Give the IUPAC name for each compound.

(a)
Interpretation:
The IUPAC nomenclature of Alkanes having complex structures and having lots of branches needs to be determined.
Concept Introduction:
Every organic compound has its own unique name which is followed by the IUPAC (International Union of Pure and Applied Chemistry). For alkane also, there are particular rules for IUPAC nomenclature. The rules are −
- The name has to end with suffix "−ane" for all alkanes.
- The longest continuous carbon chain that contains the functional group will be treated as the main chain.
- Number the carbons in the longest carbon chain. As per as the no of carbon present in the longest chain, "Pent (for 5)", "Hept (for 6)" this word will be added before "ane".
- The branched groups present in the longest carbon chain beside the main chain should be named by the number of carbon atoms present in the branched group. These groups will end with "-yl" (e.g.- ethyl, methyl) at their end.
- The position of the group on the main carbon chain and it should be mentioned (e.g. 2,3 -).
- Numbering of the main chain should start from the side having smallest distance with most priority brunch (priority depends on alphabetical order of the brunch).
- The brunched groups should be listed before the name of the main carbon chain in alphabetical order (ignoring prefix like di/tri).
- Combine the elements of the name into a single word in the following order:
- Branched groups in alphabetical order (ignoring prefixes).
- Prefix of main chain
- End with "-ane" suffix.
Answer to Problem 46P
2,3-dimethylpentane
Explanation of Solution
The longest carbon chain has 5 carbons.
And there are two methyl groups in the no 2 and 3 carbon. As the branch group has only one CH3, it is called 'Methyl'.
So, following by the above mention rule, the name of the alkane is 2,3 (the position of the branch) + dimethyl (the name of the branch, 'di' as two methyl are present )+ pent ( 5 carbon in main chain)+ "-ane"= 2,3-dimethylpentane

(b)
Interpretation:
The IUPAC nomenclature of Alkanes having same group in same carbon needs to be determined.
Concept Introduction:
Every organic compound has its own unique name which is followed by the IUPAC (International Union of Pure and Applied Chemistry). For alkane also, there is a particular rule for IUPAC nomenclature. The rules are −
- The name has to end with suffix "−ane" for all alkanes.
- The longest continuous carbon chain that contains the functional group will be treated as the main chain.
- Number the carbons in the longest carbon chain. As per as the no of carbon present in the longest chain, "Pent (for 5)", "Hept (for 6)" this word will be added before "ane".
- The branched groups present in the longest carbon chain beside the main chain should be named by the number of carbon atoms present in the branched group. These groups will end with "-yl" (e.g.- ethyl, methyl) at their end.
- The position of the group on the main carbon chain and it should be mentioned (e.g. 2,3 -).
- Numbering of the main chain should start from the side having smallest distance with most priority brunch (priority depends on alphabetical order of the brunch).
- The brunched groups should be listed before the name of the main carbon chain in alphabetical order (ignoring prefix like di/tri).
- Combine the elements of the name into a single word in the following order:
- Branched groups in alphabetical order (ignoring prefixes).
- Prefix of main chain
- End with "-ane" suffix.
Answer to Problem 46P
3,3-dimethylhexane
Explanation of Solution
The longest carbon chain has 6 carbons.
And there are two methyl groups in the no. 3 carbon. As the branch group has only one CH3, it is called 'Methyl'.
So, following by the above mention rule, the name of the alkane is 3,3 (the position of the branch) + dimethyl (the name of the branch, 'di' as two methyl are present)+ hex ( 6 carbon in main chain)+ "-ane"= 3,3-dimethylhexane.

(c)
Interpretation:
The IUPAC nomenclature of Alkanes focusing mainly on finding the longest carbon chain needs to be determined.
Concept Introduction:
Every organic compound has its own unique name which is followed by the IUPAC (International Union of Pure and Applied Chemistry). For alkane also, there is a particular rule for IUPAC nomenclature. The rules are −
- The name has to end with suffix "−ane" for all alkanes.
- The longest continuous carbon chain that contains the functional group will be treated as the main chain.
- Number the carbons in the longest carbon chain. As per as the no of carbon present in the longest chain, "Pent (for 5)", "Hept (for 6)" this word will be added before "ane".
- The branched groups present in the longest carbon chain beside the main chain should be named by the number of carbon atoms present in the branched group. These groups will end with "-yl" (e.g.- ethyl, methyl) at their end.
- The position of the group on the main carbon chain and it should be mentioned (e.g. 2,3 -).
- Numbering of the main chain should start from the side having smallest distance with most priority brunch (priority depends on alphabetical order of the brunch).
- The brunched groups should be listed before the name of the main carbon chain in alphabetical order (ignoring prefix like di/tri).
- Combine the elements of the name into a single word in the following order:
- Branched groups in alphabetical order (ignoring prefixes).
- Prefix of main chain.
- End with "-ane" suffix.
Answer to Problem 46P
5-ethylnonane
Explanation of Solution
The longest carbon chain has 9 carbons.
And there are one ethyl groups in the no. 5 carbon. As the branch groups have CH2CH3, it is called 'ethyl'.
So, following by the above mention rule, the name of the alkane is 5 (the position of the branch) + ethyl (the name of the branch) + non( 9 carbon in main chain)+ "-ane"= 5-ethylnonane

