's really
easy for me so this is propane let me
draw propane three carbon right so we've
been wrong 1 2 & 3 carbon this is my
propane but this then we number this
different color now it is two hydroxy so
let me add one which group here hydroxy
is done now I have at 1 2 3 I have cut
trial carboxylic acid right at 1 2 & 3 I
have carboxylic acids let me add so here
I'll add cos group that is this is my
cos me here and one at T also I have to
add Cu age group so let me add at this
guy
1 CH blue here and I do also I have to
add see what use here I have to add once
see oh this is my bond line formula so
with this I can create just condensed
form a very easy this is my H right and
then I have over here right then I have
see here this is my co h complete and
then I have this hair this is ch2
ch2 this is my first carbon actually and
then I have a hair one two three four
all carbon bonds are there so I have
only one carbon here right one one
carbon
leave your draw once again this is coh
so let me write CE OS here right this
part is done and then I have ch2 here
first carbon this part is done and then
I have one carbon here with this carbon
I have an H group attached and one coh
group attached right and then I have
this carbon see the two hydrogen and
then I have one more cox-2 the last
correct so my carbon is one carbon this
second carbon and this is $4
so apart from three carbons I had three
carboxylic acid okay
the next is hexane die else let us draw
hexane one two three four five six five
so I have six carbon let me number them
so these are my six carbons because it's
hexane die L and since there's no
numbers given that means the RDI group
is attached at the end right that means
one aldehyde group is attached here so
the hydrogen here then over here
similarly here also there is one early
high group iodized this means that the
aldehyde groups are attached at the end
this is my kind instructions with that I
can easily write so this is CH o CH o
and then I have 4 CH 2 then I have CH
this is my structure correct thank you
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thanks and screen
chemistry basic part 14 is what Q Bar
Exam feel calm no more fear from exam
this question
we should have following it represents
the correct IUPAC name the first one is
to diamond tread painted or to develop
eaten since it is dimethyl you should
have two numbers here this is wrong
right so this is correct to to dimethyl
pentane is correct so for a the answer
is to die tile in day let's take the
second one be two four seven time is
high octane and two five seven
try me trial opt-ins let's draw this
octane workup eight carbon three four
five six seven eight
right and this is number one two three
four five six seven eight
so it says that try me tied at second I
have a metal rod at fourth I have a
methyl group and in eighth seventh I
have type if you have second number the
same thing in the reverse way it will be
one two three four five six seven eight
the same thing I can write s257 triumph
accountant right but as I told the
Friday will be given to the lower number
so this is correct this is not correct
right because this is correct
this is not going because it is two five
seven this is two four seven for the
scene come back the next is to
chloroform methyl painting or 402 methyl
Benton again there is a conclusion here
because 2n 4 we can give any number
names so 2 will be 4 see this is correct
right because the told if you have the
conflict 2 plus 4 is equal to 6 in both
the cases so 2 will be for the lower
alphabetically-ordered group that is
close this is correct to Pluto is
correct and flu is wrong
similarly in this case you have all and
ein over and I'm so out of all lines are
of Alka alcohol and I'll kind I'll call
it a functional group so this would come
in the last this is wrong and this is
or me understand this so let me repeat
this this is dimethyl first one by
methane so it has to be two - it can't
be - this is correct in this case tri
methyl obtained we drizzle diagram and
we saw that both to 475 impossible so
two four seven we took because it has
lower number in this case we have two
one for commenting Lauro and methyl
since chloro comes first will give the
lower number to glow and in this case we
have a kind group and alcohol group this
is the primary function look this should
come in the last so this is correct so
we have to keep the condensed bond line
structure formula and also identify the
functional group for these three let's
start with the first so let's start - -
for triangle painting so we have paint
pains let us draw five carbon let's
number it one two three four five at
second position I have two methyl groups
at third position I have one with high
group that's all I have and let's other
thing I have to add carbon here this
carbon has only one bonds let me add
seeing would want this governor has one
two three four all complete this carbon
has only one bond so let me add skin one
bonds this carbon also has only one bond
so let me add three more bonds this
carbon has one two three what it we have
no hydrogen here this carbon has one
bonds let me add three more two CS three
this I have to add two more hydrogen and
this I can make it a ch3 so now this is
my formula I have to write condensed
formula right so let me write condensed
formula so it will be ch3 here one more
siesta como si señor 3 CS 3 actually
then I have one carbon here right and
then I have
ch3 and one carbon here I have this one
I have sorry one before this this is the
CSD is here here I have a reaction
correct so I have ch2 here right again I
have C H and two CS d this is my
condensed formula cut so I have cs3 here
one ch3 2 ch3 ch3 ch3 here good then I
have carbon then I have ch2 there then I
have C 1 hydrogen and two CS t let me
draw the bond length formula for this I
have five so let me first draw it is
painting so let me draw a 5 year 1 1 2 e
1 2 3 4 5 this pentane at first second I
had two methyl fail here 3 4 I have a
methyl some Hey right this is my one he
whoa
so I do I have methyl groups and for I
haven't there also if you see let me
number this ingredients that you know in
this chain matter
correct
so at the fall this carbon second Garmin
has if you see freedom diagnostic so I
have see a steep see there and then I
have one carbon right this is second
carbon act in this third carbonyls would
plugin and it is 15 things this is how
it is correct so let us draw the other
structure that is two hydroxy one two
three
propane a trick carboxylic acid let me
draw the bond line for my it's really
easy for me so this is propane let me
draw propane three carbon right so we've
been wrong 1 2 & 3 carbon this is my
propane but this then we number this
different color now it is two hydroxy so
let me add one which group here hydroxy
is done now I have at 1 2 3 I have cut
trial carboxylic acid right at 1 2 & 3 I
have carboxylic acids let me add so here
I'll add cos group that is this is my
cos me here and one at T also I have to
add Cu age group so let me add at this
guy
1 CH blue here and I do also I have to
add see what use here I have to add once
see oh this is my bond line formula so
with this I can create just condensed
form a very easy this is my H right and
then I have over here right then I have
see here this is my co h complete and
then I have this hair this is ch2
ch2 this is my first carbon actually and
then I have a hair one two three four
all carbon bonds are there so I have
only one carbon here right one one
carbon
leave your draw once again this is coh
so let me write CE OS here right this
part is done and then I have ch2 here
first carbon this part is done and then
I have one carbon here with this carbon
I have an H group attached and one coh
group attached right and then I have
this carbon see the two hydrogen and
then I have one more cox-2 the last
correct so my carbon is one carbon this
second carbon and this is $4
so apart from three carbons I had three
carboxylic acid okay
the next is hexane die else let us draw
hexane one two three four five six five
so I have six carbon let me number them
so these are my six carbons because it's
hexane die L and since there's no
numbers given that means the RDI group
is attached at the end right that means
one aldehyde group is attached here so
the hydrogen here then over here
similarly here also there is one early
high group iodized this means that the
aldehyde groups are attached at the end
this is my kind instructions with that I
can easily write so this is CH o CH o
and then I have 4 CH 2 then I have CH
this is my structure correct thank you
visit exam viacom to watch free
education videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more
thanks and screen
About this Video
Oct 17, 2025
Last updated
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