(d)
Interpretation:
The IUPAC nomenclature of Alkanes having skeletal structure needs to be determined.
Concept Introduction:
Every organic compound has its own unique name which is followed by the IUPAC (International Union of Pure and Applied Chemistry). For alkane also, there is a particular rule for IUPAC nomenclature. The rules are −
- The name has to end with suffix "−ane" for all alkanes.
- The longest continuous carbon chain that contains the functional group will be treated as the main chain.
- Number the carbons in the longest carbon chain. As per as the no of carbon present in the longest chain, "Pent (for 5)", "Hept (for 6)" this word will be added before "ane".
- The branched groups present in the longest carbon chain beside the main chain should be named by the number of carbon atoms present in the branched group. These groups will end with "-yl" (e.g.- ethyl, methyl) at their end.
- The position of the group on the main carbon chain and it should be mentioned (e.g. 2,3 -).
- Numbering of the main chain should start from the side having smallest distance with most priority brunch (priority depends on alphabetical order of the brunch).
- The brunched groups should be listed before the name of the main carbon chain in alphabetical order (ignoring prefix like di/tri).
- Combine the elements of the name into a single word in the following order:
- Branched groups in alphabetical order (ignoring prefixes).
- Prefix of main chain
- End with "-ane" suffix.
Answer to Problem 46P
4-ethyl-2,6-dimethyloctane
Explanation of Solution
The longest carbon chain has 8 carbons.
And there are two methyl groups in the no. 2 and 6 carbon, one ethyl group in no. 4 carbon. As the branch groups have CH3and CH2CH3, they are called 'methyl' and 'ethyl' simultaneously.
So, following by the above mention rule, the name of the alkane is 4 (the position of the branch having alphabetically order 1st) + ethyl (the name of the branch) + 2,6 (position of 2nd brunch) + dimethyl( name of the brunch and 'di' as two methyl are present)+ oct ( 8 carbon in main chain)+ "-ane"= 4-ethyl-2,6-dimethyloctane

(e)
Interpretation:
The IUPAC nomenclature of Alkanes needs to be determined.
Concept Introduction:
Every organic compound has its own unique name which is followed by the IUPAC (International Union of Pure and Applied Chemistry). For alkane also, there is a particular rule for IUPAC nomenclature. The rules are −
- The name will always end with the suffix "−ane" for alkanes.
- Find the longest continuous carbon chain that contains the functional group.
- Number the carbons in the longest carbon chain. Depends on the no. of carbon in the longest chain, "Pent (for 5)", "Hept (for 6)" this word will be added before "ane" in the nomenclature.
- Look for branched groups present in the longest carbon chain beside the main chain. These groups are named by counting the number of carbon atoms in the branched group. These groups will all end in "-yl" (e.g.- propyl, pentyl)
- Note the position of the group on the main carbon chain. If there is more than one of the same types of branched group then both numbers must be listed (e.g. 2,3 -).
- Numbering of the main chain should start from the side having smallest distance with most priority brunch (priority based on alphabetical order).
- The branched groups should be listed before the name of the main carbon chain in alphabetical order (ignoring di/tri/tetra).
- Combine the elements of the name into a single word in the following order:
- Branched groups in alphabetical order (ignoring prefixes).
- Prefix of main chain
- End with "-ane" suffix.
Answer to Problem 46P
4-ethyl-2,2-dimethylhexane
Explanation of Solution
The longest carbon chain has 6 carbons.
And there are two methyl groups in the no. 2 carbon, one ethyl group in no. 4 carbon. As the branch groups have CH3and CH2CH3, they are called 'methyl' and 'ethyl' simultaneously.
So, following by the above mention rule, the name of the alkane is 4 (the position of the branch having alphabetically order 1st) + ethyl (the name of the branch) + 2 (position of 2nd brunch) + dimethyl( name of the brunch and 'di' as two methyl are present)+ hex ( 6 carbon in main chain)+ "-ane"= 4-ethyl-2,2-dimethylhexane

(f)
Interpretation:
The IUPAC nomenclature of Alkanes having lots of brunch in skeletal structure needs to be determined.
Concept Introduction:
Every organic compound have their own unique name which is followed by the IUPAC (International Union of Pure and Applied Chemistry). For alkane also, there is a particular rule for IUPAC nomenclature. The rules are −
- The name has to end with suffix "−ane" for all alkanes.
- The longest continuous carbon chain that contains the functional group will be treated as the main chain.
- Number the carbons in the longest carbon chain. As per as the no of carbon present in the longest chain, "Pent (for 5)", "Hept (for 6)" this word will be added before "ane".
- The branched groups present in the longest carbon chain beside the main chain should be named by the number of carbon atoms present in the branched group. These groups will end with "-yl" (e.g.- ethyl, methyl) at their end.
- The position of the group on the main carbon chain and it should be mentioned (e.g. 2,3 -).
- Numbering of the main chain should start from the side having smallest distance with most priority brunch (priority depends on alphabetical order of the brunch).
- The brunched groups should be listed before the name of the main carbon chain in alphabetical order (ignoring prefix like di/tri).
- Combine the elements of the name into a single word in the following order:
- Branched groups in alphabetical order (ignoring prefixes).
- Prefix of main chain
- End with "-ane" suffix.
Answer to Problem 46P
5-ethyl-2,6-dimethyloctane
Explanation of Solution
The longest carbon chain has 8 carbons.
And there are two methyl groups in the no. 3 and 6 carbon and one ethyl in no 5 carbon. As the branch groups have CH3and CH2CH3, they are called 'methyl' and 'ethyl' simultaneously.
So, following by the above mention rule, the name of the alkane is 5 (the position of the branch having alphabetically order 1st) + ethyl (the name of the branch) + 2,6 (position of 2nd brunch) + dimethyl( name of the brunch and 'di' as two methyl are present)+ oct ( 8 carbon in main chain)+ "-ane"= 5-ethyl-2,6-dimethyloctane.
Want to see more full solutions like this?
Chapter 12 Solutions
GENERAL,ORGANIC, & BIOLOGICAL CHEM-ACCES
- Could you please turn this into a complete Lewis dot structure formula for me so I can visualize it more clearly? and then do the explaining for the question.arrow_forwardplease solve. If the answer is "no error" and it asks me to type something, and i typed a-helix, its always wrong.arrow_forwardCan you please solve and explain this for me in a simple way? I cant seem to comprehend this problem.arrow_forward
- Part I. Problem solving. Include all necessary calculations 13 provide plots and graphs. Complexation wl diphenyl carbazide (OPC) in acidic media is another type of sensitive photometric method used for the analysis of aqueous. hexavalent chromium. At 540nm the cherry-red complex as a result of DPC reaction w/ chromium can be photometrically measured. at this wavelength. - a 25mL The UV-vis analysis for the determination of nexavalent chromium in ground water sample is given below. The experiment was based on external calibration method w/ each measurement sample prepared are as follows lab sample analysis contained the standard 100 ppb croy cor groundwater sample, volumes used as indicated below), 12.50 mL of 0.02 M H2Soy and 5.50 ml of 100 ppm DPC (wi water to adjust final volume to 25-ml). The main stripping method was square wave voltammetry, following the conditions set in the main ASV experiment. Standard 100 Volumetric Groundwater H2SO4 0.20 M, flask Sample, mL ppb CrO4*, 100…arrow_forwardplease helparrow_forwardPredict the products of the following reactions. Draw mechanism arrows for each step for a, b, and c. a.) HBr b.) HI H₂O H2SO4 d.) C12 HO H2SO4 1.) BH3 2.) H2O2, NaOHarrow_forward
- K for the following reaction is 0.11 at constant temperature. If the equilibrium concentration of HCl is 0.5 M, what is the equilibrium concentration of NH3. NH4CI(s) ⇌ NH3(g) + HCI(g)arrow_forwardplease help by Draw the following structures (Lewis or line-angle drawing).arrow_forwardplease helparrow_forward
- Consider the reaction: 2 A (aq) ⇌ B(aq) Given the following KC values and starting with the initial concentration of A = 4.00 M, complete ICE diagram(s)and find the equilibrium concentrations for A and B.A) KC = 4.00B) KC = 200C) KC = 8.00 x10-3arrow_forward5) Consider the reaction: Cl2 (g) + F2 (g) ⟷ 2 ClF (g) KP=? The partial pressure of 203 kPa for Cl2 and a partial pressure of 405 kPa for F2. Upon reaching equilibrium, thepartial pressure of ClF is 180 kPa. Calculate the equilibrium concentrations and then find the value for KP.arrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forward
